ca.analysis and odes - accelerated convergence to the mean using quadratic weights

If the sequence $x_1,x_2,dots$ is periodic, the unweighted averages $(sum_{i=1}^n x_i)/n$ converge to the asymptotic average of the $x_n$'s with error $O(1/n)$, but the weighted averages $(sum_{i=1}^n i(n+1-i)x_i)/(n(n+1)(n+2)/6))$ converge even more quickly, with error $O(1/n^2)$.

This fact is easy to prove (e.g. first prove it for $(x_n) = (zeta^n)$ with $zeta$ an arbitrary root of unity and then appeal to linearity), but it's something I stumbled upon on my own, and I don't really understand what's going on. Can anyone provide a context for this fact? My guess is that it must be well-known to people who study series-convergence (and acceleration thereof), and also well-known to Fourier analysts, though possibly in disguised form. (Speaking of disguises: This question is related to my earlier question A specific Dedekind-esque sum ; in my earlier post, the relevant sequence is almost-periodic rather than periodic, and the discrepancy goes down like $O((log n)/n^2)$ rather than $O(1/n^2)$.)

I suspect that $O(1/n^2)$ is the end of the line, in the sense that no weighted average of $x_1,dots,x_n$ with fixed coefficients will differ from the asymptotic average of the $x_n$'s by $O(1/n^c)$ for any $c>2$, and I might even try to give a proof using the geometry of numbers, but I suspect this is old stuff and would appreciate some pointers.

Thanks!

Jim Propp

gr.group theory - Quotient of manifolds by groups and embeddings

Let $f:X_1to X_2$ be a closed submanifold. Let $rho:G_1to G_2$ be a closed Lie subgroup. Let $G_1$ acts on $X_1$ and $G_2$ on $X_2$ and suppose $f$ is $rho$-equivariant. I would like to get a morphism $overline{f}:X_1/G_1to X_2/G_2$ and conditions for $overline{f}$ to be also a closed embedding. Do you know a reference where I can find this situation treated, in the category of analytic manifolds(or algebraic varieties)?

Resources on Invariant Theory

Hi,

So my question is pretty much summed up by the summary - basically I've run into a need to teach myself some of the basics of invariant theory and was looking for a good place to start. I'd prefer online / freeish resources if anyone knows of any that can be trusted - otherwise if someone can recommend a good introductory book on the subject that would be most appreciated.

(Currently considering getting or perhaps convincing my library to order for me P. Olivers Classical Invariant Theory - but I thought I'd ask around first).

As a second question - two in one if you will - and perhaps this will help to let you know how little I know about invariant theory - can anyone quickly summarize the difference between Classical and Geometric Invariant Theory? Presently I am thinking what I need is the classical kind - but again more information is always helpful.

P.S. Geometric invariant theory may not have been the best tag for this question, but apparently new users can't create new tags, so I picked the closest one - anyone that has the power to change the tag to something more appropriate please do so.

discrete geometry - Number of spanning trees in a grid

Expanding on Steve Huntsman's answer, call the product which appears in A007341 f(n). That is,

$$f(n) = prod_{k=0}^{n-1} {prod_{l=0}^{n-1}}^prime left(2 - cos {pi k over n} - cos {pi l over n } right)$$

where the $prime$ on the second product indicates that we start at $l=1$ in the case $k = 0$. The number of interest here is $a(n) = 2^{n^2-1} f(n)/n^2$ .

The product is the exponential of a sum, so

$$log f(n) = sum_{k=0}^{n-1} {sum_{l=0}^{n-1}}^prime log left(2 - cos {pi k over n} - cos {pi l over n } right).$$

This sum is, in turn, $n^2$ times a Riemann sum for the integral

$$ C = int_0^1 int_0^1 log(2-cos xpi - cos ypi) : dx : dy $$

which I believe converges, although actually evaluating it numerically is tricky. If you believe that, then $log f(n) sim Cn^2$ as $n to infty$, and $log a(n) sim (C+log 2) n^2$ as $n to infty$. From evaluating $f(n)$ for various $n$, it appears that $C$ is near $0.473$, $e^C$ is near $1.605$ and so we have

$$ a(n) approx 3.21^{n^2} $$

where I write $p(n) approx q(n)$ for $log p(n)/log q(n) to 1 $ as $n to infty$, i. e. $log p(n) sim log q(n)$.

Irreducible Polynomials in UFD and corresponding Quotient Field

Hello,

"Let $D$ be a UFD and let $F$ be its quotient field. Further let
$f$ be a primitive polynomial of positive degree in $Dleft[xright]$.
From this it follows that that $f$ is irreducible in $Dleft[xright]$ if and only
if $f$ is irreducible in $Fleft[xright]$."

I've shown the forward direction, i.e. irreducible in UFD $Rightarrow$ irreducible in quotient field, but am struggling to understand how the converse direction would go.

In particular, it strikes me that it might be productive to try and show that $f$ is prime in the UFD since this is equivalent to irreducibility, but I don't know how to show this. Also, $f$ should have no "denominators" in $Fleft[xright]$ since it's also in $Dleft[xright]$.

Any advice on how to proceed?

abstract algebra - Injective modules and Pontrjagin duals

Let me sketch a proof of the result (which I would guess is the one Lang gives) with some motivation that I hope resonates with your question.

There is a general adjunction, if $A to B$ is a ring homomorphism and $M$ and $N$ are $A$- and $B$-modules respectively, of the form
$Hom_A(N,M) = Hom_B(N,Hom_A(B,M)).$
This shows that if $M$ is injective over $A$, then $Hom_A(B,M)$ is injective as a $B$-module,
and also that if $N hookrightarrow M$ is an embedding of $A$-modules, then the induced map
$N to Hom_A(B,M)$ is also an embedding.

Thus it suffices to construct enough injectives in the category of $mathbb Z$-modules;
the above considerations then produce enough injectives for modules over any ring.

Now $I$ is injective if and only if $Hom(text{--},I)$ is exact, which is already a kind of Pontrjagin duality property: it gives an exact functor reversing arrows. Furthermore,
we always have a double duality map $M to Hom(Hom(M,I))$, and another feature of Pontrjagin duality is that this map should be an isomorphism. Let's here consider the weaker property that it is injective (so we at least have a non-degenarate duality, although not necessarily perfect).

Suppose then that $I$ is some injective module over $mathbb Z$, with the property that
the double duality map $M to Hom(Hom(M,I),I)$ is injective for all $M$ (i.e. with the property that it gives rise to some kind of Pontrjagin duality). Then
given this $I$, if we take a free module surjecting onto $Hom(M,I),$ and then apply
$Hom(text{--},I)$ and compose with the double duality map, we get an embedding from
$M$ into a product of copies of $I$, and hence an embedding of $M$ into an injective.

Thus constructing enough injectives is reduced to finding such $I$, and here we are
guided by the idea that we are looking for an object that establishes some kind of
(at least weak form of) Pontrjagin duality.

Now in order for the double duality to be injective, one needs that for any $m in M$,
there is a map from $langle m rangle$ to $I$ that is non-zero on $m$ (since this will extend to
all of $M$ by injectivity of $I$). So we need $I$ to be injective, i.e. divisible,
and to contain torsion elements of arbitrary order. Thus we could take $I = {mathbb Q}/{mathbb Z}$, as in Lang, or also the circle group $S^1$. (But the latter has infinite order elements which are not necessary, so ${mathbb Q}/{mathbb Z}$ is more economical.
Since it also has a more algebraic/less analytic feel than the circle, this is probably why it is more commonly used in such arguments.)

ac.commutative algebra - Converse to Hilbert basis theorem?

Dear Zev,

There are some sources which give the converse. See e.g. pp. 64-65 of

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

For that matter, see also pp. 32-33 of loc. cit. for the Chinese Remainder Theorem and its converse. (And I am not the only one to do this...)

Note that in both cases the converse is left as an exercise. I think (evidently) that this is the right way to go: it may not be so easy for the journeyman mathematician to come up with the statement of the converse, but having seen the statement it is a very valuable exercise to come up with the proof. (In particular, I believe that in a math text or course at the advanced undergraduate level and beyond, most exercises should indeed be things that one could find useful later on, and not just things which are challenging to prove but deservedly forgettable.)

Finally, by coincidence, just yesterday in my graduate course on local fields I got to the proof of the "tensor product theorem" on the classification of norms in a finite-dimensional field extension (which came up in a previous MO answer). The key idea of the proof -- which I found somewhat challenging to write; I certainly admit to the possibility of improvements in the exposition -- seems to be suspiciously close to the valuation-theoretic analogue of the converse of the Chinese Remainder Theorem! See pp. 18-19 of

http://math.uga.edu/~pete/8410Chapter2.pdf

if you're interested.

nt.number theory - An Expectation of Cohen-Lenstra Measure

Let me spell out the cokernel description of the Cohen-Lenstra distribution in more detail, as my answer will depend on it.

A map $(mathbb{Z}_p)^N to (mathbb{Z}_p)^N$ is given by an $N times N$ matrix of $p$-adic integers. Choose such a map by picking each of the digits of each integer uniformly at random from ${ 0, 1, ..., p-1 }$; this is the same as using the additive Haar measure on $mathrm{Hom}((mathbb{Z}_p)^N, (mathbb{Z}_p)^N)$. With probability $1$, this map does not have determinant $0$, so its cokernel is a finite abelian $p$-group. Let $mu_N$ be the probability measure on isomorphism classes of abelian $p$-groups which assigns each $p$-group the probability that it arises as this cokernel.

The Cohen-Lenstra distribution is the limit as $N to infty$ of $mu_N$. As shown in several of the references you link to, it is given by the formula
$$lim_{N to infty} mu_N(G) = frac{1}{ |mathrm{Aut}(G)|} prod_{i=1}^{infty} (1-1/p^i) .$$

For notational convenience, it will help to distinguish between the domain and range of a map in $mathrm{Hom}((mathbb{Z}_p)^N, (mathbb{Z}_p)^N)$. I will call the former $U^N$ and the latter $V^N$.

---

Now, to answer your question. Let $A$ be a fixed finite abelian $p$-group. Let $e_N(A)$ be the expected number of surjections from an abelian $p$-group $G$, picked according to measure $mu_N$, to $A$. Ignoring issues about interchanging limits, we want to show that $lim_{N to infty} e_N(A)=1$.

Lets start by considering $H_N(A) := mathrm{Hom}(V^N, A)$. The set $H_N(A)$ has cardinality $|A|^N$, as any map is specified by giving the image of a basis for $V^N$. Inside this set, let $S_N(A)$ be the surjective maps and $C_N(A)$ the nonsurjective maps.

For any map $f in S_N(A)$, let's consider the possibility that it extends to the cokernel of a random map $U^N to V^N$. This will occur if and only if the $N$ generators of $U^N$ land in the kernel of $f$. Since $f$ is in $S_N(A)$, its kernel has index $|A|$. So the probability that $U^N$ is mapped into the kernel of $f$ is $1/|A|^N$.

We want to compute
$$e_N(A) = |S_N(A)| cdot (1/|A|^N) = 1 - |C_N(A)|/A^N.$$

If $A$ can be generated by $r$ elements, then $|C_N(A)|/A^N leq (1-1/|A|^r)^{lfloor N/r rfloor}$, so the second term drops out as $N to infty$. (To see this bound, group the basis elements of $V^N$ into $N/r$ groups of size $r$; the probability that these $r$ basis elements are not sent to the $r$ generators of $A$ is $(1-1/|A|^r)$. This bound is probably much weaker than the true rate of convergence.)

division algebras - Why don't quaternions have an overall phase?

There is a unique copy of $mathbb R$ in $mathbb H$ and it is the center.
On the other hand, for every quaternion $q$ not in $mathbb R$, the generated $mathbb R$-algebra with unit is a copy of $mathbb C$, but none is central.

I like to view quaternions as $q=(x,X)in mathbb Rtimes mathbb R^3$ with multiplication
$$
(x,X).(y,Y) = (x.y - langle X,Yrangle, X times Y + x.Y + y.X)
$$
This uses oriented Euclidean $mathbb R^3$, and it turns
http://en.wikipedia.org/wiki/Quaternion#Quaternions_and_the_geometry_of_R3
into a definition.
One can then take any oriented orthonormal basis $i,j,k$ for the basic imaginary quaternions.

Edit: Note that $Xtimes Y$ is a Lie bracket and $-langle X,Yrangle$ the corresponding Killing form ($mathfrak smathfrak o(3,mathbb R)$ up to a constant), one can repeat this construction for every real Lie algebra. Only for 3 of them one obtains an associative algebra.

Second edit: The reference for the first edit is:

Peter W. Michor, Wolfgang Ruppert, Klaus Wegenkittl: A connection between Lie algebras and general algebras. Rendiconti Circolo Matematico di Palermo, Serie II, Suppl. 21 265--274, (1989)(pdf)

Moreover, a similar construction as the above on $mathbb Ctimes mathbb C^3$ using also complex conjugation (see "Greub: Multilinear algebra, 2n ed. 1978" page 289) leads to octonians.

ag.algebraic geometry - Concerning the dimension of a complex variety modulo a prime

To make more precise the answer of Felip. You have a scheme $X=Spec(A)$ over $mathbb Z$, where $A$ is a finitely generate $mathbb Z$-algebra such that its generic fiber $X_{mathbb Q}$ (just consider your polynomials as polynomials with rational coefficients) gives $V$ by field extension $mathbb{C}/mathbb{Q}$. Of course, $X_{mathbb Q}$ has the same dimension as $V$, and $X_{mathbb Q}$ is smooth if and only $V$ is smooth.

Questions 1-2. You want to compare $dim X_p$ with $dim X_{mathbb Q}$. In general $dim X_pge dim X_{mathbb Q}$. The equality holds under some flatness condition at $p$. Namely, there is rather general result: if $f: Yto Z$ is a flat morphism of finite type between noetherian schemes and suppose that $Z$ is integral and universally catenary (e.g.any scheme of finite type over a noetherian regular scheme, so any open subset of $Spec(mathbb Z)$ is OK), then all [EDIT: non-empty] fibers $Y_z$ of $f$ have the same dimension [EDIT: if the generic fiber of $f$ is equidimensional (i.e. all irreducibles components have the same dimension)].

Problem: your $X$ is not necessarily flat over $mathbb Z$. But the flatness over $mathbb Z$ (or any Dedekind domain) is easy to detect: it is equivalent to be torsion free. So consider the ideal $I$ equal to
$$ { ain A mid ka=0 text{ for some non zero } k in mathbb Z }$$
(don't know how to type "{" and "}"). Then $A/I$ is flat over $mathbb Z$ and defines a closed scheme of $X$ which coincides with $X$ over an open subset of $mathbb Z$. Actually, as $I$ is finitely generated, there exists a positive integer $N$ such that $NI=0$. Then $I=0$ over $Spec(mathbb Z[1/N])$. So for any prime number $p$ prime to $N$, $X_p$ will have dimension $dim V$. Now you have to compute such a $N$... For hypersurface, $N$ is just the gcd of the coefficients (see comments in Felip's answer). I don't know whether efficient methodes exist in general. Of course if a prime $p$ does not divid any polynomial in the definning ideal of $X$, then this $p$ is OK. But you have to test this property for all polynomials and not just a set of generators (Example: the ideal generated by $T_1+pT_2, T_1$, then $p$ is bad).

Question 3. Suppose $V$ is smooth (and connected for simplicity), and you are looking for the $p$ such that $X_p$ is also smooth. You first proceed as in Question 1 to find an open subset $Spec(mathbb Z[1/N]$ over which $X$ is flat. Let $d=dim X_{mathbb Q}=dim X_p$ for all $p$ prime to $N$. Write
$$ X=Specbig(mathbb Z[T_1, ldots T_n]/(F_1,ldots, F_m)big)$$
Then $X_p$ is smooth is and only if the Jacobian matrix of the $F_i$'s mod $p$ has rank $n-d$ at all points of $X_p$. Equivalently, the ideal $Jsubseteq mathbb Z[T_1,ldots, T_n]$ generated by $F_1,...,F_m$ and the rank $n-d$ minors of the Jacobian matrix is the unit ideal mod $p$. Therefore, the computation consists in determining the ideal $J$. As $X_{mathbb Q}$ is supposed to be smooth, $J$ is generated by a positive integer $M$. Now for all $p$ prime to $MN$, $X_p$ is smooth of dimension $dim V$.

[EDIT]: The assertion on the dimensions of the fibers needs some hypothesis on the generic fiber of $Yto Z$. We must assume it is equidimensional. Otherwise $Y_z$ has dimension equal to that of the generic fiber for $zin Z$ belonging to the image of all irreducible components of $X$. For the initial question, it is better to assum the original variety $V$ is irreducible.

gr.group theory - projections of finitely presented groups

I think this is true. Let $G= langle g_1, ldots, g_n; R_1, ldots, R_m rangle$ be a finite presentation for $G$. Let $p:Gto G$ be a projection (or retract). Then $p(G)< G$ is generated by $p(g_1), ldots, p(g_n)$. Let $F_n$ be the free group on $n$ generators, and $Rin F_n$ a relator of $p(G)$, so $R(p(g_1), ldots, p(g_n))=1$. I claim that $p(R_1),ldots, p(R_m)$ give finitely many relations for $p(G)$. Give a new presentation for $G$, $$langle g_1, ldots, g_n , y_1, ldots, y_n ; y_1^{-1} p(g_1) , ldots, y_n^{-1} p(g_n), R_1, ldots, R_m rangle.$$ With respect to this presentation, the homomorphism $p$ is given by $p(g_i)=p(g_i), p(y_i)=p(g_i)$. Then we must have the relation $R(y_1, ldots, y_n)=1 in G$, and therefore $R(y_1, ldots, y_n) $ may be written as a product of conjugates of the relators. Then $p(R(y_1,ldots,y_n)) = R(p(g_1),ldots,p(g_n))$ is a product of conjugates of projections of the relators of the second presentation of $G$. This shows that $p(G)$ is finitely presented, and since the first $n$ relators project to the trivial relator, we see that the relators are given by $p(R_1),ldots, p(R_m)$.

Here's a reformulation:

Let $G$ be a finitely presented group, with a projection $p:Gto G$, so $p^2=p$.

Then there is a free group $F = langle g_1, ldots, g_nrangle$ and an epimorphism $phi:Fto G$ such that $ker(phi)$ is finitely normally generated. We may write the presentation as $$ langle g_1,ldots, g_n; R_1,ldots, R_mrangle.$$ The kernel of $p$ is normally generated by $phi(g_i) p(phi(g_i))^{-1}$ (exercise)*. Let $p': Fto F$ be a lift of $p$, which exists by the universal property of $F$, so that $pcirc phi = phi circ p'$. Then $ker(pcirc phi) = phi^{-1}( ker(p)) $ is normally generated by elements projecting to the normal generators of $ker(p)$ togther with $ker(phi)$. We may take $g_i p'(g_i)^{-1}$ together with the finitely many normal generators of $ker(phi)$ to get normal generators of $ker(pcirc phi)$. This gives a finite presentation of $p(G)$, given by $$langle g_1,ldots,g_n ; R_1,ldots, R_m, g_1 p'(g_1)^{-1},ldots, g_n p'(g_n)^{-1} rangle.$$

In my previous "argument", I attempted to project this presentation by $p'$ to $p'(F)$, which is a free subgroup of $F$, but this is unnecessary, and I was incorrect that the new relators project to trivial relators in $p'(F)$.

*(exercise) Here's the details to show the kernel is normally generated by $phi(g_i) p(phi(g_i))^{-1}$. For simplicity of notation, I'll denote the generators as $g_i$ instead of $phi(g_i)$, and assume that they are semigroup-generators of $G$. Then $ker(p)$ is
normally generated by $g_i p(g_i)^{-1}$. Call this normal subgroup $K$. Notice that it suffices to show that $g p(g)^{-1} in K$, for all $gin G$, since if $gin ker(p)$, then $g p(g)^{-1} = g in K$. We may show this by induction on the length of the word representing $g$. Clearly it's true if $g= g_i$. Now, suppose it is true for products of $k$ generators, $kin mathbb{N}$. Then if $g= g_{i_1} cdots g_{i_{k+1}}$, we have
$$g p(g)^{-1}= g_{i_1} cdots g_{i_{k+1}} p(g_{i_{k+1}})^{-1} cdots p(g_{1})^{-1} = g_{i_1} p(g_{i_1})^{-1} ( p(g_{i_1}) ( g_{i_2} cdots g_{i_{k+1}} p(g_{i_{k+1}})^{-1} cdots p(g_{i_2})^{-1} ) p(g_{i_1})^{-1}) in K$$ by induction and the fact that $K$ is closed under products and conjugations.

pr.probability - How to partially uniformize a deck by partial shuffles?

This question is a variant of a previous one; it was originally a posed as an edit of this former question, but I came to think it could be more suitable to pose it anew.

Assume I have a deck of cards that I would like to shuffle. Unfortunately, the deck is so big that I cannot hold it entirely in my hands. Let's say that the deck contains M cards, and that the operation I can perform are: 1. cut (deterministically) a deck into any number of sub-decks, without looking at the cards but remembering for all i where the i-th card from top of the original deck has been put; 2. gather several decks into one deck in any order (but assume that we do not intertwin the various decks, nor change the order inside any of them); 3. shuffle any deck of at most n cards. Assume moreover that such a shuffle consist in applying an unknown random permutation drawn uniformly.

Due to arithmetic arguments, it is not possible to achieve uniform distribution over all permutations of the original deck by such shuffles (see David Speyer's answer to the question above). However, one can can consider partial uniformity as follows.

Call a random permutation of ${1,ldots,M}$ $r$-uniform if the law of $((a_1),ldots,(a_r))$ is uniform for all tuples $(a_1,ldots,a_r)$ of ${1,ldots,M}$. Then, what is the maximal such $r$ that can be achieved by a process as described above? Given a $r$, how many shuffles are needed to achieve $r$-uniformity?

Note that this measure of uniformity makes sense for the game Race for the galaxy, since some subsets of cards play well together, and one wants to break the artificial weight given when playing a game to the event that such cards are close.

Here is a first thing one can say: if $M=2n$ and $n$ is even, then it is possible to achieve $1$-uniformity in $4$ partial shuffles. Simply cut the deck in two equal sub-decks, shuffle both, gather them, divide the result into a sub-deck containing the first and third quarters and another containing the second and fourth quarters, shuffle both, and gather them. At the end, each individual card has random distribution.

inequalities - Is there a good reason why a^{2b} + b^{2a}

Fixed now. I spent some time looking for some clever trick but the most unimaginative way turned out to be the best. So, as I said before, the straightforward Taylor series expansion does it in no time.

Assume that $a>b$. Put $t=a-b=1-2b$.

Step 1:
$$
begin{aligned}
a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)left[frac{1}2b^2+frac{1+t}{3!}b^3+frac{(1+t)(2+t)}{4!}b^4+dotsright]
\
&le 1-b(1-t)-t(1-t)left[frac{b^2}{1cdot 2}+frac{b^3}{2cdot 3}+frac{b^4}{3cdot 4}+dotsright]
\&
=1-b(1-t)-t(1-t)left[blogfrac 1{a}+b-logfrac {1}aright]
\
&=1-b(1-t^2)+(1-b)t(1-t)logfrac{1}a=1-bleft(1-t^2-t(1+t)logfrac 1aright)
end{aligned}
$$
(in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)

Step 2.
We need the inequality $e^{ku}ge (1+u)(1+u+dots+u^{k-1})+frac k{k+1}u^{k+1}$ for $uge 0$.
For $k=1$ it is just $e^uge 1+u+frac{u^2}{2}$. For $kge 2$, the Taylor coefficients on the left are $frac{k^j}{j!}$ and on the right $1,2,2,dots,2,1$ (up to the order $k$) and then $frac{k}{k+1}$. Now it remains to note that $frac{k^0}{0!}=1$, $frac{k^j}{j!}ge frac {k^j}{j^{j-1}}ge kge 2$ for $1le jle k$, and $frac{k^{k+1}}{(k+1)!}ge frac{k}{k+1}$.

Step 3:
Let $u=logfrac 1a$. We've seen in Step 1 that $a^{2b}le 1-b(1-tmu)$ where $mu=u+(1+u)t$. In what follows, it'll be important that $mulefrac 1a-1+frac 1a t=1$ (we just used $logfrac 1ale frac 1a-1$ here.

We have $b^{2a}=b(a-t)^t$. Thus, to finish, it'll suffice to show that $(a-t)^tle 1-tmu$. Taking negative logarithm of both sides and recalling that $frac 1a=e^u$, we get the inequality
$$
tu+tlog(1-te^u)^{-1}ge log(1-tmu)^{-1}
$$
to prove.
Now, note that, according to Step 2,
$$
begin{aligned}
&frac{e^{uk}}kge frac{(1+u)(1+u+dots+u^{k-1})}k+frac{u^{k+1}}{k+1}
gefrac{(1+u)(mu^{k-1}+mu^{k-2}u+dots+u^{k-1})}k+frac{u^{k+1}}{k+1}
\
&=frac{mu^k-u^k}{kt}+frac{u^{k+1}}{k+1}
end{aligned}
$$
Multiplying by $t^{k+1}$ and adding up, we get
$$
tlog(1-te^u)^{-1}ge -ut+log(1-tmu)^{-1}
$$
which is exactly what we need.

The end.

P.S. If somebody is still interested, the bottom line is almost trivial once the top line is known. Assume again that $a>b$, $a+b=1$. Put $t=a-b$.

$$
begin{aligned}
&left(frac{a^b}{2^b}+frac{b^a}{2^a}right)^2=(a^{2b}+b^{2a})(2^{-2b}+2^{-2a})-left(frac{a^b}{2^a}-frac{b^a}{2^b}right)^2
\
&le 1+frac 14{ [sqrt 2(2^{t/2}-2^{-t/2})]^2-[(1+t)^b-(1-t)^a]^2}
end{aligned}
$$
Now it remains to note that $2^{t/2}-2^{-t/2}$ is convex on $[0,1]$, so, interpolating between the endpoints, we get $sqrt 2(2^{t/2}-2^{-t/2})le t$. Also, the function $xmapsto (1+x)^b-(1-x)^a$ is convex on $[0,1]$ (the second derivative is $ab[(1-x)^{b-2}-(1+x)^{a-2}]$, which is clearly non-negative). But the derivative at $0$ is $a+b=1$, so $(1+x)^b-(1-x)^age x$ on $[0,1]$. Plugging in $x=t$ finishes the story.

elliptic curves - Quadratic Twist of Legendre Form

To amplify on what Kevin Buzzard said: the Legendre form of an elliptic curve is

$E_{lambda} : y^2 = x(x-1)(x-lambda)$

Since it has only one parameter, it has much less freedom than $y^2 = x^3 + ax + b$. Strictly speaking, an elliptic curve over a field $K$ is a twist of another elliptic curve over $K$, if they are not isomorphic over $K$ but they are over an extension $L/K$. So it doesn't make sense to ask when the quadratic twist (meaning that $K=mathbb{Q}$ and $L=mathbb{Q}(sqrt{d})$ is a quadratic extension) is isomorphic to the curve. However, I think that you mean the curve

$E_{lambda,d}: d y^2 = x(x-1)(x-lambda)$ rewritten into Legendre form. However, we do have

$j = 256 frac{(lambda^2 - lambda + 1)^3}{lambda^2 (lambda-1)^2}$

and there are six "versions" of $lambda$ which yield isomorphic curves in Legendre form:

$lambda,1-lambda,1/lambda,1/(1-lambda),lambda/(lambda-1),(lambda-1)/lambda$.

This seriously restricts the possible $d$'s for which $E_{lambda,d}$ can be written in Legendre form to a at most six values, namely if $pm d lambda, pm d, pm d (1-lambda)$ are squares.

[Added later]:
It might make more sense to think of $Y^2 = X(X-a)(X-b)$ as a "generalized Legendre form". In that line of thought, the "standard" Weierstrass form $Y^2 = X^3 + a X + b$ is associated with the moduli space $X(1)$, and $a$ is a modular form for weight 4, and b is a modular form of weight 6 for $Gamma(1)$ (suitably normalized Eisenstein forms). The "generalized Legendre form" is associated with $X(2)$ which parametrizes elliptic curves along with the full 2-torsion subgroup (satisfying a normalization of the Weil Pairing). In that case $a$ and $b$ are modular form of weight 2 for $Gamma(2)$

fa.functional analysis - Bibliography for topologies defined by a family of seminorms

I just had a look at

Topological Vector Spaces,
Distributions and Kernels

by Francois Treves. It is divided into three parts:

I Topological Vector Spaces. Spaces of Funtions

• covering: basic material about locally convex spaces and Frechet spaces (with a lot of examples)

II Duality, Spaces of Distributions

• topologies on Duals, transposes of linear maps, convolution, barreled spaces

III Tensor Products. Kernels

• injective and projective tensor products and their relation to bilinear forms, nuclear spaces, nuclear mappings, Schwartz kernel theorem and applications

From the first sight, this looks like a good place to start if you are already familiar with functional analysis on Banach and Hilbert spaces.

ca.analysis and odes - Differentiate under integral sign for iterated integral?

As usual when differentiating something with respect to a variable that appears twice. The chain rule for partial derivatives.

For example, consider function $z = f(u,v)$. Suppose we want $(d/dt)f(t,t)$. Let $u=v=t$ and use
$dz/dt = (partial z/partial u)(du/dt) + (partial z/partial v)(dv/dt)$.

Thus...
$$
frac{d}{dt}int_0^tint_0^t f(x,y)\,dx\,dy =
int_0^t f(t,y)\,dy + int_0^t f(x,t)\,dx
$$

By the way, why did you write $partial/partial t$ to differentiate a function of the single variable $t$? It's not wrong, just confusing to students.

prime numbers - Computing the Mertens function

I wonder if anybody can help me with this problem.

I'm trying to compute the Mertens function for large $n$. The most obvious algorithm is just to compute all primes up to $sqrt{n}$ and then to sieve. That takes at least an order of $nlog n$ operations, and really even more.

The most recent article that I could find that discusses methods to compute the function directly is dated 1994, and it proposes to do exactly that.

Are there any known algorithms that let you compute Mertens faster than by sieving? I know that $pi(n)$ can be computed in $O(n^{2/3})$, I looked into that algorithm but it does not seem to be easily adaptable to my task.

Alternatively, I could use an algorithm to compute $M(n+dn)-M(n)$ for $dnll n$ (say $dnsim sqrt{n}$ ) in $O(sqrt{n})$ time or less.

at.algebraic topology - How does $pi_1(SO(3))$ relate exactly to the waiters trick?

(This is a bit long for a comment but isn't intended as an answer.)

I've broken glasses in the past demonstrating this trick so I strongly advise doing it with empty plastic mugs until you have it down pat. For those who can't quite work it out, here's a simpler way to do it: take an elastic band and two rods that have distinguishable "up" and "down". Loop the band around the two rods and keep it fairly taut (just so it doesn't fall off). Ascii picture:

   |     |
 /-|-----|-   
/  |     |     
  |     |  /   
 ---------/   
   |     |   

Now turn one of the rods upside-down. By moving the rod through space but without turning it (or the system falling apart), try to untangle the elastic band.

Can't do it? Good. Can? Whoops! Either you've done something wrong or Whitehead did.

Start again from the beginning. Now turn one of the rods through a full twist (so it's right-side-up again). Now try to untangle the elastic band as before.

Can't do it? Try again! It's possible.

So in mathematical terms, what you are looking for is the position of the second rod plus it's "up-down"ness. In terms of the waiter's arm, you want the position of the hand and the expression of agony on his face: if his face is in agony, his arm is twisted; if his face is calm, his arm isn't twisted.

Mnemonic: twisting someone's arm twice gets you nowhere.

the group law for an elliptic curve

Let $varphi(u)$ be holomorphic in the neighborhood of the origin of the complex plane. One says that $varphi(u)$ admits an algebraic addition theorem if it satisfies a functional equation of the form $G[varphi(u). varphi(v), varphi(u+v)]=0,$ where where G(X,Y,Z) is a non vanishing polynomial in the three variables X,Y,Z with complex constant coefficients, while $u,v, u+v$ are in the domain of $varphi(u)$. Examples are the rational functions, the exponential function, the trigonometric functions, and the elliptic functions. Then it can be proved that $varphi(u)$ and $varphi'(u)$ are connected by an algebraic equation $A[varphi(u)$,$varphi'(u)]=0$, which defines the ELLIPTIC CURVE parameterized by $varphi(u)$. It is known that ANY elliptic curve can be realized in this way.

If one finds the greatest common divisor of G(X,Y,Z) and $X'frac{dG}{d Y}-Y'frac{dG}{d X},$, where $X'$ means the derivative with respect to $u$, etc., we obtain an irreducible polynomial $D(Z,X,X',Y,Y')$, which, when put equal to zero, gives the GROUP LAW for A(X,X')=0. If the degree of D in Z is equal to one, then the group law is rational in X,X',Y,Y'. Here is the question:

Prove: the degree of D in Z is one iff $varphi(u)$ is uniform (i.e., has no branch points)

The proof must not use the properties of $varphi(u)$ as rational, trigonometric, or elliptic functions...rather, only the uniformity of $varphi(u)$.

I have not seen a proof...it is desirable that one be found. There ARE proofs using the properties of the rational, trigonometric, and elliptic functions.

This gives an elementary algorithm for finding the group law for any elliptic curve without detouring through the Weierstrass $wp$ function.

dg.differential geometry - Walking around Santa Cruz, track around the soccer field

This may be enough to get you started.

In building a model railroad track, to make a complete circle, I need 12 curved pieces.
Assuming I want a more elaborate layout, but want to keep things on the same level and
allow no branches or crossovers ( i.e. I have only straight and curved pieces and no
risers to lift track off my flat building surface), I need many curved pieces. If
I start at one place, go counterclockwise around the track, and count pieces that
curve left and pieces that curve right, I get 12 more pieces that curve left than curve
right after I complete one loop and return to my starting place.

Is your scenario similar to building railroad track?

Gerhard "Ask Me About System Design" Paseman, 2010.08.15

lie groups - What's the classification of the algebraic subgroups of Sp(4,R)?

On the one hand, I could not find a published answer with a cursory search. On the other hand, as Ben says, you could work out the answer "by hand". Instead of writing down a sheer list, which might be complicated (and I haven't done the work), I'll write down the main ingredients.

A Zariski-closed subgroup $H$ of any connected semisimple Lie group $G$ has three pieces: (1) finite, (2) connected semisimple, and (3) connected solvable. The Zariski topology forces $H$ to have only finitely many components; if $H_0$ is the connected subgroup, then $H/H_0$ is the finite piece. Then the Lie algebra of $H_0$ has a Levi decomposition, so that you get the other two pieces. The way to analyze the question is to chase down the possibilities for all three pieces.

I think that the finite part always lifts to a slightly larger finite subgroup of $H$. This is not true for groups in general, but I think that it is true in context. Then this finite group is contained in a maximal compact group of $G$. Happily, the compact core of $text{Sp}(4,mathbb{R})$ is $text{SU}(2)$, and the finite subgroups are classified by simply laced Dynkin diagrams.

A semisimple, connected subgroup of $G$ corresponds to a semisimple Lie subalgebra, and that complexifies. The Lie algebra $text{sp}(4,mathbb{C})$ does not have very many inequivalent semisimple subalgebras. From looking a rank, they are isomorphic to $text{sl}(2,mathbb{C})$ or $text{sl}(2,mathbb{C}) oplus text{sl}(2,mathbb{C})$. I am confusing myself a little with the possible positions of the former, although I know there are only a few. The latter embeds in only one way. Then you would work backwards to get the real forms of these complex subalgebras; again there wouldn't be very many.

Finally the solvable part also complexifies and I think that it is contained in a Borel subalgebra at the Lie algebra level.

As for the more general question, for $text{Sp}(2n,mathbb{R})$, there is a tidy converse answer that also shows you that you can't expect a tidy answer for all fixed $n$. Namely, if $G$ is any algebraic group, you can classify its anti-self-dual (or symplectically self-dual) representations. Every algebraic group will have some, because every algebraic group has representations in $text{GL}(n,mathbb{R})$. A more interesting case is when $G$ has an irreducible symplectically self-dual representation. For that purpose, you check that the irreducible representation is real, and then check the Frobenius–Schur indicator.

ag.algebraic geometry - Ramification in morphisms of surfaces

Let me expand jvp's answer, giving a picture of the situation in the case of a $general$ flat triple cover $f colon X to Y$.

Let $R subset Y$ be the ramification divisor and $B subset Y$ the branch divisor, that is $B = f(R)$. Then $R$, $B$ are both reduced and irreducible, and $B$ has only a finite number of ordinary cusps $q_1, ldots, q_t$ as singularities. These cusps are exactly the points over which $f$ is $totally$ $ramified$. Moreover $R$ is isomorphic to the normalization of $B$, in particular it is $smooth$.

One has the equality of divisors

$f^*(B)=2R + R'$,

where $R'$ is another irreducible curve, isomorphic to $R$, which meets $R$ in a finite number of points $p_1, ldots, p_t$. Notice that $R'$ is $not$ a component of the ramification locus, since the latter consists of $R$ alone.

Moreover

• $R$ and $R'$ are tangent at $p_1, ldots, p_t$;


• $p_1, ldots ,p_t$ are the preimages of the cusps $q_1, ldots, q_t$.

Summing up, in this case your $S$ is the set whose elements are the points $p_1, ldots ,p_t$. They correspond to the points where the ramification divisor $R$ meets the curve $R'=f^*(B) setminus R$. In other words, they come from the singular points of the branch divisor $B$ (whereas the ramification divisor $R$ is smooth).

This is easy to see; a good reference is Miranda's paper "Triple covers in algebraic geometry".

Anyway, the crucial fact here is that a general triple cover is not a Galois cover, so over the branch locus $B$ there are both points where $f$ is ramified (the curve $R$) and points where it is not (the curve $R'$).

If you consider instead any Galois cover, say with group $G$, then every preimage of a branch point is a ramification point (and the stabilizers of points lying on the same fibre are conjugated in $G$). In this case there are formulae relating the ramification number of a point on $X$ with the ramification numbers of the components of the ramification locus passing through it.

See Pardini's paper "Abelian covers of algebraic varieties" for more details.

dg.differential geometry - Divisor Intersections and Chern Class Products

I'm not sure what you mean by a line bundle on a real algebraic variety and its Chern classes, but for smooth complex analytic manifolds the Chern class of a line bundle corresponding to a divisor is Poincare dual to the homological class of the divisor, as explained e.g. in Griffiths-Harris, Chapter 1, Chern classes of line bundles. So the $c_2([D_1]oplus [D_2])=c_1[D_1]c_1[D_2]$ is indeed the indeed Poincare dual to the intersection class of $D_1$ and $D_2$. If the manifold is a surface, then we get the a number.

Maybe this not what you meant, but in that case you should really add some more details to your question.

dg.differential geometry - Divisors and vector bundles in various categories

I'm taking a first course on complex manifolds, and am trying to square what I hear with what I know of (real) differential geometry. Please forgive me if this question is misguided!

Here are two examples of ways of making vector bundles from codimension-one submanifolds.

• (The tensor powers of its associated line bundle) In a complex manifold $M$ with a codimension-1 complex submanifold $D$, take as an atlas a system $(U_alpha)$ of slice-coordinate charts for $V$, together with other charts $(V_beta)$ covering $Msetminus D$. For each $nin mathbb{Z}$, define a line bundle via the following transition functions: $phi_{beta_1beta_2}:V_{beta_1}cap V_{beta_2}to GL_1(mathbb{C})$ is uniformly =1; $phi_{alpha_1beta_2}:U_{alpha_1}cap V_{beta_2}to GL_1(mathbb{C})$ is $z_1^n$, and $phi_{alpha_1alpha_2}:U_{alpha_1}cap U_{alpha_2}to GL_1(mathbb{C})$ is $z_1^n/w_1^n$, where $z_1$ and $w_1$ are the coordinates whose vanishing determines $D$ on $U_{alpha_1}$ and $U_{alpha_2}$ respectively.

Comment: This also seems to work fine if we replace "complex" by "smooth (real)" throughout. However, the family of line bundles isn't so interesting: the even ones are all trivial; the odd ones are mutually isomorphic.

• (Vector bundles on spheres) For each homotopy class of maps $S^{n-1}to GL_k(mathbb{R})$, we can construct a vector bundle of rank $k$ on $S^n$, by using a representative of this class to define a transition function on the intersection of the "north" and "south" stereographic projection charts (which has $S^{n-1}$ as a retract).

I'd like to know: are these indeed analogous? Are they special cases of, say, a general method for constructing a smooth (respectively, complex) rank-$k$ vector bundle on a smooth (resp., complex) manifold out of a map from a codimension-one submanifold into $GL_k(mathbb{R})$ (resp., $GL_k(mathbb{C})$?

soft question - Magic trick based on deep mathematics

The following trick uses some relatively deep mathematics, namely cluster algebras. It will probably impress (some) mathematicians, but not very many laypeople.

Draw a triangular grid and place 1s in some two rows, like the following except you may vary the distance between the 1s:

1   1   1   1   1   1   1
  .   .   .   .   .   .
.   .   .   .   .   .   .
  .   .   .   .   .   .
.   .   .   .   .   .   .
  .   .   .   .   .   .
1   1   1   1   1   1   1

Now choose some path from the top row of 1s to the bottom row and fill it in with 1s also, like so:

1   1   1   1   1   1   1
  1   .   .   .   .   .
.   1   .   .   .   .   .
  1   .   .   .   .   .
.   1   .   .   .   .   .
  .   1   .   .   .   .
1   1   1   1   1   1   1

Finally, fill in all of the entries of the grid with a number such that for every 2 by 2 "subsquare"

  b
a   d
  c

the condition $ad-bc=1$ is satisfied, or equivalently, that $d=frac{bc+1}{a}$. You can easily do this locally, filling in one forced entry after another. For example, one might get the following:

1   1   1   1   1   1   1
  1   2   3   2   2   1
.   1   5   5   3   1   .
  1   2   8   7   1   .
.   1   3   11  2   1   .
  .   1   4   3   1   .
1   1   1   1   1   1   1

The "trick" is that every entry is an integer, and that the pattern of 1s quickly repeats, except upside-down. If you were to continue to the right (and left), then you would have an infinite repeating pattern.

This should seem at least a bit surprising at first because you sometimes divide some fairly large numbers, e.g. $frac{5cdot 11+1}{8} = 7$ or $frac{7cdot 3+1}{11} = 2$ in the above picture. Of course, the larger the grid you made initially, the larger the numbers will be, and the more surprising the exact division will be.

Incidentally, if anyone can provide a reference as to why this all works, I'd love to see it. I managed to prove that all of the entries are integers, and that they're bounded, and so there will eventually be repetition. However, the repetition distance is actually a simple function of the distance between the two rows of 1, which I can't prove.

sp.spectral theory - $Spin^c$-Dirac-operator on the 3-torus

Here is a possible unpromising start that hints at probable headaches. Square the Dirac to get

$$ D_alpha^2= Delta+ c(dalpha)$$

where $c(dalpha)$ denotes the Clifford multiplication by the $2$-form $dalpha$. Note that

$$ {rm spec}(D_alpha^2)= bigl(; mathrm{spec}(D_alpha);bigr)^2 $$

To find ${rm spec}(D_alpha^2)$ you need to understand spectrum of ordinary differential operators of the form

$$ -partial^2_theta + A(theta) $$

acting on functions $u: S^1 to mathbb{C}^2$ where $A(theta)$ is a $2times 2$ complex hermitian matrix depending smoothly on $thetain S^1$. I don't know how to find the spectrum of such an operator but maybe you can find something in the literature.

rt.representation theory - Conjugacy classes of reductive groups defined over local commutative rings

Background: I'm trying a problem on representations of reductive groups over various finite rings towards which this is very relevant (what I want to do is a very specialized case of this problem, and I want to know what background theory has been done for this situation in the literature and what is known about this). In characteristic $0$ over an algebraically closed field, and over finite fields, classifying conjugacy classes in reductive groups over field is a very well-known and well-studied problem (sometimes in preparation for studying representations of these).

Question: Let $R$ be a local commutative ring, either in characteristic $0$, algebraically closed, or in characteristic $p$ (algebraically closed OR finite field). If you want to complicate matters, and have an answer for non-commutative rings as well, I would be happy to see it, but I think the problem is non-trivial enough as is - and as far as I know, only $GL$ can be easily defined over non-commutative rings). Edit: Also specify that $R$ is an algebra over its residue field, and has an identity (which I believe is necessary to make the following argument work; again if you don't need this restriction feel free to not use it).

Let $G$ be a reductive group (if you want, feel free to restrict to just the classical groups, $GL$, $SL$, $Sp$, and $SO$) defined over $R$. What can be said about classifying conjugacy classes in $G$? What is clear is the Levi decomposition of $G$, as the semi-direct product of the reductive group defined over the residue field of $R$, and $N$, the set of all matrices that are congruent entry-wise to the identity matrix, modulo the maximal ideal of $R$ (the latter is the normal subgroup). Using the semi-direct product, one can say something implicit about the conjugacy classes; first by studying the conjugation action of the reductive group on the unipotent algebraic group $N$, then studying the conjugacy classes in $N$, then extending this to the whole group.

Are there any special cases of this problem that have been studied in the literature? Is there something more than can be said in general (further to what I have said above about the semi-direct product)?.

ac.commutative algebra - When is a commutative ring the limit of its local rings?

The map from A to the inverse limit of all its localizations is always injective. This boils down to the fact that the global sections of the structure sheaf O on Spec A are just A. The map from A to the global sections of O just takes A to the section which is the image of a in A_p on each stalk. So it is the map from
A to the product of all its localizations, and this is therefore injective, so the map to the inverse limit will also be.

But it does not always have to be surjective. Indeed, we can just take a commutative von Neumann regular ring that is not a product of fields. The reason this will work is that every prime ideal in a commutative VNR ring is maximal, and every localization is a field. So the inverse limit of the localizations will be a product of fields.

Here is a commutative VNR that is not a product of fields; take the subring of a countably infinite product of copies of a fixed field k consisting of sequences that are eventually constant. I got this from Lam, Lectures on Modules and Rings, Example 7.54p. 263. It is easy to see this is VNR, and surely it is not a product of fields.

ag.algebraic geometry - Endomorphisms of bundles associated to codimension 2 subvarieties

The bundle constructed from the subvariety $Z subset X$ comes in exact triple
$$
0 to L to E to J_Z to 0,
$$
where $L$ is a line bundle on $X$ extending $det N_{Z/X}$. (In case $X = P^2$ and $Z$ is a set of points, $L$ can be chosen to be arbitrary (since each line bundle on $Z$ is trivial)). So, you can use this triple to compute any cohomological invariant of $E$. For example, if you are interested in $Gamma(P^2,End E)$ you can use the spectral sequence
with the first term having the following form
$$
begin{array}{ccccc}
Hom(L,J_Z) & to & Ext^1(J_Z,J_Z) oplus Ext^1(L,L) & to & Ext^2(J_Z,L) cr
& & Hom(J_Z,J_Z) oplus Hom(L,L) & to & Ext^1(J_Z,L) cr
& & & & Hom(J_Z,L)
end{array}
$$
and converging to $Ext^i(E,E) = H^i(P^2,End E)$. So, you see that the contributions to $Gamma(P^2,End E)$ come

1) from $Hom(J_Z,L)$;

2) from $Ker(Hom(J_Z,J_Z) oplus Hom(L,L) to Ext^1(J_Z,L))$; and

3) from $Ker(Hom(L,J_Z) to Ext^1(J_Z,J_Z) oplus Ext^1(L,L))$ (here one should also take into account the $d_2$ differential).

So, everything can be computed.

ct.category theory - Image of composite morphisms

This cannot happen in a regular category. Below I give a proof using the sequent calculus of subobjects in a regular category. It can be deciphered using the book 'Sketches of an Elephant Volume 2' by Peter T. Johnstone, in particular chapter D1.

I write $beta:=I$ and $gamma:=J$. I hope the definition of image given in your book is the same as mine, namely the image of a subobject (~mono) $S$ under a morphism $phi$ is the least subobject of the codomain of $phi$ through which $phicircoverline{S}$ factors, where $overline{S}in S$.

Assume we know
$exists x(alpha(x)wedge f(x)=y) dashvvdash_{y:Y}quad beta(y)$ and $exists y(beta(y)wedge g(y)=z) dashvvdash_{z:Z}quad gamma(z)$. We then want to prove two things. The first is that $exists x(alpha(x)wedge g(f(x))=z)vdash_{z:Z}quad gamma(z)$, the second that $gamma(z)vdash_{z:Z} quad exists x(alpha(x)wedge g(f(x))=z)$.

For the first we have the following.
$alpha(x)wedge g(f(x))=z$
$vdash_{x:X,z:Z} quadalpha(x)wedge g(f(x))=z wedge f(x)=f(x)$
$vdash_{x:X,z:Z}quad alpha(x)wedge g(f(x))=z wedge beta(f(x))$
$vdash_{x:X,z:Z}quad gamma(g(f(x)))$. Therefore $alpha(x)wedge g(f(x))=zvdash_{x:X,z:Z}quad gamma(z)$ and hence $exists x(alpha(x)wedge g(f(x))=z)vdash_{z:Z}quad gamma(z)$.

The second also holds. First note that $beta wedge g(y)=z$
$vdash_{y:Y,z:Z}quad exists x(alpha(x)wedge f(x)=y)wedge g(y)=z$
$vdash_{y:Y,z:Z}quad exists x(alpha(x)wedge f(x)=ywedge g(y)=z)$
$vdash_{y:Y,z:Z}quad exists x(alpha(x)wedge g(f(x))=z)$ from which we may conclude that $gamma(z)vdash_{z:Z} quad exists y(beta(y)wedge g(y)=z)vdash_{z:Z} quad exists x(alpha(x)wedge g(f(x))=z)$.

ag.algebraic geometry - Regular functions on affine schemes

I think a good reference might be Eisenbud-Harris, Geometry of Schemes. They construct the structure sheaf $mathcal{O}$ by specifying it on principal open subsets (viz. the 'important' ones) and extending it uniquely to other open subsets.

On a given ring $R$, you have a basis of open sets of Spec $R$ consisting of the $text{D}(f)$'s.

($D(f) = Spec R - V(Rcdot f)$, where $Rcdot f$ stands for the ideal generated by $f$).

With each $D(f)$ we associate the localization $R_f$.

With a general open subset $U$ we associate the inverse limit of the $R_f$, for $D(f) subseteq U$.

More concretely, if $U = Spec R - V(I)$, then $D(f) subseteq U$ if and only if $V(I) subseteq V(Rcdot f)$ if and only if $f in sqrt{I}$. So $mathcal{O}(U)$ is the inverse limit of the rings $R_f$, for $f in sqrt{I}$.

http://en.wikipedia.org/wiki/Inverse_limit

Under what conditions does the second cohomology preserve injectivity?

Sasha's comment is correct. In your case you have the trivial action on $p$-power order cyclic modules. So let me write the map $B to B/A$ as $B = mathbb Z/p^n to
mathbb Z/p^m = B/A,$ where $m leq n$, and the map is the natural one (reduce a mod $p^n$ class to a mod $p^m$ class).

Now $H^1$ of $G$ against a trivial module is just homs of $G^{ab}$ into this module, so
we have to look at
$$Hom(G^{ab},mathbb Z/p^n ) to Hom(G^{ab},mathbb Z/p^m).$$
Now $G^{ab}$ is (in your setting) itself an abelian $p$-power order group, so is a product
of cyclic $p$-power order groups, so (since $Hom$ from a finite product is the product
of the individual $Hom$s) we see that it is enough to consider whether
$$Hom(mathbb Z/p^r,mathbb Z/p^n) to Hom(mathbb Z/p^r,mathbb Z/p^m)$$ is surjective.

Assuming that $m < n$ (i.e., in the original terms of the problem, that $A$ is non-zero,
so that the question is non-trivial), then this map is surjective if and only
if $n leq r.$ (This is not hard to check; see below for a careful explanation.)

Putting this together for all $r$, we get the following: assuming that $A$ is non-zero, that $B$
is cyclic of $p$-power order, and that $G$ is a $p$-group, then the map of $H^2$ is injective precisely when each cyclic direct summand of $G^{ab}$ has order at least
that of $B$.

Proof of surjectivity fact:
The $Hom$ space $Hom(mathbb Z/p^r,M)$ is equal to $M[p^r],$ the $p^r$-torsion subgroup of $M$,
for any abelian group $M$ (just look at the image of $1$ mod $p^r$). So we have to consider
the surjectivity (or non-surjectivity) of
$(mathbb Z/p^n)[p^r] to (mathbb Z/p^m)[p^r],$ which
is the map
$$p^{max(0,n-r)}mathbb Z/p^nmathbb Z to p^{max(0,m-r)}mathbb Z/p^n mathbb Z.$$
This is surjective if $n - r leq 0,$ or if $m = n$, but otherwise is not.

knot theory - Difference between Alexander polynomial and Blanchfield pairing

The Blanchfield pairing has many formulations, I like to think of it as a sesquilinear form:

$$ A otimes A to Lambda / mathbb Z[t^pm] $$

where $A$ is the Alexander module and $Lambda$ is the field of fractions of $mathbb Z[t^pm]$. This pairing has to be a duality isomorphism, ie: the adjoint

$$ overline{A} to Hom_{mathbb Z[t^pm]} (A, Lambda/mathbb Z[t^pm]) $$

is an isomorphism of $mathbb Z[t^pm]$-modules. $overline{A}$ is $A$ but given the opposite action of $mathbb Z[t^pm]$ (you substitute $t longmapsto t^{-1}$ before multiplication by a polynomial)

The Blanchfield pairing can be anything of that form. So you take the Alexander module, and soup it up with such an isomorphism between $overline{A}$ and its ``Ext dual'' $Hom_{mathbb Z[t^pm]} (A, Lambda/mathbb Z[t^pm]) $. That is the extra information in the S-equivalence class.

edit: the pairing has a nice geometric interpretation. $A$ is $H_1(tilde C)$ where $tilde C to C$ is the universal abelian cover of the knot complement. Since $A$ is $mathbb Z[t^pm]$-torsion, given any $[x] in A$ let $p$ be such that $px = partial X$. Then you define the pairing $langle x, yrangle = (sum_i (X cap t^{i}y)t^i)/p$ provided $X$ and $y$ are transverse representatives when projected to $C$ (in any way that that makes sense). Here $cap$ is the standard algebraic intersection number of transverse chains.

set theory - On statements independent of ZFC + V=L

There are numerous examples of such statements. Let me organize some of them into several categories.

First, there is the hierarchy of large cardinal axioms that are relatively consistent with V=L. See the list of large cardinals. All of the following statements are provably indpendent of ZFC+V=L, assuming the consistency of the relevent large cardinal axiom.

• There is an inaccessible cardinal.




• There is a Mahlo cardinal.




• There is a weakly compact cardinal.




• There is an indescribable cardinal.




• and so on, for all the large cardinals that happen to be relatively consistent with V=L.

These are all independent of ZFC+V=L, assuming the large cardinal is consistent with ZFC, because if we have such a large cardinal in V, then in each of these cases (and many more), the large cardinal retains its large cardinal property in L, so we get consistency with V=L. Conversely, it is consistent with V=L that there are no large cardinals, since we might chop the universe off at the least inaccessible cardinal.

Second, even for those large cardinal properties that are not consistent with V=L, we can still make the consistency statement, which is an arithmetic statement having the same truth value in V as in L.

• Con(ZFC)




• Con(ZFC+`there is an inaccessible cardinal')




• Con(ZFC+`there is a Mahlo cardinal')




• Con(ZFC+`there is a measurable cardinal')




• Con(ZFC+`there is a supercompact cardinal').




• and so on, for any large cardinal property. Con(ZFC+large cardinal property).

These are all independent of ZFC+V=L, assuming the large cardinal is consistent with ZFC, since on the one hand, if W is a model of ZFC+Con(ZFC+phi), then L^W is a model of ZFC+V=L+Con(ZFC+phi), as Con(ZFC+phi) is an arithmetic statement. And on the other hand, by the Incompleteness theorem, there must be models of ZFC+¬Con(ZFC+phi), and the L of such a model will have ZFC+V=L+¬Con(ZFC+phi).

Third, there is an interesting trick related to the theorem of Mathias that Dorais mentioned in his answer. For any statement phi, the assertion that there is a countable well-founded model of ZFC+phi is a Sigma¹₂ statement, and hence absolute between V and L. And the existence of a countable well-founded model of a theory is equivalent by the Lowenheim-Skolem theorem to the existence of a well-founded model of the theory. Thus, the truth of each of the following statements is the same in V as in L.

• There is a well-founded set model of ZFC. This is equivalent to the assertion: there is an ordinal α such that L_α is a model of ZFC.




• There is a well-founded set model of ZFC with ¬CH. (This is also equivalent to the previous statement.)




• There is a well-founded set model of ZFC with Martin's Axiom.




• and so on. For all the statements known to be forceable, you can ask for a well-founded set model of the theory.




• There is a well-founded set model of ZFC with an inaccessible cardinal.




• There is a well-founded set model of ZFC with a measurable cardinal.




• There is a well-founded set model of ZFC with a supercompact cardinal.




• and the same for any large cardinal notion.

These are all independent of ZFC=V=L, since they are independent of ZFC, and their truth is the same in V as in L. I find it quite remarkable that there can be a model of V=L that has a transitive model of ZFC+'there is a supercompact cardinal'. The basic lesson is that the L of a model with enormous large cardinals has very different properties and kinds of objects in it than a model of V=L arising elsewhere. And I believe that this gets to the heart of your question.

Since all these statements are studied very much in set theory, and are very interesting, and are independent of ZFC+V=L, I find them to be positive instances of what was requested.

However, how does this relate to Shelah's view in Dorais's excellent answer? He seems there to dismiss the entire class of consistency strength statements as combinatorics in disguise. What does he mean exactly? Since we set theorists are very interested in these statements, I don't think that he means to dismiss them as silly tricks with the Incompleteness theorem. Perhaps he means something like: to the extent that we believe that a large cardinal property LC is consistent, then we don't really want to consider the theory ZFC+V=L, but rather, the theory ZFC+V=L+Con(LC). That is, we aren't so interested in models having the wrong arithmetic theory, so we insist that Con(LC) if we are comitted to that. And none of the examples I have given exhibit independence from that corresponding theory.

gr.group theory - Size of the smallest group not satisfying an identity.

To make the question a little less open-ended while (I hope) retaining its spirit, let me interpret the question as asking for the rate of growth of the function $mu(k)$ defined as the maximum of $M(w)$ over all nontrivial words $w$ of length up to $k$ in any number of symbols, where length is the number of symbols and their inverses multiplied together in $w$. (E.g., $x^5 y^{-3}$ has length $8$.) George Lowther observed that $M(x^{n!})>n$, so $mu(n!)>n$. One can replace $n!$ by $operatorname{LCM}(1,2,ldots,n)$, which is $e^{(1+o(1))n}$, so this gives $mu(k) > (1-o(1)) log k$ as $k to infty$.

I will improve this by showing that $mu(k)$ is at least of order $k^{1/4}$.

Let $C_2(x,y):=[x,y]=xyx^{-1}y^{-1}$. If $C_N$ has been defined, define $$C_{2N}(x_1,ldots,x_N,y_1,ldots,y_N):=[C_N(x_1,ldots,x_N),C_N(y_1,ldots,y_N)].$$ By induction, if $N$ is a power of $2$, then $C_N$ is a word of length $N^2$ that evaluates to $1$ whenever any of its arguments is $1$.

Given $m ge 1$, let $N$ be the smallest power of $2$ such that $N ge 2 binom{m}{2}$. Construct $w$ by applying $C_N$ to a sequence of arguments including $x_i x_j^{-1}$ for $1 le i < j le m$ and extra distinct interdeterminates inserted so that no two of the $x_i x_j^{-1}$ are adjacent arguments of $C_N$. The extra indeterminates ensure that $w$ is not the trivial word. If $w$ is evaluated on elements of a group of size less than $m$, then by the pigeonhole principle two of the $x_i$ have the same value, so some $x_i x_j^{-1}$ is $1$, so $w$ evaluates to $1$. Thus $M(w)>m$. The length of $w$ is at most $2N^2$, which is of order $m^4$. Thus $mu(k)$ is at least of order $k^{1/4}$.

I have a feeling that this is not best possible$ldots$

lo.logic - What is lambda calculus related to?

I think, based on Your interests (programming, LISP), that it could work if You followed the path of a Haskell programmer:

Haskell wiki

Subpage Learning Haskell is a good place to start (see it in the leftmost column, second below Learning header).

Despite of Haskell being a LISP-successor, still, there will be a learning curve (Haskell is very clean, compared to LISP, and it reaches back to several deep logical foundations), but this learning curve can be distributed well along a larger time span, and it will yield fun during the whole time. Even the first minutes will yield pleasure (total lack of side effects, the beauty of currying, the extremely clean economy of concepts). The deeper details will come later automatically (category theory, lambda calculus, combinatory logic, type theory).