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Answer by algori for A_infinity structure on cohomology and the weight filtration

2013-04-24T20:44:58Z

This was meant to be a comment before it got too long.

One of the ways to define a minimal $A_\infty$ model of a cdga (more generally, an $A_\infty$-algebra) $A$ is Merkulov's recipe (see Merkulov http://arxiv.org/pdf/math/9809172.pdf or Chuang and Lazarev http://arxiv.org/pdf/0802.3507.pdf for a summary). All we need is а decomposition $A=W\oplus K$ into subcomplexes with the differential of $A$ vanishing on $W$ (so $W$ can be identified with $H^*(A)$ ), $K$ acyclic, and a contracting homotopy $h:K\to K$ satisfying $h^2=0$. This decomposition is called the Hodge decomposition since it becomes one when $A=\mathcal{E}^*(M)$ , the algebra of complex-valued differential forms on a K\"ahler manifold $M$, in which case $W$ is the subspace of harmonic forms. Note that $W$ needn't be a subalgebra. Note also to find such a decomposition it suffices to find a splitting $A$ into cocycles and some complement and then splitting off the cohomology inside the cocycles, see Chuang, Lazarev, ibid., p. 5.

So if $A$ is some kind of cochain algebra associated to a complex algebraic variety and we take any mixed Hodge structure on $K$ compatible with $h$ and the differential and such that the multpplication $A\otimes A\to A$ is a map of mixed Hodge structures (the latter condition is probably too strong and may be relaxed in some way). Applying Merkulov's formula we get higher multiplications $m_n$ on $H^*(A)$ which are maps of mixed Hodge structures of degree $0$. These maps will be linear combinations of compositions of maps constructed from maps of the form $(0,h)$, multiplications in $A$ and projections to $W$ along $K$. In particular the weight of $m_n(x_1,\ldots,x_n)$ will be the sum of the weights of $x_1,\ldots, x_n$. Of course there is no reason this $A_\infty$ structure should be compatible with morphisms of varieties. I would guess it is possible to choose the splittings so that it is but this would require some more work.

Constructible sheaves and dg-modules

2011-09-08T22:23:28Z

Let $M$ be a smooth manifold, $A_M$ the de Rham algebra of $M$, $D_{A_M}$ the derived category of the category of differential graded (dg) $A_M$-modules and $D^+_c(M)$ the bounded below constructible derived category of sheaves of real vector spaces on $M$. The category $D^+_c(M)$ knows a lot about the topology of $M$; for example, it allows one to compute the real cohomology of $M$ together with all Massey products. However, unsurprisingly, this category is also quite complicated. Informally, the question I'd like to ask is: can one describe at least some pieces of $D^+_c(M)$ in terms of dg-modules (which is something much more manageable)?

In "Equivariant sheaves and functors", 12.3, J Bernstein and V. Lunts construct two mutually inverse equivalences $\gamma_M:\langle \mathbb{R}_M\rangle\to \langle A_M\rangle$ and $\mathcal{L}_M: \langle A_M\rangle\to \langle \mathbb{R}_M\rangle$. Here $\langle\cdot\rangle$ stands for the full triangulated subcategory generated by $\cdot$, $\mathbb{R}_M$ is the constant sheaf on $M$ and $\langle \mathbb{R}_M\rangle$ and $\langle A_M\rangle$ are subcategories of $D^+_c(M)$ and $D_{A_M}$ respectively.

The functors are defined as follows: $\gamma_M$ takes a sheaf, multiplies it by the de Rham complex and then takes the global sections; $\mathcal{L}_M$ takes a module, replaces it with a $\mathcal{K}$-projective resolution and multiplies the result by the de Rham complex. (A complex of dg $A_M$-modules is $\mathcal{K}$-projective, if $Hom$ from it to an acyclic complex is acyclic.)

Notice that $\gamma_M$ is in fact defined on the whole of $D^+(M)$ . I would like to ask: is there a subcategory $D$ of $D^+_c(M)$ larger than the one generated by the constant sheaf such that $\gamma_M$ restricted to $D$ is fully faithful? In particular, if $i:N\subset M$ is a submanifold and $M,N,M\setminus N$ are all simply connected, what happens if we take $D=\langle\mathbb{R}_M, i_*\mathbb{R}_N\rangle$ ?

Here is a related result. Suppose we fix a stratification of $M$ with all strata and their closures simply-connected. Consider the subcategory $D\subset D^+_c(M)$ formed by complexes which are constructible with respect to the chosen stratification and let $I^*$ be an injective resolution of the direct sum of the constant sheaves on the strata. Then, due to a result by B. Keller (Deriving dg-algebras, Ann ENS, 1994, no 1, 63-102) by taking $C^*\to Hom(C^*,I^*)$ we get a fully faithful functor from $D$ to $D_{End(I^*)}$ where $End(I^*)$ is the (global) endomorphism algebra of $I^*$ . However, this is not exactly what I'm looking for since the endomorphism algebra is still quite difficult to describe explicitly in the example I'm interested in.

Algorithmically unsolvable problems in topology

2012-06-01T22:16:45Z

This question is inspired by a paper by B. Poonen that appeared on the arxiv some time ago: http://arxiv.org/abs/1204.0299. The paper gives a sample of algorithmically unsolvable problems from various areas of mathematics.

The topology part however contains only two such problems: the homeomorphism problem for 4-manifolds, which was shown to be undecidalbe by Markov in 1958, and the problem of recognizing $S^n,n\geq 5$ up to homeomorphism. The indecidability in both cases basically boils down to the undecidability of the group isomorphism problem.

Note that both the above problems become decidable if one restricts one's attention to simply-connected PL-manifolds. This follows in the first case from the fact that simply-connected PL 4-manifolds are determined up to homeomorphism by the integral cohomology and in the second case from the generalized Poincare conjecture.

This makes one wonder what happens if one imposes some natural topological restrictions like simple connectedness. So I would like to ask if the following problems are decidable for simply-connected finite simplicial complexes, maybe under some further restrictions (e.g. for those those simplicial complexes that are homeomorphic to smooth or PL-manifolds):

the homeomorphism problem
the homotopy equivalence problem
the rational (or mod a prime $p$) homotopy equivalence problem

Personally, I do not hold out much hope that any of these turns out to be algorithmically decidable. For instance, the rational homotopy type of a space $X$ can be seen as an infinite collection of maps $H^{\otimes n}\to H$ of degrees $2-n$ (where $H=H^*(X,\mathbb{Q})$ ) subject to some condition, up to an equivalence relation, and it looks plausible that all the components in this collection matter. However, it is not completely clear to me how to prove this.

Answer by algori for A problem on convex geometry

2012-12-12T22:06:14Z

Since Petya answered question 2, let me answer question 1. Set $S=\partial K$ and assume at first that $S$ is smooth. Let $f,g:S^{n-1}\to S^{n-1}$, where $S^{n-1}$ is the unit sphere, be the maps that take an element $x$ to the outward unit normal to $S$ at the points of intersection of $S$ with the half-lines starting at the origin and going in the direction of $-x$ and $x$ respectively. These maps are homeomorphisms and they are homotopic to minus identity and identity respectively. So the map $x\mapsto g^{-1}(-f(x))$ has degree 1, and so it has a fixed point by the Lefschetz theorem (here we use the fact that $n$ is odd, so $n-1$ is even). Any such fixed point gives you a pair of opposite points on $S$ with parallel tangent planes.

upd: the above argument was essentially given in the posting. Let me explain how one can deduce the general case from it. One can approximate (e.g., with respect to the Hausdorff metric) an arbitrary compact convex $K$ with a sequence $(K_i)$ of smooth convex bodies that all have the origin in their interiors. For each $i$ let $a_i,b_i\in\partial K_i$ be opposite points with parallel tangent planes. By choosing a subsequence we may assume that there are $a=\lim a_i,b=\lim b_i$, which will again be opposite points in $\partial K$. By choosing a subsequence again we may assume that $\lim T_{a_i}\partial K_i$ and $\lim T_{b_i}\partial K_i$ also exist. These will be the required support planes.

Answer by algori for Pullback map in homology

2012-12-08T04:41:12Z

6672 -- the wrong way maps in homology are in fact less mysterious than they look. Suppose $f:X\to Y$ is a continuous map of oriented compact topological manifolds. Then there is a map $$H_i(Y,\mathbb{Z})\to H^{\dim Y-i}(Y,\mathbb{Z})\to H^{\dim Y-i}(X,\mathbb{Z})\to H_{\dim X-(\dim Y-i)}(X,\mathbb{Z})$$ where the arrows on the left and on the right are the Poincar\'e duality and the one ine the middle is the usual pullback in cohomology. We need the orientations to specify the Poincar\'e isomorphisms. Here $\dim$ stands for the dimensions over the reals.

There is a version of this in the case when $X,Y$ are oriented but no longer compact, in which case one has to use the Borel-Moore homology instead of the ordinary homology.

Also note that in a similar way one can construct the wrong way cohomology map $H^i(X,\mathbb{Z})\to H^{i-\dim X+\dim Y}(Y,\mathbb{Z})$, which is nothing but the Gysin map when $f$ is a closed embedding.

Answer by algori for A cohomology computation request.

2012-12-02T03:36:02Z

Reladenine -- your $X$ is the complement in the affine 3-space of the union of two hypersurfaces $Y$ and $Z$, the first given by $x=0$, the second by $yz=x$. The intersection $Y\cap Z$ is the union of two intersecting affine lines. Moreover, both $Y$ and $Z$ are isomorphic to $\mathbb{C}^2$ (note that $Z$ is the graph of a function). So the Borel-Moore homology of $Y\cup Z$ is given by $H^{BM}_i(Y\cup Z)=\mathbb{C}^2$ if $i=4,3$ and $\mathbb{C}$ if $i=2$. So, by the Alexander duality $\tilde H^j(X)\cong H^{BM}_{6-1-j}(Y\cup Z)$ where $\tilde H$ stands for the reduced cohomology, $H^i(X)=\mathbb{C}$ if $i=0,3$ and $\mathbb{C}^2$ if $i=1,2$. Applying the Poincar\'e duality $H^i(X)\cong H^{6-i}_c(X)$ one gets the answer you give in your posting.

Answer by algori for Non-stably trivial bundle with trivial charactertic classes

2012-11-18T22:44:03Z

Let $E=\gamma^1\otimes\mathbb{C}$ be the complexified tautological bundle over $X=\mathbb{P}^6(\mathbb{R})$, and set $F=4E=E\oplus E\oplus E\oplus E$. It is not hard to check that $c(E)=1+a$ and $w(E_\mathbb{R})=1+\bar a$ with $a\in H^2(X,\mathbb{Z})\cong\mathbb{Z}/2$ and $\bar a \in H^2(X,\mathbb{Z}/2)\cong\mathbb{Z}/2$ being the non-zero element. So all imaginable characteristic classes of $F$ and $F_\mathbb{R}$ vanish by the Whitney sum formula (note that $H^*(X,\mathbb{Z})=\mathbb{Z}[a]/2a=a^4=0$ ).

However, the group $\tilde K^0(X)$ of the classes of complex vector bundles on $X$ up to stable equivalence is cyclic of order 8 and is generated by $E-1$, see e.g. Karoubi, K-theory, corollary 6.47 from Chapter IV. So $4(E-1)=F-4\in \tilde K^0(X)$ is non-zero, and so $F$ is not stably trivial.

Answer by algori for Does every orientable surface embed in $\mathbb{R}^{3}$

2012-11-18T00:18:29Z

Expanding slightly on my comment above: here is how one can get the embedding theorem from the classification theorem.

The classification theorem for non-compact surfaces (theorem 3 in http://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf; by the way, the comment links to a version that tries to charge one $30 unless one's reading this from a university account; sorry about that; this has been fixed now) states that

Every surface is homeomorphic to a surface formed from a sphere $\Sigma$ by first removing a closed totally disconnected set $X$ from $\Sigma$, then removing the interiors of a finite or infinite sequence $D_1,D_2,\ldots$ of nonoverlapping closed discs in $\Sigma - X$, and finally suitably identifying the boundaries of these discs in pairs. (It may be necessary to identify the boundary of one disc with itself to produce an odd "cross cap.'')[...]

To get a proper embedding in $\mathbb{R}^3$ first note that one may assume that $X$ is non-empty; otherwise the surface will be compact (since we are throwing away something open and then identifying something), in which case everything is clear. Let $f:\Sigma\to [0,\infty)$ be a smooth function such that $f^{-1}(0)=X$.

Take an $x_0\in X$ and identify $\Sigma\setminus \{x_0\}=S^2\setminus \{x_0\}$ with $\mathbb{R}^2\subset\mathbb{R}^3$. First we properly embed $\Sigma-X=\mathbb{R}^2\setminus X$ in $\mathbb{R}^3$ as the graph of the function $\frac{1}{f}$.

Let $D_i,D_j$ be two disks whose boundaries are to be identified. There is an $\varepsilon(D_i,D_j)>0$ such that one can join $D_i$ and $D_j$ with a curve $\gamma(D_i,D_j)\subset \Sigma$ that misses $f^{-1}([0,\varepsilon(D_i,D_j)))$. Now delete the portion of the graph over the interiors of $D_i$ and $D_j$, attach vertical tubes ("chimneys") that reach at least as high as $1/\varepsilon(D_i,D_j)$ to the resulting boundaries, and then connect the tops of the chimneys with a horizontal tube along $\gamma(D_i,D_j)$.

If $D_k,D_l$ are two other disks that we must throw away and then glue the boundaries we proceed in a similar way but this time we may have to make the chimneys higher so that they miss the first tube. And so on. It is not hard to see that each ball centered at the origin intersects only finitely many tubes.

Answer by algori for Is it true that every projective curve is a union of two affine curves?

2012-11-17T19:15:21Z

One way to see this would be as follows: if $X\subset\mathbb{P}^n$ is a curve one can find hyperplanes $H_1,H_2$ such that $X\cap H_1\cap H_2=\varnothing$. Then $X$ will be the union of two affine curves $X\setminus H_1$ and $X\setminus H_2$.

Answer by algori for Leray Spectral Sequence

2012-11-16T23:13:20Z

Ru -- the Leray spectral sequence exists for any map $f:X\to Y$ of arbitrary topological spaces and any sheaf $F$ on $X$ and its second term is $$E_2^{p,q}=H^p(Y,R^q f_*F)$$ where $R^q f_*F$ are the sheaves on $Y$ that are obtained by sheafifying the presheaves $U\mapsto H^q(f^{-1}(U),F)$. Here are some remarks that might help:

If $f$ is a locally trivial fibration and $F$ is constant then all $R^q f_*F$ are locally constant; if in addition $Y$ is simply-connected then the sheaves are constant and we can express $E_2$ in terms of the constant cohomology of $Y$.
It may happen that all fibers $f^{-1}(y),y\in Y$ are homeomorphic but some or all $R^q f_*F$ are non-constant; take e.g. $X=(\mathbb{R}\setminus \{ 0\})\sqcup \{ 0\}, Y=\mathbb{R},f$ the identity map.
Nevertheless, if $f:X\to X/G$ where $G$ is a connected Lie group that acts nicely on $X$ (say so that the quotient is Hausdorff) with finite stabilizers, and $F$ is a constant sheaf with stalk $\mathbb{Q}$ (or $\mathbb{R}$ or $\mathbb{C}$) then any sheaf $R^q f_*F$ is constant with stalk $H^q(G,\mathbb{Q})$ (resp., $H^q(G,\mathbb{R})$ and $H^q(G,\mathbb{C})$).

Two possible references (which means, to be honest, that there may be better references but that's where I first learned this from) are Godement, Topologie alg\'ebrique et th\'eorie des faisceaux, the very end of chapter 4, and Griffiths-Harris, the very end of vol.1

Answer by algori for Can we get the HOMFLY polynomial for a torus knot from the Kauffman Polynomial?

2012-11-16T18:23:18Z

John -- a formal answer is yes: The 2-variable Kauffman polynomial of a torus $(m,n)$-knot determines the Jones polynomial, which allows one to determine the $m$ and $n$ from the formula

$$V(K_{m,n})=\frac{t^{{(m-1)(n-1)}/2}(1-t^{m+1}-t^{n+1}+t^{m+n})}{1-t^2},$$

which allows one to compute the HOMFLY polynomial. Now, this is not really an answer, as you are presumably interested in how to find and rigorously prove an explicit formula for the Kauffman and Homfy polynomials for torus knots or links. I'm sure this has been done but can't think of a reference. If I find one, I'll edit my posting to include it.

A small remark: the HOMFLY polynomial wasn't discovered by Jones. It was discovered by the 6 authors the first letters of whose names make up HOMFLY, and also 2 other the first letters of whose names don't: Hoste, Ocneanu, Millett, Freyd, Lickorish, Yetter, Przytycki and Traczyk. They were certainly inspired by Jones's work though.

Answer by algori for Compact homogeneous spaces that admit a self map of degree >1

2012-11-16T15:18:28Z

If $X$ is a simply connected compact manifold, then one sufficient condition for the existence of maps $X\to X$ of any sufficiently divisible degree is formality: there is a commutative differential graded algebra (cdga) $A$ and cdga maps $A\to H^*(X,\mathbb{R}),A\to \Omega^*(X)$ (the de Rham complex) that both induce isomorphisms in cohomology, see Sullivan, Infinitesimal computations in topology, \S 12, in particular, Theorem 12.2.

Lie groups are formal, and so are their classifying spaces and also many homogeneous spaces, in particular symmetric spaces, see e.g. F\'elix, Halperin, Thomas, Rational homotopy theory, \SS 12 and 15.

Answer by algori for Cohomology of a fiber bundle with fiber $H$ and base space $BG$

2012-11-13T00:20:16Z

A typical example would be the case when $G$ is a subgroup of $H$. Then $(EH\times H)/G$ (diagonal action) is 1. homotopy equivalent to $H/G$, and 2. fibered over $BG=EH/G$ with fiber $H$. Note that this works both in the Lie case and the discrete case but in the latter case what we get is not very interesting since the fiber of our fibration is a potentially infinite discrete space.

[upd: There is one thing one can extract from this though: the $i$-th cohomology group of $G$ with coefficients in the infinite product $\Pi_{h\in H}\mathbb{Z}_{(h)}$ is $\Pi_{h G\in H/G} \mathbb{Z}_{(hG)}$ when $i=0$ and is 0 otherwise; this may be of some use when $G$, or its index in $H$, is finite.]

On the other hand, if $G$ is normal in $H$ one can go a bit further: $BH=EH/H$ is the quotient of $BG=EH/G$ by a free action of $H/G$. So, as above, we construct a fibration over $B(H/G)$ with fiber $BG$ and total space $BH$. If we now take an $H$-module $M$ (i.e., a local system on $BH$) we get the Hochschild-Serre spectral sequence

$$E_2^{pq}=H^p(H/G,H^q(G,M))\Rightarrow H^{p+q}(H,M).$$

There are lots of references where this is discussed. One could take a look e.g. at the original paper by Hochschild and Serre (Cohomology of group extensions, Transactions AMS 1953).

Eichler-Shimura isomorphism and mixed Hodge theory

2009-12-23T06:50:22Z

Let $Y(N),N>2$ be the quotient of the upper half-plane by $\Gamma(N)$ (which is formed by the elements of $SL(2,\mathbf{Z})$ congruent to $I$ mod $N$). Let $V_k$ be the $k$-th symmetric power of the Hodge local system on $X(N)$ tensored by $\mathbf{Q}$ (the Hodge local system corresponds to the standard action of $\Gamma(N)$ on $\mathbf{Z}^2$).

$V_k$ is a part of a variation of polarized Hodge structure of weight $k$. So the cohomology $H^1(Y(N),V_k)$ is equipped with a mixed Hodge structure (the structure will be mixed despite the fact that $V_k$ is pure because $Y(N)$ is not complete). The complexification $H^1(Y(N),V_k\otimes\mathbf{C})$ splits

$$H^1(Y(N),V_k\otimes\mathbf{C})=H^{k+1,0}\oplus H^{0,k+1}\oplus H^{k+1,k+1}.$$

There is a natural way to get cohomology classes $\in H^1(Y(N),V_k\otimes\mathbf{C})$ from modular forms for $\Gamma(N)$. Namely, to a modular form $f$ of weight $k+2$ one associates the secion

$$z\mapsto f(z)(ze_1+e_2)^k dz$$

of $$Sym^k(\mathbf{C}^2)\otimes \Omega^1_{\mathbf{H}}.$$

Here $\mathbf{H}$ is the upper half plane and $(e_1,e_2)$ is a basis of $\mathbf{C}^2$ coming from a basis of $\mathbf{Z}^2$. This pushes down to a holomorphic section of $V_k\otimes \mathbf{C}$.

Deligne had conjectured (Formes modulaires et repr\'esentations l-adiques, Bourbaki talk, 1968/69) that the above correspondence gives a bijection between the cusp forms of weight $k+2$ and $H^{k+1,0}$. (This was before he had even constructed the Hodge theory, so strictly speaking this can't be called a conjecture, but anyway.) Subsequently this was proved by Zucker (Hodge theory with degenerating coefficients, Anns of Maths 109, no 3, 1979). See also Bayer, Neukirch, On automorphic forms and Hodge theory, (Math Ann, 257, no 2, 1981).

The above results concern cusp forms and it is natural to ask what all modular forms correspond to in terms of Hodge theory. It turns out that all weight $k+2$ modular forms give precisely the $k+1$-st term of the Hodge filtration on $H^1(Y(N),V_k\otimes\mathbf{C})$ i.e. $H^{k+1,0}\oplus H^{k+1,k+1}$.

The proof of this is not too difficult but a bit tedious. So I would like to ask: is there a reference for this?

upd: The original posting contained non-standard notation; this has been fixed.

Conjuring phantoms by hand?

2012-09-25T22:05:48Z

A map $f:X\to Y$ of CW-complexes is called a phantom if $f$ restricted to the $n$-skeleton of $X$ is contractible for all $n$. The first non-trivial example of such a map, with $X=\Sigma\mathbb{P}^\infty(\mathbb{C})$ and $Y$ an infinite wedge of 4-spheres, was constructed by J. F. Adams and G. Walker in 1964. Subsequently, it was shown that many other examples exist. In particular, in 1966 B. Gray showed that there are continuously many non-homotopy equivalent phantoms $K(\mathbb{Z},2)\to S^3$.

I would like to ask if there is a way to construct a non-trivial phantom map, or at least to prove such maps exist, by hand (say, using the material covered in Hatcher's Algebraic topology).

The motivation is this: a colleague of mine has gone away for a while and has asked me if I could replace him during his topology problem classes. I agreed but the instructions given to me were rather vague ("just show them some cool examples..."). I have enough examples to fill all the sessions (a couple of those found on MO by the way), but still I was wondering if I could construct a phantom by hand, and learn myself how to do this. The students seem to be pretty smart but haven't seen much beyond the basic cohomology and homotopy theory, not yet anyway.

Answer by algori for Sets with compatible right and left G-actions

2012-09-03T17:59:30Z

Suppose we have a group $G$ acting on the right and on the left as above on a set $X$. These actions give rise to a left action of $G\times G$ on $X$ by setting $(g_1,g_2)\cdot x=g_1\cdot x \cdot g_2^{-1}$. Conversely, if we have a left action of $G\times G$ on a set $X$, we can construct compatible left and right actions by setting $g\cdot x= (g,e)\cdot x$ and $x\cdot g= (e,g^{-1})\cdot x$. The compatibility follows from the fact that the factors of $G\times G$ commute.

Of course, instead of left actions of $G\times G$ one can also consider right actions.

Answer by algori for Second lowest weight piece of the cohomology of an algebraic variety

2012-08-30T18:19:45Z

Here is a geometric description of $W_{k+1}H^k(U)/W_k H^k(U)$. Set $d=\dim U$. Suppose $X$ is a compactification of $U$ such that the complement $X\setminus U$ is the union of normal crossing divisors, or, more generally, union of smooth varieties $D_i$ all of whose intersections are smooth. Take submanifolds $Y_i\subset D_i$ (in the $C^\infty$ sense) of dimension $2d-k-1$ such that $\sum [Y_i]$ vanishes in $X$. Then $\bigcup Y_i$ bounds a chain $c$ in $X$. It is useful to think of $c$ as a $C^\infty$ submanifold with boundary in $X\setminus U$. If we intersect this submanifold with $U$ we get a submanifold without boundary. The Poincar\'e dual of it will be in $W_{k+1}$ and all elements of $W_{k+1}$ arise in this way.

Notice that $c$ is not uniquely defined: there may be several chains that bound $\bigcup Y_i$. But the difference between resulting classes $\in H^k(U)$ will be the Poincar\'e dual class of a $C^\infty$ submanifold of $X$, and so will be of weight $k$, as expected.

t-structures and higher categories?

2010-07-04T17:59:11Z

I'm curious to find out where the viewpoint of higher categories may be useful so here is a somewhat vague question (which may or may not have a reasonable answer).

Given a triangulated category, one can consider the set of all possible t-structures on it. Simple examples where one can compute things by hand indicate that this is something complicated but not hopelessly so. See for example the paper http://arxiv.org/abs/0909.0552 by Jon Woolf which describes a three parametric family of t-structures on the constructible bounded derived category of $\mathbf{P}^1(\mathbf{C})$ stratified by a point and its complement. Some of these t-structures are more interesting than others and there is one that is the most interesting of them all since by taking the bounded derived category of its heart one gets back the triangulated category one started with. (For that t-structure the heart is the category of perverse sheaves on $\mathbf{P}^1(\mathbf{C})$.)

On the other hand, the set of t-structures on a triangulated category is interesting since there lurks somewhere the conjectural motivic t-structure whose existence implies Grothendieck's standard conjectures. See the recent paper http://arxiv.org/abs/1006.1116 by Beilinson.

On the triangulated categories page of the n-category lab website it says "Therefore, all the structure and properties of a triangulated category is best understood as a 1-categorical shadow of the corresponding properties of stable (infinity,1)-categories". See http://ncatlab.org/nlab/show/triangulated+category. Note that this is quite a strong statement, since it is referring to all, and not just some, properties and structure of a triangulated category.

So I'd like to ask: is there a higher categorical analog of a t-structure? More generally, how does the higher categorical viewpoint help one understand the set of all (or maybe all "nice" in an appropriate sense) t-structures on a given trangulated category, provided it is the homotopy category of a stable $(\infty,1)$ category?

upd: as Mike points out in the comments, the answer to the first question is yes and it is given by proposition 6.15 of Lurie's Stable Infinity Categories. The second, more "philosophical" question remains.

Answer by algori for product of two sub-Grassmannians

2012-08-28T18:51:52Z

Here is a slightly different way to do this: the cohomology of $G(k,n)$ is generated by the Chern classes $c_1,\ldots,c_k$ of the universal bundle $\gamma (k,n)$ and the Chern classes $c'_1,\ldots,c'_{n-k}$ of the universal quotient bundle, subject to one relation,

$$(1+c_1+\cdots +c_k)(1+c_1'+\cdots +c'_{n-k})=1.$$

Using this we can express $c'_i$'s in terrms of $c_i$'s or vice versa.

Now if we embed $G(k_1,n_1)\times G(k_2,n_2)$ in $G(k,n)$ as above $\gamma(k,n)$ restricts to $p^*_1(\gamma(k_1,n_1))\oplus p^*_2(\gamma(k_2,n_2))$ where $p_1$ and $p_2$ are the projections to the first, respectively, second factor. This allows one to compute the cohomology map $H^*(G(k,n))\to H^*(G(k_1,n_1)\times G(k_2,n_2))$ induced by the embedding, and hence also the homology map in the opposite direction. This gives an explicit way of calculating the image of the fundamental class modulo the values of the Chern classes on the Schubert cells.

(If necessary I can add more details.)

Answer by algori for Covering spaces of surfaces

2012-08-15T16:39:18Z

For an explicit example of a non-cyclic group $G$ acting freely on a surface $S$ take the union $X$ of the edges of the 3-cube $[-1,1]^3\subset\mathbb{R}^3$; set $S$ to be the boundary of a small neighborhood of $X$, and take the group generated by the rotations through $\pi$about the coordinate axes in $\mathbb{R}^3$ as $G$; note that $G$ is isomorphic to Klein's group $\mathbb{Z}/2\oplus\mathbb{Z}/2$, the "simplest" non-cyclic group there is.

Answer by algori for cohomology and $j_!$

2012-08-10T18:45:45Z

If $X$ is smooth and we consider sheaves of $k$-modules, $k$ a commutative ring, then $H^*(X, j_!\mathcal{S})\cong H^*_c(U,\mathcal{S})$ ; the latter is equipped with a non-degenerate pairing $$H^*_c(U,\mathcal{S})\otimes H^{2d-*}(U,\mathcal{S}^{\vee})\to k,$$

where $\mathcal{S}^\vee$ is the Verdier dual local system (i.e., the local system that is constructed from the representation of $\pi_1(U)$ dual to the one that gives $\mathcal{S}$) and $d=\dim X$. So if $k$ is a field, we do get a statement relating the vanishing of $H^*(X, j_!\mathcal{S})$ and $H^{*}(U,\mathcal{S}^{\vee})$ .

Answer by algori for Holomorphic bundles and maps to the Grassmannian ?

2012-08-07T22:34:56Z

Here are some remarks; hope they will help.

Let us consider the algebraic case first: then we shall see what one can hope for in the analytic case. Let $X$ be a smooth compact complex algebraic variety and let $\mathcal{E}$ be a vector bundle on $X$. As Youloush points out, in general it is not true that $\mathcal{E}$ is obtained as a pullback of the universal quotient bundle over some Grassmannian: the universal quotient bundle has sections but it may happen that $\mathcal{E}$ doesn't.

Suppose however there is an ample line bundle $\mathcal{L}$ on $X$. Then, for some $n$, $\mathcal{E}\otimes\mathcal{L}^{\otimes n}$ is generated by global sections (e.g. Hartshorne, part II 7.6 and II 5.17), and as such, it is the quotient of a free sheaf of rank $k=\dim H^0(X,\mathcal{E}\otimes\mathcal{L}^{\otimes n})$ on $X$. In other words, $\mathcal{E}\otimes\mathcal{L}^{\otimes n}$ is the pullback of the universal quotient bundle on $G_{k-r}(\mathbb{C}^k),r=rank(\mathcal{E})$ under the map that takes an $x\in X$ to the subspace of $H^0(X,\mathcal{E}\otimes\mathcal{L}^{\otimes n})$ formed by the sections that vanish at $x$.

So every vector bundle on $X$, up to twisting by a power of $\mathcal{L}$, is in fact induced by a map from $X$ to a Grassmannian. However, in order for this to work, there must be at least one ample line bundle on $X$, which automatically makes $X$ projective (e.g. Hartshorne, ibid.). Something similar holds in the analytic case as well. Either there is a positive line bundle on $X$, in which case $X$ is projective by the Kodaira embedding theorem, or there isn't, in which case the above trick doesn't work, and I'm not sure there is one that does.

Mappings of mapping class groups

2009-12-03T00:12:46Z

Let $X$ be a compact non-orientable surface, maybe with boundary, and let $\tilde X$ be the orienting cover of $X$. If I understand correctly, any smooth automorphism of $X$ lifts naturally to an automorphism of $\tilde X$ (since $\tilde X$ can be viewed as the space of couples (a point $x$ of $X$, an orientation of $T_xX$)). Moreover, composition lifts to composition and isotopic automorphisms lift to isotopic automorphisms. So we get a map $MCG(X)\to MCG(\tilde X)$ where $MCG$ stands for the mapping class group. (In the definition of the mapping class groups we do not require that maps or isotopies should be the identity on the boundary.)

Is there a simple way to describe the kernel of the above map?
Is it true that the mapping class group of a of non-orientable surface injects into the mapping class group of some orientable surface (not necessarily the orienting cover)?
(a slightly unrelated question which nonetheless involves mapping class groups) Is it true that the mapping class group of an orientable surface without boundary injects into the mapping class group of some orientable surface with boundary?

Answer by algori for Convergence of analytic covering maps to a covering map

2012-02-20T01:54:19Z

If $f_n:U\to\mathbb{C}$ are holomorphic and $f=\lim_{n\to\infty} f_n$ uniformly on each compact subset of $U$, then $f'$ has no zeros, provided none of the $f'_n$'s have any and $f$ is not constant. Indeed, then $f'_n\to f'$ uniformly on all compact subsets of $U$ and if $f'(z)=0$, then we can apply Rouch\'e's theorem to $f',f'-f'_n$ and a disk $D\subset U$ such that $z\in D$ , $f'$ has no zeros on $\partial D$ and $sup_{\partial D}|f'-f_n'|<inf_{\partial D}|f'|$ to conclude that $f'_n$ has a zero in the interior of $D$ as well.

On the other hand, it may happen that $f$ is constant: take $U$ to be the unit disk and $f_n(z)=z/n$.

Answer by algori for Strong Kodaira vanishing

2012-02-16T01:41:47Z

Re question 1: yes, see Deligne, Cohomologie des intersections compl`etes, SGA 7 II, th\'eor`eme 1.1. The proof is by "force brutale" as Deligne himself puts it, so I'm not sure this generalizes.

Answer by algori for intersections of $SO_2^n, SL_2^n$ with $SO_{2n}, Sp_{2n}$

2012-01-30T19:56:09Z

[upd: this answers the old version of the question, which has since been changed.]

The intersection is $SO(2)^n=SO(2)\times SO(2)\times\cdots\times SO(2)$, as perhaps expected. The inclusion $SO(2)^n\subset (SL_2)^n\cap SO(2n)$ is clear. The other inclusion follows from this observation: If a linear map that preserves a vector subspace is to be orthogonal, it must preserve the metric restricted to the subspace.

Answer by algori for Description of $GL_3/U$

2012-01-31T21:29:27Z

As already mentioned, the quotient $GL_n(\mathbb{C})/B$ is the space $Fl$ of all flags ($0\subset$ a line $\subset$ a 2-plane $\subset\cdots\subset \mathbb{C}^n$). There are line bundles $L_i , i=1,\ldots, n$ on this space where $L_i$ is obtained by pulling back the tautological bundles from Grassmannians of $i$- and $i-1$-planes in $\mathbb{C}^n$ and quotienting one by the other.

Let $L_i^0$ be the total space of $L_i$ minus the zero section and set $L^0$ to be the fibered product $$L_1^0\times_{Fl}L_2^0\times_{Fl}\cdots\times_{Fl}L^0_n.$$ The group $GL_n(\mathbb{C})$ acts transitively on the space $L^0$ and each stabilizer is conjugate to the group of unipotent matrices in $GL_n(\mathbb{C})$. So $GL_n(\mathbb{C})/\mbox{unipotent matrices}\cong L_0$.

Answer by algori for References: Infinite dimensional Lie algebras

2012-01-31T15:38:27Z

I think the following references might be useful (copied from mathscinet)

MR0874337 (88b:17001) Fuks, D. B.(2-MOSC) Cohomology of infinite-dimensional Lie algebras. Translated from the Russian by A. B. Sosinskiĭ. Contemporary Soviet Mathematics. Consultants Bureau, New York, 1986. xii+339 pp. ISBN: 0-306-10990-5

MR1756408 Feigin, B. L.(J-KYOT-R); Fuchs, D. B.(1-CAD) Cohomologies of Lie groups and Lie algebras [MR0968446 (90k:22014)]. Lie groups and Lie algebras, II, 125–223, Encyclopaedia Math. Sci., 21, Springer, Berlin, 2000. 22E60 (17B45 17B56 22E41)

There is a chance the first reference treats the question you are interested in (I don't have the book at hand). The second one is a very readable survey of Lie group and Lie algebra cohomology.

Answer by algori for Are there any known formulas about the Hodge-Deligne structure of quotients by actions of groups?

2012-01-30T21:45:08Z

The rational cohomology of the quotient of a variety $X$ by an action of a finite group $G$ is the $G$-invariant part of $H^*(X,\mathbb{Q})$; this is a Hodge substructure if $G$ acts by automorphisms. So one can deduce the Hodge polynomial of the quotient from the $G$-equivariant Hodge polynomial of $X$.

Answer by algori for Direct limit of sheaves over paracompact spaces

2012-01-26T18:47:17Z

In general not: take $X$ to be $\mathbb{Z}_{\geq 0}$ and $\mathcal{F}_i$ the constant sheaf supported at $X\cap [0,i]$. The direct limit will be the constant sheaf on $X$ and will have many more sections than there are in the direct limit of the sections of the $\mathcal{F}_i$'s.

However, if you consider compactly supported cohomology (or if $X$ itself is compact), then everything is fine. See e.g. Godement, theorem 4.12.1.

Comment by

2013-04-24T21:52:09Z

Dan -- you are right: multiplication is a map of MHS's of degree 0!

Comment by

2013-02-22T13:53:31Z

Thanks, Marek, this looks interesting.

Comment by

2013-02-22T13:48:44Z

quid -- I agree that dg-algebras would be a more appropriate tag.

Comment by

2013-01-08T03:17:48Z

Anton -- I'm casting the final vote to close, as the question needs a serious revision, starting from the phrase "It specifies some area". If you revise it and explain what exactly you are interested in, I'll vote to reopen. And, as Misha says, it wouldn't do any harm if you LaTeXed the formulae.

Comment by

2013-01-07T01:49:04Z

asdfasdf -- I can't tell you what to do, as I don't know your situation well enough. But in any case, it's a good idea to try and come up with some specific questions and go and see the teacher about them. Just think of the first thing you don't really understand. Then the second. And so on.

Comment by

2013-01-06T22:01:26Z

.. but you could try math.stackexchange.com

Comment by

2013-01-05T23:13:59Z

Joseph -- what is slightly confusing is that it says "convergence a.e." in the title but then in the posting the OP asks about the convergence in measure.

Comment by

2013-01-05T19:56:06Z

Leon -- I see, so you don't allow reflections. In this case the space, call it $X$, is an open cone over the set of all triangles with the sum of the sides equal 1. The latter space, call it $Y$, is the union of two triangles with sides of one glued to the sides of the other in a bijective way, i.e., $Y=S^2$, which makes $X$ homeomorphic to $\mathbb{R}^3$.

Comment by

2013-01-05T19:29:32Z

Todd -- so you did but I didn't see it when I started writing mine.

Comment by

2013-01-05T19:08:23Z

unknown google -- if I understand you correctly, your quotient space is juet the space of all triangles in $\mathbb{R}^2$ with ordered vertices; it includes degenerate triangles, in which one of the vertices lies in the interior of the segment that joins the other two. This space is indeed a subspace of $\mathbb{R}^3$: each triangle is determined, up to a composition of rotations, reflections and translations, by the lengths of the edges (these are ordered, as the vertices are). If you do not allow reflections, then a triangle is determined by the lenghts of the sides plus orientation.

Comment by

2012-12-30T14:35:14Z

Hiro -- I'm afraid I still can't follow your notation. Do you mean that $E_n$ is the $n$-th term of the spectral sequence, i.e., $E_n=H_*(E_{n-1},d_{n-1})$? In that case one would normally expect to see $E^n$, since you write $E^1_{a,b}$, with 1 on top. Also, the term a spectral sequence converges to is denoted $E^\infty$.

Comment by

2012-12-29T17:05:42Z

Hiro -- what exactly are $E_n, f_n$ and $F_n$? If $E_n=\bigoplus_{a+b=n}E^1_{a,b}$ and similarly for $F_n$ then, under your conditions, where can infinitely many terms appear from?

Comment by

2012-12-20T18:12:22Z

Chen -- regarding this and your other question: math.stackexchenge.com is a more suitable place for that.

Comment by

2012-12-16T20:44:47Z

mohammad-83 -- your last integral is the Laplace transform of some function, so I would try to take look in some book on integral transforms. But it is not very clear why you expect that the result can be written in a nice form.

Comment by

2012-12-16T20:32:23Z

chewbacca -- you can delete the question if you wish. Pietro -- this doesn't look a very safe bet to me.