User eric wofsey - MathOverflow

Answer by Eric Wofsey for sufficient conditions for rational homotopy equivalence

2013-04-16T16:17:50Z

Yes. By a generalization of the Hurewicz theorem (which can be formulated more generally in terms of Serre classes), if $X$ is simply connected and has trivial rational homology below dimension $D$, then $\pi_i(X)\otimes\mathbb{Q}=\tilde{H}_i(X,\mathbb{Q})$ via the Hurewicz map for all $i\leq D$. In particular, for $i=D$ this implies we can find elements of $\pi_D(X)$ whose Hurewicz images form a basis for $\tilde{H}_D(X,\mathbb{Q})$. These elements together give a map from a wedge of $D$-spheres to $X$ which induces an isomorphism on rational homology and is hence a rational equivalence.

Answer by Eric Wofsey for How would you say that a small category is embedded into functors from a large $C'$ to abelian groups?

2013-04-08T04:48:39Z

In your example, there likely exists a cardinal $\kappa$ such that $C'$ is $\kappa$-accessible, as are the functors $C'\to Ab$ associated to each object of $C$ (i.e., they preserve $\kappa$-filtered colimits). In this case, you can consider $C$ embedded into the category of $\kappa$-accessible functors from $C'$ to $Ab$, which is only "large" rather than "very large". This is equivalent to restricting to the subcategory of $\kappa$-compact objects of $C'$.

Answer by Eric Wofsey for Why does the naive choice of homogeneous coordinate ring of a product of projective schemes not work?

2013-04-06T19:01:18Z

A ($\mathbb{Z}$-)grading on a ring $S$ is equivalent to an action of the group $\mathbb{G}_m$ on $U=\operatorname{Spec} S$. This is an exercise in algebra; the statement is that the symmetric monoidal category of comodules over the Hopf algebra $A[t,t^{-1}]$ is equivalent to the category of graded $A$-modules. The construction $\operatorname{Proj} S$ then corresponds to taking the quotient of $U\setminus U_0$ by the $\mathbb{G}_m$ action, where $U_0$ is the subscheme corresponding to the "irrelevant ideal". This is really just the familiar fact that $\mathbb{P}^n$ is the quotient of $\mathbb{A}^{n+1}\setminus 0$ by rescaling.

What happens when we take a product of two such schemes $U$ and $V$ with corresponding graded rings $S$ and $T$? I'm going to ignore $U_0$ and $V_0$ in this discussion to simplify notation (i.e., I'm going to pretend that $\operatorname{Proj} S$ is just $U/\mathbb{G_m}$). Well, $U\times V$ will naturally have an action of $\mathbb{G}_m\times \mathbb{G}_m$; this corresponds to the natural bigrading on $S\otimes T$. By taking the diagonal $\mathbb{G}_m\to\mathbb{G}_m\times\mathbb{G}_m$, we get an action of $\mathbb{G}_m$ on the product, and this corresponds to the "total" grading on $S\otimes T$. But if we mod out $U\times V$ by the diagonal action of $\mathbb{G}_m$, we don't get $U/\mathbb{G}_m\times V/\mathbb{G}_m$; to get that, we would need to mod out the entire action of $\mathbb{G}_m\times \mathbb{G}_m$. Alternatively, instead of modding out the whole action of $\mathbb{G}_m\times\mathbb{G}_m$ at once, we could first mod out the action of the antidiagonal $\alpha:\mathbb{G}_m\to\mathbb{G}_m\times\mathbb{G}_m$ given by $\alpha(t)=(t,t^{-1})$ and then mod out the action of the diagonal; this works since the antidiagonal and the diagonal together generate the whole group. What do we get when we mod out the antidiagonal? Well, it turns out that in this case the quotient will still be affine, and its coordinate ring can be found by taking invariants of the action on $S\otimes T$. If $x\in S_n$ and $y\in T_m$, then the the antidiagonal action of $t\in\mathbb{G}_m$ sends $x\otimes y$ to $t^{n-m}x\otimes y$, so the invariants are exactly $\bigoplus S_n\otimes T_n$. Thus the product $U/\mathbb{G}_m\times V/\mathbb{G}_m$ can be obtained as $\operatorname{Proj}(\bigoplus S_n\otimes T_n)$.

Answer by Eric Wofsey for How to call a simplicial set where every boundary of a simplex can be filled?

2013-04-03T06:00:23Z

This is just a contractible Kan complex. It's equivalent to the same condition where you replace the pairs $(\Delta^k,\partial\Delta^k)$ with all pairs $(A,B)$ where $B\subset A$, since $A$ can be built from $B$ by iteratively filling in simplices whose boundaries are already filled in. In particular, any such $X$ will satisfy the Kan condition, and applying the condition to the pair $(X\times \Delta^1, X\times\partial\Delta^1)$ gives a contraction of $X$.

Conversely, given a contractible Kan complex $X$ and a map $\partial \Delta^k\to X$, by contractibility it extends over the cone on $\partial\Delta^k$. But that cone is just a $(k+1)$-horn, and filling in the horn gives a $k$-simplex in $X$ extending the original map.

Answer by Eric Wofsey for is there an anyon structure analogous to spin structure for rank 2 bundle?

2013-04-03T03:34:54Z

I know nothing about the physics you have in mind, but I can tell you about the topology. The classifying space $BSO(2)$ is a $K(\mathbb{Z},2)$, so oriented 2-plane bundles are in bijection with classes in $H^2(B,\mathbb{Z})$. For any $n$, the $n$-sheeted cover $SO(2)\to SO(2)$ corresponds to multiplication by $n$ on $H^2(B,\mathbb{Z})$, so a bundle lifts to that cover iff the corresponding cohomology class divisible by $n$, which is equivalent to the vanishing of the mod $n$ reduction of the class in $H^2(B,\mathbb{Z}/n)$ (generalizing the second Stiefel-Whitney class for $n=2$). When a lift exists, the set of lifts is a torsor over $H^1(B,\mathbb{Z}/n)$, since the homotopy fiber of the map $n:BSO(2)\to BSO(2)$ is a $K(\mathbb{Z}/n,1)$.

The universal cover is contractible, so a bundle lifts to the universal cover iff the bundle is trivial. The set of such lifts is a torsor over $H^1(B,\mathbb{Z})$.

Answer by Eric Wofsey for Cup-squares in the mod 2 cohomology of a cell complex embedded in 3-space

2013-04-02T22:17:00Z

It is true. Let's first assume that $X$ is a finite complex. Then by Alexander duality, $H^2(X;\mathbb{Z})=H_0(S^3\setminus X,\mathbb{Z})$ must be torsion-free. Now the squaring map $H^1(X;\mathbb{Z}/2)\to H^2(X;\mathbb{Z}/2)$ coincides with the Bockstein map. This means that if the squaring map is nonzero, then $H_1(X;\mathbb{Z})$ must have $\mathbb{Z}/2$ as a direct summand. But then $H^2(X;\mathbb{Z})$ would have $\mathbb{Z}/2$ as a summand, and in particular cannot be torsion-free.

Now let $X$ be arbitrary, and write $X=\bigcup X_n$ as a union of finite complexes; from now on we only use $\mathbb{Z}/2$ coefficients. Then $H^1(X_n)$ is finite for all $n$, so $\lim^1 H^1(X_n)=0$. It follows that the map $H^2(X)\to\lim H^2(X_n)$ is an isomorphism. Now for any $x\in H^1(X)$, the image of $x^2$ in $H^2(X_n)$ must be 0 for all $n$ by the previous paragraph. Thus $x^2=0$.

Answer by Eric Wofsey for Are subfunctors of left exact functors also left exact?

2013-04-01T17:32:28Z

Here's a counterexample with additive functors on abelian categories. If $A$ is an abelian group, let $F(A)$ denote the subgroup of elements that are divisible by $2$. It is easy to see that $F:Ab\to Ab$ is an additive functor, and $F$ is a subfunctor of the identity. But $F$ is not left exact because it does not preserve kernels. For instance, $F$ sends the exact sequence $$0\to \mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}$$ to $$0\to2\mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z},$$ which is not exact.

Answer by Eric Wofsey for Generalizations and limitations of Quillen's F-isomorphism theorem

2013-03-23T04:20:00Z

This might be a rather different direction than you're interested in, but analogues of Quillen's theorem have been studied for graded cocommutative Hopf algebras. These generalizations were originally motivated by computations in stable homotopy theory that used finite subalgebras of the Steenrod algebra. Wilkerson ("The cohomology algebras of finite dimensional Hopf algebras") proved that an analogue of Quillen's theorem holds for all finite subalgebras of the mod 2 Steenrod algebra, but gave counterexamples for more general connected graded cocommutative Hopf algebras, including subalgebras of the mod $p$ Steenrod algebra for odd $p$. Palmieri ("A note on the cohomology of finite dimensional Hopf algebras") gave a generalization of the theorem that holds for any finite-dimensional connected graded cocommutative Hopf algebra, but Palmieri uses a very general class of algebras in the place of elementary abelian subgroups, to the point that the theorem essentially becomes purely formal.

Answer by Eric Wofsey for Are Thom spectra MU, MSO and K-theory spectra KU, KO modules over some truncations of the sphere spectrum?

2013-03-13T02:11:34Z

For any $k$ and any $0<n<\infty$ , $K(n)\wedge \pi_{\leq k}S=0$. Indeed, this is true for any spectrum with finitely many homotopy groups, since $K(n)\wedge H\mathbb{Z}=0$ and any such spectrum has a finite filtration into Eilenberg-MacLane spectra. If a spectrum $M$ admits a $\pi_{\leq k}S$-module structure, then $M$ is a retract of $\pi_{\leq k}S\wedge M$, and so $K(n)\wedge M$ is a retract of $K(n)\wedge\pi_{\leq k}S\wedge M=0$ and hence $K(n)\wedge M=0$. But this is not true for $n=1$ and all of your examples (for $MSO$, you must work at an odd prime).

Answer by Eric Wofsey for Pullbacks as manifolds versus ones as topological spaces

2013-03-12T13:38:12Z

Here's a counterexample. Let $X=Y=\mathbb{R}$, and let $Y'$ be a point. Let $f:\mathbb{R}\to\mathbb{R}$ be a smooth map such that $f^{-1}(\{0\})=\{1,1/2,1/3,\dots\}\cup\{0\}$ , and let $f'$ map $Y'$ to $0$. Then the topological pullback is just the space $W=\{1,1/2,1/3,\dots\}\cup\{0\}$ . But since manifolds are locally connected, any map from a manifold to $W$ factors through $Z$, a space with the same points as $W$ but with the discrete topology. Thus $Z$ is the pullback in manifolds, but the canonical map $Z\to W$ is not a homeomorphism.

An interesting question is whether this is the only way to get a counterexample. That is, can you prove the following: A pair of smooth maps has a pullback in manifolds iff every path component of the topological pullback is a manifold.

EDIT: I believe you can get a path-connected counterexample by taking a pair of maps for which the topological pullback is a topologist's sine curve which has been connected up so that if you "unglued" the two halves of it, it would be homeomorphic to $\mathbb{R}$. That is, it should look kind of like a letter P, except that as the loop of the P approaches the vertical line, it wobbles up and down infinitely like a topologist's sine curve. Then any map from a manifold in should factor through $\mathbb{R}$, since manifolds are locally path-connected.

Answer by Eric Wofsey for Is there a conceptual reason why topological spaces have quotient structures while metric spaces don't?

2013-03-06T12:48:34Z

You can define a (pseudo)metric on a quotient of a metric space. Let $X$ be a metric space with metric $d$ and an equivalence relation $\sim$. Say that a chain between two points $x,y\in X$ is a sequence of points $x=a_0\sim b_0$, $a_1\sim b_1$, $\ldots$ $a_n\sim b_n=y$, and define the length of such a chain to be $\sum d(b_i,a_{i+1})$. We can now define the distance $d([x],[y])$ between two equivalence classes to be the infimum of all lengths of chains from $x$ to $y$.

It's easy to see that this is a pseudometric on $X/\sim$ (a metric where the distance between two points might be $0$). This descends to a true metric on the quotient $Y=X/\sim'$, where $x\sim' y$ if $d([x],[y])=0$. Furthermore, $Y$ can be characterized by the following universal property: distance-decreasing maps from $Y$ to a metric space $Z$ are naturally in bijection with distance-decreasing maps from $f:X\to Z$ such that $f(x)=f(y)$ whenever $x\sim y$.

More generally, a similar construction shows that the category of metric spaces and distance-decreasing maps has all connected colimits (colimits over connected diagrams). If you generalize metrics to allow the distance between two points to be infinite, you can construct all colimits, and also all limits (use the sup metric on products).

Answer by Eric Wofsey for Can sine be made into a homomorphism?

2013-02-26T23:53:27Z

If you require the group structure to be continuous, this is impossible. Indeed, in that case, the image $[-1,1]$ would have to be a group as well. But $[-1,1]$ is not homogeneous, so it cannot be a topological group.

Without requiring continuity, it is possible. Let me first give a construction where you restrict from $\mathbb{R}$ to $X=[-\pi/2,\pi/2]$; I will write $f:X\to X$ for the sine function. In that case, the (set-theoretic) dynamical system given by $f$ has a particularly simple structure: $f$ is injective with a unique fixed point 0, $X_n=f^n(X)\setminus f^{n+1}(X)$ has cardinality $2^{\aleph_0}$ for all $n$, and $\bigcap f^n(X)=\{0\}$ . Such a dynamical system is completely determined by giving the sequence of sets $X_n$ for $n\geq 0$ with bijections $X_n\to X_{n+1}$. Let $Y$ be the free $\mathbb{Q}$-vector space on $X$; then it is easy to see that the induced homomorphism $\tilde{f}:Y\to Y$ will have the same properties. Choosing compatible bijections $X_n\to Y_n$, we get a group structure on $X$ for which $f$ is a homomorphism.

Extending this to all of $\mathbb{R}$ is now easy: identify $\mathbb{R}$ with $X\times \mathbb{Z}$ in such a way that the projection $X\times\mathbb{Z}\to X$ sends each $x\in \mathbb{R}$ to the unique $y\in [-\pi/2,\pi/2]$ such that $\sin x=\sin y$. Then the sine map $\mathbb{R}\to \mathbb{R}$ can be identified with the projection to $X$ followed by $f:X\to X$.

This construction is quite flexible if you want the group structure to have various properties. For instance, you could replace $Y$ with the free group on $X$, or the free abelian group on $X$, or the free $\mathbb{F}_p$-vector space on $X$, or many other constructions; you could also replace $X\times\mathbb{Z}$ with any semidirect product of $X$ with a countable group.

Answer by Eric Wofsey for do behavior of gimel or GCH determine all infinte products of cardinals?

2013-02-22T23:31:16Z

Arbitrary products are determined by exponentiation even without GCH. As in the second part of Toink's answer, I assume the $\kappa_i$ are nondecreasing and unbounded with limit $\kappa$.

First, note that $cf(\delta)=cf(\kappa)$. Choose a cofinal sequence $(i_\alpha)$ in $\delta$ of length $\gamma=cf(\kappa)$. I claim it is possible to choose this sequence such that $f(\alpha)=|[i_\alpha,i_{\alpha+1})|$ is eventually a nondecreasing function of $\alpha$. First, if $f(\alpha)<\gamma$ for all sufficiently large $\alpha$, by regularity of $\gamma$ we can replace the $i_\alpha$ with a subsequence for which $f(\alpha)$ is either eventually always the predecessor of $\gamma$ (if $\gamma$ is successor) or eventually an increasing sequence of cardinals approaching $\gamma$ (if $\gamma$ is limit).

If $f(\alpha)\geq \gamma$ for arbitrarily large $\alpha$, let $\eta=\limsup f(\alpha)$. If $f(\alpha)=\eta$ for arbitrarily large $\alpha$, we can choose a subsequence of the $i_\alpha$ such that $f(\alpha)=\eta$ for all sufficiently large $\alpha$. Otherwise, if $cf(\eta)<\gamma$, we can similarly find a subsequence to make $f(\alpha)=\eta$ for all sufficiently large $\alpha$ (since $f(\alpha)$ must get arbitrarily close to $\eta$ over and over again as $\alpha$ approaches $\gamma$). Finally, if $cf(\eta)=\gamma$, we can pick a subsequence such that $f(\alpha)$ is eventually an increasing sequence of cardinals approaching $\eta$.

Now $$\prod \kappa_i=\prod_{\alpha<\gamma} \left(\prod_{i_\alpha}^{i_{\alpha+1}} \kappa_{\beta}\right),$$ and by induction on $\delta$ we can compute each of these smaller products. By our choice of the sequence $i_\alpha$, these smaller products are also eventually nondecreasing; splitting off an initial segment and again using induction, we may assume they are actually nondecreasing. As long as they are not eventually constant, the cofinality of their supremum will be $\gamma$. Thus we have reduced the problem to the case $\delta=\gamma=cf(\kappa)$.

Now by choosing a bijection $\delta=\delta^2$ we can split the product into $\delta$ many products, each of size $\delta$. Each one of these subproducts must be at least $\sup \kappa_i=\kappa$, so the entire product is at least $\kappa^\delta$. But the product is also clearly at most $\kappa^\delta$, so we have $\prod \kappa_i=\kappa^\delta$.

Answer by Eric Wofsey for Left/right exact functor "in nature" which is not a right/left adjoint

2013-02-16T16:20:47Z

I would disagree that the hypotheses of the adjoint functor theorem are much stronger than exactness. Left exactness is equivalent to preserving all finite limits, and the hypotheses of the adjoint functor theorem are existence of all limits, preserving all limits, and a smallness condition that usually is easy to verify. Furthermore, to know that a left exact functor preserves all limits, it suffices to know that it preserves arbitrary products (since any limit can be expressed as a kernel of an appropriate map between products). So in most typical applications, the only difference between being left exact and having a left adjoint is whether a functor preserves infinite products.

This also shows how to find a counterexample: find a left exact functor that does not preserve infinite products. For instance, if $M$ is any flat module over a commutative ring $R$, tensoring with $M$ is left exact, but will not preserve infinite products unless $M$ has nice finiteness properties (if $R$ is Noetherian, the condition is that $M$ is finitely generated). In particular, if $R=\mathbb{Z}$ you could take $M=\mathbb{Q}$, or if $R$ is a field you could take $M$ to be any infinite-dimensional vector space.

As Todd notes in his comment, you can similarly get an example for right exactness instead of left exactness by Homming out of a projective module that is not finitely generated.

You can get a more artificial sort of counterexample by taking abelian categories with a size restriction on their objects that prevents the adjoint from existing (because you don't have all (co)limits). For instance, take the category of countable-dimensional vector spaces over some field, and consider the endofunctor given by tensoring with a countably infinite dimensional space $V$. This is right exact, and it ought to have a right adjoint given by $\mathrm{Hom}(V,-)$. But this right adjoint is undefined because (for instance) $\mathrm{Hom}(V,V)$ is uncountable-dimensional and so it doesn't exist in our category.

Answer by Eric Wofsey for (Homotopy theory) When does the 2 of 3 property not imply 2 of 6?

2013-02-05T16:47:11Z

Here's a rather tautological example. Consider the category $$X\rightarrow Y\rightarrow Z\rightarrow A.$$ That is, $X$, $Y$, $Z$, and $A$ are the only objects, and the only morphisms are those appearing in the diagram (and their composites). Then let $W$ consist of the identity maps, the map $X\to Z$, and the map $Y\to A$. Then this satisfies 2 out of 3 but not 2 out of 6.

Answer by Eric Wofsey for Is an additive category a balanced category?

2013-01-25T09:29:26Z

An additive category need not be balanced. Consider the full subcategory of abelian groups consisting of all torsion-free groups. Then for any $n\neq0$, the map $n:\mathbb{Z}\to\mathbb{Z}$ is monic and epic, but it is not an isomorphism unless $n=\pm1$. The only nontrivial part of this is that it is epic, and this is simply the statement that a map from $\mathbb{Z}$ to a torsion-free group is uniquely determined by its value at $n$.

Answer by Eric Wofsey for Is every finite subcomplex of a contractible simplicial complex contained in a finite contractible subcomplex?

2013-01-21T23:12:05Z

Let $X$ be any finite complex such that any map $X\to X$ homotopic to the identity is surjective and for which there is a surjective but nullhomotopic PL map $f:X\to X$ (for instance, $X$ could be $S^n$ for any $n>0$). Let $K$ be the mapping telescope obtained by iterating $f$. Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $X$ is $K$ itself. Indeed, in order to be able to contract the first copy of $X$, you must take a subcomplex that contains the entire second copy, and to contract the second copy, you must contain the third copy, and so on.

For $X=S^1$, you can even visualize this example pretty easily. Take a cylinder, and deform it so that one end of it looks like a crescent, and then collapse the crescent to a circle by gluing together the two "C" shapes that make up the crescent and gluing the two endpoints of the "C". Glue that circle to one end of another cylinder, and then deform the other end of that cylinder similarly. Repeating this infinitely gives the telescope $K$.

Answer by Eric Wofsey for Why all irreducible representations of compact groups are finite-dimensional ? [EDIT: Subtleties: AC,etc]

2013-01-20T19:55:12Z

Here's a pretty easy direct argument. Let $X$ be a unitary representation of a compact group $G$. We note that for any finite-rank operator $T$ on $X$, $g\mapsto gTg^{-1}$ is a norm-continuous map $G\to B(X)$. This is because $T$ is a sum of operators of the form $\langle -, u\rangle v$, which conjugate to $\langle -, gu\rangle gv$, and $g\mapsto gv$ is norm-continuous for any fixed $v$ (the map $G\to U(X)$ is strong operator continuous).

Now let $T$ be any finite-rank positive operator on $X$. By averaging the conjugates of $T$ over $G$, we obtain an invariant positive operator $\tilde{T}$. By the continuity noted above and compactness of $G$, $\tilde{T}$ can be approximated in norm by finite "Riemann sums" of conjugates of $T$, and is thus compact.

If $X$ is irreducible, $\tilde{T}$ has to be a multiple of the identity. Since $\tilde{T}$ is compact, it follows that $X$ is finite-dimensional. More generally, eigenspaces of $\tilde{T}$ give finite-dimensional subrepresentations of any representation, and it follows easily that every unitary representation is a sum of irreducible representations (which are finite-dimensional).

Answer by Eric Wofsey for Automorphism of finite groups and Hurwitz spaces

2013-01-20T00:27:39Z

No, this isn't even true if $G$ is $S_n$ itself: the symmetric group $S_6$ has an outer automorphism.

Answer by Eric Wofsey for continuous R^2xR^2xR^2/E^+(2) -> R^3 injection?

2013-01-05T19:54:10Z

By translation, fix the first point to be at the origin. Consider the other two points as complex numbers, and take their quotient. As long as the other two points are not both at the origin, this continuously gives an element of $\mathbb{CP}^1\cong S^2$. Up to rotation, the other two points are determined by this quotient together with a scale parameter, such as the sum the sides of the triangle. A point on the sphere together with a nonzero scale parameter gives you a point in $\mathbb{R}^3\setminus\{0\}$ . Thus we have a continuous injection (in fact, homeomorphism) from your space (except for the point where all three points coincide) to $\mathbb{R}^3\setminus\{0\}$ .

What about the case when all three points coincide? Well, a sequence in your space will converge to the case when all three points coincide iff the scale parameter converges to 0. Thus we can continuously extend the map to send that point to $0\in\mathbb{R}^3$. We thus get that your space is actually homeomorphic to $\mathbb{R}^3$.

Answer by Eric Wofsey for Irreducible cohomology theories

2013-01-04T16:43:35Z

The sphere spectrum, representing stable cohomotopy, is irreducible (you can see this, for instance, from the fact that its homology is irreducible and any summand would be connective and thus would have to have nontrivial homology). But the associated cohomology of a point is the stable homotopy groups of spheres, which are certainly not irreducible in general.

Answer by Eric Wofsey for A lost lemma about periodicity in a grid of long exact sequences?

2013-01-03T20:38:05Z

This has a simple interpretation in terms of spectral sequences. Think of the top left 2x2 square of the original square as a triple complex. Call the 3 dimensions $x$ (horizontal), $y$ (vertical), and $z$ ($C_{ij}$ differential). By using either double complex spectral sequence, we see that the total cohomology of the $xy$-plane is just $C_{33}$. Thus the total cohomology of the triple complex is $H^*(C_{33})$.

On the other hand, we can also compute the total cohomology of the triple complex by a spectral sequence that first takes the $z$-cohomology and then takes the $xy$-cohomology. A pair $([\alpha],[\beta])$ in your lemma gives a class that survives this spectral sequence: $g([\alpha])-u([\beta])$ is the $d_1$ differential, and the $d_2$ differential will vanish for degree reasons. The operation taking $([\alpha],[\beta])$ to $[\gamma]$ is just the isomorphism between the limit of this spectral sequence and the total cohomology $H^*(C_{33})$.

Note that in your proof, $\chi$ is only defined up to a cocycle in $C_{22}$, and so $[\gamma]$ will only be defined modulo the image of $vh:H^{k-1}(C_{22})\to H^{k-1}(C_{33})$. This indeterminacy reflects exactly the fact that $([\alpha],[\beta])$ corresponds to an element of the associated graded of a filtration on $H^{k-1}(C_{33})$ (whose first term is the image of $vh$), rather than an element of $H^{k-1}(C_{33})$ itself.

Answer by Eric Wofsey for Rational Morava E-theory of cyclic groups

2013-01-01T19:33:13Z

Your guess is correct, and I think this is a confusion pretty much everyone has when they first see these computations. What's going on here is essentially just the simple algebraic fact that $$\mathbb{Z}[p^{-1}][[x]]\neq\mathbb{Z}[[x]][p^{-1}].$$

In computation (1), you can work directly with a Gysin sequence for $E_n$ , without going through $K(n)$ and Bocksteins. The Gysin sequence tells you that $$E_n^*(B\mathbb{Z}/p)=E_n^*[[x]]/([p](x)),$$ where $[p](x)$ is the $p$-series of the formal group law. More generally, this holds for any complex-oriented theory $E$ such that $[p](x)\in E^*[[x]]$ is a nonzero divisor.

Since the formal group law for $E_n$ has height $n$, the first coefficient of $[p](x)$ which is a unit is the $x^{p^n}$ coefficient. It follows that $[p](x)$ differs from a monic polynomial of degree $p^n$ by a unit in $E_n^*[[x]]$ , so $E_n^*(B\mathbb{Z}/p)$ is free of rank $p^n$ over $E_n^*$ .

On the other hand, if we do the same computation with $E_n\otimes \mathbb{Q}$ instead of $E_n$ , $$\frac{[p](x)}{x}=p+\dots\in(E_n\otimes \mathbb{Q})^*[[x]]$$ is now a unit. Thus $(E_n\otimes\mathbb{Q})^*(B\mathbb{Z}/p)$ is free of rank 1 over $(E_n\otimes\mathbb{Q})^*$ . However, $[p](x)/x$ is only a unit in $(E_n\otimes \mathbb{Q})^*[[x]]$ , not in $E_n^*[[x]]\otimes\mathbb{Q}$ . This is because to construct a power series inverse for it, we need coefficients with arbitrarily large powers of $p$ in the denominator. Thus we find that $(E_n\otimes \mathbb{Q})^*[[x]]/([p](x))$ and $E_n^*[[x]]\otimes\mathbb{Q}/([p](x))$ look quite different from each other, and this is exactly the discrepancy between your two computations.

As for a computation along the lines of (2) that gives the answer from (1), I don't know of anything like that. You could make an AHSS computation for $E_n$ rather than $E_n\otimes\mathbb{Q}$ and only tensor with $\mathbb{Q}$ afterwards, which would give the right answer. However, the AHSS $$H^*(\mathbb{Z}/p,E_n^*)\implies E_n^*(B\mathbb{Z}/p)$$ is quite subtle: I don't know how to compute the differentials in it without already knowing the final answer, and the filtration of the spectral sequence distorts almost all of the structure of $E_n^*(B\mathbb{Z}/p)$ (for instance, the associated graded you get from the spectral sequence consists mostly of $p$-torsion, whereas the actual answer is torsion-free).

Answer by Eric Wofsey for Is a conservative finite limit preserving functor of (infinity,1)-categories homotopically faithful?

2012-12-31T17:47:02Z

Here's a counterexample. Let $\mathcal{C}=\mathcal{D}$ be the stable $(\infty,1)$-category of perfect complexes over $\mathbb{Z}_{(p)}$, and let $F(X)=X\otimes \mathbb{Z}/p$ (the derived tensor product). Then $F$ is exact, and does not send any nonzero objects to zero and hence reflects equivalences. But $F$ is not faithful, since it kills the multiplication by $p$ map on any object.

Answer by Eric Wofsey for Is every functor inducing a homotopy equivalence a composition of adjoint functors?

2012-12-28T13:01:22Z

The answer is no. Let $C$ be a category such that the unique map from $C$ to the terminal category is a composition of $n$ adjoints. Then $C$ has an object $x_0$ such that every other object of $C$ can be connected to $x_0$ by a zigzag of length at most $n$; this is easy to prove by induction on $n$.

In particular, let $R$ be the "infinite zigzag", the unique poset such that $|BR|$ is homeomorphic to $\mathbb{R}$. Then $BR$ is contractible, but the unique map from $R$ to the terminal category cannot be a composition of adjoints. Note, however, that this map is a transfinite composition of adjoints (of length $\omega$). It seems plausible to me that any functor between (small) categories which is an equivalence on nerves could be a transfinite composition of adjoints.

Answer by Eric Wofsey for Compact MU or BP Modules

2012-12-14T23:55:55Z

No nonzero finite spectrum admits an $MU$-module structure. Indeed, suppose $F$ is a finite spectrum with an $MU$-module structure. Then for all $n$, $F$ has a map $v_n:\Sigma^{2p^n-2}F\to F$, which induces an isomorphism on $K(n)_*F$ (there's a subtlety here in that it's not obvious that the $v_n$ map on $F$ and the $v_n$ map on $K(n)$ give rise to the same map on $K(n)\wedge F$; see eg the end of the proof of Lemma 7 of http://math.harvard.edu/~lurie/252xnotes/Lecture33.pdf). Thus for each $n$, $F/v_n$ is $K(n)$-acyclic. But $F/v_n$ is finite, so this implies it is also $K(m)$-acyclic for all $m<n$ , and so $v_n$ is also an isomorphism on $K(m)_*F$. But by finiteness of $F$, for any $m$ sufficiently large we can find $n>m$ for which $v_n$ must be $0$ on $K(m)_*F$ just for reasons of degree (by the AHSS for $K(m)_*F$). Thus $K(m)_*F=0$ for all sufficiently large $m$, which implies $F=0$.

I would also add that even if you did have a finite spectrum with an $MU$-module structure, it could not possibly be compact as an $MU$-module. Indeed, if it were, after smashing with $H\mathbb{Z}$ it would be a compact $H\mathbb{Z}\wedge MU$-module. But $\pi_*(H\mathbb{Z}\wedge MU)$ is a polynomial ring on infinitely many generators, and so all but finitely many of those generators have to act non-nilpotently on any compact module (basically, any "finite presentation" of a compact module can only involve finitely many of the polynomial generators). Since $\pi_*(H\mathbb{Z}\wedge F)=H_*(F)$ vanishes in all but finitely many degrees for $F$ finite, this is impossible.

Answer by Eric Wofsey for This is not a category. What is it?

2012-12-14T19:23:54Z

It's called a groupoid. Given an object $A$, call the degenerate edge from $A$ to itself the identity map at $A$. Given an edge $f:A\to B$, let $f^{-1}:B\to A$ denote the unique edge that fills in a 2-simplex whose other two edges are $f$ and the identity. Given $f:A\to B$ and $g:B\to C$, define $gf$ to be the unique edge that fills in a 2-simplex whose other two edges are $f^{-1}$ and $g$. The condition for $n=3$ using the maps $f$, $1$, and $f$ gives that $(f^{-1})^{-1}=f$. It is now easy to see that the identity maps are units for our composition operation, and $f^{-1}f=ff^{-1}=1$.

The condition for $n=3$ now says that $(hf^{-1})(gf^{-1})^{-1}=hg^{-1}$. Setting $g=1$ tells us that $(hf^{-1})f=h$, and so we have cancellation on the right. Setting $h=1$, $f=x$, and $g=yx$ gives $x^{-1}y^{-1}=(yx)^{-1}$. Setting $f=y$, $g=z^{-1}$, and $h=xy$ now says that $x(z^{-1}y^{-1})^{-1}=(xy)z$. Since $(z^{-1}y^{-1})^{-1}=yz$, this says our composition operation is associative. It follows that composition defines a groupoid, and it is now easy to see the simplicial set is the nerve of this groupoid.

Answer by Eric Wofsey for Impossibility of continuously picking k independent rows from a rank k matrix

2012-12-11T05:09:38Z

As Angelo points out, this statement is trivial if by "rows" you mean rows with respect to a fixed basis. A generalization would be to allow the "rows" to come from any basis. Up to some duality, this is equivalent to asking to be able to continuously choose a set of $k$ linearly independent vectors whose span is disjoint from the kernel of your linear map.

Here's a simple way to see you can't do this. The Grassmannian $G_{n,k}$ of $k$-planes in $F^n$ ($F=\mathbb{R}$ or $\mathbb{C}$) embeds in the space of rank $k$ matrices by sending a $k$-plane to the orthogonal projection onto it. If we could continuously choose $k$ linearly independent vectors whose span is disjoint from the kernel of such a projection, then by applying the projection to these vectors we could continuously choose bases for all $k$-planes. That is, we would have a trivialization of the tautological vector bundle on $G_{n,k}$. But the tautological bundle is certainly not trivial if $0<k<n$ (you can use characteristic classes, or use the universal property of the Grassmannian and simply give an example of any nontrivial rank $k$ bundle generated by $n$ sections).

Answer by Eric Wofsey for Generators of Thick Subcategories

2012-12-01T01:03:18Z

When $R$ is the Eilenberg-MacLane spectrum of a Noetherian ring, thick subcategories are in bijection with specialization-closed subsets of $\mathrm{Spec}\ \pi_0(R)$. Such a thick subcategory is generated by a single compact object iff the specialization-closed subset is actually Zariski-closed (and in that case a generator is given by a Koszul complex for generators of an ideal corresponding to the closed set). To say that this holds for all thick subcategories imposes a rather strong condition on the ring $\pi_0(R)$; I believe it's actually equivalent to $\mathrm{Spec}\ \pi_0(R)$ having only finitely many points. (Note that although this condition does hold for $p$-local spectra and the corresponding "Spec" has infinitely many points, one for each height, this situation is badly non-Noetherian.)

The same story holds more generally if the graded ring $\pi_*(R)$ is Noetherian and stratifies the category of $R$-modules, in the sense of Benson-Iyengar-Krause. For instance, this automatically holds if $\pi_*(R)$ is a regular ring concentrated in even degrees.

Here's a sketch of a proof that if $R$ is a Noetherian ring such that any specialization-closed subset of $\mathrm{Spec}\ R$ is closed, then $\mathrm{Spec}\ R$ is finite. A specialization-closed set is just a union of closed sets, so this implies that any set consisting only of closed points is closed. But then quasicompactness implies there can only be finitely many closed points. We can now look at prime ideals of dimension (coheight) 1, and by a similar compactness argument there can be only finitely many of them. Continuing by induction on dimension, we get that for each dimension, $R$ can only have finitely many primes of that dimension. But $\mathrm{Spec}\ R$ has to be finite-dimensional (since, say, there are only finitely many maximal ideal and the localization at any maximal ideal is finite-dimensional), so it can only have finitely many points in total.

Something that might be more reasonable to ask for is that any thick subcategory is a join of irreducible thick subcategories, and that any irreducible thick subcategory is generated by a single object. Here I say a thick subcategory is irreducible if it is not the join of two smaller thick subcategories. In the stratified case, irreducible thick subcategories correspond to irreducible closed subsets of $\mathrm{Spec}\ \pi_*(R)$ , and all of these thick subcategories are indeed generated by a single object.

Answer by Eric Wofsey for Number of II${}_1$ factors

2012-11-29T21:19:01Z

Your argument is correct. An alternate and more "intrinsic" argument is to look at the predual, which is a separable Banach space. There are only continuum many separable Banach spaces, since they are determined by the metric on a countable dense subset that is a $\mathbb Q[i]$-vector space. Thus there are only continuum many separable von Neumann algebras, when considered just with their structure of dual Banach spaces. As Nik points out in the comments, the dual Banach space structure may not determine the algebra structure. However, given any such dual Banach space, you can fix a countable weak*-dense subset, and then any algebra structure will be determined by what it does on that countable set, so you again have only continuum many possibilities.

Comment by Eric Wofsey

2013-05-02T05:26:24Z

"Quasiprojective" looks more correct than "quasi-projective" to me.

Comment by Eric Wofsey

2013-04-08T05:16:32Z

This may be easier to understand if you think about additive functors on $C$ as "modules" over $C$, generalizing the case when $C$ has a single object and is a ring.

Comment by Eric Wofsey

2013-04-08T05:13:51Z

If $C$ contains less than $\kappa$ morphisms, then it should be easy to write any $C$-shaped diagram in $Ab$ as a $\kappa$-filtered colimit of diagrams in which all the abelian groups have size less than $\kappa$. Given any element of any of the groups of the diagram, just take the subdiagram that it "generates".

Comment by Eric Wofsey

2013-04-08T03:22:58Z

Why not just take the Yoneda embedding on $C$ itself, or some appropriate small subcategory of $C'$ that contains $C$?

Comment by Eric Wofsey

2013-04-08T02:54:14Z

If $B$ is projective over $A$, it is easy to see that it commutes. I conjecture that the converse holds: If $B$ is not projective, there is some chain of modules for which it does not commute. Any sort of general description of $\bigcap M_i\otimes B$ is likely to involve $\lim^1$, which tends to be pretty nasty to compute (when it's nontrivial).

Comment by Eric Wofsey

2013-04-07T08:23:29Z

This question seems awfully broad and vague. What is your motivation? Is there some example you have in mind? Do you want to know when taking the intersection commutes with tensor products?

Comment by Eric Wofsey

2013-04-07T01:18:15Z

Degree 2 maps of chain complexes only correspond to homotopies between homotopies if the endpoints of the (second) homotopies are required to be fixed in place throughout the (first) homotopies. So the lack of degree 2 maps tells you that the degree 1 maps are not homotopic as homotopies whose endpoints are fixed, but if one endpoint is allowed to move they may still be homotopic. To get the space of maps equipped with a homotopy to a given map (so one endpoint but not the other is fixed), you have to do something more subtle using a simplicial structure on the category of chain complexes.

Comment by Eric Wofsey

2013-04-07T00:35:03Z

If you think of $f$ as modding out an equivalence relation on $X$, then this takes $x$ to its equivalence class. Other than that, I've never seen a term for this.

Comment by Eric Wofsey

2013-04-07T00:23:33Z

It is certainly not true that the space of maps homotopic to a given equivalence is contractible; I'm not sure what led you to that conclusion. You may be confusing it with the space of maps equipped with a homotopy to a given map, which is contractible.

Comment by Eric Wofsey

2013-04-05T02:09:48Z

Some of the answers to mathoverflow.net/questions/915/… may be relevant

Comment by Eric Wofsey

2013-04-02T18:53:55Z

It turns out that $f_!$ is still left exact. This follows from the fact that $f_*$ is left exact and for any subsheaf $\mathcal{F}\subset\mathcal{G}$, $f_!\mathcal{G}\cap f_*\mathcal{F}=f_!\mathcal{F}$ as subsheaves of $f_*\mathcal{G}$.

Comment by Eric Wofsey

2013-04-01T17:38:47Z

It is obvious that a subfunctor of a left exact functor sends short exact sequences to sequences that are exact on the left (i.e., it sends injections to injections), but it is neither obvious nor true that it preserves exactness in the middle.

Comment by Eric Wofsey

2013-03-28T18:19:34Z

I don't understand what you're asking for. You could take $J=I$, since $I$ itself is directed.

Comment by Eric Wofsey

2013-03-28T01:34:53Z

I don't have time to give a proper answer right now, but under suitable hypotheses, a statement like this comes from the theory of Bousfield localization.

Comment by Eric Wofsey

2013-03-27T23:32:54Z

Complex line bundles are classified by $H^2$ (with coefficients in $\mathbb{Z}$), so you can get extra line bundles on a product that come from $H^1(X)\otimes H^1(Y)$. The simplest example of this is for $X=Y=S^1$: every complex line bundle on $X$ or $Y$ is trivial, but $H^2(X\times Y)=\mathbb{Z}$.