User neil strickland - MathOverflow

Answer by Neil Strickland for Can the fact that the square of an integer is a natural number be categorified?

2013-05-23T18:27:23Z

Probably the most natural thing to ask for Theorem 1 is as follows. Let $\mathcal{A}$ be the category whose objects are pairs of finite sets, and whose morphisms are pairs of bijections. Let $\mathcal{B}$ be the category of finite sets and functions. (Allowing more morphisms in $\mathcal{A}$ or fewer morphisms in $\mathcal{B}$ would make the problem harder.) We have functors $F,G:\mathcal{A}\to\mathcal{B}$ given by $F(A,B)=(A\times B)\uplus(B\times A)$ and $G(A,B)=(A\times A)\uplus(B\times B)$. We then ask whether there is a natural injective map $j:F\to G$. I think the answer is negative. Indeed, the subset $j^{-1}(A\times A)\subseteq F(A,B)$ would have to be preserved by the action of $\text{Aut}(A)\times\text{Aut}(B)$ on $F(A,B)$, so it would have to be $(A\times B)$ or $(B\times A)$, wlog the former. Now $j$ gives an $\text{Aut}(A)\times\text{Aut}(B)$-equivariant map from $A\times B$ to $A\times A$, which consists of two equivariant maps $p,q:A\times B\to A$. For fixed $b$ we have an $\text{Aut}(A)$-equivariant map $p(-,b):A\to A$, and (provided that $|A|>2$) the only possibility is $p(a,b)=a$. Similarly $q(a,b)=a$, and we see that $j$ cannot be injective.

[Added later]

This whole story may be related to Thomason's paper "Beware the phony multiplication on Quillen's $\mathcal{A}^{-1}\mathcal{A}$". Here $\mathcal{A}$ is a symmetric bimonoidal category (like the category of finite-dimensional vector spaces, under $\oplus$ and $\otimes$). From this Quillen constructed a new category $\mathcal{A}^{-1}\mathcal{A}$. The objects are pairs $(A,B)$, where $A$ and $B$ are objects of $\mathcal{A}$. I won't spell out the morphisms except to say that $(A,B)$ is supposed to represent the "formal difference" $A-B$, and everything is guided by that. It is reasonable to hope that there should be a tensor product on $\mathcal{A}^{-1}\mathcal{A}$ compatible with the original tensor product on $\mathcal{A}$. However, Thomason showed that several purported constructions of such a tensor product contain subtle errors, and that a wide class of approaches are doomed to fail.

Answer by Neil Strickland for Why don't more mathematicians improve Wikipedia articles?

2013-05-22T11:19:02Z

In general, I find the wikipedia maths pages to be very comprehensive and useful, and I have not found any serious errors. However, they are often not very well organized or clearly explained. I agree that it would be good if more professional mathematicians tried to help with this.

I rewrote the article http://en.wikipedia.org/wiki/Homology_theory a few years ago. I thought about working on some other related ones. However, it seemed to me that it would be best to do a general reorganization, deleting some pages, renaming others and redistributing material in a more coherent way. I did not know how to go about presenting such a proposal to the relevant community, and how to ensure that there was enough consensus that a lot of work would not go to waste, and how to make the changes offline so as not to leave an inconsistent mess while the work was going on. So I did not end up doing anything.

Answer by Neil Strickland for How much of character theory can be done without Schur's lemma or the Artin-Wedderburn theorem?

2013-05-21T18:53:20Z

It might be useful here to think about the Morava $K$-theory rings $K(n)^*(BG)$ (for finite groups $G$). For the trivial group you get the graded ring $k=\mathbb{Z}/p[v_n,v_n^{-1}]$, where $|v_n|=2p^n-2$; for general groups you get a finitely generated graded module over $k$ (and all such modules are free). There is a ring structure, induction and restriction maps, an inner product and so on, making $K(n)^*(BG)$ closely analogous to $R(G)$. However, one can find examples where $K(n)^*(BG)$ has no basis that is permuted by $Aut(G)$, which means that we have nothing analogous to irreducible characters. One way to make your question more precise would be to restrict attention to methods that also work in this context.

If you don't like the fact that $k$ has characteristic $p>0$, there are naturally occurring lifted versions where the trivial group gives you the ring of $p$-adic integers, or a formal power series ring over the $p$-adics or something a bit larger than that. If we let $E$ denote one of these variants, it works out that there is a kind of character theory due to Hopkins, Kuhn and Ravenel. Instead of conjugacy classes of elements of $G$, you need to consider the set $C_n$ of conjugacy classes of $n$-tuples of mutually commuting elements of $p$-power order. There is then a certain ring $L$ that is an algebraic extension of $\mathbb{Q}\otimes E^*(\text{point})$ , and a natural isomorphism $L\otimes E^*(BG)\to Map(C_n,L)$, analogous to the description of $\mathbb{C}\otimes R(G)$ by class functions.

Convergence at the radius of convergence

2013-05-14T20:07:02Z

Suppose I have (roughly speaking) a multivalued meromorphic function $f(z)$ on all of $\mathbb{C}$ that is single-valued and holomorphic on the open unit disc and has some branch points of finite order (but no poles) on the unit circle. Does the Taylor series always converge uniformly to $f(z)$ on the closed unit disc? This seems very likely but I do not think I have ever seen a proof.

I know an argument that works for the simplest possible case, namely the function $f(z)=\sqrt{1-z}$.

Is this basis of simplex polynomials known?

2013-05-03T13:30:46Z

Put $R_n=\mathbb{R}[t_0,\dotsc,t_n]/(\sum_it_i-1)$ (the ring of polynomial functions on the $n$-simplex). Consider a monomial $t^a=t_0^{a_0}\dotsb t_n^{a_n}$. Let $(b_0,\dotsc,b_n)$ be the sequence $(a_0,\dotsc,a_n)$ arranged in nondecreasing order. I'll say that $t^a$ is admissible if $b_n=b_{n-1}+1$. One can show that the admissible monomials form a basis for $R_n$ with various convenient properties. In the case $n=1$, you just get the monomials $t_0^{i+1}t_1^i$ and $t_0^it_1^{i+1}$, for example (so $1$ has to be expressed as $t_0+t_1$). Does this appear in the literature, and if so, under what name?

Answer by Neil Strickland for Examples of applications of the Freyd-Mitchell embedding theorem.

2013-04-20T20:02:52Z

An interesting class of examples comes from the following construction, also due to Freyd. Let $\mathcal{T}$ be a small triangulated category. Form a new category $\mathcal{A}$, with one object $I(u)$ for each morphism $u$ in $\mathcal{T}$. The morphisms from $I(u:A\to B)$ to $I(v:C\to D)$ are the pairs $(f:A\to C,g:B\to D)$ such that $gu=vf$, modulo those for which $gu=vf=0$. There is a functor $J:\mathcal{T}\to\mathcal{A}$ given by $J(X)=I(1_X)$. One can show that $\mathcal{A}$ is abelian, that $J$ is full and faithful, and that the essential image of $J$ is the subcategory of projective objects, which is the same as the subcategory of injective objects.

I think that the embedding theorem is much less obvious for these abelian categories than it is for the more usual examples.

As a special case, we can take $\mathcal{T}$ to be the category of finite spectra in the sense of stable homotopy theory. The embedding theorem, combined with the above construction, tells us that $\mathcal{T}$ embeds in the category of $R$-modules for some $R$ (which is not unique). Freyd also conjectured more specifically that in this case the stable homotopy functor gives a full and faithful embedding of $\mathcal{T}$ in the category of modules over the ring of stable homotopy groups of spheres (and he proved that many interesting consequences would follow from that). This conjecture is still wide open half a century later.

Answer by Neil Strickland for Homotopy classes of maps to Lie groups

2013-04-16T07:24:35Z

If the dimension of $M$ is low relative to that of $G$ then the calculation of $[M,G]$ typically reduces to stable homotopy theory or generalised cohomology, for which many methods are known. For example, if $\dim(M)<2n$ then $$[M,U(n)]\simeq [M,U(\infty)] \simeq [\Sigma M,BU(\infty)] \simeq K(\Sigma M) $$ (where $K$ denotes complex $K$-theory). This can often be understood using explicit constructions with vector bundles, or using the Atiyah-Hirzebruch spectral sequence $H^*(M;K^*)\Longrightarrow K^*(M)$ . Similar methods work for $[M,O(n)]$, $[M,SU(n)]$, $[M,Sp(n)]$ and so on, provided that $n$ is large enough. If you want to consider small $n$ then it may be possible to work back from large $n$ using fibrations like $U(n)\to U(n+1)\to S^{2n+1}$.

Is $\mathbb{H}P^\infty_{(p)}$ an H-space?

2012-10-12T09:29:08Z

Put $X=\mathbb{H}P^\infty$ (so $X$ classifies quaternionic line bundles, and $\Omega X=S^3$). There is no obvious reason for $X$ to be an H-space, because the tensor product of quaternionic vector spaces is not naturally a quaternionic vector space. Below I will prove that there is no nonobvious H-space structure. However, the obstruction that I use has order $12$ and so vanishes if we localise at a prime $p>3$. My guess is that $X_{(p)}$ is not an H-space for any prime $p$; does anyone know a proof of that?

Note that $H^*(X)=\mathbb{Z}[y]$ with $|y|=4$, and this has a Hopf algebra structure given by $\psi(y)=y\otimes 1+1\otimes y$, which is compatible with all Steenrod operations. Thus, there do not seem to be any primary obstructions.

However, if $X$ were an H-space then $S^3=\Omega X$ would have two commuting binary operations with the same identity and so (by a standard argument) they would be the same and would be commutative. However, it is known that $S^3$ is not homotopy commutative: the commutator map $S^6=S^3\wedge S^3\to S^3$ is the standard generator $\nu'$ of $\pi_6(S^3)\simeq\mathbb{Z}/12$.

Answer by Neil Strickland for A formal group law over oriented bordism

2013-04-12T20:17:14Z

MSO is certainly complex-oriented
The resulting formal group law satisfies $[-1]_F(x)=-x$
Let $R_*$ be the universal example of a graded ring with a formal group with the above property, so there is a natural map $R_*\to MSO_*$ . I think this is injective, and it becomes an isomorphism after inverting $2$.
Let $U$ be the set of all positive integers not of the form $2^j-1$, let $U_0$ be the subset of all even integers, and put $U_1=U\setminus U_0$. One can show that $R_*$ has a polynomial generator $a_i$ of degree $2i$ for each $i\in U$, and the only relations are $2a_i=0$ for $i\in U_1$.
The additive structure of $MSO_*$ is known by old work of Wall and Atiyah. I don't know a really convincing interpretation in terms of formal groups. All the torsion is killed by $2$, and there is a lot of stuff in odd degrees.

Answer by Neil Strickland for Proving that a space cannot be delooped.

2013-04-12T19:50:05Z

I'll assume for simplicity that $X$ is connected and of finite type.

The most basic cohomological criterion (which I suspect that others have considered too elementary to mention) is that if $X$ is a loop space then $H^*(X;\mathbb{Q})$ has a Hopf algebra structure, and so (by a theorem of Milnor and Moore) it is a tensor product of polynomial algebras and exterior algebras over $\mathbb{Q}$. Thus, if the ring structure of $H^*(X;\mathbb{Q})$ is any more complicated than that, then $X$ cannot be a loop space. Similarly, $H^*(X;\mathbb{Z}/p)$ must be a tensor product of polynomial algebras, exterior algebras, and truncated polynomial algebras of the form $(\mathbb{Z}/p)[x]/x^{p^m}$ for various $m$.

In another MO question I asked about the case $X=\mathbb{H}P^\infty_{(p)}$, which I suspect is not a loop space for any prime $p$, although it satisfies the obvious primary cohomological tests. I still don't know a convincing answer for that case.

Polynomial maps between noncommutative groups

2012-12-14T11:34:57Z

Below I will give some definitions. My question is: do these appear in the literature, and if so, under what name?

Let $G$ and $H$ be groups that may not be commutative. For $y\in G$, define $R_y:G\to G$ by $R_y(x)=xy$. Let $f$ be a function from $G$ to $H$. Define $\delta_n(f):G^n\to H$ by \begin{align*} \delta_0(f) &= 1 \\ \delta_1(f)(x) &= f(1) f(x)^{-1} \\ \delta_2(f)(x,y) &= f(1) f(x)^{-1} f(xy) f(y)^{-1} \\ \delta_3(f)(x,y,z) &= f(1) f(x)^{-1} f(xy) f(y)^{-1} f(yz) f(xyz)^{-1} f(xz) f(z)^{-1} \end{align*} and in general $$ \delta_{n+1}(f)(x_1,\dotsc,x_{n+1}) = \delta_n(f)(x_1,\dotsc,x_n) \delta_n(f\circ R_{x_{n+1}})(x_1,\dotsc,x_n)^{-1}. $$ This has one term for each subset $J\subseteq\{1,\dotsc,n\}$ , with exponent $(-1)^{|J|}$. The order of the terms corresponds to the Binary Reflected Gray Code (see Wikipedia, for example). One could imagine using other orders such as lexicographic, but the BRGC order seems to do the right thing for the examples that I am considering.

I'll say that $f$ is polynomial of degree at most $n$ if $\delta_{n+1}(f)$ is the constant function with value $1$. Clearly $f$ is polynomial of degree at most $0$ iff it is constant, and it is polynomial of degree at most $1$ iff it is a constant times a homomorphism. The commutative case is fairly well-known, and is consistent with the usual meaning of 'polynomial' for maps $\mathbb{Z}^p\to\mathbb{Z}^q$. I know of a 1971 paper by Andreas Dress, but it would not surprise me if there were earlier references. However, I have never seen the noncommutative case.

Even in the commutative case, it takes some work to prove that any composite of polynomial maps is polynomial. I do not know whether that holds in the noncommutative case.

OCR for handwritten mathematics

2012-04-30T07:03:15Z

I am in the process of scanning a large collection of handwritten notes. They consist of diagrams and formulae with a relatively small proportion of actual words. Of course it would be hopeless to get an OCR program to digest the diagrams or formulae, but it would be useful if I could get one to find and transcribe enough of the words to build an index. Has anyone tried this kind of thing?

Answer by Neil Strickland for Interesting examples of a 4-torsion X in a triangulated category such that $2 End(X/2X)\neq 0$?

2012-11-14T08:58:00Z

If $\text{Hom}(S,S/2)$ refers to maps of degree zero, then that group is $\mathbb{Z}/2$. However, $\text{Hom}(S[2],S/2)$ is $\mathbb{Z}/4$, as is $\text{Hom}(S/2,S/2)$. (My sign convention for the shift is such that $S[n]$ is the sphere $S^n$.) On the other hand, $\text{Hom}(S/2,S/2[i])$ will be zero for $i>1$ but nonzero for most $i\leq 1$.

Answer by Neil Strickland for Equivariant K-theory of $S^1$-action on $S^2$

2012-11-04T10:14:09Z

Let $L$ denote $\mathbb{C}$ with $S^1$ acting by multiplication, and let $\mathbb{C}$ denote $\mathbb{C}$ with trivial $S^1$-action. Then the projective space $P(L\oplus\mathbb{C})$ is homeomorphic to $S^2$, and the natural $S^1$-action is the one that you mentioned. Thus, your problem is a special case of calculating $K_G(PV)$, where $V$ is a complex representation of a compact Lie group $G$. There is an evident map from $R(G)=K_G(\text{point})$ to $K_G(PV)$, and the tautological bundle $T$ also gives an element of $K_G(PV)$, so the polynomial ring $R(G)[T]$ maps to $K_G(PV)$. Put $f(t)=\sum_{k=0}^{\text{dim}(V)}(-1)^k\Lambda^k(V^*)t^k$. The constant bundle with fibre $V$ splits as $T\oplus T^\perp$, and using this one can check that $f(T)=0$ in $K_G(PV)$. With more work it can be shown that $K_G(PV)=R(G)[T]/f(T)$. This is stated as Proposition 3.9 in Segal's "Equivariant K-Theory"; the proof relies on a result that Segal states as Proposition 3.8, but does not prove; for that, see Proposition 4.9 of Atiyah's "Bott periodicity and the index of elliptic operators". A more direct argument is possible for the case that you mention, but the result above gives the general context.

Answer by Neil Strickland for Is the stable homotopy category idempotent complete?

2012-10-20T19:48:49Z

Yes, this is a standard fact. Given a self-map $e\colon X\to X$, we write $e^{-1}X$ for the telescope of the sequence $X\xrightarrow{e}X\xrightarrow{e}X\xrightarrow{e}\dotsb$ (constructed as the cofibre of a suitable self-map of $\bigvee_{i=0}^\infty X$). If $e$ is idempotent, one can check that the natural map $X\to e^{-1}X\vee (1-e)^{-1}X$ is an equivalence, and that $e$ acts as the identity on the first factor and as zero on the second; in other words, we have a splitting of $e$.

Answer by Neil Strickland for Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex?

2012-10-15T21:01:03Z

This is not really an answer, but a comment about an interesting special case. Suppose that $G$ acts smoothly on $S^2$. By averaging we can choose a $G$-invariant Riemannian metric. This gives $S^2$ a conformal structure, making it a Riemann surface. Any Riemann surface homeomorphic to $S^2$ is conformally equivalent to the standard Riemann sphere. Thus, we can reduce to the case where $G$ acts on $\mathbb{C}\cup\{\infty\}$ by conformal and anticonformal maps, which must have the form $z\mapsto (az+b)/(cz+d)$ or $z\mapsto (a\overline{z}+b)/(c\overline{z}+d)$. I think it even works out here that the quotient $(\mathbb{C}\cup\{\infty\})/G$ is always either a sphere or a disc. Thus, one cannot get any local pathology in this context. This contrasts with other settings where smooth functions can generate topological pathology: for example, any closed subset of $\mathbb{R}^n$, however fractal, can be expressed as the zero set of a smooth function $f\colon\mathbb{R}^n\to\mathbb{R}$.

Along somewhat similar lines, I think one can show that when $X$ is a one-dimensional CW complex with continuous action of a finite group $G$, then $X/G$ is again a one-dimensional CW complex (up to homeomorphism, not just homotopy equivalence).

Answer by Neil Strickland for Is a representation sphere dualizable inside naive G-spectra?

2012-07-13T10:30:50Z

Let $G$ be of order $2$, and let $V$ be the nontrivial one-dimensional representation. I'll write $S(V)$ for the unit sphere in $V$ (which is just $G$); the one-point compactification of $V$ is then the unreduced suspension of $G$, which I'll call $S^V$. We can only form spectra from $G$-spaces with fixed basepoint, so we need to consider $S(V)_+=G_+$ rather than $S(V)$ itself. There is a cofibration $G_+\to S^0\to S^V$ , and $S^0$ is its own dual, so the question for $S^V$ is equivalent to the question for $G_+$ .

In the category of genuine $G$-spectra, $G_+$ is its own dual, which means that there are unit and counit maps $S^0\xrightarrow{\eta}G_+\wedge G_+\xrightarrow{\epsilon}S^0$ making certain diagrams commute. In the naive category, $\{S^0,G_+\wedge G_+\}^G$ is the colimit of the unstable mapping sets $[S^n,S^n\wedge G_+\wedge G_+]^G$, which are trivial because $S^n\wedge G_+\wedge G_+$ is free away from the basepoint. Thus, we cannot construct the required map $\eta$, and $G_+$ is not self-dual. The same applies if we replace one of the $G_+$ 's by something else, so $G_+$ is not dualisable.

Answer by Neil Strickland for Koszulness of the cohomology ring of moduli of stable genus zero curves

2012-06-15T17:17:56Z

Here is an alternative presentation of the cohomology, taken from the unpublished PhD thesis of my student Daniel Singh. It has the disadvantage that one marked point is treated specially, so some symmetry is lost, but otherwise has many pleasant properties.

Put $S=\{1,\dotsc,n-1\}$ .

For each subset $T\subseteq S$ with $|T|>1$ we have a generator $x_T$ in degree two.
For each pair of sets $T,U$ with $T\cap U\neq\emptyset$ we have $(x_{T\cup U}-x_T)(x_{T\cup U}-x_U)=0$.
Now consider a set $T$ as before, and disjoint subsets $U_1,\dotsc,U_r\subseteq T$, again with $|U_i|>1$. Put $m=(|T|-1)-\sum_i(|U_i|-1)$. Then $x_T^m\prod_i(x_T-x_{U_i})=0$.
Moreover, there are no more generators or relations.

One can also give a basis for the cohomology consisting of monomials in the generators $x_T$.

Consider a monomial $y=\prod_Tx_T^{n_T}$. The shape of $y$ is $\{T : n_T>0\}$ .
We say that a collection $\mathcal{F}$ of subsets of $S$ is a forest if all elements have size at least two, and any two elements are either disjoint or nested.
Given a forest $\mathcal{F}$ and a set $T\in\mathcal{F}$, let $U_1,\dotsc,U_r$ be the maximal elements of $\{V\in\mathcal{F}:V\subset T\}$ , and then put $m(\mathcal{F},T)=(|T|-1)-\sum_i(|U_i|-1)$.
We say that our monomial $y$ is admissible if $\text{shape}(y)$ is a forest and $n_T\lt m(\text{shape}(y),T)$ for all $T\in\text{shape}(y)$.

It can be shown that the admissible monomials form a basis for the cohomology.

I do not know whether the algebra is Koszul, but I think that this presentation is well-adapted for investigating that question.

Answer by Neil Strickland for The Constructions of Davis and Januszkiewicz.

2012-06-14T09:56:27Z

The DJ construction works with a simplicial complex $K$ and a subtorus $W\leq\prod_{v\in V}S^1$ (where $V$ is the set of vertices of $K$). People tend to be interested in the case where $|K|$ is homeomorphic to a sphere, but that isn't really central to the theory. However, it is important that we have a simplicial complex rather than something with more general polyhedral structure. It is also important that we have a subtorus, which gives a sublattice $\pi_1(W)\leq\prod_{v\in V}\mathbb{Z}$, which is integral/rational information. I don't think that the DJ approach will help you get away from the rational case.

I like to formulate the construction this way. Suppose we have a set $X$ and a subset $Y$. Given a point $x\in\prod_{v\in V}X$, we put $\text{supp}(x)=\{v:x_v\not\in Y\}$ and $K.(X,Y)=\{x:\text{supp}(x) \text{ is a simplex}\}$ . The space $K.(D^2,S^1)$ is a kind of moment-angle complex, and $K.(D^2,S^1)/W$ is the space considered by Davis and Januskiewicz; it has an action of the torus $T=\left(\prod_{v\in V}S^1\right)/W$. Generally we assume that $W$ acts freely on $K.(D^2,S^1)$. There is a fairly obvious complexification map $K.(D^2,S^1)/W\to K.(\mathbb{C},\mathbb{C}^\times)/W_{\mathbb{C}}$. Under certain conditions relating the position of $W$ to the simplices of $K$, one can check that $K$ gives rise to a fan, that the complexification map is a homeomorphism, and that both $K.(D^2,S^1)/W$ and $K.(\mathbb{C},\mathbb{C}^\times)/W_{\mathbb{C}}$ can be identified with the toric variety associated to that fan.

Answer by Neil Strickland for Properties of endmorphism rings of E(n),K(n)-localized spheres

2012-05-31T18:13:12Z

Firstly, using the universal property of localisation we see that $[L_nS,L_nS]_*=\pi_*(L_nS)$ and similarly for $L_{K(n)}$, so we can just talk about the homotopy rings.

Next, $\pi_*(L_1S)$ is closely related to the image of $J$ and is described completely in papers by Bousfield and Ravenel. There is a copy of $\mathbb{Q}/\mathbb{Z}_{(p)}=\mathbb{Z}/p^\infty$ in degree $-2$, so the annihilators of powers of $p$ give a strictly increasing chain of ideals, proving that $\pi_*(L_1S)$ is not Noetherian. In $\pi_*(L_{K(1)}S)$ there is no $\mathbb{Z}/p^\infty$ but there are copies of $\mathbb{Z}/p^k$ for all finite $k$ so again the ring is not Noetherian. At height $2$ there are voluminous calculations by Shimomura and coauthors, and Behrens has recently explained an easier way to organise the answers, although they are still very complicated. The same line of argument shows that they are very far from being Noetherian. At larger heights I do not think that anything very significant is known.

Answer by Neil Strickland for Cup products of connected sum

2012-05-30T13:30:42Z

The natural description is like this. There are augmentations $\epsilon_X:H^*(X)\to\mathbb{Z}$ (for $X\in\{M,N\}$) and orientation classes $u_X\in H^d(X)$. Put $$R'=\{(a,b)\in H^*(M)\times H^*(N):\epsilon_M(a)=\epsilon_N(b)\} = H^*(M\vee N)$$ and $R=R'/(u_M,-u_N)$. Then one can use the cofibration $$ S^{d-1} \to M \# N \to M\vee N $$ to identify $H^*(M\#N)$ canonically with $R$. Thus, if $i+j=d$ with $i,j>0$ then the product of $H^i(M)$ with $H^j(N)$ in $H^d(M\# N)$ is zero, but the product of $H^i(M)$ with $H^j(M)$ is the same as in the original manifold $M$.

Answer by Neil Strickland for A homotopyish Landweber exact functor theorem

2012-05-22T19:10:57Z

Here are three methods that I know:

In the case $M_*=(MU_*/I)[S^{-1}]$ (where $I$ is generated by a regular sequence) there is a more direct construction by reducing to the cases $M_*=MU_*/a$ and $M_*=MU_*[a^{-1}]$ . My paper 'Products on MU-modules' is probably the sharpest version, but there are many earlier versions in a similar spirit.
In the case $M_*=MU_*[x_1,\dotsc,x_r]$ with $|x_i|=0$ you can use $MU\wedge\Sigma^\infty_+\mathbb{N}^r$ (and this has an $E_\infty$ structure).
In the case $M_*=MU_*[n^{-1}]$ (for some $n\in MU_0=\mathbb{Z}$) you can note that there are natural $E_\infty$ maps $$ MU\xleftarrow{f}\Sigma^\infty_+DS^0\xrightarrow{}\Sigma^\infty_+QS^0,$$ where $f$ has degree $n$ on the bottom cell. The smash product $$ MU\wedge_{\Sigma^\infty_+DS^0}\Sigma^\infty_+QS^0$$ then has the required property.

There are some fairly obvious ways to combine these methods and generalise them slightly.

Under the general conditions of the Landweber theorem, I know of several people including myself who have looked quite hard for a more direct construction, but without success.

Modular representations with unequal characteristic - reference request

2012-05-16T10:09:03Z

Let $G$ be a finite group, and let $K$ be a finite field whose characteristic does not divide $|G|$. I am interested in the theory of finitely generated modules over $K[G]$. Of course many problems are not present here because $K[G]$ is semisimple and all modules are projective. My case is partly covered by Section 15.5 of Serre's book "Linear Representations of Finite Groups". However, Serre likes to assume that $K$ is "sufficiently large", meaning that it has a primitive $m$'th root of unity, where $m$ is the least common multiple of the orders of the elements of $G$. I do not want to assume this, so some Galois theory of finite extensions of $K$ will come into play. I do not think that anything desperately complicated happens, but it would be convenient if I could refer to the literature rather than having to write it out myself. Is there a good source for this?

[UPDATED]:

In particular, I would like to be able to control the dimensions over $K$ of the simple $K[G]$-modules. As pointed out in Alex Bartel's answer, these need not divide the order of $G$. I am willing to assume that $G$ is a $p$-group for some prime $p\neq\text{char}(K)$.

[UPDATED AGAIN]:

OK, here is a sharper question. Put $m=|K|$ (which is a power of a prime different from $p$) and let $t$ be the order of $m$ in $(\mathbb{Z}/p)^\times$. Let $L$ be a finite extension of $K$, let $G$ be a finite abelian $p$-group, and let $\rho:G\to L^\times$ be a homomorphism that does not factor through the unit group of any proper subfield containing $K$. Then $\rho$ makes $L$ into an irreducible $K$-linear representation of $G$, and every irreducible arises in this way. If I've got this straight, we see that the possible degrees of nontrivial irreducible $K$-linear representations of abelian $p$-groups are the numbers $tp^k$ for $k\geq 0$. I ask: if we let $G$ be a nonabelian $p$-group, does the set of possible degrees get any bigger?

Answer by Neil Strickland for Open source LaTeX lecture notes/slides/books

2012-05-11T19:00:45Z

I have lots of undergraduate notes in PDF form on my web pages (http://shef.ac.uk/nps). No one has ever asked me for the source files. Nonetheless, I have thought about releasing them explicitly in LaTeX form under a creative commons license. This has not happened for fairly mundane reasons:

Some of the notes are partially based on earlier notes by colleagues, and I have not discussed things with them.
Most courses have a fairly elaborate setup with separate files for notes, problem sheets and lecture slides, auxiliary files that are \included in the main files, Maple worksheets used to generate jpeg diagrams and so on. I have not worked out a good way to package everything.
All my courses have detailed solutions to all the problems, which I release to students a week or so after the problems have been assigned. I have not decided what would be the best thing to do with such solutions if I were to make the source files freely available. At the moment I have things set up so that the solutions are in the same files as the problems, with LaTeX macros etc to switch them on and off.

Answer by Neil Strickland for Do you use the Mathematics Subject Classification (MSC) when searching for literature?

2012-05-11T16:08:49Z

In the UK we have a group called MAGIC (http://maths-magic.ac.uk/index.php) which runs graduate level courses in mathematics shared by video across 19 universities. In the early planning stages for that project (about 2005) I downloaded from MathSciNet lists of BiBTeX entries for all papers published by the participating universities in the previous decade, and sorted them by MSC code to get a systematic overview of the research activity across the network. That was quite a useful exercise.

Answer by Neil Strickland for Do you use the Mathematics Subject Classification (MSC) when searching for literature?

2012-05-11T15:01:28Z

I get emails from the AMS listing new publications in my area (http://www.ams.org/membership/individual/benefits/e-cmp) and that works by MSC code (I get everything under classification 55 = algebraic topology). This is still of some use, but less so than the equivalent arxiv emails (I get everything under math.at). This is the only use that I make of MSC codes.

Answer by Neil Strickland for What is / are the softwares to use to draw surfaces of the form of a two or three-holed torus , or torus, or torus with cusps attached to it?

2012-05-11T09:32:39Z

The solutions that people have mentioned are good if you are happy with a two-dimensional line drawing. It would be better to have three-dimensional equations that could be plotted using Maple, or something like that, but that seems to be surprisingly hard. The equation $$ 3x_3^2x_4-2(x_1^2+x_2^2)x_4-2x_4^3+2(x_1^2-x_2^2)x_3 = 0 $$ defines a highly symmetric surface of genus 2 embedded in $S^3$, and one can project stereographically into $\mathbb{R}^3$ to get a nice picture like this:

(There's a lot to be said about this example; I will have an undergraduate working on it over the summer.) However, I do not know similarly nice equations for surfaces of higher genus, or with the two tori in the same plane rather than at right angles, or with cusps.

Answer by Neil Strickland for Is it true that all real projective space $RP^n$ can not be smoothly embedded in $R^{n+1}$ for n >1

2012-05-10T19:23:30Z

I'll use cohomology with coefficients $\mathbb{Z}/2$ everywhere.

Suppose that the space $P=\mathbb{R}P^{n-1}$ embeds in $S^{n}$ (where $n>2$). Recall that $$ H^*(P)=(\mathbb{Z}/2)[x]/x^{n} = (\mathbb{Z}/2)\{1,x,\dotsc,x^{n-1}\} $$ By examining the top end of the long exact sequence of the pair $(S^{n},P)$ we find that $H^{n}(S^{n},P)$ has rank two. Lefschetz duality says that this group is isomorphic to $H_0(S^{n}\setminus P)$, so we see that $S^{n}\setminus P$ has two connected components. (I don't need any orientation conditions here as I am working mod 2.) Let $A$ and $B$ be the closures of these components, so $A\cap B=P$ and $A\cup B=S^{2n}$. Lefschetz duality also gives $H^{n}(A)\times H^{n}(B)=H^{n}(S^{n}\setminus P)=H_0(S^{n},P)=0$.

We now have a Mayer-Vietoris sequence relating the cohomology groups of $A$, $B$, $P$ and $S^{n}$. As $H^1(S^{n})=H^2(S^{n})=0$ this gives an isomorphism $H^1(A)\times H^1(B)\to H^1(P)=\{0,x\}$. After exchanging $A$ and $B$ if necessary, we can assume that $H^1(B)=0$ and that there is an element $a\in H^1(A)$ that maps to $x$ in $H^1(P)$. It follows that $a^{n-1}$ maps to $x^{n-1}$, which generates $H^{n-1}(P)$, so the Mayer-Vietoris connecting map $H^{n-1}(P)\to H^{n}(S^{n})=\mathbb{Z}/2$ must be zero. This contradicts exactness at the next stage, because $H^{n}(A)\times H^{n}(B)=0$.

Answer by Neil Strickland for When do two maps between groups give the same map between representation rings?

2012-05-10T13:19:31Z

Your two questions are not just closely related but identical. If $R(f_1)=R(f_2)$ then $\chi(f_1(x))=\chi(f_2(x))$ for all characters $\chi$, and standard representation theory allows you to deduce that $f_1(x)$ is conjugate to $f_2(x)$.

For a basic example where pointwise conjugacy is different from conjugacy, let $G$ be elementary abelian of order $4$ with generators $a$ and $b$, and let $H$ be the symmetric group on six letters. Define \begin{align*} f_1(a) &= (1\;2)(5\;6) \\ f_1(b) &= (3\;4)(5\;6) \\ f_1(ab) &= (1\;2)(3\;4) \\ f_2(a) &= (1\;2)(3\;4) \\ f_2(b) &= (1\;3)(2\;4) \\ f_2(ab) &= (1\;4)(2\;3). \end{align*} Note that $5$ and $6$ are fixed by the image of $f_2$, but no point is fixed by the image of $f_1$, so $f_1$ and $f_2$ are not conjugate. However, for each $x$ we see that $f_1(x)$ and $f_2(x)$ have the same cycle type and so are conjugate.

Answer by Neil Strickland for Categorical description of the restricted product (Adeles)

2012-05-09T18:02:25Z

Personally I think that the restricted product description should be avoided. It is best to define $\widehat{\mathbb{Z}}$ to be the inverse limit of the system of all quotients $\mathbb{Z}/n$ (without gratuitously factoring $n$ as a product of primes) and then put $\mathbb{A}=(\mathbb{Q}\otimes\widehat{\mathbb{Z}})\times\mathbb{R}$. We can topologise this by giving $\mathbb{R}$ the usual topology, and $\mathbb{Q}\otimes\widehat{\mathbb{Z}}$ the topology for which the sets $q\otimes\widehat{\mathbb{Z}}$ form a basis of neighbourhoods of zero. Now the adeles for any number field $K$ can be defined as $\mathbb{A}\otimes K$. Any $\mathbb{Q}$-basis for $K$ identifies $\mathbb{A}\otimes K$ with $\mathbb{A}^d$ and thus gives a topology on $\mathbb{A}\otimes K$, which is easily seen to be independent of the choice of basis. The connection with primes/valuations for $K$ should be a theorem, not a definition.

Comment by Neil Strickland

2013-05-24T13:30:15Z

I wish you luck with your project, but your request is not really on topic for this site; please see the FAQ. It would be better to ask at math.stackexchange.com, where I am sure people will be happy to help.

Comment by Neil Strickland

2013-05-23T19:18:43Z

@André: if you replace general linear groups by symmetric groups, that's more or less my argument below.

Comment by Neil Strickland

2013-05-21T17:20:55Z

Your question is not really appropriate for this site; please see the FAQ. It would be better to ask at math.stackexchange.com.

Comment by Neil Strickland

2013-05-21T11:32:57Z

mathoverflow.net/howtoask

Comment by Neil Strickland

2013-05-19T17:11:26Z

@Dylan: if I remember rightly, $L[1/2]$ is equivalent to an Eilenberg-MacLane spectrum, whereas $L_{(2)}$ is equivalent to $kO_{(2)}$, so the $K(n)$-localizations are not hard. However, these are not $E_\infty$ equivalences (or at least not obviously so).

Comment by Neil Strickland

2013-05-17T07:54:44Z

@David: $\pi_{31}(S^2)$ is well out of the stable range. There is a homomorphism to the 29th stable group, but it is unlikely to be injective or surjective.

Comment by Neil Strickland

2013-05-14T21:50:27Z

@Tommaso: I don't think we can switch the sum and the derivative here. The usual argument for that involves a contour integral, and the contour would have to wind around the branch point, which would break the proof.

Comment by Neil Strickland

2013-05-14T21:47:06Z

@Alexandre: how does this follow from Abel's theorem? That tells us that if we have a point $u$ on the unit circle and the Taylor series converges at $u$, then the Taylor series converges uniformly on the line segment $[0,u]$. It is not obvious that the hypothesis is satisfied, and even if it is, the conclusion is weaker than uniform convergence on the whole closed disk.

Comment by Neil Strickland

2013-05-13T20:40:54Z

Your question is not really appropriate for this site; please see the FAQ. It would be better to ask at math.stackexchange.com or stats.stackexchange.com instead.

Comment by Neil Strickland

2013-05-07T10:49:06Z

Please read mathoverflow.net/howtoask

Comment by Neil Strickland

2013-05-04T09:24:32Z

In the case where $F'$ is a single point, you are essentially asking whether the fibration $E\to B$ has a section. It must be possible to set up a kind of obstruction theory for that, but I don't think it would be very tractable in general. If you have a particular fibration in mind, you will probably need to find an approach tailored to that case.

Comment by Neil Strickland

2013-04-21T15:50:57Z

@Aaron: see ams.org/mathscinet-getitem?mr=209333 and the other papers by Freyd in the same proceedings volume. Unfortunately I do not think that the full text is online anywhere.

Comment by Neil Strickland

2013-04-20T20:10:21Z

This is clear if $A$ is Noetherian and $M$ and $N$ are finitely generated, so presumably you are thinking of some more general context. Are you willing to assume any finiteness conditions at all?

Comment by Neil Strickland

2013-04-19T13:46:02Z

mathoverflow.net/howtoask

Comment by Neil Strickland

2013-04-18T07:41:03Z

mathoverflow.net/howtoask