Let G be a real reductive group, and P any parabolic subgroup. In the paper 'Canonical extensions of Harish-Chandra modules to representations of $G$' by Casselman, a result says that if we begin with a smooth representation of moderate growth of P, then the smooth induced representation of G induced from P is also of moderate growth.(Proposition 4.1 in that paper)
Casselman didn't give details of the proof. Does anybody know any proof of this result in other places?
I am sorry for this rather dumb-sounding question. But I am thinking of it for the last two days and am unable to find an answer.
Let $K, L$ be an algebraic number fields, ie a finite extensions of $mathbb Q$. Let $F = KL$ be their compositum. Let $mathcal O_K, mathcal O_L, mathcal O_F$ be the rings of integers of these fields. Let $n(K), n(L), n(F)$ be the degrees of these number fields.
Now, we have a triplet $(K, mathcal O_K, i_{K} : mathcal O_K hookrightarrow mathbb{R}^{n(K)})$, where $i_K$ is the embedding $x mapsto (x^{(1)}, x^{(2)}, ldots , x^{(n(K))})$, where $x^{(1)}, x^{(2)}, ldots , x^{(n(K)}$ are the conjugates of $x$.
What bothers me in this situation is that the conjugates depend on the number field. To alleviate this situation, I go ahead as follows.
We consider the similar triplets $(L, mathcal O_K, i_{L} : mathcal O_L hookrightarrow mathbb{R}^{n(L)})$ and $(F, mathcal O_F, i_{F} : mathcal O_F hookrightarrow mathbb{R}^{n(F)})$. There is a natural embedding $mathcal O_K hookrightarrow mathcal O_F$ and similarly $mathcal O_L hookrightarrow mathcal O_F$ and these respect the inclusions of the number rings into their respective Euclidean spaces. So we can "include" the triplets for $K$ and $L$ into $F$.
Now, we order all number fields via inclusion. This is a directed set. And the set of all triplets considered above of the form $(K, mathcal O_K, i_{K} : mathcal O_K hookrightarrow mathbb{R}^{n(K)})$, is a directed system of such triplets. So we take the direct limit. The result should be some embedding of the ring of all algebraic integers into a countable dimension Euclidean space.
Question:
Does this embedding give a lattice?
I would be grateful for answers. Again I am sorry if this is a stupid question.
Such orthonormal bases do exist, as proved in:
Bourgain, J. A remark on the uncertainty principle for Hilbertian basis. J. Funct. Anal. 79 (1988), no. 1, 136--143 (MathSciNet link).
The theorem says that for each $rho>1/2$ there is an orthonormal basis for $L^2(mathbb{R})$ such that all of the variances of the basis elements and their Fourier transforms are less than $rho$. After the statement Bourgain remarks:
Thus Balian’s strong uncertainty principle does not hold for a nonperiodic
basis.
It is remarkable that this appears to have been discovered (rediscovered?) well after von Neumann's time. Powell proved more recently the result that Matt Hastings mentioned, namely that in such a case the sequence of means of the orthonormal basis is unbounded.
My old answer, posted before reading Matt Hastings's comment led me to the correct question, was to the question of whether all of the variances can be finite. It was this:
Yes, because you can take an orthonormal basis in the Schwartz space by applying Gram-Schmidt to a countable $L^2$ dense subset of the Schwartz space.
V. A. Gorbatov claims that he solved 4-color problem in 1964, and he repeats this claim in his book which appears in 2000. (The paper and the book are in Russian; the book is also translated to Spanish --- see refs below.) Obviousely, I doubt that it is true, but I do NOT want to read the book...
Question: Did anyone checked this proof?
Why do I care. This claim appears in Russian wikipedia and I would be happy to have a good reason to remove it.
(At the moment ru.wikipedia states Gorbatov's claim and says that nobody checked his proof.)
Summary of answers: I have got few answers, roughly stating that there is no sign that Gorbatov is a serious mathematician. I agree with that, but it is not an answer to my question...
This is an excellent and interesting question! You are asking whether the 2-step iteration of a normal measure μ on a measurable cardinal κ is uniquely factored by the steps of the iteration itself.
The answer is Yes.
Let me denote κ0 just by κ and j02 by j. Since
μ1 is a measure in M1, it has the
form j01(m)(κ), where m =
(να | α < κ). Since you
have said that μ1 is not in
ran(j01), we may choose the
να to be all different, and different
from μ0. In this case, there is a partition
of κ as the disjoint union of Xα,
with Xα in να and none
in μ0. Let x = (Xα | α
< κ). Note that κ is not in
j01(Xα) for any α <
κ, and similarly κ1 is not in
j(Xα). But κ is in
j01(x)(β) for some β <
κ1, since this is a partition of
κ1. Apply j12 to conclude that
κ1 is in j(x)(β) for this β.
Thus, there is some β in the interval [κ,
κ1) having the form β =
j(f)(κ1) for the function f that picks the
index. From this, it follows from normality of
μ0 that we can write κ =
j(g)(κ1) for some function g, since any
β < κ1 generates κ via
j01. In my favored terminology, the seed
κ1 generates κ via j and in fact
generates all β in [κ,κ1) via
j.
Similarly, suppose that δ is in the interval
[κ1,j(κ)). We know δ =
j12(f)(κ1) for some function f
on κ1 in M1. We also know f =
j01(F)(κ) for some F in V. Thus, δ =
j(F)(κ, κ1). In Y, let
(α,β) be the smallest pair with δ =
j(F)(α,β). It cannot be that both are below
κ1, since this would be inside
ran(j12) and so the least pair must have β
= κ1. Thus, δ generates
κ1, which we already observed generates
κ.
To summarize, every ordinal in the interval
[κ1,j(κ)) generates
κ1, which generates all the ordinals
β in [κ,κ1), any of which
generate κ and all the other such β.
This is enough to answer your question. The k " N in your
question is just an arbitrary elementary substructure of
M2 containing ran(j), so suppose we have Y
elementary in M2 and ran(j) subset Y. The case Y
= ran(j) is one of your cases. Otherwise, Y has something
not in ran(j). Every object in M2 has form
j(h)(κ,κ1) for some function h, so
by looking at the smallest pair of ordinals to generate a
given object with j(h), we see that there must be ordinals
below j(κ) in Y. If Y contains any ordinal δ in
the interval [κ1,j(κ)), then it will
contain both κ and κ1, since we
observed that any such δ generates these ordinals. In
this case, Y = M2, since those two ordinals
generate everything. So we assume that Y contains no such
δ. In this last case, Y must contain some ordinal
β in the interval [κ,κ1). Since
any such β generates κ, Y contains all such
ordinals. It follows that ran(j12) subset Y and
in fact = Y, since if Y contained anything more it would
have to have an additional ordinal δ in
[κ1,j(κ)).
So we've seen that your three cases are the only
possibilities. And like your previous question, there is no
need to assume that Y or N is somehow internally definable.
By the way, this was a problem that I had solved many years
ago for my dissertation, although perhaps other people had
also thought about it. I was interested in understanding
which pairs of ordinals (α,β) generate product
measures via an embedding j, and this question is very much
related to that.
(Click the edit history to see my previous answer, which was just about the case when μ1 is in the range of j01, a case for which the answer is no.)
To supplement David's answer: there is a standard "local-global" criterion for determining whether a quadratic field $K/mathbb{Q}$ can be embedded in a rational quaternion algebra $B/mathbb{Q}$.
For this recall that $B$ is said to be ramified at a prime number $p$ if $B_p = B otimes_{mathbb{Q}} mathbb{Q}_p$ is a division algebra. Moreover, we say that $B$ ramifies "at infinity" if $B_{infty} = B otimes_{mathbb{Q}} mathbb{R}$ is a division algebra.
It is known that a rational quaternion algebra $B$ is determined up to isomorphism by the set of ramified places $p leq infty$, that this set of places is finite and of even cardinality, and conversely for any finite set of even cardinality there is a rational quaternion algebra ramifying at these places.
Now, let $B$ be a rational quaternion algebra, $K$ a quadratic field, and $p leq infty$ a ramified place of $B$. Suppose that we have an embedding $K hookrightarrow B$. Then tensoring with $mathbb{Q}_p$ we get $K_p := K otimes mathbb{Q}_p hookrightarrow B_p$.
Now, if $p$ is inert or ramified in $K$, then $K_p$ is a quadratic field extension of $mathbb{Q}_p$, and it turns out that every such quadratic extension does indeed embed in $B_p$. However, if $p$ is split in $K$, then $K_p cong mathbb{Q}_p times mathbb{Q}_p$ has nontrivial idempotent elements, so cannot embed in $B_p$ if the latter is a division algebra. (If $p = infty$, then we say that $p$ is split in $K$ iff $K$ is a real quadratic field.)
In summary, this gives a necessary local criterion for the embeddability of $K$ into $B$: each ramified prime $p leq infty$ of $B$ is nonsplit in $K$. By the local-global theory of quaternion algebras over $mathbb{Q}$, it turns out that this necessary condition is also sufficient. In particular, the quadratic fields which embed into a given quaternion algebra are precisely those which are determined by finitely many splitting conditions.
It follows easily from this that there are infinitely many quadratic fields which embed in $B$, for instance any imaginary quadratic field $mathbb{Q}(sqrt{D})$ where $D$ is divisible by each finite ramified prime $p$ of $B$. Moreover, one can see that the set of such quadratic fields has, in some natural sense, positive density (as does its complement, unless $B cong M_2(mathbb{Q})$ in which case we recover the result that every quadratic field embeds, as one sees much more easily by a Cayley's Theorem / regular representation style argument).
Dear roger123,
This is largely a response to your question aksed as a comment below Charles Siegel's answer, but it won't fit in the comment box. Since $mathcal N_{mathbb P^n/X}$ is a sheaf of modules over the sheaf of rings $mathcal O_{mathbb P^n}$ (the structure sheaf of projective space),
its global sections $mathcal N_{mathbb P^n/X}(mathbb P^n)$ are a module over the ring $mathcal O_{mathbb P^n}(mathbb P^n)$, which in turn are just $k$ (the ground field).
In short, the global sections of $mathcal N_{mathbb P^n/X}$ form a $k$-vector space.
Maybe you are being confused by the fact that $mathcal N_{mathbb P^n/X}$ is a sheaf on $mathbb P^n$ that is supported on $X$, so that people often simultaneously regard it as a sheaf on either $X$ or $mathbb P^n$. This is okay, because if $U$ is any open in $mathbb P^n$ and $mathcal F$ is a sheaf supported on $X$, then the sections over an open subset $U$ of $mathbb P^n$
(when it is regarded as a sheaf on $mathbb P^n$) will coincide with the sections over
$Ucap X$ (when it is regarded as a sheaf on $X$).
In particular, one has the equation $mathcal N_{mathbb P^n/X}(X) =
mathcal N_{mathbb P^n/X}(mathbb P^n)$ (an abuse of notation if taken literally; however
one is supposed to regard $mathcal N_{mathbb P^n/X}$ as a sheaf on $X$ on the left-hand side,
and as a sheaf on $mathbb P^n$ on the right-hand side).
One more thing: If the residue field of the Hilbert scheme at the point $P$ is $k(P)$,
then $P$ is a $k(P)$-valued point of the Hilbert scheme,
and so the corresponding closed subscheme $X$ lies in $mathbb P^n_{k(P)}$. Thus, in the
above discussion, $k$ can (and should) be taken to be $k(P)$. Thus the above discussion explains
why $mathcal N_{mathbb P^n/X}(X)$ is a $k(P)$-vector space, as it should be.
There are a couple of ways to define an action of $pi_1(X)$ on $pi_n(X)$. When $n = 1$, there is the natural action via conjugation of loops. However, the picture seems to blur a bit when looking at the action on higher $pi_n$. All of them have the flavor of the conjugation map, but are more geometric than algebraic, and in some cases work is needed to show the map is well defined. Here are a couple I have seen:
There is a homotopy equivalence $f : S^n to S^n vee I$. taking the basepoint of $S^n$ to the endpoint of the unit interval "far away" from $S^n$. Given a path $alpha$ from $x_0$ to $x_1$, one can get a basepoint changing homomorphism $pi_n(X,x_0) to pi_n(X,x_1)$ by taking $g : S^n to X$ and mapping it to $(g vee alpha) circ f$. If $alpha$ is a loop this gives an action of $pi_1$
Another way to proceed may be to look at elements of $pi_n(X,x_0)$ as homotopy classes of maps $I^n to X$ that send $partial I^n$ to $x_0$. Then a base change homomorphism could be obtained by using a path $alpha$ to define a map $I^n cup (partial I^n times I) to X$, which can be filled in to a map $I^{n+1} to X$. Then the action would be to take the face opposite the original $I^n subset I^{n+1}$.
These both define the same standard action of $pi_1$ on $pi_n$, but lose the algebraic flavor of the group action and instead have this stronger geometric feel, which can make working with the action a bit cumbersome. Are there other ways of looking at this action that are more algebraic?
Perhaps, can something be done wherein $pi_0(Y)$ acts on $pi_n(Y)$, where $Y$ is some sufficiently nice space like $Omega X$, and does this coincide with the above defined actions? Is this a useful way of viewing the action?
OK, I guess, your first question is addressed to me. The answer is: fixed point localization. In my paper "Elliptic Gromov-Witten invariants and the mirror conjecture", a formula is found for the genus-1 (no descendants) potential of a semisimple target. It is a theorem, discovered and proved by fixed point localization when a torus acts on the target with isolated fixed points, and the GW-invariants are understood as equivariant ones. Since the answer is expressed in genus-0 data making sense for any semisimple Frobenius manifold, the conjectural extension to all such manifold is immediate. (The conjecture was proved by Dubrovin-Zhang in the sense that they showed my formula being the only candidate that would satisfy Getzler's relation.) The paper of mine you are asking about, "Semisimple Frobenius structures at higher genus", does exactly the same that the elliptic paper, but for higher genus GW-invariants, first without, and then with gravitational descendants.
After the fact, there is a more satisfying description of how that formula could have been invented. Dubrovin's connection of a semisimple Frobenius manifod allows for an asymptotical fundamental solution (which looks like the complete stationary phase asymptotics of oscillating integrals on the mirror theory). It's construction ("the $R$-matrix") is contained in the key lemma in that elliptic paper I've mentioned. Another way to interpret this solution is to say - in terms of overruled Lagrangian cones in symplectic loop spaces as the objects that describe genus-0 theory in lieu of Frobenius structures - that the overruled Lagrangian cone of a semisimple Frobenius manifold is isomorphic to the Cartesian product of several such cones corresponding to the one-point target space, and moreover, the isomorphism is accomplished by transformations from the twisted loop group: $$ L = M (L_{pt}times cdots times L_{pt}).$$ The "mysterious" conjectural higher genus formula simply says that the same relation persists for the total descendant potentials of higher genus theory:$$D sim hat{M} (D_{pt}otimescdotsotimes D_{pt}),$$ where the elements of the loop group are quantized, and the equality is replaced by proportionality up to a non-zero "central constant".
This is a request for references about a peculiar categorical
construction I've run into in some work I've been doing, and about which I'd
like to learn as much as I can.
Let $mathrm{Cat}$ be the category of small categories, and let
$mathrm{PSh}(C)$ be the category of presheaves of sets on a category $C$.
Suppose we
are given a "reasonable" endofunctor $Xicolon mathrm{Cat}to
mathrm{Cat}$. I want to consider a certain "intertwining" functor
$$
Vcolon Ximathrm{PSh}(C) to mathrm{PSh}(Xi C)
$$
defined by the formula
$$
(VX)(gamma) = mathrm{Hom}_{Ximathrm{Psh}(C)}(Agamma, X),
$$
where $X$ is an object of $Ximathrm{PSh}(C)$, $gamma$ is an object
of $Xi C$, and $Acolon Xi Cto
Ximathrm{PSh}(C)$ is the functor obtained by
applying $Xi$ to the Yoneda functor $Cto mathrm{PSh}(C)$.
Note: it's unreasonable to expect for a randomly chosen $Xi$ that the
category $Xi mathrm{PSh}(C)$ is even defined, since
$mathrm{PSh}(C)$ is a large
category, and $Xi$ is given as a functor on small categories. And even if it is defined, it's unreasonable to expect that $V$ is
well-defined, since $(VX)(gamma)$ may not be a set. But here are
some reasonable examples:
Let $Xi C= Ctimes C$. Then $Vcolon mathrm{PSh}(C)times
mathrm{PSh}(C)to mathrm{PSh}(Ctimes C)$ is the "external
product" functor, which takes a pair of presheaves $(X_1,X_2)$ on $C$ to
the presheaf $(c_1,c_2) mapsto X_1(c_1)times X_2(c_2)$ on $C^2$.
You can generalize this by considering $Xi C= mathrm{Func}(S,C)$,
where $S$ is a fixed small category.
Let $Xi C = C^{mathrm{op}}$. Then $Vcolon
mathrm{PSh}(C)^{mathrm{op}} to mathrm{PSh}(C^{mathrm{op}})$ is
a sort of "dualizing" functor, which sends a presheaf $X$ on $C$
to the presheaf $cmapsto mathrm{Hom}_{mathrm{PSh}(C)}(X, Fc)$ on
$C^mathrm{op}$; here $Fcolon Cto mathrm{PSh}(C)$ represents the
Yoneda functor.
Let $Xi C=mathrm{gpd} C$, the maximal subgroupoid of $C$.
Then $Vcolon mathrm{gpd}\,mathrm{PSh}(C)to
mathrm{PSh}(mathrm{gpd}C)$ is such that $(VX)(c)$ is the set of
isomorphisms between $X$ and the presheaf represented by $c$.
The sorts of questions I have include the following.
What makes a functor $Xi$ reasonable? Is it enough if it's
accessible?
I think $V$ should be the left Kan extension of the Yoneda
functor $Bcolon Xi Cto mathrm{PSh}(Xi C)$ along $A$. Is this
true? When can I expect to have $VAapprox B$?
How does $V$ of a composite $Xi Psi$ relate to the composite of the
$V$s of each term?
Given a functor $fcolon Cto D$, you get a bunch of functors
between the associated presheaf categories. How does $V$ interact
with such functors?
There's really only one or two examples of $Xi$ that really I need to
understand this for, and I don't want to spend time working out a general theory of this thing. It would be most convenient if someone can
point me to a reference which talks about this construction. Even one
that deals with particular instances of it would be helpful.
This is not an answer to your question. Rather, it is a "less mysterious" version of the Milnor-Stasheff construction of the Stiefel-Whitney classes, which doesn't refer explicitly to Steenrod operations. (I think I learned about this from my thesis advisor when I was a lad ... it was so long ago ...)
Let $Vto X$ be a real vector bundle. Let $S^infty=bigcup S^n$, the infinite dimension sphere. Taking product with $S^infty$ gives a vector bundle $Vtimes S^inftyto Xtimes S^infty$. I produce a vector bundle $V'to Xtimes RP^infty$ by dividing out by an action of the cyclic group of order $2$ on both base and total space:
The Euler class $e(V')$ of $V'$ is an element of degree $n$ in $H^*(Xtimes RP^infty; Z/2) = H^*(X;Z/2)[t]$. The following formula holds:
$$ e(V') = t^n + w_1(V)t^{n-1}+cdots + w_n(V).$$
So if you have an Euler class, then you can use this as the definition of the Stiefel-Whitney classes. The mod-2 Euler class is fairly easy to define from Milnor-Stasheff's point of view: $e(V')$ is the pullback along the $0$ section of the orientation class in the cohomology of the Thom space of $V'$.
It's easy to check the axioms for this guy. It's certainly natural, since $Vmapsto V'$ and $e$ are functorial. Whitney sum follows from $(Voplus W)'approx V'oplus W'$ and the Whitney sum formula for the Euler class. If $Rto *$ is the trivial bundle, then $R'to RP^infty$ is the canonical line, so $e(R')=t$ and so $w_0(R)=1$ and $w_1(R)=0$. You can use this to show that $w_0(V)in H^0X$ is equal to $1$ for any bundle over $X$, by pulling back $V$ over any point of $X$ (where it becomes trivial). If $Lto RP^infty$ is the canonical line, then $L'to RP^inftytimes RP^infty$ is $L_1otimes L_2$, the tensor product of the canonical line bundles over each factor. So $e(L')=s+t= 1cdot t^1 + scdot t^0$, giving $w_0(L)=1$ and $w_1(L)=s$ (where $sin H^1RP^infty$ is the generator).
Added later. I wrote the above while I was a bit feverish :). It didn't occur to me when describing it that it's a pretty standard way to construct characteristic classes; the variant which gives chern classes is probably more familiar.
I also said that it's a "version" of the Steenrod operation construction of SW classes, so let me try to explain that. I'll sketch a "direct" proof that the Steenrod operation definition of SW classes is equivalent to the one I gave above (i.e., without refering the axioms that M-S give for SW classes).
Steenrod operations come from an "extended square" construction on cohomology classes (see my answer in Why does one think to Steenrod squares and powers?). If $X$ is a space, let $DX=(Xtimes X times S^infty)/(Z/2)$, where I divide by the involution $(x_1,x_2,y)to (x_2,x_1,-y)$. The "extended square" is a function
$$P: H^n(X) to H^{2n}(DX).$$
Cohomology is with mod-2 coefficients. If you restict along the "diagonal" embedding $d: Xtimes RP^infty to DX$, you get Steenrod squares:
$$d^*(P(a)) = t^{n}Sq^0(a) + t^{n-1}Sq^1(a) + cdots + Sq^n(a).$$
There's a relative version of this: if $Vto X$ is a vector bundle, so is $DVto DX$; write $T(V)$ for the Thom space of $V$, and write $f: T(V) to T(DV)$ for the map induced by diagonal inclusion. If $uin H^nT(V)$ is the orientation class, then
$$f^*(P(u))= t^{n}Sq^0(u)+t^{n-1}Sq^1(u)+cdots +Sq^n(u).$$
According to Milnor-Stasheff, $Sq^i(u)=u,w_i(V)$.
The neat fact is that $P(u)in H^{2n}T(DV)$ has to be the orientation class $u'$ of $DVto DX$! So as long as I can describe the orientation class, I don't need to know about Steenrod opeartions! Thus, $f^*(u')in H^*TV[t]$ is the polynomial whose coefficients are the SW classes. To get the formula I gave originally, observe that $f^*(u')=u, e(V')$; this is because the pullback of the bundle $TVto TX$ along $d: Xto DX$ is the same as the bundle $V+V' to X$.
Why is $P(u)$ the orientation class of $DV$? The orientation class of a bundle in ordinary cohomology mod-2 is the unique element which restricts to the fundamental class of the sphere when you restrict to each fiber, so you just have to check that $P(u)$ has this property. And this is pretty easy (the operation $P$ is natural, and it's easy to understand how $P$ works when you have a discrete space, or a bundle over a discrete space.)
I'm not quite sure the best way to ask this, so bear with me: Does anyone know of a subset of integers such that, for any odd prime p, the subset only occupies (p-1)/2 equivalence classes mod p (and does so uniformly)?
For example, take the subset of squares. Elementary number theory shows that they (as quadratic residues) occupy (p+1)/2 equivalence classes mod p. But the answer to the above is not to take the non-residues since being a non-residue is a local property, not a property of an integer.
It is possible to construct such a set of integers one element at a time in an ad hoc manner using some initial members, a whole lot of CRT, and making a somewhat arbitrary choice at each step. But is there a more ``well-known'' set that has this property?
I am looking for a good reference on composite residues of multi-variable contour integrals (something better and more explicit than Griffiths and Harris or Tsikh). This means I want to evaluate $oint frac{d^nz}{p(z)}$ where $p(z)$ is a homogeneous degree $n$ polynomial in $n$ variables, and I want the residue at $z = 0$.
I understand that this is a difficult problem in general, but in my particular case one can regard the $z$ variables as entries in a matrix, and $p(z)$ is a product of various minor determinants of this matrix. I know what the answer should be ahead of time, and I know that it's pretty, but I can't prove it, so I'd be interested in references where similar residues have been computed.
Thanks,
Jared
Even worse! ;) You might have a Hausdorff countably compact space
which is not locally countably compact in the stronger sense that every
point has a neighborhood base of countably compact sets (of course, for Hausdorff spaces, the strong and weak forms of "locally compact" are equivalent; but you need $T_3$ to get the corresponding equivalence for "locally countably compact").
Consider $ omega_2$, the third infinite cardinal, with the usual order topology, and make the topology finer by adding $A= { alpha in omega_2 | alpha text{ has cofinality } omega_1 }$ as an open set. So, a base for the topology is given by the open intervals together with the sets of the form $L cap A$, for $L$ an open interval.
With this topology, $ omega_2$ remains countably compact, since every infinite
subset has a complete accumulation point. It obviously remains Hausdorff. Notice that, with this topology,
any ordinal of cofinality $ omega_1$ is an isolated point,
provided it is not a limit of a set of ordinals of cofinality $ omega_1$.
If you consider instead an ordinal $alpha$ of cofinality $ omega_1$ such that $alpha$ is also a limit of ordinals of cofinality $ omega_1$, then $A$ is a neighborhood of $alpha$, but no neighborhood $U$ of
$alpha$ contained in $A$ is countably compact, since any such $U$ contains
a closed infinite discrete set, by the above remark (I mean, the infinite discrete set is closed in U).
I am sure that this example or something similar appears in the literature, but I do
not recall where. Probably there are also simpler counterexamples (in the counterexample above, too, you can start with an ordinal much smaller than $ omega_2$, of course). I just found that N. Noble (Two examples on preimages of metric spaces , Proc. Amer. Math. Soc. 36 (1972), 586-590) mentions that the existence of
such counterexamples is relevant in constructing other counterexamples about $k$-spaces.
A lot of time has gone by, but I think it might be useful to point this out.
This question is about the space of all topologies on a
fixed set X. We may order the topologies by refinement, so
that τ ≤ σ just in case every τ open set is open in σ.
Equivalently, we say in this case that τ is coarser
than σ, that σ is finer than τ or that
σ refines τ. (See wikipedia on comparison of
topologies.)
The least element in this order is the indiscrete topology and the largest topology is the discrete topology.
One can show that the collection of all topologies on a fixed set is a complete lattice. In the downward direction, for example, the intersection of any collection of
topologies on X remains a topology on X, and this intersection
is the largest topology contained in them all. Similarly,
the union of any number of topologies generates a smallest
topology containing all of them (by closing under finite
intersections and arbitrary unions). Thus, the collection of all topologies on X is a complete
lattice.
Note that the compact topologies are closed downward in
this lattice, since if a topology τ has fewer open sets than
σ and σ is compact, then τ is compact.
Similarly, the Hausdorff topologies are closed upward,
since if τ is Hausdorff and contained in σ, then
σ is Hausdorff. Thus, the compact topologies inhabit
the bottom of the lattice and the Hausdorff topologies the
top.
These two collections kiss each other in the compact
Hausdorff topologies. Furthermore, these kissing points,
the compact Hausdorff topologies, form an antichain in the
lattice: no two of them are comparable. To see this,
suppose that τ subset σ are both compact
Hausdorff. If U is open with respect to σ, then the
complement C = X - U is closed with respect to σ and
hence compact with respect to σ in the subspace
topology. Thus C is also compact with respect to τ in
the subspace topology. Since τ is Hausdorff, this
implies (an elementary exercise) that C is closed with respect to τ, and so U is
in τ. So τ = σ. Thus, no two distinct compact Hausdorff topologies are comparable, and so these topologies are spread out sideways, forming an antichain of the lattice.
My first question is, do the compact Hausdorff topologies
form a maximal antichain? Equivalently, is every topology
comparable with a compact Hausdorff topology? [Edit: François points out an easy counterexample in the comments below.]
A weaker version of the question asks merely whether every
compact topology is refined by a compact Hausdorff
topology, and similarly, whether every Hausdorff topology
refines a compact Hausdorff topology. Under what
circumstances is a compact topology refined by a unique
compact Hausdorff topology? Under what circumstances does a
Hausdorff topology refine a unique compact Hausdorff
topology?
What other topological features besides compactness and
Hausdorffness have illuminating interaction with this
lattice?
Finally, what kind of lattice properties does the lattice
of topologies exhibit? For example, the lattice has atoms,
since we can form the almost-indiscrete topology having
just one nontrivial open set (and any nontrivial subset
will do). It follows that every topology is the least upper
bound of the atoms below it. The lattice of topologies is
complemented.
But the lattice is not distributive (when X has at least
two points), since it embeds N5 by the
topologies involving {x}, {y} and the topology generated by
{{x},{x,y}}.
So I was wondering: are there any general differences in the nature of "what every mathematician should know" over the last 50-60 years? I'm not just talking of small changes where new results are added on to old ones, but fundamental shifts in the nature of the knowledge and skills that people are expected to acquire during or before graduate school.
To give an example (which others may disagree with), one secular (here, secular means "trend over time") change seems to be that mathematicians today are expected to feel a lot more comfortable with picking up a new abstraction, or a new abstract formulation of an existing idea, even if the process of abstraction lies outside that person's domain of expertise. For example, even somebody who knows little of category theory would not be expected to bolt if confronted with an interpretation of a subject in his/her field in terms of some new categories, replete with objects, morphisms, functors, and natural transformations. Similarly, people would not blink much at a new algebraic structure that behaves like groups or rings but is a little different.
My sense would be that the expectations and abilities in this regard have improved over the last 50-60 years, partly because of the development of "abstract nonsense" subjects including category theory, first-order logic, model theory, universal algebra etc., and partly because of the increasing level of abstraction and the need for connecting frameworks and ideas even in the rest of mathematics. I don't really know much about how mathematics was taught thirty years ago, but I surmised the above by comparing highly accomplished professional mathematicians who probably went to graduate school thirty years ago against today's graduate students.
Some other guesses:
These are all just conjectures. Contradictory information and ideas about other possible secular trends would be much appreciated.
The definition of $U'$ doesn't depend on the Topology on $Y$.
Given any map $f:Xrightarrow Y$, we can consider the equivalence relation $x'sim x$, iff $f(x)=f(x')$ on $X$ and factor the map $f$ as $Xrightarrow X/sim quad rightarrow Y$. Call the first $f'$. Note that,the second map is injective. Hence we get the same set $U'$ for any set $U$, if we replace $f$ by $f'$.
So we might assume without restriction, that $f:Xrightarrow X/sim $ is a quotient map. Then the upper statement is equivalent to saying, that $f$ is closed.
Suppose $f$ is closed. Then one gets for the complements $U'^c=f^{-1}(f(U^c))$. And hence $U'^c$ must be closed.
There is the closed map lemma, which shows, that under the nice conditions mentioned in the comments above properness is also sufficient.
Now suppose $f:Xrightarrow X/sim$ is not closed and a quotient map. Then there is a closed set $Asubset X$, such that its image is not closed. By definition of the quotient topology, a subset of $X/sim$ is closed, if and only if its preimage is closed. So $f^{-1}(f(A))$ is not closed and hence $U=A^c$ gives the desired counterexample.
So we get: A map $f:Xrightarrow Y$ has the property, if and only if the induced map $Xrightarrow X/sim$ is closed.
This question may end up [closed], but I'm going to ask and let the people decide. It's certainly the kind of question that I'd ask people at tea, and it's not one whose answer I've been able find with Google.
TeX, I have heard, is Turing complete. In theory, this means that we can do modular arithmetic with LaTeX programs. I'd like to know how this can be done in practice.
Background: I've been using the foreach command in TikZ to draw NxN arrays of nodes, indexed by pairs of integers (m,n). I'd like to be able to use modular arithmetic and an ifthenelse statement to put different decorations on the nodes, depending on the value of (m+n) mod p. Obviously, one can just do this by hand. But that's not the world I want to live in.
So, a basic set-up in modern enumerative geometry is that you have some object $X$ (say, a "nice" stack, for whatever definition of "nice" you need) and then you want to "count" the curve of genus $g$ intersecting a bunch of cohomology classes and of degree $beta$, so you look at $bar{mathcal{M}}_{g,n}(X,beta)$, then pull back the classes, intersect, and get a Gromov-witten Invariant. Famously, this gives the Kontsevich formula counting rational curves in the projective plane passing through $3d-1$ points. And though GW-invariants can be negative and rational, there are nice cases where they do count something legitimate, such as the genus $0$, $ngeq 3$ case into a homogeneous space.
So, enough background, here's my question (and this is largely idle curiosity, so no specific motivation): can we do this in higher dimensions? For instance, given a (smooth?) variety $V$ and marking a bunch of subvarieties $W_1,ldots,W_n$ (maybe restricting them to points?) can we form $bar{mathcal{M}}_{V,(W_1,ldots,W_n)}(X,beta)$ a moduli space of stable mappings of varieties deformation equivalent to the one we started with into our space, represented by a given cohomology class $beta$ and with $W_1,ldots,W_n$ intersecting some cohomology classes, so that we can get something that can be called higher dimensional Gromov-Witten invariants? If this has been studied, under what conditions does it actually count subvarieties? For instance, if $X$ is $mathbb{P}^N$ and $V=mathbb{P}^2$, and maybe if we loosen things to just needing rational maps, could we use something like this to count rational surfaces, satisfying some incidence conditions?
EDIT: since I'm not an expert on Banach spaces, I feel I shouldn't say anything more, but anyway; an essential ingredient is an exact formula in Hilbert spaces for $|x + y|^2$. Just an idea (perhaps I am being stupid): maybe if you have a Banach space where $| x + y | = F(|x|, |y |, g(x,y))$ for some reasonably simple functions $F$, $g$ then something can be said.
If you examine the proof for Hilbert spaces, it makes essential use of the scalar product; so it's not really surprising that it doesn't work for general Banach spaces. The norm is a nice enough structure to do a lot of things, but not that nice.
It also demonstrates that Banach spaces have far more detailed structure than just ordinary vector spaces, but that Hilbert spaces have even more structure again. In fact, there are many properties Hilbert spaces have which general Banach spaces don't (and many which even characterise Hilbert space uniquely).
With luck, this will be blunder-free (and if it's not, please tell me!):
Consider the short exact sequence
$$1 to mu_2 to overline{mathbb Q}^{times} to overline{mathbb Q}^{times}
to 1,$$
with the third arrow being squaring, and with trivial $G_k$-action. Passing to cohomology,
the sequence of $H^0$s is just this same sequence, and the sequence of higher cohomology
becomes
$$0 to Hom(G_K,mu_2) to Hom(G_K,overline{mathbb Q}^{times}) to
Hom(G_K,overline{mathbb Q}^{times}) to H^2(G_K,mu_2),$$
with the last arrow being the obstruction you asked about.
Added: See Brian Conrad's comment below for a cleaner point of view, showing
that $H^2(G_K,mu_2)$ is the precise obstruction space.
Yes, the κth fixed point will have the same cofinality as κ, since the κ many earlier fixed points are cofinal in κ.
More generally, for any function f from the ordinals to the ordinals, it is not difficult to see that the collection of ordinals α for which f " α subset α, that is, the closure points of f, will form a closed unbounded class C. And for any limit ordinal β, the β-th element of C has the same cofinality as β, since it is the limit of the β many smaller elements of C.
In particular, if we consider the beth function, where α maps to Bethα, then β is a closure point of this function if and only if Bethα < β for all α < β, and this implies Bethβ = β. So the club C in this case consists of the Beth fixed points.
I recently heard the following puzzle: There are three nails in the wall, and you want to hang a picture by wrapping a wire attached to the picture around the nails so that if any one nail is removed the picture still stays but if any two nails are removed then the picture falls down. An answer, schematically, is given by abca-1b-1c-1 (i.e., wrap it clockwise around each of the three nails in some order, then counterclockwise around each of the three nails in the same order).
This reminded me very much of the Borromean rings, where three rings are linked but when any one ring is removed the other two become unlinked. So I was trying to figure out if there might be some way to transform one situation into the other. My first instinct was to put the rings in S3 and have one of them pass through ∞, but that isn't really right. What seems to be tripping me up is that with the picture there's an extra object (the wire) that doesn't show up with the Borromean rings, but I have a vague idea that perhaps we could change the former situation by saying that we make the loop in the complement of three unlinked rings, and then perhaps "pulling really hard on the wire" would somehow thread the rings together. Maybe my issue is that what's really going on with the picture is just that we're making a loop in the complement of three points in plane, and I'm confounding phenomena of different dimensions...
Does anyone have any ideas?
My undergraduate advisor said something very interesting to me the other day; it was something like "not knowing quantum mechanics is like never having heard a symphony." I've been meaning to learn quantum for some time now, and after seeing it come up repeatedly in mathematical contexts like Scott Aaronson's blog or John Baez's TWF, I figure I might as well do it now.
Unfortunately, my physics background is a little lacking. I know some mechanics and some E&M, but I can't say I've mastered either (for example, I don't know either the Hamiltonian or the Lagrangian formulations of mechanics). I also have a relatively poor background in differential equations and multivariate calculus. However, I do know a little representation theory and a little functional analysis, and I like q-analogues! (This last comment is somewhat tongue-in-cheek.)
Given this state of affairs, what's my best option for learning quantum? Can you recommend me a good reference that downplays the historical progression and emphasizes the mathematics? Is it necessary that I understand what a Hamiltonian is first?
(I hope this is "of interest to mathematicians." Certainly the word "quantum" gets thrown around enough in mathematics papers that I would think it is.)
Is there any efficient algorithm for counting number of digits in n! without actually calculating n!?
[Addendum -- PLC]: I voted to close the question as "no longer relevant" because of Gerry's answer that one could just use Stirling's formula, as supplemented by a comment which referred to a formula of Kamenetsky given on the online journal of integer sequences. It seems now that it is an open question whether Kamenetsky's formula always (rather than just "most of the time") gives exactly the right answer, so there is more here than I had realized. A followup question which provides more context has been asked here:
How good is Kamenetsky's formula for the number of digits in n-factorial?
I can't speak for these authors, but what I understand by a "Fourier-Mukai" transform between Fukaya categories is the functor between extended Fukaya categories associated with a Lagrangian correspondence. I expect these will appear a good deal in the next few years, in symplectic topology and its applications to low-dim topology and mirror symmetry (cf. these papers of Auroux and Abouzaid-Smith).
A Lagrangian correspondence from $X$ to $Y$ is an embedded Lagrangian submanifold of $Xtimes Y$ with symplectic form $(-omega_X)oplusomega_Y$. One can take the graph of a symplectomorphism, for instance, or the vanishing moment-map locus $mu^{-1}(0)$ as a correspondence from a Hamiltonian $G$-manifold $M$ to the quotient $mu^{-1}(0)/G$. According to Wehrheim-Woodward, a generalised Lagrangian in $X$ is a sequence of symplectic manifolds $X_0=pt., X_1, X_2,dots,X_d=X$ and Lagrangian correspondences $L_{i,i+1}$ from $X_i$ to $X_{i+1}$. Generalised Lagrangians (subject to the usual sorts of restrictions and decorations) form objects in the extended Fukaya category $F^{sharp}(X)$, whose $A_infty$-structure is under construction by Ma'u-Wehrheim-Woodward. If it happens that two adjacent Lagrangian correspondences have a smooth, embedded composition, say $L' = L_{i+1,i+2}circ L_{i,i+1}$, then deleting $X_{i+1}$ from the chain and substituting $L'$ for its two factors results in an isomorphic object; see arxiv:0905.1368.
The geometric mechanism behind this is the idea of "pseudo-holomorphic quilts", see arXiv:0905.1369, arxiv:math.SG.0606061.
Whilst $F^sharp(X)$ seems even less tractable than the Fukaya category $F(X)$, we expect that in many cases the embedding $F(X)to F^sharp(X)$ induces a quasi-isomorphism of module categories.
A Lagrangian correspondence from $X$ to $Y$ induces a "Fourier-Mukai" $A_infty$-functor between extended Fukaya categories. Even better, we expect that in this way one gets an $A_infty$-functor $F(X_-times Y)to hom(F^{sharp}(X),F^{sharp}(Y))$.
Interpolated paragraph, in response to Kevin's query: The definition of the $A_infty$-functor associated with a correspondence $C$ from $X$ to $Y$ is very simple, at least on objects. An object of $F^sharp(X)$ is a chain of Lagrangian correspondences, beginning at the 1-point manifold and ending at $X$. We just tack $C$ to the end of this sequence. The clever thing about this formalism is that when there's a nice geometric way to pass Lagrangians through a correspondence (for instance taking the preimage of $Lsubset mu^{-1}(0)/G$ in $mu^{-1}(0)subset M$) the geometric and formal approaches give quasi-isomorphic objects in $F^sharp(Y)$. (Usually you can't pass a Lagrangian submanifold through a correspondence without making a terrible mess - hence the formal approach.)
For me, all this is exciting because we can at last compute Floer cohomology using its naturality properties, rather than by direct attack on the equations.
[In the new paper that inspired your question, the authors suggest that their F-M kernels should be coisotropic branes in the sense of Kapustin-Orlov. Floer cohomology for such branes is supposed to be some weird mixture of pseudo-holomorphic discs and Dolbeault cohomology over a holomorphic foliation - but there is no concrete proposal on the table
and for the moment this is just an intriguing idea. For the purposes of homological mirror symmetry, idempotent endomorphisms in the Fukaya category apparently provide the enlargement that K-O observed was necessary.]
I have read in the literature about quasi-trigonal curves. Such a curve C is a hyperelliptic curve X with two points p,q identified (basically a pinch). They seem to be pretty important in the theory of Prym varieties, since the prym of a double cover of a q-t curve is the jacobian of an hyperelliptic curve B.
There are a few things that are not completely clear to me:
first, can the two points p,q identified be one fiber of the hyperelliptic pencil on X?
second, can the hyperelliptic curve B be reconstructed (say, find its weierstrass points) starting from the data of the weierstrass points of X plus p and q?
Third, on the other hand the canonical model of a q-t curve sits on a rational normal cone, the singular point coinciding with the vertex of the cone.
Do the ruling of the cone induce the hyperelliptic map on C? Is there a model of C on a Hirzebruch surface or one just needs to blow up the cone in the vertex?
Primes of that form are very rare for n > 2 so you can't expect that they fill an arithmetic progression. Heath-Brown proved that there are infinitely many primes which are sums of two cubes. The proof is hard.
See corrections below...
EDIT, Will Jagy. Evidently Felipe posted from an ipod and was not able to edit at length. He will be able to fix everything when we he settles down his weary feet from his current travels. He meant the bit about such primes being rare. From his comment, it seems most likely that he meant to say originally that there are infinitely many primes of shape $$ x^3 + 2 y^3.$$
Moving on to Victor's comment, primes of shape $ a^3 + b^3$ actually solve
$ p = 3 a^2 - 3 a + 1.$ This means that the number of such primes up to some large positive $M$ is at most $$ c_4 sqrt M$$ which is asymptotically much less than the number of primes up to the same $M$ in some arithmetic progression, that being $$ c_5 frac{M}{log M} $$
So, infinitely many primes, indeed a set of full density, fail to have such an expression.
Finally, it would be a remarkable thing if it were known that there were infinitely many such primes. We certainly do not know the status for primes $n^2 + 1.$ Indeed, I am moved to ask: is there any polynomial in one variable, integer coefficients, and degree at least two that is known to represent infinitely many primes?
He then gives a link to the Friedlander-Iwaniec theorem, maybe I have it right.
http://en.wikipedia.org/wiki/Friedlander%E2%80%93Iwaniec_theorem
For Felipe, Whenever We May Find Him: $$ $$
http://www.youtube.com/watch?v=buMH_tAu2Zg
It is perhaps instructive to note that there is an open subset of $mathbb{R}$ which has finite measure and contains all rational numbers. Just number the rational numbers as $mathbb{Q}={q_1,q_2,ldots,}$ and let $$U=bigcup_{i=1}^infty (q_i-2^{-i-1},q_i+2^{-i-1}).$$ The measure of $V$ is less than 1 (you can make it less than some prescribed $epsilon$ easily enough). And the measure of open sets seems easier to understand, since any open set is the disjoint union of open intervals and the measure is the sum of the length of these intervals. (But note that the intervals defining $U$ above are not disjoint; there will be a great deal of overlap, so overlapping intervals join together to make bigger intervals …)
This is a bit counterintuitive of course, since there doesn't seem to be a whole lot of room outside of $U$, yet the complement of $U$ has infinite measure. If you think about it, it must mean that even though $U$ is the countable union of open intervals, the complement of $U$ has an uncountable number of components, each consisting of just one point. (Any connected subset of $mathbb{R}$ is an interval, but an interval containing no rational point must be a singleton.)
I have the following setup:
$X, Y$ are topological spaces (if it helps, they can both be $T_1$ and normal. They can even be countably paracompact. They can't be assumed paracompact). $V$ is a normed space (it can be Banach if you like). $f : X to Y$ is a perfect surjection.
I have continuous and bounded $g : X to V$ and given $epsilon > 0$ would like to find continuous $h : Y to V$ such that $d(h(x), g(f^{-1}(x))) < epsilon$
Is there some sort of selection theorem that will let me do this? I've used the Michael selection theorem to good effect elsewhere, but it doesn't apply here due to the lack of convexity of the target sets (even if they were convex the hypotheses don't apply due to potential non-paracompactness of Y, but one might be able to work something out using countable paracompactness and compactness of the targets).
I found Teruyoshi Yoshida's exposition of the subject very helpful:
http://www.dpmms.cam.ac.uk/~ty245/Yoshida_2003_introDL.pdf
As JT commented, the curve you wrote down is really the Deligne-Lusztig variety for SL_2, not GL_2. Ben is also right about the curve being $mathbf{P}^1 - mathbf{P}^1(mathbf{F}_q)$, only he is using a different definition of DL variety from you I would presume. The way Ben has it, the DL variety is a subvariety of G/B, but the curve you want is a subvariety of G/U, where U is the unipotent radical. One formulation is a cover of the other with galois group equal to the rational points on a twist of the torus T. We'll take the G/U point of view here.
So let's start with $G=text{SL}_2$ over the field $mathbf{F}_q$. We'll let $B$ be the usual Borel and $U$ its unipotent radical. We can then identify $G/B$ with $mathbf{P}^1$ and $G/U$ with $mathbf{A}^2$. The latter identification sends $(a,b,c,d)$ to $(a,c)$.
Let $w=(0,1,1,0)$ be the nontrivial Weyl element. We let $X_w$ be the subvariety of $G/B$ consisting of elements $x$ for which $x$ and $F(x)$ are in relative position $w$, where $F$ is the Frobenius map. This is $mathbf{P}^1 - mathbf{P}^1(mathbf{F}_q)$ as Ben says.
For cosets $x,yin G/B$ in relative position $w$, and a coset $gUin G/U$ for which $gB=x$, we are going to define a new coset $w_{x,y}(gU)in G/U$ as follows. First find a $g'in G$ for which $g'B=x$ and $g'wB = y$. We may further take $g'$ so that $g'U=gU$. (This can be done because of the Bruhat decomposition of $G/B times G/B$--wait a moment to see how this plays out for $text{SL}_2$.) Then define $w_{x,y}(gU) = g'wU$. (Pardon the abuse of notation of the symbol $w$.) The Deligne-Lusztig variety $Y_w$ is defined as the set of $gUin G/U$ for which $F(gU)=w_{gB,F(gB)}(gU)$.
When does a point $(x,y)inmathbf{A}^2=G/U$ lie in $Y_w$? We need to calculate $w_{gB,F(gB)}(gU)$, where $g=(x,*,y,*)in G$. We have $gB=gcdotinfty=x/y$ and $F(gB)=(x/y)^q$. So we must now find $g'in G$ with $g'U=gU$ and $g'wB=F(g)wB$. The first condition means that $g'=(x,*,y,*)$ and the second means that $g'cdot 0=(x/y)^q$. Thus $g'=(x,ux^q,y,uy^q)$, where $u$ must satisfy $u(xy^q-x^qy)=1$. We find that $w_{gB,F(gB)}(gU)=g'wU=(ux^q,uy^q)$. The condition that $(x,y)in Y_w$ is exactly that $(x^q,y^q)=(ux^q,uy^q)$, which implies that $u^{-1}=x^qy-xy^q=1$. So that's the equation for the Deligne-Lusztig variety.
The equation for the DL variety for the longest cyclic permutation in the Weyl group of $text{SL}_n$ is $det(x_i^{q^j})=1$, where $0leq i,jleq n-1$.
I believe Lusztig calculated the zeta functions of his varieties in a very general setting, but I was never able to trudge through it all. There must be a simple answer for the behavior of the zeta functions for the $text{GL}_n$ varieties--if you ever write it up I'd certainly love to read it! I can start you off: for $text{SL}_2$ over $mathbf{F}_q$, the DL curve has a compactly supported $H^1$ of dimension $q(q-1)$, and the $q^2$-power Frobenius acts as the constant $-q$. (The behavior of the $q$-power Frobenius might be a little subtle--I suspect it has to do with Gauss sums.)
Good luck!
Let $a_1,cdots, a_n$ be integers such that $a_igeq 2$ for all $i$ and $k>0$ another integer.
I am interested in whether there exist integers $x_1,cdots, x_n$ with $0<x_i<a_i$ satisfying:
$$ frac{x_1}{a_1}+ cdots + frac{x_n}{a_n}=k (*)$$
For example, if $n=2,k=1$, there exist a solution in the specified range iff $text{gcd}(a_1,a_2) neq 1$, in other words $text{lcm}(a_1,a_2)-a_1a_2 neq 0$.
Question: For what pairs $n,k$ is there a polynomial $F$ whose inputs are $text{lcm}$ of various subsets of ${a_i}_{1leq ileq n}$ such that (*) has a solution in the specified range if and only if $Fneq 0$ (see some examples I have in mind below)?
Some observations: If $kgeq n$ then $Fequiv 0$ works, trivially.
Some calculations for small values of $(n,k)$ suggest the following strange function: for each subset $I subset {1,cdots,n}$ define
$$f_I(a_1,cdots,a_n):= frac{prod_{iin I}a_i}{text{lcm}{a_i}_{iin I}}$$
For $I=varnothing$ we set $f_I=1$. Let:
$$F(a_1,cdots,a_n):= sum_{Isubset {1,cdots,n}}(-1)^{|I|}f_I(a_1,cdots,a_n) $$
(You can replace $F$ by a polynomial of the $text{lcm}$s which vanishes at the same time). This works for $(n,k)=(2,1)$ and probably $(3,1)$ (which would imply $(3,2)$). It does not work for $(4,1)$ but seems to work for some sequence with $(n,k)= (4,2)$ (may be the function depends on $n/k$?). Have anyone seen this kind of formula before in other contexts?
PS: I am not sure what tags should be used. Please feel free to re-tag.
EDIT: Apparently this question is related to existence of lattice points on a polytope. I checked through some of the references pointed to in this question on MO, but could not find the exact answer to what I wanted.
I would like to state something about the existence of solutions $x_1,x_2,dots,x_n in mathbb{R}$ to the set of equations
$sum_{j=1}^n x_j^k = np_k$, $k=1,2,dots,m$
for suitable constants $p_k$. By "suitable", I mean that there are some basic requirements that the $p_k$ clearly need to satisfy for there to be any solutions at all ($p_{2k} ge p_k^2$, e.g.).
There are many ways to view this question: find the coordinates $(x_1,dots,x_n)$ in $n$-space where all these geometric structures (hyperplane, hypersphere, etc.) intersect. Or, one can see this as determining the $x_j$ necessary to generate the truncated Vandermonde matrix $V$ (without the row of 1's) such that $V{bf 1} = np$ where ${bf 1} = (1,1,dots,1)^T$ and $p = (p_1,dots,p_m)^T$.
I'm not convinced one way or the other that there has to be a solution when one has $m$ degrees of freedom $x_1,dots,x_m$ (same as number of equations). In fact, it would be interesting to even be able to prove that for finite number $m$ equations $k=1,2,dots,m$ that one could find $x_1,dots,x_n$ for bounded $n$ (that is, the number of data points required does not blow up).
A follow on question would be to ask if requiring ordered solutions, i.e. $x_1 le x_2 le dots le x_n$, makes the solution unique for the cases when there is a solution.
Note: $m=2$ is easy. There is at least one solution = the point(s) where a line intersects a circle given that $p_2 ge p_1^2$.
Any pointers on this topic would be helpful -- especially names of problems resembling it.
The answer is YES if $n$ is even. But if $n$ is odd, then the answer is NO since $-I$ is not a sum of two squares.
See
Griffin and Krusemeyer, Matrices as sums of squares, Linear and Multilinear Algebra 5 (1977/78), no. 1, 33-44
for the proofs of these facts and generalizations.
I'm currently learning about knot theory, so please correct me if I'm saying something senseless. I'll try to describe the things just as I think they are.
First, suppose we have constructed the reduced Burau representation
$psi_n^r: B_ntotext{GL}_{n-1}({mathbb Z}[t^{pm 1}])$
of the Artin Braid Group on $n$ strands.
(1) From this representation, one can construct the Alexander-Polynomial $nabla$ as the knot invariant corresponding to the Markov function
(X) $betamapsto (-1)^{n+1}frac{s^{-langlebetarangle} (s-s^{-1})}{s^n-s^{-n}} text{det}(psi^r_n(beta) - I_{n-1}))$
(here $langlebetaranglein{mathbb Z}$ is the image of $beta$ under $B_nto B_n/[B_n,B_n]={mathbb Z}$ and $s=t^{1/2}$.)
Now the Alexander polynomial satisfies the Skein relation $nabla(L_+) - nabla(L_-) = (s^{-1} - s)nabla(L_0)$, and this suffests to look at the quotient of ${mathbb Z}[s^{pm 1}][B_n]$ by the relation $sigma_i - sigma_i^{-1} = (s^{-1}-s)cdot 1$, because the Markov function above factors through this quotient. This was the first motivation for me to study the Hecke algebra - just take some knot invariant and mod out every relation in the group algebra of $B_n$ which is satisfied by the invariant; in fact, viewing it in this way, I'd rather say "Hecke-Algebra of the Alexander Polynomial".
(2) On the other hand, one could start from a more representation theoretic viewpoint and define the Hecke-algebra ${mathcal H}_n^s$ to be the quotient of ${mathbb Z}[t^{pm 1}][B_n]$ by the relation $T_i^2 = (t-1) T_i + tcdot 1$ in order to study those representations of $B_n$ where the representing matrices of the $T_i$ satisfy one fixed quadratic relation. The representing matrices in the reduced Burau representation do satisfy the above quadratic equation, and so one gets a representation of the Hecke algebra ${mathcal H}_n^t$.
These are two quite different ways which lead to the study of Hecke algebras -- can somebody tell me what the relation between these two constructions is? I'd also like to get some geometric intuition for (X), if there is one (the homological construction of the Bureau representation is very natural to me, but in the definition of the Markov function (X) I'm struggling to see the motivation - I'd like to "see" that this definition is the right one in order to get a Markov function, without just doing a huge calculation).
I know that this is not a very precise question, but I'd just like to hear about what do you think is the "right" way to think about and motivate the study of the Hecke algebra.
Thank you.
I have to agree strongly with Ame's answer, in part. Weibel is a great place to go for the formalism. Once you have a little bit of the formalism, though, where to go depends on interests. To really see a lot of the power of derived functors, Hartshorne chapter 3 has some good theorems and exercises using them (though he does bounce back and forth with Cech cohomology...) and Huybrecht's "Fourier Mukai Transforms in Algebraic Geometry" book is very clear (at least to me) in the first few chapters, where he discusses derived categories (Note: I do disagree with Harry, this is a SECOND step, not a first for most) and makes extensive use of them, as he's interested in talking about derived categories, and they're the most natural thing in the universe there.
Also, the following papers might help:
Fourier Mukai Transforms and Applications to String Theory talks about how FM transforms (which are compositions of derived functors) help in string theory
Derived categories for the working mathematician This is a wonderful paper, very clear about what the derived category is, and might help in conjunction with the books above.
Yes. The isomorphism in $mathrm{Ho}(mathcal{M})$ is represented by a morphism in $mathcal{M}$ from a cofibrant replacement for $A$ to a fibrant replacement for $B$. The "converse to the Whitehead lemma" states that a map in a model category is a weak equivalence iff its image in the homotopy category is an isomorphism. Combining this with the definition of (co)fibrant replacement, we see that $A$ and $B$ are connected by a 3-step zig-zag of weak equivalences.
If $kgeq0$ and $f$ is a polynomial in $k+1$ variables with integer coefficients, let $$X(f)={ninmathbb Z:exists x_1,dots,x_kinmathbb Z:f(n,x_1,dots,x_k)=0}.$$ This is the typical diophantine subset of $mathbb Z$.
Clearly $X(f)cup X(g)=X(h)$ with $h(n,x_1,dots x_{k+l})=f(n,x_1,dots,x_k)g(n,x_{k+1},dots,x_{k+l})$. It follows that a finite union of diophantine subsets of $mathbb Z$ is diophantine. It follows that to show that a finite set is diophantine, then, it suffices to check that for all $ainmathbb Z$ the set ${a}$ is diophantine.
Let then $ainmathbb Z$ and let $f(n)=n-a$. Then obviously ${a}=X(f)$.