set theory - Dispensing with the notion of infinity for the sake of coverings

I don't quite know what you mean by "coverings of topology", but it is possible to formalize a notion of size for infinite sets which relies on the part-whole conception, rather than the bijective correspondence conception. These two views are mutually exclusive, in the sense that size for finite sets satisfies both properties, infinite sets can only support one of the two. But either choice can be made to work!

So, if you require the notion of size to equate two bijective sets, then the even numbers are equal in size to the natural numbers (this is the traditional Cantorian view). You could also take a mereological view, and say that one set is smaller than another if every element of the first set is a member of the second set. In this interpretation, the even numbers are smaller than the natural numbers.

A recent issue of the Review of Symbolic Logic had an article about these issues, including both some history of mathematics and more recent logical systems which formalize the mereological view. See Paolo Mancosu's "Measuring the Size of Infinite Collections of Natural Numbers: Was Cantor's Theory of Infinite Number Inevitable?"

ag.algebraic geometry - Is Hartshorne's definition of the category of varieties natural?

I would just comment, but I'm a new user so I can't.

I believe you are referring to Hartshorne's definition on p. 15:

"A variety over $k$ is any affine, quasi-affine, projective, or quasi-projective variety as defined above."

The reason he defines things this way is that in section I.1 he defined affine and quasi-affine varieties; in section I.2 he defined projective and quasi-projective varieties. You are right that he hasn't made very clear the relation between the two. Exercise I.2.9 on projective closures hints at the relation, but it hasn't been made completely precise (he just uses the terminology "identify"). To make this precise, we need to say what the morphisms between any two different varieties are.

Suffice it to say, the identification of Exercise I.2.9 given by projective closure is in fact an isomorphism in the sense of section I.3, so all varieties are isomorphic to quasi-projective varieties.

If this is your first encounter with algebraic geometry, however, I'm not sure I would recommend chapter I of Hartshorne very highly. I made the mistake of thoroughly studying chapter I myself, and I think there are far better ways to spend your time.

gr.group theory - Field and Group Isomorphisms

Let me see if I can clear up your confusion. It is true that any algebraic extension of a field $F$ embeds in a given algebraic closure, but this embedding is not unique; you can compose any given embedding with an element of the Galois group of the normal closure, and if the extension isn't itself Galois then you'll even end up with embeddings that have different images. Also, it's usually considered bad form to say "the" algebraic closure of a field because algebraic closures are only unique up to isomorphism and the choice of an algebraic closure cannot be made canonically (to my knowledge). Anyway, here's what I think you wanted to say:

"If I have an extension $E/F$ and an isomorphism $sigma : F to F'$, I should be able to find a corresponding extension $E'/F'$ and an isomorphism $tau : E to E'$ which restricts to $sigma$."

This statement is true, but again, $tau$ is not unique. (And I wouldn't call this an "isomorphism theorem." Those refer to a specific set of theorems.) I also don't know why you give a statement about isomorphisms and then ask for a statement about homomorphisms (in the group setting). What kind of statement, exactly, are you looking for?

ag.algebraic geometry - Colimits of schemes

This is not an answer to the question, but is intended to show that Martins's concern about the possible distinction between the colimit in the category of schemes vs. in the category of locally ringed spaces is a valid one. Indeed, if I haven't blundered below, then it seems that in some circumstances at least the direct limit (colimit) of schemes exists, but does not coincide with the direct limit in the category of locally ringed spaces.

For example, suppose that $X_n$ is the Spec of $k[x]/(x^n),$ for some field $k$ (and the transition maps are the obvious ones). Then the direct limit of the $X_n$ in the category of locally ringed spaces is a formal scheme which is not a scheme, whose underlying topological space is a point, and whose structure sheaf (which is in this context simply a ring, namely the stalk at the unique point) is $k[[x]]$.

On the other hand, suppose given compatible maps from the $X_n$ to a scheme $S$. These must all map the common point underlying the $X_n$ to some point $s in S$, which lies in some affine open Spec $A$. Thus the maps from the $X_n$ factor through Spec $A$, and correspond to compatible maps $A rightarrow k[x]/(x^n),$ i.e. to a map $A rightarrow k[[x]].$ This in turn gives a map Spec $k[[x]] rightarrow$ Spec $Asubset X,$ and so we see that the natural compatible maps from the $X_n$ to Spec $k[[x]]$ identify Spec $k[[x]]$ with the direct limit of the $X_n$ in the category of schemes.

EDIT: As is noted in a comment of David Brown's attached to his answer, this example
generalizes, e.g. if $I$ is an ideal in a ring $A$, then the direct limit in the category of schemes of Spec $A/I^n$ coincides with Spec $hat A$, where $hat A$ is the $I$-adic completion of $A$.

FURTHER EDIT: I am no longer certain about the claim of the previous paragraph. If $A/I$ (and hence each $A/I^n$) is local then for any scheme $S$ the maps Spec $A/I^n to S$ factor through an affine open subscheme, so one reduces to computations in the category of rings, and hence finds that indeed the direct limit of the Spec $A/I^n$ equals Spec $hat{A}$. More generally, I'm currently unsure ... .

ho.history overview - Galois theory timeline

EDIT. Here is the part of the answer that has been rewritten:

We give below a short proof of the Fundamental Theorem of Galois Theory (FTGT) for finite degree extensions. We derive the FTGT from two statements, denoted (a) and (b). These two statements, and the way they are proved here, go back at least to Emil Artin (precise references are given below).

The derivation of the FTGT from (a) and (b) takes about four lines, but I haven't been able to find these four lines in the literature, and all the proofs of the FTGT I have seen so far are much more complicated. So, if you find either a mistake in these four lines, or a trace of them the literature, please let me know.

The argument is essentially taken from Chapter II (link) of Emil Artin's Notre Dame Lectures [A]. More precisely, statement (a) below is implicitly contained in the proof Theorem 10 page 31 of [A], in which the uniqueness up to isomorphism of the splitting field of a polynomial is verified. Artin's proof shows in fact that, when the roots of the polynomial are distinct, the number of automorphisms of the splitting extension coincides with the degree of the extension. Statement (b) below is proved as Theorem 14 page 42 of [A]. The proof given here (using Artin's argument) was written with Keith Conrad's help.

Theorem. Let $E/F$ be an extension of fields, let $a_1,dots,a_n$ be distinct generators of $E/F$ such that the product of the $X-a_i$ is in $F[X]$. Then

• the group $G$ of automorphisms of $E/F$ is finite,




• there is a bijective correspondence between the sub-extensions $S/F$ of $E/F$ and the subgroups $H$ of $G$, and we have
  $$
  Sleftrightarrow Hiff H=text{Aut}(E/S)iff S=E^H
  $$
  $$
  implies[E:S]=|H|,
  $$
  where $E^H$ is the fixed subfield of $H$, where $[E:S]$ is the degree (that is the dimension) of $E$ over $S$, and where $|H|$ is the order of $H$.

PROOF

We claim:

(a) If $S/F$ is a sub-extension of $E/F$, then $[E:S]=|text{Aut}(E/S)|$.

(b) If $H$ is a subgroup of $G$, then $|H|=[E:E^H]$.

Proof that (a) and (b) imply the theorem. Let $S/F$ be a sub-extension of $E/F$ and put $H:=text{Aut}(E/S)$. Then we have trivially $Ssubset E^H$, and (a) and (b) imply
$$
[E:S]=[E:E^H].
$$
Conversely let $H$ be a subgroup of $G$ and set $overline H:=text{Aut}(E/E^H)$. Then we have trivially $Hsubsetoverline H$, and (a) and (b) imply $|H|=|overline H|$.

Proof of (a). Let $1le ile n$. Put $K:=S(a_1,dots,a_{i-1})$ and $L:=K(a_i)$. It suffices to check that any $F$-embedding $phi$ of $K$ in $E$ has exactly $[L:K]$ extensions to an $F$-embedding $Phi$ of $L$ in $E$; or, equivalently, that the polynomial $pinphi(K)[X]$ which is the image under $phi$ of the minimal polynomial of $a_i$ over $K$ has $[L:K]$ distinct roots in $E$. But this is clear since $p$ divides the product of the $X-a_j$.

Proof of (b). In view of (a) it is enough to check $|H|ge[E:E^H]$. Let $k$ be an integer larger than $|H|$, and pick a
$$
b=(b_1,dots,b_k)in E^k.
$$
We must show that the $b_i$ are linearly dependent over $E^H$, or equivalently that $b^perpcap(E^H)^k$ is nonzero, where $bullet^perp$ denotes the vectors orthogonal to $bullet$ in $E^k$ with respect to the dot product on $E^k$. Any element of $b^perpcap (E^H)^k$ is necessarily orthogonal to $hb$ for any $hin H$, so
$$
b^perpcap(E^H)^k=(Hb)^perpcap(E^H)^k,
$$
where $Hb$ is the $H$-orbit of $b$. We will show $(Hb)^perpcap(E^H)^k$ is nonzero. Since the span of $Hb$ in $E^k$ has $E$-dimension at most $|H| < k$, $(Hb)^perp$ is nonzero. Choose a nonzero vector $x$ in $(Hb)^perp$ such that $x_i=0$ for the largest number of $i$ as possible among all nonzero vectors in $(Hb)^perp$. Some coordinate $x_j$ is nonzero in $E$, so by scaling we can assume $x_j=1$ for some $j$. Since the subspace $(Hb)^perp$ in $E^k$ is stable under the action of $H$, for any $h$ in $H$ we have $hxin(Hb)^perp$, so $hx-xin(Hb)^perp$. Since $x_j=1$, the $j$-th coordinate of $hx-x$ is $0$, so $hx-x=0$ by the choice of $x$. Since this holds for all $h$ in $H$, $x$ is in $(E^H)^k$.

[A] Emil Artin, Galois Theory, Lectures Delivered at the University of Notre Dame, Chapter II, available here.

PDF version: http://www.iecn.u-nancy.fr/~gaillapy/DIVERS/Fundamental.theorem.of.galois.theory/

---

Here is the part of the answer that has not been rewritten:

Although I'm very interested in the history of Galois Theory, I know almost nothing about it. Here are a few things I believe. Thank you for correcting me if I'm wrong. My main source is

http://www-history.mcs.st-and.ac.uk/history/Projects/Brunk/Chapters/Ch3.html

Artin was the first mathematician to formulate Galois Theory in terms of a lattice anti-isomorphism.

The first publication of this formulation was van der Waerden's "Moderne Algebra", in 1930.

The first publications of this formulation by Artin himself were "Foundations of Galois Theory" (1938) and "Galois Theory" (1942).

Artin himself doesn't seem to have ever explicitly claimed this discovery.

Assuming all this is true, my (probably naive) question is:

Why does somebody who makes such a revolutionary discovery wait so many years before publishing it?

I also hope this is not completely unrelated to the question.

inequalities - How often are irrational numbers well-approximated by rationals?

$mathcal{Q}(x,lambda)$ has positive relative density if and only if $lambdale 1$.
This follows from Weyl's Theorem on Uniform Distribution. (There is a nice concise proof in Cassels' "Diophantine Approximation".)

Weyl's Theorem: Let $Isubset mathbb{R}$ be an interval of length $epsilon le 1$. Let $S_N(I)$ be the set of all integers $q$ in the interval $[1,N]$ such that for some integer $p$, it holds that $xq-pin I$. Then

$$frac{Card(S_N(I))}{N} to epsilon
text{ as } Ntoinfty.$$

Here's a proof-sketch, using Weyl's Theorem, that if $lambda > 1$ then $mathcal{Q}(x,lambda)$ has relative density zero:

Fix $epsilon > 0$, and take $I$ (in Weyl's Theorem) to be the interval $(-epsilon,epsilon)$. Suppose $lambda>1$. Let $qin mathcal{Q}(x,lambda)$; so for some $pin mathbb{Z}$, $$|xq-p| < q^{1-lambda}.$$

There is an integer $M$, depending only on $epsilon$, such that $|xq-p| < epsilon$ whenever $p$ and $q$ satisfy the above inequality and $qge M$. Therefore $$mathcal{Q}(x,lambda)cap [M,N]subset S_N(I).$$ It follows from Weyl's Theorem that the relative density of $mathcal{Q}(x,lambda)$ does not exceed $2epsilon$. Since $epsilon$ is arbitrary, the relative density of $mathcal{Q}(x,lambda)$ must be zero.

This can be proved in a more elementary but laborious way using the "Ostrowski Number System", which is explained in the Rockett and Szusz book on continued fractions.

pr.probability - Inequality constraints, probability distributions, and integer partitions

Whether this is a valid approach for calculating the probability will depend on what assumptions you are making on the probabilities of the various partitions. In particular, this calculation seems to require for its validity that all partitions be equally likely. Do you actually want (10,0,0) to be as likely as (4,3,3)?

Edit: Concerning asymptotic behavior, take the case where the number of balls equals the number of urns; call it $n$. One can ask for the proportion of balls in the first urn as a function of $n$, which is the same as asking for the average size of the biggest part in a partition of $n$ (which is also the same as asking for the average number of parts in a partition of $n$). This was studied by Kessler and Livingston, The expected number of parts in a partition of $n$, Monatshefte Math 81 (1976) 203-212. Let $P(n)$ be the total number of parts in all the partitions of $n$ (which is the same as the sum of the largest parts in all the partitions of $n$), and let $p(n)$ be the number of partitions of $n$. Then $${P(n)over p(n)}=sqrt{3nover2pi}(log n+2gamma-log(pi/2))+O(log^3n)$$ where $gamma$ is Euler's constant. More information at sequence A006128 at the Online Encyclopedia of Integer Sequences.

EDIT 15 July: Let $f_{j,k}(n)$ be the sum of the $j$-th biggest parts over all partitions of $n$ into at most $k$ parts, $$f_{j,k}(n)=sum_{eqalign{a_1gedotsge a_k&ge0cr a_1+dots+a_k&=ncr}}a_j,$$ and let $p_k(n)$ be the number of partitions of $n$ into at most $k$ parts. You are asking for the asymptotic behavior of $f_{j,k}(n)/np_k(n)$ as $k$ and $n$ increase, but it is not clear how you want them to increase (that is, what relation they should bear to each other as they increase), nor is it clear what you want $j$ to be doing while $k$ and $n$ are increasing. Perhaps you are only interested in the case $j=1$, which is the case in your example.

In any event, you need asymptotics for $p_k(n)$. From the generating function $$sum_np_k(n)x^n={1over(1-x)(1-x^2)cdots(1-x^k)}$$ one can get, for fixed $k$, $p_k(n)=C_kn^{k-1}+O(n^{k-2})$, where $C_k=(k!(k-1)!)^{-1}$ (I hope someone checks my calculations here).

Now I can tell you something about $f_{k,k}(n)$ - this is the case when you're interested in the $it last$ urn. Let's write $g_k(n)=f_{k,k}(n)$. It's not hard to prove that $g_k(n+k)-g_k(n)=p_k(n)$, from which it follows that $$sum_ng_k(n)x^n={x^kover(1-x)cdots(1-x^{k-1})(1-x^k)^2}$$ and then you get, again for fixed $k$, $g_k(n)=D_kn^k+O(n^{k-1})$, where $D_k=k^{-1}(k!)^{-2}$. So, the probability that the ball is from the last urn approaches $k^{-2}$ as $n$ increases with $k$ fixed.

I didn't do so well with the first urn. Let's just take the case $k=3$. We want $$h_3(n)=f_{1,3}(n)=sum_{eqalign{age bge c&ge0cr a+b+c&=ncr}}a.$$ Starting with $n=1$, this produces the sequence $1,3,6,11,17,27,37,52,69,90,113,144dots$. I didn't find this sequence in the Online Encyclopedia of Integer Sequences. I think, but can't prove, that the generating function is $$sum_nh_3(n)x^n={x+3x^2+4x^3+3x^4over(1-x)^2(1-x^3)^2}.$$ This gives $h_3(n)=(11/216)n^3+O(n^2)$. Together with $p_3(n)=(1/12)n^2+O(n)$, you get the probability of the ball coming from the first of the three urns converging to 11/18, as $n$ increases (and conditional on my unproved formula for the generating function).

EDIT 8 February 2011: There's a typo in the last display, it should be $$sum_nh_3(n)x^n={x+3x^2+4x^3+3x^4over(1-x^2)^2(1-x^3)^2}.$$ George Andrews wrote out a proof of this formula (and his method should apply more generally to $sum_nf_{j,k}(n)x^n$). Let $a=r+s+t$, $b=s+t$, $c=t$, then the generating function is $sum_{r,s,t}(r+s+t)x^{r+2s+3t}$, which can be split up into three pieces each of which is two geometric progressions times a sum of the shape $sum_uuy^u$. That last sum is well-known, and after some algebra the conjectured formula pops out.

tag removed - Coloring Points in the Plane

This is the well-known unit distance graph problem. If we call $U=U(mathbb R^2)$ as the unit distance graph on the plane; that is, the vertices are the points on the plane and the edges are the pairs at distance one from each other.

It is well-known that $$4 leq chi(U(mathbb R^2))leq 7.$$

The lower bound is found by drawing a finite unit distance subgraph of $U$ which has chromatic number 4 while the upper bound is found by coloring the plane with 7 colors after dividing it into hexagons of a fixed,small diameter.

medskip

Recently, I came across the study of the chromatic index of a supergraph of $U$ called odd-distance graphs. Moreover, I think that this problem is equivalent to finding a measure of some sort but that is all I remember.

We once tried to use Hammel Basis of the plane to come up with a proof which at least improves the above bounds but it does not seem to work(well, we were able to prove that $mathbb Z$ is an integral domain... funny...).

characteristic p - Restriction theorems over finite fields

Not directly. The key connection between restriction and Kakeya in Euclidean settings is that thanks to Taylor expansion, a surface in Euclidean space looks locally flat, and so the Fourier transform of measures on that surface are a superposition of Fourier transforms of very flat measures, which by the uncertainty principle tend to propagate along tubes. The arrangement of these tubes is then governed by the Kakeya conjecture.

In finite fields, there is no notion of Taylor expansion, unless one forces it into existence by working on the tangent bundle of a surface, rather than a surface itself. So there is a connection between restriction and Kakeya on the former object, but not the latter. (This is discussed at the end of my paper with Mockenhaupt.)

Ref request: A graph G contains H as a minor iff it contains one of finitely many graphs as a topological minor

For definitions of graph minors and topological minors, see wikipedia's article on graph minors.

Theorem: For every graph H, there is a finite set of graphs, say S(H), such that G contains H as a minor if and only if G contains some graph from S(H) as a topological minor.

Can anyone point me to a paper/book where this is proved? (I know how to prove it, I just want a reference to cite.)

ag.algebraic geometry - Square of an elliptic curve and projective plane

Let's assume one takes $E = mathbb{C}^* / langle p rangle$ an elliptic (Tate) curve over the complex field ($p = e^{2 pi i tau}$ where $1, tau$ are the 2 periods in additive notation; $Im tau > 0$). On this take points $u_1, u_2, u_3$ such that $u_1 u_2 u_3 = 1$ and then mod out by the action of the symmetric group $S_3$. So we essentially have a hypersurface in $E^3$ - a copy of $E^2$ with coordinates $(u_1, u_2)$ and we mod out by permuting $u_1, u_2$ and $1/u_1 u_2$ (the $u_i$'s are zeros and their reciprocals poles of an elliptic function - essentially the only one up to constant with these zeros and poles).

The question: is this quotient space $mathbb{P}^2$? I believe the answer is yes, but I can't see a way of using theta functions or other gadgets to explicitly give the isomorphism (whereby a theta function I mean $$theta_p(x) = prod_{l ge 0}(1-p^l x)(1-p^{l+1}/x)$$ which reduces to the Jacobi theta via the triple product identity).

Finally, does this work over other fields (reals, finite fields, other reasonable fields)?

spectral sequences in number theory

I posit the following example, in response to your ambiguous question:

The coniveau spectral sequence seems to play an important role in 'arithmetic geometry'. One instance is in class field theory for schemes:

From W. Raskind's nice survery article "Abelian class field theory of arithmetic schemes" [AMS, 1992, pgs. 100-101]:

Let $X$ be an arithmetic scheme, $n>0$ invertible on $X$. Then there is a coniveau spectral sequence (in the etale site):

$$E^{p,q}_1 = bigoplus _{ xin X^{p} } H^{q-p} (k(x), ;mathbb{Z}/n ; (j-p)) Rightarrow H^{p+q} (X, mathbb{Z}/n ;(j)) $$

Without going into more details, this sequence plays an important role in defining a reciprocity map from a class group of $X$ to abelian fundamental groups.

That's all I will say for now in hopes that the above provides for motivation to delve further into studying coniveau, etc.

Finally, one of the best articles I have seen on coniveau is by Colliot-Thélène, Hoobler, and Kahn, "The Bloch-Ogus-Gabber theorem" which can be found at:
http://www.math.jussieu.fr/~kahn/preprints/prep.html

It might be nice to have others' remarks/comments on coniveau, but I don't have any precise questions yet.

big list - Books you would like to see translated into English.

Probably most of the works from Oskar Perron. It has been mentioned already Die Lehre von den Kettenbrüchen, both volumes, but we could also ask for Irrationalzahlen or any of the other works from Perron.
Also worth being mentioned, for applied mathematicians, are the works of Grigory Isaakovich Barenblatt, previous to 1994; this is because Barenblatt has consistently worked about scaling phenomena, but from about the beginnings of the 1990's he began to do it on his own, whereas earlier work includes the participation of other marvelous mathematicians, like Z'eldovich; or even works on his own, but it is interesting to compare the evolution of his ideas. So, the name of books with his participation previous to the 1990's, and to my knowledge, not translated into English:
* Ja, B Zeldovich, G. I. Barenblatt, V. B. Librovich, G. M. Maxvikadze "Matematicheskaja teorija gorenija i vsriva", 1980
* G. I. Barenblatt, "Podobie, avtomodelnoct, promezhutochnaja asimptotika: teorija i prilozhenija k geofizicheskoi gidrodinamike", 1982
* A. P. Licitsin, G. I. Barenblatt "gidrodinamika i osadkoobrasovanie", 1983
* G. I. Barenblatt, V. N. Entov, V. M. Rizhik, "Dvizhenie zhidkocteii i gazov v prirodnix plactax" 1984
* G. I. Barenblatt, "Analiz razmernosteii" . Uch. pos. M.: MFTI, 1987. 168 с. (I think this last work made it to English under the translation as "Dimensional Analysis", but in that case I saw it only once, at the library of the Department of Applied Mathematics and Theoretical Physics -DAMTP-of Cambridge, UK, many years ago and is likely out of print anyway, plus the edition, to my knowledge was not revised; on top of that, DAMTP changed from Silver Street to Wilberforce road, and I have no idea if that book survived the moving, if indeed was at that library).

Notice also, that in the Nachlass (the collection of manuscripts, left after the death of an academician, and of course in particular a mathematician) of people like Bernhard Riemann or Ernst Zermelo, there might be still some untranslated documents, but then again they also need to be interpreted in a way that could be meaningful, and this because they are not finished, published or even unpublished works, but sketches of something not fully developed.

fa.functional analysis - Naive questions about "matrices" representing endomorphisms of Hilbert spaces.

This is a very basic question and might be way too easy for MO. I am learning analysis in a very backwards way. This is a question about complex Hilbert spaces but here's how I came to it: I have in the past written a paper about (amongst other things) compact endomorphisms of $p$-adic Banach spaces (and indeed of Banach modules over a $p$-adic Banach algebra), and in this paper I continually used the notion of a "matrix" of an endomorphism as an essential crutch when doing calculations and proofs. I wondered at the time where more "conceptual" proofs existed, and probably they do, but I was too lazy to find them.

Now I find myself learning the basic theory of certain endomorphisms of complex separable Hilbert spaces (continuous, compact, Hilbert-Schmidt and trace class operators) and my instinct, probably wrong, is to learn the theory in precisely the same way. So this is the sort of question I find myself asking.

Say $H$ is a separable Hilbert space with orthonomal basis $(e_i)_{iinmathbf{Z}_{geq1}}$. Say $T$ is a continuous linear map $Hto H$. Then $T$ is completely determined by its "matrix" $(a_{ij})$ with $Te_i=sum_ja_{ji}e_j$. But are there completely "elementary" conditions which completely classify which collections of complex numbers $(a_{ij})$ arise as "matrices" of continuous operators?

I will ask a more precise question at the end, but let me, for the sake of exposition, tell you what the the answer is in the $p$-adic world.

In the $p$-adic world, $sum_na_n$ converges iff $a_nto 0$, and life is easy: the answer to the question in the $p$-adic world is that $(a_{ij})$ represents a continuous operator iff

(1) For all $i$, $sum_j|a_{ji}|^2<infty$ (equivalently, $a_{ji}to 0$ as $jtoinfty$), and

(2) there's a universal bound $B$ such that $|a_{ij}|leq B$ for all $i,j$.

[there is no inner product in the $p$-adic case, so no adjoint, and the conditions come out being asymmetric in $i$ and $j$].
See for example pages 8--9 of this paper of mine, although of course this isn't due to me---it's in Serre's paper on compact operators on $p$-adic Banach spaces from the 60s---see Proposition 3 of Serre's paper. In particular, in the $p$-adic world, one can identify the continuous maps $Hto H$ (here $H$ is a $p$-adic Banach space with countable ON basis $(e_i)$) with the collection of bounded sequences in $H$, the identification sending $T$ to $(Te_i)$.

In the real/complex world though, the analogue of this result fails: the sequence $(e_1,e_1,e_1,ldots)$ is a perfectly good bounded sequence, but there is no continuous linear map $Hto H$ sending $e_i$ to $e_1$ for all $i$ (where would $sum_n(1/n)e_n$ go?).

Let's consider the finite rank case, so $T$ is a continuous linear map $Hto H$ with image landing in $mathbf{C}e_1$. Then by Riesz's theorem, $T$ is just "inner product with an element of $H$ and then multiply by $e_1$". Hence we have an additional condition on the $a_{ij}$, namely that $sum_j|a_{ij}|^2<infty$. Furthermore a continuous linear map is bounded, as is its adjoint.

This makes me wonder whether the following is true, or whether this is still too naive:

Q) Say $(a_{ij})$ $(i,jinmathbf{Z}_{geq1})$ is a collection of complex numbers satisfying the following:

There is a real number $B$ such that

1) For all $i$, $sum_j|a_{ij}|^2leq B$

2) For all $j$, $sum_j|a_{ij}|^2leq B$

Then is there a unique continuous linear map $T:Hto H$ with $Te_i=sum_ja_{ji}e_i$?
My guess is that this is still too naive. Can someone give me an explicit counterexample? Or, even better, a correct "elementary" list of conditions characterising the continuous endomorphisms of a Hilbert space?

On the other hand, it clearly isn't a complete waste of time to think about matrix coefficients. For example there's a bijection between Hilbert-Schmidt operators $T:Hto H$ and collections $(a_{ij})$ of complexes with $sum_{i,j}|a_{ij}|^2<infty$, something which perhaps the experts don't use but which I find incredibly psychologically useful.

lo.logic - What is a topos?

Let us concentrate on Grothendieck topoi. As mentioned in earlier posts, these are those topoi which arise as the category of sheaves for a category equipped with a Grothendieck topology. These are those topoi which "behave the most like sheaves of sets on a topological space". Let me try to explain to what extent Grothendieck topoi are topological in nature.

First, given a continuous map f:X->Y, it produces a geometric morphism Sh(X)->Sh(Y). If X and Y are sober, then there is a bijection between Hom(X,Y) and Hom(Sh(X),Sh(Y)) (where the later is again geometric morphisms). This means, if we restrict to sober spaces, we get a fully faithful functor Sh:SobTop->topoi. (Recall via stone duality that the category of sober spaces is equivalent to the category of locales with enough points).

More generally, if G is a topological groupoid (a groupoid object in Top), we can construct its classifying topos. This can be defined as follows: Take the enriched nerve of G to obtain a simplicial space, applying the functor "Sh" (viewing the nerve as a diagram of space) and obtain a simplicial topos. Now take the (weak) colimit of the diagram to obtain a topos BG. This topos can be described concretely as equivariant sheaves over G_0.

Geometrically, BG is a model for the topos of "small sheaves" over the topological stack associated to G. In fact, on etale topological stacks* (this include all orbifolds), we also have an equivalence between maps of stacks and geometric morphisms between their categories of sheaves, so, there is a subcategory (sub-2-category) of Grothendieck topoi which is equivalent to etale topological stacks. These Grothendieck topoi are called topological etendue.

*(over sober spaces)

It turns out that a large class of topoi can be obtained as BG for some topological groupoid. In fact, every Grothendick topos "with enough points" is equivalent to BG for some topological groupoid. The more general statement is that EVERY Grothendieck topos is equivalent to BG for some localic groupoid (a groupoid object in locales). Since locales are a model for "pointless topology", we see in some sense, every Grothendieck topos is "topological". You can make sense of the statement that every Grothendieck topos is equivalent to the category of small sheaves on a localic stack.

soft question - What are your favorite instructional counterexamples?

The Warsaw circle $W$ http://en.wikipedia.org/wiki/Continuum_%28topology%29 is a counterexample for quite a number of too naive statements.

The Warsaw circle can be defined as the subspace of the plane $R^2$ consisting of the graph of $y = sin(1/x)$, for $xin(0,1]$, the segment $[−1,1]$ in the $y$ axis, and an arc connecting $(1,sin(1))$ and $(0,0)$ (which is otherwise disjoint from the graph and the segment).

Some observations: $W$ is weakly contractible (because a map from a locally path connected space cannot ''go over the bad point'').

Let $I$ denote the segment $[−1,1]$ in the $y$ axis. Then $W/Icong S^1$ is just the usual circle, and thus we have a natural projection map $g:W to S^1$. The point-preimages of $g$ are either points or, for a single point on $S^1$, a closed interval.

Thus the assumptions of the Vietoris-Begle mapping theorem hold for $g$, proving that $g$ induces an isomorphism in Cech cohomology. Thus the Cech cohomology of $W$ is that of $S^1$, but it has the singular homology of a point, by Hurewicz.

Since $Ito W$ is an embedding of compact Hausdorff spaces, we have an induced long exact sequence in (reduced) topological $K$-theory (see, for example, Atiyah's $K$-theory Proposition 2.4.4). Since $I$ is contractible, we get that $W$ and $S^1$ also have the same topological $K$-theory.

Note that the Warsaw circle is a compact metrizable space, being a bounded closed subspace of $R^2$. By looking on points on $I$ one sees that $W$ is not locally path-connected (and, in particular, not locally contractible).

The above observations imply:

1. A map with contractible point-inverses does not need to be a weak homotopy equivalence, even if both, source and target, are compact metric spaces. Assuming that the base and the preimages are finite CW complexes does not help.




2. The Vietoris-Begle Theorem is false for singular cohomology (in particular, the wikipedia version of that Theorem is not quite correct).




3. The embedding $Ito W$ cannot be a cofibration in any model structure on $Top$, where the weak equivalences are the weak homotopy equivalences and the interval $I$ is cofibrant. Because then we would have a cofiber sequence $Ito Wto S^1$ and thus also a long exact sequence in singular cohomology.




4. $W$ does not have the homotopy type of a CW complex (since it is not contractible).




5. Even though the map $g$ is trivial on fundamental groups, it does not lift to the universal cover $p: mathbb{R} to S^1$, because $g$ cannot be nullhomotopic. Thus the assumption of local path connectivity in the lifting theorem is necessary.

teaching - "Homotopy-first" courses in algebraic topology

I was a heavily involved TA for such a graduate course in 2006 at UC Berkeley.

We started with a little bit of point-set topology introducing the category of compactly generated spaces. Then we moved into homotopy theory proper. We covered CW-complexes and all the fundamental groups, Van-Kampen's Theorem, etc. From this you can prove some nice classical theorems, like the Fundamental Theorem of Algebra, the Brauwer Fixed Point Theorem, the Borsuk-Ulam Theorem, and that $R^n neq R^m$ for $n neq m$. I felt like this part of the course went fairly well and is sufficiently geometric to be suitable for a first level graduate course (you can draw lots of pictures!).

At this point you can take the course in a couple different directions which all seem to have their own disadvantages and problems. The main problem is lack of time. A very natural direction is to discuss obstruction theory, since it is based off of the same ideas and constructions covered so far. However this is not really possible since the students haven't seen homology or cohomology at this point!

Instead, for a bit we discussed the long exact sequences you get from fibrations and cofibrations. You could then try to lead into the definition of cohomology as homotopy classes of maps into a $K(A,n)$. But this definition is fairly abstract and doesn't show one of the main feature of homology/cohomology: It is extremely computable. Still, I could imagine a course trying to develop homology and cohomology from this point of view and leading into CW homology and the Eilenberg-Steenrod axioms.

Another direction you can go is into the theory of fiber bundles (this is what we tried). The part on covering space theory works fairly well and you have all the tools at your disposal. However when you want to do general fiber bundle theory it can be difficult. A natural goal is the construction of classifying spaces and Brown's representability theorem.
The problem is that the homotopy invariance of fiber bundles is non-trivial to prove. You should expect to have to spend fair amount of time on this. It is really more suited for a second course on algebraic topology.

The main problem with all of these approaches is that it is difficult to cover the homotopy theory section and still have enough time to cover homology/cohomology properly. You know this has to be the case since it is hard to do the reverse: cover homology and cohomology, and still have enough time to cover homotopy theory properly.

What this means is that you'll be in the slightly distasteful situation of having bunch of students who have taken a first course on algebraic topology, but don't really know about homology or cohomology. This is fine if you know that these students will be taking a second semester of algebraic topology. Then any gaps can be fixed. However, in my experience this is not a realistic expectation. As you well know, you will typically have some students who end up not being interested in algebraic topology and go into analysis or algebraic geometry or some such. Or you might have some students who are second or third year students in other math fields and are taking your course to learn more about homology and cohomology. They would be done a particular disservice by a course focusing on "homotopy first".

ag.algebraic geometry - Is there an example of a variety over the complex numbers with no embedding into a smooth variety?

A really neat well known example is as follows:

Choose a conic $C_1$ and a tangential line $C_2$ in $mathbb{P}^2$ and asssociate to a point $P$ on $C_1$ the point of intersection $Q$ of $C_2$ and the tangent line to $C_1$ at $P$. This gives a birational isomorphism from $C_1$ to $C_2$. Identify the curves by this map to get the quotient variety $phi:mathbb{P}^2rightarrow{X}$ with $C:=phi(C_1)$. Now if there was an embedding of $X$ in a smooth scheme then, there would surely exist an effective line bundle on $X$, say $L$ whose pull back to $mathbb{P}^2$ will obviously be effective. Let us see why this is a contradiction. Let $L'$ be the pullback of $L$ to $mathbb{P}^2$. Note that the degrees of $L'|C_1$ and $L'|C_2$ both coincide with the degree of $L|C$ and are therefore equal. But $L'congmathcal{O}(k)$ and therefore the degrees in question are $2k$ and $k$ respectively for $C_1$ and $C_2$. Therefore $k=0$ and $L'congmathcal{O}$, which is non-effective! A contradiction!

In view of VA's comment, I give a complete proof here for constructing $X$ as a scheme.

In our special case it is a trivial pushout construction: Here I am thinking of $Y$ as $C_1amalg{C_2}$, $Y'=C$ (the quotient by the birational isomorphism above), and $Z=mathbb{P}^2$, but the argument is more general provided any finite set of closed points in $Z$ is contained in an affine open set. $X$ will denote the quotient.

Claim: Suppose $j:Yrightarrow{Z}$ is a closed subscheme of a scheme $Z$, and $g:Yrightarrow{Y'}$ is a finite surjective morphism which induces monomorphism on coordinate rings. Then there is a unique commutative diagram (which I don't know how to draw here, but one visualize it easily): $Yxrightarrow{j}{Z}$, $Yxrightarrow{g}Y'$, $Y'rightarrow{X}$, $Zxrightarrow{h}X$ where $X$ is a scheme, $h$ is finite and induces monomorphisms on coordinate rings and $Y'rightarrow{X}$ is a closed immersion.

Proof: First assume that $Z$ is affine, in which case $Y,Y'$ are both affine too. Let $A,A/I,B$ be their respective coordinate rings. Then $Bsubset{A/I}$ in a natural way. Let's use $j$ again to denote the natural map $Arightarrow{A/I}$. Put $A'=j^{-1}(B)$ and $Spec(A')=X$. The claim is clear for $X$. Also if $Z$ is replaced by an open subset $U$ such that $g^{-1}g(Ucap{Y})=Ucap{Y}$, $X$ would be replaced by $U'=h(U)$ which is an open subset.

Now this guarantees the existence of $X$ once it has been shown that $Z$ can be covered by affine open subsets $U$ such that $g^{-1}g(Ucap{Y})=Ucap{Y}$. But this is obvious in our example. For our example it is also clear from the construction of $X$ that it is actually reduced and irreducible. QED.

I hope this is satisfactory.

dg.differential geometry - Diameter of a circle in an embedded Riemannian manifold

This question was inspired by an answer to the "Magic trick based on deep mathematics" question. I wanted to post it as a comment, but I ran out of characters! I'm sure there must be a collection of standard results related to this question, but I don't know where to start looking.

First, a quick definition. The diameter of a set $S in mathbb{R}^n$ is $sup{d(x, y) mid x, y in S}$.

A sheet of paper is a good physical example of a Riemannian 2-manifold with boundary, and a table is a good physical model of (a subset of) $mathbb{R}^2$. Embed the paper isometrically in $mathbb{R}^2$ by laying flat on a table.

Draw the outline of a circular cup on the paper. It seems obvious that no matter how you embed the paper in $mathbb{R}^2$, the outline of the cup will always be a metric circle, and it will always have the same diameter $D$.

Now, lift the paper into the air, embedding it isometrically in $mathbb{R}^3$. If you let the paper flop around, the outline of the cup might not be a metric circle anymore... but will it still have diameter $D$?

Finally, cut along the outline of the cup, removing an open disk from the sheet of paper. The paper now has a second boundary component, and it's no longer simply connected. The paper has also gained a surprising property: you can bend it around in midair (that is, embed it isometrically in $mathbb{R}^3$) so that the outline of the cup has diameter greater than $D$! What's the important property of the paper that we changed to make this possible?

---

Comments

I don't think you need to cut along the outline of the cup to make this work... you could probably just cut out any disk contained within the outline of the cup. So maybe simply-connectedness is the important property?

My gut tells me that if you draw two dots on the sheet of paper, the distance between the dots is maximized when the paper is flat on the table. When you bend the paper around in midair, the dots can get closer together, but they can never get farther apart. I think this is equivalent to the statement that if $delta$ is the natural distance function on the paper, $d$ is the distance function in $mathbb{R}^3$, and $F$ is an isometric embedding of the paper in $mathbb{R}^3$, $d(Fx, Fy) le delta(x, y)$ for all points $x$ and $y$ on the paper.

fa.functional analysis - Convergence of Affine Transformations

Hi,

could you perhaps specify what kind of space your transformations are acting on? Before you do that, let me try to still share some things...

If it's a Riemannian (sub)manifold, e.g., Euclidean plane, then your problem fits well within the framework of dynamical systems, in particular "discrete-time" dynamical systems as your transformations are indexed by a countable set. Depending on the transformations, you might end up having an attracting set, and the limiting operator would amount to a kind of "dynamical projection" of the entire space to that attracting set. If transformations are all equal, i.e., a_k = a_l, for all k,l, then you have an "affine, time-invariant (autonomous) system". If not, you have an "affine, time-varying (non-autonomous) system".

Most of dynamical systems will not be phrased exactly as you presented it, rather, the sequence of transformations will be generated as a solution of a difference/differential equation, especially at the introductory level. The language you are using is more common in ergodic theory, which I think is the appropriate setting for types of questions you are asking. It deals with limiting processes for (semi)groups of operators, however, the operators in question are often considered to be linear (they are composition operators on spaces of functions/distributions). Perhaps more advanced texts do generalize to non-linear operators. Additionally, reapplication of affine transformation (if it's the same one), translates to a reapplication of a linear operator + a series generated by reapplication of linear operator to the offset vector, in a handwavy way. :) I therefore believe there is hope for your problem within the context of ergodic theory. As an intro text, perhaps you'd want to look into Silva: Invitation to ergodic theory (recently published by AMS). For a more advanced treatment, you'd have to look for something more advanced, e.g., Petersen to start with, but perhaps going to Cornfeld, Fomin, Sinai (I have still to even parse through that).

Hope this helped.

ag.algebraic geometry - Criteria for a map of schemes to be an isomorphism

Let $ X, Y $ be separated finite type schemes over an algebraically closed field $ k $. Assume that $ Y $ is reduced. Let $ phi : X rightarrow Y $ be a morphism of schemes. Suppose that $ phi $ gives a bijection on $ k $-points and an injection on $ S$-points for all $k$-schemes $ S$. Prove or disprove that $ phi $ is an isomorphism (or add some extra hypotheses to ensure that $ phi $ is an isomorphism).

When $ k = mathbb{C} $, $ X $ is normal, an $ Y $ is normal and irreducible, then I have a proof which uses the following crazy fact:
If $ X $ and $ Y $ are irreducible varieties over $mathbb{C} $ and $ Y $ is normal, then a morphism $ phi $ inducing a bijection on $ mathbb{C} $-points is an isomorphism.
My original question follows from this fact via a small tweaking of the usual Yoneda argument.

If anyone can give me a proof or reference for this last fact, I would be grateful too. I read it in the appendix of Kumar's book on Kac-Moody groups.

Edit: In light of some counterexamples, let me assume that X is irreducible and Y is normal and irreducible.

ac.commutative algebra - Isomorphism between direct sum of modules

There are even counterexamples in the case $A = {mathbb Z}$: at the end of B. Jónsson’s paper “On direct decompositions of torsion-free abelian groups,” Math. Scand. 5 (1957), 230–235, an example is given of torsion-free, finite-rank abelian groups $B notcong C$ such that $B oplus B cong C oplus C$.

A further counterexample, which I believe has been pointed out independently by L. S. Levy, R. Wiegand, and R. G. Swan: let $A$ be the coordinate ring of the real 2-sphere and ${}_AM$ the module for the tangent bundle; then $M oplus M$ is free of rank $4,$ but $M$ is not free of rank $2$.

In the positive direction, K. R. Goodearl has proved (“Direct sum properties of quasi-injective modules,” Bull. Amer. Math. Soc. 82 (1976), no. 1, 108–110, Theorem 3) that if $M$ and $N$ are quasi-injective modules over a ring (commutative or not), then $M^n cong N^n$ implies $M cong N$ for any positive integer $n$.

Your question is related to an important open problem in noncommutative ring theory, the “separativity” problem for von Neumann regular rings: if $R$ is a von Neumann regular ring (or more generally an exchange ring), and $A$ and $B$ are finitely generated projective left $R$-modules with the property that $A oplus A cong A oplus B cong B oplus B$, must we have $A cong B$?  An affirmative answer would resolve several major open problems, as explained in P. Ara, K. R. Goodearl, K. C. O’Meara, and E. Pardo’s paper “Separative cancellation for projective modules over exchange rings,” Israel J. Math. 105 (1998), 105–137.

ct.category theory - Lax Functors and Equivalence of Bicategories?

First of all, for any two bicategories A and B, there is a bicategory $Fun_{x,y}(A,B)$ where x can denote either strong, lax, or oplax functors, and y can denote either strong, lax, or oplax transformations. There's no problem defining and composing lax and oplax transformations between lax or oplax functors, and the lax/oplax-ness doesn't even have to match up. It's also true that two x-functors are equivalent in one of these bicategories iff they're equivalent in any other one. That is, any lax or oplax transformation that is an equivalence is actually strong/pseudo.

Where you run into problems is when you try to compose the functors. You can compose two x-functors and get another x-functor, but in general you can't whisker a y-transformation with an x-functor unless x = strong, no matter what y is, and moreover if y isn't strong, then the interchange law fails. Thus you only get a tricategory with homs $Fun_{x,y}(A,B)$ if x=y=strong. (In particular, I think this means that there isn't a good notion of "equivalence of bicategories" involving lax functors.)

For a fixed strong functor $Fcolon Ato A'$, you can compose and whisker with it to get a functor $Fun_{x,y}(A',B) to Fun_{x,y}(A,B)$ for any x and y. However, the same is not true for transformations $Fto F'$, and the answer to your question is (perhaps surprisingly) no! The two bicategories are not equivalent.

Consider, for instance, A the terminal bicategory (one object, one 1-morphism, one 2-morphism) and A' the free-living isomorphism, considered as a bicategory with only identity 2-cells. The obvious functor $A' to A$ is an equivalence. However, a lax functor from A to B is a monad in B, and a lax functor from A' to B consists of two monads and a pair of suitably related "bimodules". If some lax functor out of A' is equivalent to one induced by composition from A (remember that "equivalence" doesn't depend on the type of transfomation), then in particular the two monads would be equivalent in B, and hence so would their underlying objects. But any adjunction in B whose unit is an isomorphism gives rise to a lax functor out of A', if we take the monads to be identity 1-morphisms, the bimodules to be the left and right adjoint, and the bimodule structure maps to be the counit and the inverse of the unit. And of course can have adjunctions between inequivalent objects.

By the way, I think your meaning of "equivalence" for bicategories is becoming more standard. In traditional literature this sort of equivalence was called a "biequivalence," because for strict 2-categories there are stricter sorts of equivalence, where you require either the functors to be strict, or the two composites to be isomorphic to identities rather than merely equivalent to them, or both. These stricter notions don't really make much sense for bicategories, though. For instance, in a general bicategory, even identity 1-morphisms are not isomorphisms, so if "equivalence" were to demand that FG be isomorphic to the identity, a general bicategory wouldn't even be equivalent to itself!

sheaf theory - Equivalence of ordered and unordered cech cohomology.

I don't know if this is in SGA IV.5, but that's a good place to look for questions about Cech cohomology.

As I described here, the Cech cohomology with respect to a cover is the same as the sheaf cohomology in the sieve associated to that cover. If $mathcal{U}$ is a cover of $X$, let $R$ be the category whose objects are maps $V rightarrow X$ that factor through some object in $mathcal{U}$. Then

$check{H}^p(mathcal{U}, F) = varprojlim^{(p)}_{R} F = Ext^p(mathbf{Z}_R, F)$

where $varprojlim^{(p)}$ is the $p$-th derived functor $varprojlim$. This can be calculated by taking a projective resolution of $mathbf{Z}_R$. Here are two ways to do it:

$displaystyle K_p = sum_{i_1 < i_2 < cdots < i_p} mathbf{Z}_{U_{i_1} cap cdots cap U_{i_p}}$

$displaystyle L_p = sum_{i_1, ldots, i_p} mathbf{Z}_{U_{i_1} cap cdots cap U_{i_p}}$.`

One must check, of course, that these are indeed resolutions. (I don't have a slick explanation of why they are resolutions. The best I can do is to say that these complexes are associated via the Dold--Kan correspondence to simplicial resolutions of the final presheaf on $R$.) Taking $Hom$ into $F$ yields the two Cech complexes in question.

tag removed - Algorithms for laying out directed graphs?

If the x-coordinates are compatible with the acyclic structure of your DAG (that is, for an edge u->v, the x coordinate of u should always be less than that of v) then this is a standard problem in graph drawing, known as Sugiyama-style layered drawing. (Usually it is the y coordinates that are fixed but that makes no difference.) Some versions of the problem (e.g. finding the exact minimum number of edge crossings) can be NP-hard but effective heuristics are known. See e.g. chapter 9 of Di Battista, Tamassia, and Eades, "Graph Drawing: Algorithms for the Visualization of Graphs", Prentice-Hall, 1999.

Searching Google scholar for "layered graph drawing" should also turn up some more recent references, but be careful that some of them (including mine) which use the term to mean something unrelated.

fa.functional analysis - Regular borel measures on metric spaces

When teaching Measure Theory last year, I convinced myself that a finite measure defined on the Borel subsets of a (compact; separable complete?) metric space was automatically regular. I used the Borel Hierarchy and some transfinite induction. But, typically, I've lost the details.

So: is this true? Are related questions true? What are some good sources for this sort of questions? As motivation, a student pointed me to http://en.wikipedia.org/wiki/Lp_space#Dense_subspaces where it's claimed (without reference) that (up to a slight change of definition) the result is true for finite Borel measures on any metric space.

(I'm normally only interested in Locally Compact Hausdorff spaces, for which, e.g. Rudin's "Real and Complex Analysis" answers such questions to my satisfaction. But here I'm asking more about metric spaces).

To clarify, some definitions (thanks Bill!):

• I guess by "Borel" I mean: the sigma-algebra generated by the open sets.


• A measure $mu$ is "outer regular" if $mu(B) = inf{mu(U) : Bsubseteq U text{ is open}}$ for any Borel B.


• A measure $mu$ is "inner regular" if $mu(B) = sup{mu(K) : Bsupseteq K text{ is compact}}$ for any Borel B.


• A measure $mu$ is "Radon" if it's inner regular and locally finite (that is, all points have a neighbourhood of finite measure).

So I don't think I'm quite interested in Radon measures (well, I am, but that doesn't completely answer my question): in particular, the original link to Wikipedia (about L^p spaces) seems to claim that any finite Borel measure on a metric space is automatically outer regular, and inner regular in the weaker sense with K being only closed.

soft question - japanese/chinese for mathematicians?

This may not be the answer you're looking for, but I thought I'd share my experience as someone who was born in Japan but was transplanted quickly into the United States. My Japanese is not nearly as good as it should be, but is certainly good enough to read math.

A beautiful part of reading Japanese/Chinese math is that you can grasp the meaning without knowing how to pronounce anything. I don't know any technical Chinese, but in Japanese,

写像

is the word for "mapping" or "function", and the literal meaning of its characters hints at this. Let me explain.

The first character means to transcribe, to picture, or to give a visual form--poetically, it can mean to simply give an abstract form to something, rather than a visual one. (For instance, the word 写真 means photograph, where the second character in this particular word means "truth". It might be silly to think the word for photograph is to "picture something truly/in its true form", but that's a beautiful translation to ponder on another occasion.)

The final character in 写像 means figure, or image, or an embodiment. For instance, the word 画像 means "image" in the computer sense of file type. In fact the character 像 alone can mean "image" in the sense of mathematics, as in the image of something under a map.

In short, the word for "function" or "map" can be literally and clumsily translated back into English as "forming an image" or "creating a figure" or "realizing a form", most abstractly. I doubt any Japanese person ever thinks in these terms, no more than we think of the word "projection" deeply in terms of its Latin roots. But to harzard a guess at the meanings of these words can be a beautiful experience, and one unique to those weirdos who know the meanings of things without knowing how to say them.

So it may be a really interesting experience to simply learn the meaning of each commonly occurring (math) character---I'll list a few below---and to get a feel for the mathematical meanings of their combinations via intuition. When I've read Japanese math books, the feeling of knowing the meaning on a page without knowing how to pronounce a word has been the most rewarding and beautiful part. If you choose to do this, the best tip I have is to simply write: Make sure you copy and write the characters over and over again, so you begin to distinguish subtle differences between them.

For the enjoyment of some, here are examples of Japanese math words and the meanings of their constituent characters. I'll list some irrelevant meanings of some characters--though characters often only take on one of many meanings based on context, I still think it's fun to know their other possible meanings.

空間 (space)

空 = sky, emptiness, space, air

間 = between, the space between, an interval of time

位相 (topology)

位 = rank (as in seniority or importance in an organization), a word for counting dead souls, decimal place, position. As a verb, it can mean to locate--i.e., to determine the position of.

相 = form, shape, appearance, the relationship of one thing to another.

Strangely enough, 位相　can also mean the phase of something, as in the angle or phase of a complex number or a wave. It also mean the phase of something as in "solid/liquid/gaseous". I would assume that the term first came to use to describe the states of matter, was tangentially used to describe the phase of wave-like phenomena since the English term "phase" was used in both instances.

微分 (derivative, to take the derivative of)

微 = infinitesimal, tiny, slight

分 = to divide, an amount of something.

In learning language so much emphasis is placed on the sounds of things, rather than on the abstract units of meaning. I suppose Chinese characters were developed exactly to avoid this aural emphasis, but it is always a joy to have zero verbal understanding with a Chinese or Korean person, but to be able to communicate by writing characters in the air.

Well, perhaps this was not helpful in the least, but maybe it will at least entertain some non-Japanese-speakers. (By the way, I'd be very curious to hear if the Chinese technical terms are the same, as almost all technical terms in Japanese utilize kanji, or Chinese characters.)

gt.geometric topology - Periodic mapping classes of the genus two orientable surface

In the paper listed below there is a calculation of all the finite group actions on a genus 2 surface. There are 20 of them, with the groups ranging from order 2 to order 48. Nine of the actions are of cyclic groups, of orders 2,2,3,4,5,6,6,8,10 respectively. The paper also does the genus 3 case. The techniques are mostly algebraic. It is an interesting exercise to try to find nice geometric pictures of all the actions.

S.A.Broughton, Classifying finite group actions on surfaces of low genus, J.Pure Appl.Alg. 69 (1991), 233-270.

problem solving - Math Puzzle: calculating the dimensions of variable rectangles in a fixed square

I've got the following problem,

I've got a fixed size square and within there are a fixed number of rectangles to be contained within it. I want the rectangles to cover the maximum amount of space within the square. The size of the rectangles is determined by a weighting. The higher the value of an individual weight of a rectangle, the bigger its surface area.

Assuming I have a predetermined weighting for all 9 rectangles, how do I derive at the coordinates (for their position in the square) and their dimensions (width and length).

this has had me puzzled, hope someone can help me out...thanks!!

lo.logic - A few questions on model theory, especially model theory of rings

"The complex numbers are easy to deal with, whereas for the integers it is much harder..."

What you might have heard about is that the complete first-order theory of the complex numbers with just the ring operations -- which is the same as the theory of algebraically closed fields of characteristic zero -- is decidable (i.e. "computable"), as was first proved by Tarski. So in principle one could write a computer program where you could input any first-order sentence in the language of rings, and in a finite amount of time it would tell you whether or not this sentence is true in the complex numbers.

However, if you look at the ring of integers, no such thing is true; the first-order theory of the integers is undecidable. This is a theorem of Alonzo Church, and is closely related to Goedel's famous incompleteness theorem.

The negative answer to Hilbert's Tenth Problem is a different issue -- this doesn't follow immediately from Church's Theorem, and was proved much later, by Davis, Putnam, Julia Robinson, and Matiyasevich.

I think of these things as more "logical folklore" than model theory, per se -- any reasonable introductory book on mathematical logic (e.g. Enderton's A Mathematical Introduction to Logic) will have a lot to say about them.

soft question - Best online mathematics videos?

My good friend Professor Elvis Zap has the "Calculus Rap," the "Quantum Gravity Topological Quantum Field Theory Blues," a vid on constructing "Boy's Surface," "Drawing the hypercube (yes he knows there is a line missing in part 1)," A few things on quandles, and a bunch of precalculus and calculus videos. In order to embarrass all involved, he posted the series "Dehn's Dilemma" that was recorded in Italy last summer.

gauge theory - Casson's invariant and the trivial connection contribution to witten's 3-manifold invariant

This question might turn out not to make any sense, but here it is: Witten's (and Reshetikhin and Turaev's) 3-manifold invariant can be "defined" as an integral over the space of connections on the trivial SU(2) bundle over the 3-manifold M, modulo gauge transformation. In the stationary phase approximation as the level k-->infinity, we write this integral as the sum of functions f_i(k), where the sum is over the set {c_i} of critical points of the chern-simons functional (which is basically the log of the integrand). The critical points are the flat connections. In his analysis of the function f_tr(k) corresponding to the trivial flat connection (product connection), Rozansky (as well as previous authors, I believe) found the Casson invariant (as the second coefficient if you write f_i as a power series in 1/k, or something like that).

Now, at first glance it strikes me as odd that the casson invariant shows up here, because the casson invariant can, by definition, be thought of as a (signed) sum over ALL flat connections, EXCEPT the trivial connection. So my question is: what gives? Why does the contribution from the trivial connection give the casson invariant? Why isn't it instead, say, the sum of the 2nd coefficients of the contributions from all other flat connections? is there some explanation for this? Have contributions from other connections, say reducible connections for a homology 3-sphere, been analyzed? Does the casson invariant show up there as well?

If you've gotten this far, here's another question: right now, i can write |H_1(M)|*(Casson-Walker invariant of M), say for lens spaces M=L(p,q) as the sum over certain INTEGERS, one per conjugacy class of reducible flat connection on SU(2)XM, PLUS the signature of the 2-bridge knot b(p,q). I'm trying to identify this decomposition of the casson invariant with others in the literature, of which there are many, but I'm having the problem that none of these are (always) integral decompositions, even though they ADD UP to an integer. Well this is a long shot, but any ideas would be appreciated!

ag.algebraic geometry - Arbitrary products of schemes don't exist, do they?

Let me rephrase the question (and Ilya's answer). Given an arbitrary collection $X_i$ of schemes, is the functor (on affine schemes, say)

$Y mapsto prod_i Hom(Y, X_i)$

representable by a scheme? If the $X_i$ are all affine, the answer is yes, as explained in the statement of the question. More generally, any filtered inverse system of schemes with essentially affine transition maps has an inverse limit in the category of schemes (this is in EGA IV.8). The topology in that case is the inverse limit topology, by the way.

It is easy to come up with examples of infinite products of non-separated schemes that are not representable by schemes. This is because any scheme has a locally closed diagonal. In other words, if $Y rightrightarrows Z$ is a pair of maps of schemes then the locus in $Y$ where the two maps coincide is locally closed in $Y$.

Suppose $Z$ is the affine line with a doubled origin. Every distinguished open subset of an affine scheme $Y$ occurs as the locus where two maps $Y rightrightarrows Z$ agree. Let $X = prod_{i = 1}^infty Z$. Every countable intersection of distinguished open subsets of $Y$ occurs as the locus where two maps $Y rightarrow X$ agree. Not every countable intersection of open subsets is locally closed, however, so $X$ cannot be a scheme.

Since the diagonal of an infinite product of separated schemes is closed, a more interesting question is whether an infinite product of separated schemes can be representable by a scheme. Ilya's example demonstrates that the answer is no.

Let $Z = mathbf{A}^2 - 0$. This represents the functor that sends $Spec A$ to the set of pairs $(x,y) in A^2$ generating the unit ideal. The infinite product $X = prod_{i = 1}^infty Z$ represents the functor sending $A$ to the set of infinite collections of pairs $(x_i, y_i)$ generating the unit ideal. Let $B$ be the ring $mathbf{Z}[x_i, y_i, a_i, b_i]_{i = 1}^infty / (a_i x_i + b_i y_i = 1)$. There is an obvious map $Spec B rightarrow X$. Any (nonempty) open subfunctor $U$ of $X$ determines an open subfunctor of $Spec B$, and this must contain a distinguished open subset defined by the invertibility of some $f in B$. Since $f$ can involve at most finitely many of the variables, the open subset determined by $f$ must contain the pre-image of some open subset $U'$ in $prod_{i in I} Z$ for some finite set $I$. Let $I'$ be the complement of $I$. If we choose a closed point $t$ of $U'$ then $U$ contains the pre-image of $t$ as a closed subfunctor. Since the pre-image of $t$ is $prod_{i in I'} Z cong X$ this shows that any open subfunctor of $X$ contains $X$ as a closed subfunctor.

In particular, if $X$ is a scheme, any non-empty open affine contains a scheme isomorphic to $X$ as a closed subscheme. A closed subscheme of an affine scheme is affine, so if $X$ is a scheme it is affine.

Now we just have to show $X$ is not an affine scheme. It is a subfunctor of $W = prod_{i = 1}^infty mathbf{A}^2$, so if $X$ is an affine scheme, it is locally closed in $W$. Since $X$ is not contained in any closed subset of $W$ except $W$ itself, this means that $X$ is open in $W$. But then $X$ can be defined in $W$ using only finitely many of the variables, which is impossible.

Edit: Laurent Moret-Bailly pointed out in the comments below that my argument above for this last point doesn't make sense. Here is a revision: Suppose to the contrary that $X$ is an affine scheme. Then the morphism $p : X rightarrow X$ that projects off a single factor is an affine morphism. If we restrict this map to a closed fiber then we recover the projection from $Z$ to a point, which is certainly not affine. Therefore $X$ could not have been affine in the first place.

mathematics education - How should one present curl and divergence in an undergraduate multivariable calculus class?

I have taught multivariable calculus exactly once, to engineering students at Concordia University in Montreal. I found the course to be replete with expository challenges like the one you mention: namely, to explain what is going on with the various concepts of vector analysis in something like geometric terms, but of course without introducing anything like differential forms. [Conversely, it is possible to know Stokes' Theorem in the form $int_{partial M} omega = int_M d omega$ and still not have any insight into flux, divergence and other such geometric and physical notions. I myself spent about 10 years in this position.]

I thought hard and often found explanations that were much more satisfactory than the textbook, which was amazingly laconic. Or rather, I found explanations which were much more satisfactory to me. The students had a lot of trouble conceptualizing the material, to the extent that my lectures almost certainly would have been more successful if I hadn't tried to give geometric explanations and intuition but simply concentrated on the problems. Thus Gerald Edgar's comment rings true to me. But let me proceed on the happier premise that you want to give more motivation to the bright student who approaches you outside of class.

One thing which was useful for me was to read the "physical explanations" that the book sometimes gave and try to make some kind of mathematical sense of them. For context, I should say that I have never taken any physics classes at the university level and that I have rarely if ever met a mathematician who has less physics background than I. Moreover, when I took introductory multivariable calculus myself (around the age of 17), I found the physical explanations to be so vague and so far away from the mathematics as to be laughable. For instance, the geometric intuition for a curl involved some story about a paddlewheel.

So when I taught the class, I tried to make some mathematical sense out of the names "incompressible" (zero divergence) and "irrotational" (zero curl), and to my surprise and delight I found that this was actually rather straightforward once I stopped to think about it.

Let me also tell you about my one "innovation" in the course (I am sure it will be commonplace to many of the mathematicians here). It seems strange that there are two versions of Stokes' theorem in three-dimensional space (one of them is called Stokes' theorem and one of them is called Gauss' Theorem or -- better! -- the Divergence Theorem) whereas in the plane there is only Green's Theorem. Stokes' Theorem is about curl, whereas Gauss' theorem is about divergence. What about Green's Theorem?

The answer is that Green's theorem has a version for divergence -- i.e., a flux version involving normal line integrals -- and a version for curl -- a circulation version involving tangent line integrals -- but these two versions are formally equivalent. Indeed, one gets from one to the other by applying the "turning" operators L and R: L applied to a planar vector field rotates each vector 90 degrees counterclockwise, and R is the inverse operator. Then (with the convention that the curl of a planar vector field should always be pointing in the vertical direction, so we can make a scalar function out of it)

curl(L(F)) = div(F)

and by making this formal substitution one gets from one version of Green's Theorem to the other.

See

http://math.uga.edu/~pete/handoutfive.pdf

http://math.uga.edu/~pete/handouteight.pdf

http://math.uga.edu/~pete/reviewnotes.pdf

spectral sequences - Torsion in K-theory versus torsion in cohomology

Inspired by this question, I wonder if anyone can provide an example of a finite CW complex X for which the order of the torsion subgroup of $H^{even} (X; mathbb{Z}) = bigoplus_{k=0}^infty H^{2k} (X; mathbb{Z})$ differs from the order of the torsion subgroup of $K^0 (X)$, where $K^0$ is complex topological K-theory. This is the same as asking for a non-zero differential in the Atiyah-Hirzebruch spectral sequence for some finite CW complex X, since this spectral sequence always collapses rationally.

Even better, is there an example in which X is a manifold? An orientable manifold?

Tom Goodwillie's answer to the question referenced above gave examples (real projective spaces) where the torsion subgroups are not isomorphic, but do have the same order.

It's interesting to note that the exponent of the images of these differentials is bounded by a universal constant, depending only on the starting page of the differential! This is a theorem of D. Arrlettaz (K-theory, 6: 347-361, 1992). You can even change the underlying spectrum (complex K-theory) without affecting the constant.

ag.algebraic geometry - Normal bundle of a quotient map

I would like to know if there is a notion of normal bundle to a quotient map. In one specific case, let $G$ be a finite group acting on an algebraic variety $V$. Denote by the map $pi:V rightarrow V/G$ the quotient map. Is it possible to define the normal bundle to $pi$ having good properties. For example, I would like to get the projection formula. If $V/G$ is a substack of $X$ of co-dimension $r,$ do we get a vector bundle of rank $r$ on $V$ as the normal bundle to the composition $V rightarrow V/G rightarrow X$?
I am interested in the case where $V/G$ is a Deligne-Mumford stack.

I would like to know the conditions under which the equality
$$i^*(i_*(x))=x.c_r(N)$$ for $xin A^*(V)$ holds.

Here, $N$ is the desired normal bundle. What is the condition under which the Chern classes of this bundle exist?

The question may not be stated correctly and clearly. I apologize about it.

nt.number theory - Lower bounds on zeta(s+it) for fixed s

Yes, such conditional results are covered in Chapter 14 of the standard reference - the second edition of The Theory of the Riemann Zeta-Function by E. C. Titchmarsh. This edition has end-of-chapter notes by D. R. Heath-Brown bringing it up to date as of 1986.

In particular a lower bound
$$
|zeta(3/4 + it)| gg e^{-csqrt{log(t)}/loglog(t)}
$$
holds with some $c > 0$, conditionally on RH. See page 384 of the cited reference.

Such results are only known unconditionally for a region to the left of the line $sigma = 1$ that
narrows to zero width as $t rightarrow {pm}infty$. Not coincidentally, the best zero-free region known is also of this form. See page 135 of the cited reference for the best result of this kind.

ct.category theory - Higher-order, multi-sorted, non purely equational version of universal algebra ?

I have been looking around, unsuccessfully, for generalizations of universal algebra based on higher-order logic (rather than first order) and where the relations are not purely equational. Motivation: I need a "theory of syntax" for presentations of higher-order, non-equational theories. Furthermore, I want to be able to specify 'combinators' over these presentations, rigorously.

I am aware of Lawvere theories, but these are still equational (and neither particularly higher-order, though the multi-sorted generalization seems straightforward enough). There is a beginning of model theory done in a logical independent way, i.e. model theory over an institution; but that seems to concentrate on the model-theoretic aspects, rather than the universal algebra aspects. Perhaps what I am looking for are sketches?

[Edit:] From the various answer below, it seems I should be asking the question "how can I view type theory as a theory of syntax"? Somehow, that seems like an 'implementation' (as it requires a fair bit of 'encoding'); for example, to express the 'theory of categories' [i.e. (Obj, Mor, id, src, trg, $circ$) and 5-6 axioms, I need a dependent record. Plus what is a sort (and sort constructor for Mor), what is an operation, and what is in Prop? Universal algebra cleanly separates these.

A good question was asked: what theorems do I want? Well, whatever operations I make on theories, well-formedness of the results will require discharging some obligations -- these obligations should all be finitely expressible (and automatically well-formed). Furthermore, the resulting syntactic objects and their morphisms should form a finitely co-complete category. Note that I expect that deciding if a given (presentation of a ) theory has a model to be undecidable.

ac.commutative algebra - The inverse limit of locally free module

The answer is yes.

1) $A$ is $I$-adic complete implies that $I subset rad(A)$, the intersection of all maximal primes. Indeed, pick any $a in I$. Look at $1-a + a^2 -a^3 ... in A $ (this is where we use completeness). The inverse of this is $1+a$, so $1+a$ is an unit. Since this is true for any $ain I$, we have $Isubset rad(A)$ (see Section 1 Matsumura).

2) It suffices to prove that $M$ is free at any maximal ideal $m$ of $A$. Since $I$ is inside $m$, we may as well replace $A$ by $A_m$ and assume $A$ is local. By assumption then $M/I cong (A/I)^l$, Nakayama Lemma shows that $l$ is the mimimum number of generators of $M$.

Look at the beginning of a minimal resolution of M: $ N to F=A^l to M $.The last map become isomorphism when tensoring with $A/I^n$, so the first map has to become $0$. This means that $N subset I^nF$ for all $n>0$. This forces $N=0$ (use Artin-Rees lemma), thus $M cong F$.

nt.number theory - How does the order of a pole of a zeta function indicate any geometric information?

This is an expository note filling in the background between Steven Sam's comment and Felipe Voloch's answer.

If $X$ is a smooth projective variety, then the Weil conjectures (now theorems) describe the zeroes and poles of the zeta function in terms of the cohomology of $X$, and the action of Frobenius on it. In particular, the poles on the circle $|u|=1/q$ are the reciprocals of the eigenvalues of Frobenius acting on $H^2(X, mathbb{Q}_{ell})$.

In your example, $H^2$ is two dimensional. Over the algebraic closure $overline{F_q}$, your variety is isomorphic to $mathbb{P}^1 times mathbb{P}^1$. $H^2$ is spanned by the two classes $mathbb{P}^1 times { mbox{point} }$ and $ { mbox{point} } times mathbb{P}^1$.

If $-1$ is a square, Frobenius acts on this two dimensional vector space by multiplication by $q$, so you get a double pole at $1/q$. If $-1$ is not a square, then Frobenius multiplies by $q$ and switches the two generators. So the eigenvalues are $q$ and $-q$.

complex geometry - Holomorphic vector fields acting on Dolbeault cohomology

The question.

Let $(X, J)$ be a complex manifold and $u$ a holomorphic vector field, i.e. $L_uJ = 0$. The holomorphicity of $u$ implies that the Lie derivative $L_u$ on forms preserves the (p,q) decomposition and also that it commutes with $bar{partial}$. From this it follows that $u$ acts infinitesimally on the Dolbeault cohomology groups $H^{p,q}(X)$ of $X$. My question is, does anyone know of an example in which this action is non-trivial?

Some context.

To give some context, first note that the analgous action for de Rham cohomology is always trivial: If $M$ is any smooth manifold and $v$ any vector field, then the formula $L_v = d circ i_v + i_v circ d$ shows that the infinitesimal action of $v$ on de Rham cohmology is trivial. (This is an instance of the more general fact that homotopic maps induce the same homomorphisms on singular cohomology. The field $v$ generates diffeomorphisms which are by construction isotopic to the identity map.)

Returning to Dolbeault cohomology, suppose we know that each Dolbeault class is represented by a $d$-closed form. (For example, this is true if $X$ is a compact Kähler manifold, by Hodge theory.) Then the action is necessarily trivial. The proof is as follows. Let $alpha$ be a $bar{partial}$-closed (p,q)-form which is also $d$-closed. Then we know that $L_u alpha = d(i_u alpha)$ is also of type (p,q). So,
$$
L_ualpha = bar{partial}left((i_ualpha)^{p, q-1}right) + partialleft((i_u alpha)^{p-1, q}right)
$$
and the other contributions $bar{partial}((i_ualpha)^{p-1,q}$) and $partial((i_ualpha)^{p,q-1})$ vanish. Now the fact that $barpartial((i_ualpha)^{p-1,q}) = 0$ and our hypothesis imply that there is a (p-1, q-1)-form $beta$ such that
$$
(i_ualpha)^{p-1,q}+ barpartial beta
$$
is closed. Hence
$$
partial left((i_ualpha)^{p-1,q}right) = barpartial partial beta
$$
and so
$$
L_ualpha = bar partial left( (i_u alpha)^{p,q-1} + partial betaright)
$$
which proves the action of $u$ on $H^{p,q}(X)$ is trivial.

ac.commutative algebra - First-order UFD (factorial ring) condition / pre-Schreier rings

All rings in this post are commutative and with $1$.

Everyone knows the definition of a factorial ring, a. k. a. unique factorization domain (UFD). I have been wondering about some variations regarding this notion.

(a) A ring $R$ is called pre-pre-Schreier (this is my nomenclature) if and only if for any four elements $a$, $b$, $c$, $d$ of $R$ satisfying $ab = cd$, we can find four elements $x$, $y$, $z$, $w$ in $R$ such that $a = xy$, $b = zw$, $c = xz$, $d = yw$.

(b) A ring $R$ is called pre-Schreier if it is pre-pre-Schreier and an integral domain. (This is not my nomenclature.)

It is easy to see that a Noetherian ring is a UFD if and only if it is pre-Schreier; on the other hand, the condition on a ring to be pre-Schreier is a first-order logic formula (if I'm right; I'm not an expert in logic). This was actually my motivation to consider pre-Schreier rings: to first-orderize the UFD condition. (Is there a first-order logic formula that is equivalent to UFD always, no matter whether the ring is Noetherian or not?) As for pre-pre-Schreier rings, I was just trying to see what happens if we leave out the domain condition.

According to this paper (Remark 4.6. (1)), the polynomial ring $Rleft[Xright]$ over a pre-Schreier ring $R$ doesn't have to be pre-Schreier. My questions are now:

(1) If a Noetherian ring $R$ is pre-pre-Schreier, then what can be said about $Rleft[Xright]$ ?

(2) Can a pre-pre-Schreier ring contain nilpotents $neq 0$ ? I used to think I have proven that it can't if it is Noetherian, but now I see flaws in my argument.

gr.group theory - A ring of invariants in characteristic 2

It seems to me that the characteristic of the field does not play a big role in this question: here is a sketch of an argument.

Note first of all, that all the invariants are linear combinations of "symmetrized monomials": if m is a monomial in the polynomial ring, then form the sum of all the translates of m by the elements of your group. This means that every invariant in the polynomial ring comes from an invariant polynomial with coefficients in the prime field $mathbb{F}_2$ of K and that invariants with coefficients in $mathbb{F}_2$ are the reduction of invariants with integer coefficients. Thus we have translated the question over characteristic two to a question over the integers: it suffices to find generators and relations for the ring of invariants of your group over the integers to find generators and relations in any ring.

Over the integers I do not know what the answer is, but if you do know what the answer is over any field of characteristic different from two, maybe you can now fill in the argument. Thinking briefly about the set of generators, it seems like you might simply need the "symmetrized square-free monomials", with relations that are a bit tedious to write down, but that maybe can be nicely interpreted.

EDIT: The square-free monomials are not enough, but it seems that you do not have to look much further to describe explicitly a finite set of generators for the group algebra of a finite cyclic group over the integers. Indeed, let S be the set of monomials m for which there exists an integer r such that the exponents of m are the integers ${0,1,ldots,r}$. Then the product of all the variables of the polynomial ring together with the symmetrizations of all the monomials in S seems to generate the ring of invariants. "Symmetrize a monomial m" means sum over the cosets of the stabilizer in the cyclic group of m. I have not thought about relations, but there will be plenty!

If you really need to make explicit the fact that the ring of invariants is not Cohen-Macaulay, maybe you can, but maybe you do not need to do that...

nt.number theory - Euler and the Four-Squares Theorem

The post below is jointly by Rainer Dietmann and Christian Elsholtz.
We had worked on the problem since a while and had an independent asymptotic solution
to Euler's problem.
Our argument is possibly easier, but in it's current form
it does does not achieve the correct order of magnitude of the number of solutions. This seems to be a very nice feature of Lucia's approach!

We had intended to make the argument entirely explicit
in order to prove the statement for all $n$, not only for sufficienly large $n$. (See also the comments
after the second argument below).

We also had intended to prepare these results for publication.

Lucia, we would appreciate if you could contact us by email, the email
adresses (RD in Royal Holloway and CE in Graz) are easy to find.

$textbf{Theorem:}$
Let $n$ be a sufficiently large positive integer with
$n equiv 2 pmod 4$. Then $n$ can be written as the sum of two
positive integers, none of them having any prime factor
$p$ with $p equiv 3 pmod 4$.

This asymptotically answers a question of Euler.
Important partial results are due to R.D. James (TAMS 43 (1938), 296--302)
who proved the ternary case and an approximation to the binary case. Indeed the ternary case allows for an elementary proof,
based on Gaus' theorem on the sum of three triangular numbers: Any integer $k$
can be written as $k = frac{x(x-1)}{2}+ frac{y(y-1)}{2} + frac{z(z-1)}{2}$ and therefore
$$ 4k + 3 = (x^2 + (x - 1)^2) + (y^2 + (y - 1)^2) + (z^2 + (z - 1)^2).$$
Observe that $ (x^2 + (x - 1)^2)$ is a sum of two adjacent squares, and thus cannot be divisible by any prime $ p = 3 mod 4$. (Recall here and for later reference the following $textbf{Fact:}$ if $p|n = s^2 +t^2$, with $p = 3 mod 4$ prime, then $p|s$ and $p|t$.)

Using a well known result of the late George Greaves, one gets a short
proof of the Theorem.

$textbf{Proof.}$
By a result of Greaves (Acta Arith 29 (1976), 257--274),
each sufficiently large positive integer $n$ with
$n equiv 2 pmod 4$ can be written in the form
$$ n = p^2+q^2+x^2+y^2
$$
for rational primes $p, q$ and integers $x, y$, and the number of such representations is at least of order of magnitude
$n (log n)^{-5/2}$. We write $a=p^2+x^2$ and $b=q^2+y^2$
and take multiplicities into account:
namely the number of representations $r_2(a)$ of
$a$ as a sum of two squares is $r_2(a) leq d(a)ll a^{varepsilon}ll n^{varepsilon}$.
The same holds for $r_2(b)$. We therefore find that there are
at least
$$
n^{1-2varepsilon}
$$
many tuples $(a, b)$ with positive
integers $a, b$, such that $n=a+b$ and both $a$ and $b$ are the sum
of the square of a prime and the square of an integer.
Now suppose that $w$ is a prime with $w equiv 3 pmod 4$ and
$w$ divides $a=p^2+x^2$, say. Then by the `fact' above and
as $p$ is prime this implies that $p=w$ and $x$ is
divisible by $w$. Therefore, at most $O(1+n^{1/2}/w)$ many
$a$ can be divisible by $w$, and for any such $a$ there will be only
one corresponding $b$ since $a+b=n$. The same argument applies if
$w$ divides $b$. Moreover, clearly $w$ can be at most $n^{1/2}$. Summing over all such $w$
we conclude that the number of tuples $(a,b)$ with $a+b=n$
and $a, b$ of the form above,
where one of $a$ and $b$ is divisible by any prime congruent
$3 mod 4$, is at most $O(n^{1/2} log log n)$, which is of smaller order of magnitude than the expression $n^{1-2varepsilon}$ above.
This finishes the proof.

$textbf{Remark.}$
It seems likely that the number of representations $f(n)$ can be greatly improved by observing that one only needs $r_2(a)$ on average. This should produce $f(n)$
within a logarithmic factor. Moreover it seems possible to adapt Greaves's argument by replacing $p^2$ and $q^2$ by squares of integers not containing a prime $3bmod 4$, achieving a further logarithmic saving.

(Let us briefly reflect why the argument works: Greaves uses the fact
that Iwaniec's half dimensional sieve can also handle sums of two squares.
The contribution from the almost trivial 'fact' is also quite useful.)

Let us briefly sketch another possible approach, which could be more
suitable for getting a result for all positive integers:
Let $f(n)$ denote the number of representations as a sum of two integers, both not containing any prime factor $3 bmod 4$.
Let $r_{(a,b,c,d)}(n)$ denote the number of representations as sum $ax^2+by^2+cz^2+dt^2$.
We intend to show that $$n^varepsilon r(n) gg r_{(1,1,1,1)}(n)- 2 sum_{p=3 bmod 4} r_{(1,1,p^2,p^2)}(n) gg r_{(1,1,1,1)}(n).$$
Observe that $r_{(1,1,p^2,p^2)}(n)approx frac{1}{p^2}r_{(1,1,1,1)} $ and that $sum_{p =3mod 4} frac{1}{p^2}$ is a small and
finite number.

For a completely explicit result all we need is an explicit lower bound on
$r_{(1,1,1,1)}(n)$, which can be derived from Jacobi's formula,
and an explicit upper bound on expressions like $r_{(1,1,p^2,p^2)}(n)$,
which can be obtained either by the circle method using a Kloosterman
refinement, a modular forms approach, or via Dirichlet's hyperbola method. The big question then is if the
resulting numerical bounds allow the remaining finitely many cases
to be checked by a computer.

gn.general topology - Topological Rings

While looking through old questions, I came accross this one, and decided to throw my hat into the "ring."

Partial Answers and Observations:

1. If the identity element has a countable neighborhood base, then trivially by the continuity of the operation of addition, every point has a countable neighborhood base (as continuous maps take a filter base at some point x, and map this onto a filter base of the image point this is a particular example of homogeneity exhibited by topologically equipped algebraic objects.).




2. Provided, the topology defined on the ring is $T_0$, we have that the topology must also be $T_2$ and completely regular.




3. Moreover, since there is a countable dense subset $D$ by assumption, we have that given the above two assumptions, there exists a countable collection, of countable covers $U_n$, with the following property: Given any closed set $C$ from this space, and point $pnotin C$ there exists an $ninomega$ such that for some $Oin U_n$, $pin O$ and $O cap C = {}$. (We get such a collection by ordering the neighborhood base of each point in $D$ by reverse inclusion, and taking fixed 'sections of fixed height' from each to form the open covers)

Putting (1), (2), and (3) together produces a topological space which is about as close to a normal Moore space as you can get without actually being one, that is to say, under these assumptions: $R$ is a separable completely regular developable Hausdorff space, (and we haven't even invoked the ACC yet)

The interesting Part:

Ignoring for the moment the previous assumptions, intuitively, the ACC should somehow produce a covering property for this particular space. However, there is an interesting problem when it comes to the definition of subgroup/subring (which is required to get to the notion of ideal needed to apply the ACC): will they be open, closed, neither? Because of this we cannot really apply the ACC, to produce a nice covering property that might have tied everything together (like Lindelöf.)

Edit: While poking around Wiki, I came across something I felt I needed to add

However, if you mean that the space is a Noetherian topological space, then we get some gnarlly consequences ( http://en.wikipedia.org/wiki/Noetherian_topological_space ), like the fact that the space is compact! Which is exactly the thread we would want to tie everything together, and produces a normal moore space.

The Reality of the Matter:

The question is ill-posed, in that we do not have enough information to properly deduce a valid and fully general answer. My answer to this question has tried to highlight this point by giving you a case where, you can be about as close as you might ever want to be to something genuinely interesting, and then failing to make it interesting because of the incompatibility of the algebraic assumption with the topological ones. Even if we consider the other possibility we enter one of those strange and beautiful areas in topology where things being to become independent of $ZFC$.

Final Conclusion:

Because of this freedom or lack of information, we are left with an answer of Most Likely No. (Weak answer I know) But I can make this claim, because we honestly do not completely understand the notion of hereditary separability (in fact it was only in 2006 that J T Moore was able to produce a ZFC example of an L-space Article)

linear algebra - Approximately known matrix

What linear algebraic quantities can be calculated precisely for a nonsingular matrix whose entries are only approximately known (say, entries in the matrix are all huge numbers, known up to an accuracy of plus or minus some small number)? Clearly not the determinant or the trace, but probably the signature, and maybe some sort of twisted signatures? What is a reference for this sort of stuff? (numerical linear algebra, my guess for the name of such a field, seems to mean something else).

ca.analysis and odes - Trigonometry related to Rogers--Ramanujan identities

For integers $nge2$ and $kge2$, fix the notation
$$
[m]=sinfrac{pi m}{nk+1} quadtext{and}quad
[m]!=[1][2]dots[m], qquad minmathbb Z_{>0}.
$$
Consider the following trigonometric numbers:
$$
a_i=frac{[i+k-2]![n-i+k-2]!}{[k-2]![n+k-2]!},
qquad i=1,2,dots,n-1.
$$
Is it possible for any $n$ to express the quantities
$$
A_j=1-frac{[k-1]cdot [n+k-1]}{[j+k-1]cdot [n-j+k-1]},
qquad j=1,2,dots,n-1,
$$
as a product/quotient of terms of the form
$(1-text{product of some }a_i)$? If not (for $nge4$),
is it possible to prove that?

The affirmative answer is known for $n=2$ and $n=3$.
Namely, if $n=2$ so that we have only one $a_1$ and one $A_1$, then
$$
A_1=1-a_1.
$$
If $n=3$, then
$$
A_j=(1-a_j)(1-a_1a_2) quadtext{for } j=1,2.
$$
(The last formula is a nice trigonometric identity, by the way.)

My question is motivated (in a very sophisticated way) by a recent
question on Rogers--Ramanujan identities.
The latter one reminded me about the problem of possible $mathfrak{sl}_n$ generalizations of RRs
in their classical form "a $q$-sum"="a $q$-product". The only cases $n=2$ and $n=3$ are known;
these are the Andrews--Gordon identities and the Andrews--Schilling--Warnaar identities
(see [S. Ole Warnaar, Adv. in Math. 200 (2006) 403--434]).
An indirect implication of such identities is the family of (highly nontrivial)
numerical identities for the dilogarithm function; these come as the limit $qto1$ specialisation
and some multivariate asymptotics. The trigonometric identities above come into
play from these considerations for $n=2$ and $n=3$; any answer for $n>3$ can shed
some light on the existence of new RRs.

ct.category theory - Clifford algebra as an adjunction?

Background

For definiteness (even though this is a categorical question!) let's agree that a vector space is a finite-dimensional real vector space and that an associative algebra is a finite-dimensional real unital associative algebra.

Let $V$ be a vector space with a nondegenerate symmetric bilinear form $B$ and let $Q(x) = B(x,x)$ be the associated quadratic form. Let's call the pair $(V,Q)$ a quadratic vector space.

Let $A$ be an associative algebra and let's say that a linear map $phi:V to A$ is Clifford if
$$phi(x)^2 = - Q(x) 1_A,$$
where $1_A$ is the unit in $A$.

One way to define the Clifford algebra associated to $(V,Q)$ is to say that it is universal for Clifford maps from $(V,Q)$. Categorically, one defines a category whose objects are pairs $(phi,A)$ consisting of an associative algebra $A$ and a Clifford map $phi: V to A$ and whose arrows
$$h:(phi,A)to (phi',A')$$
are morphisms $h: A to A'$ of associative algebras such that the obvious triangle commutes:
$$h circ phi = phi'.$$
Then the Clifford algebra of $(V,Q)$ is the universal initial object in this category. In other words, it is a pair $(i,Cl(V,Q))$ where $Cl(V,Q)$ is an associative algebra and $i:V to Cl(V,Q)$ is a Clifford map, such that for every Clifford map $phi:V to A$, there is a unique morphism
$$Phi: Cl(V,Q) to A$$
extending $phi$; that is, such that $Phi circ i = phi$.

(This is the usual definition one can find, say, in the nLab.)

Question

I would like to view the construction of the Clifford algebra as a functor from the category of quadratic vector spaces to the category of associative algebras. The universal property says that if $(V,Q)$ is a quadratic vector space and $A$ is an associative algebra, then there is a bijection of hom-sets

$$mathrm{hom}_{mathbf{Assoc}}(Cl(V,Q), A) cong mathrm{cl-hom}(V,A)$$

where the left-hand side are the associative algebra morphisms and the right-hand side are the Clifford morphisms.

My question is whether I can view $Cl$ as an adjoint functor in some way. In other words, is there some category $mathbf{C}$ such that the right-side is
$$mathrm{hom}_{mathbf{C}}((V,Q), F(A))$$
for some functor $F$ from associative algebras to $mathbf{C}$. Naively I'd say $mathbf{C}$ ought to be the category of quadratic vector spaces, but I cannot think of a suitable $F$.

I apologise if this question is a little vague. I'm not a very categorical person, but I'm preparing some notes for a graduate course on spin geometry next semester and the question arose in my mind.