User tom church - MathOverflow

Answer by Tom Church for Orbits of a symplectic group on its Lie algebra in the finite field case

2010-07-11T15:53:21Z

It won't tell you everything about the orbits, but the decomposition of $\mathfrak{sp}_{2n}\textbf{F}_p$ as an $\text{Sp}_{2n}\textbf{F}_p$ -representation is known. This can be found in Hogeweij, "Almost-classical Lie algebras. I." Nederl. Akad. Wetensch. Indag. Math. 44 (1982), no. 4, 441-452, but it is hard to extract the answer from that paper, so I'll briefly give the argument.

If $p$ is odd, then $\mathfrak{sp}_{2n}\textbf{F}_p$ is irreducible with highest weight $2\omega_1$, where $\omega_1,\ldots,\omega_n$ are the fundamental weights for $\text{Sp}_{2n}\textbf{F}_p$ .

If $p=2$, we proceed as follows. Let $H\approx \textbf{F}_p^{2n}$ be the standard representation of $\text{Sp}_{2n}\textbf{F}_p$ . Note that as a subspace of $\mathfrak{gl}_{2n}\textbf{F}_p\approx H^*\otimes H$ , the condition defining $\mathfrak{sp}_{2n}\textbf{F}_p$ describes exactly the subspace $\text{Sym}^2 H$ inside $H\otimes H\approx H^*\otimes H$ , so we are looking at orbits of $\text{Sp}_{2n}\textbf{F}_p$ on $\text{Sym}^2 H$.

In characteristic 2 we have an embedding of $H$ into $\text{Sym}^2 H$ by $x\mapsto x\cdot x$, which is linear since $(x+y)^2 = x^2+y^2$ (in general this twists by Frobenius but we are over $\textbf{F}_2$). Since $x\cdot y=y\cdot x=-y\cdot x$, the quotient $\text{Sym}^2 H/H$ is isomorphic to $\bigwedge^2 H$. Now $\bigwedge^2 H$ has two invariant subrepresentations. One is trivial, spanned by the vector $\omega=a_1\wedge b_1+\cdots+a_n\wedge b_n$. The other is the kernel $K$ of the contraction $\bigwedge^2 H\to \textbf{F}_2$, defined by $a_i\wedge b_i\mapsto 1$, $a_i\wedge a_j\mapsto 0$, $b_i\wedge b_j\mapsto 0$, and $a_i\wedge b_j\mapsto 0$. Note that under this contraction $\omega$ is taken to $n\in \textbf{F}_2$; thus $\omega$ is contained in $K$ iff $n$ is even. Finally, $K$ is irreducible when $n$ is odd, and $K/\langle\omega\rangle$ is irreducible when $n$ is even.

If I'm not mistaken, this means the invariant subrepresentations are thus just $H$, $\langle\omega\rangle$, $H\oplus \langle\omega\rangle$, and $H+K$ (the kernel of the contraction $\text{Sym}^2 H\to \textbf{F}_2$).

Answer by Tom Church for Number of A Subset of Monomials

2010-06-09T23:10:36Z

Let $A_\ell$ be the number of monomials of degree $n$ on $\ell$ variables, which involve all $\ell$ variables and satisfy the condition (this will necessarily be 0 for $\ell>n$). The number of monomials involving all $\ell$ variables is $\binom{n-1}{\ell-1}$ by stars-and-bars. The number of monomials involving all $\ell$ variables at least twice (the invalid monomials), dividing by $x_1\cdots x_\ell$, is $\binom{n-\ell-1}{\ell-1}$. Thus $A_\ell=\binom{n-1}{\ell-1}-\binom{n-\ell-1}{\ell-1}$.

Each monomial is supported on a unique subset of the variables. For a fixed subset of size $\ell$, the monomials supported there are counted by $A_\ell$. There are $\binom{k}{\ell}$ subsets of size $\ell$. So if $N_k$ is the answer to the problem, I believe we have the formula

$$N_k=\sum_{0\leq \ell\leq k} A_\ell \binom{k}{\ell}=\sum_{0\leq \ell\leq k}\binom{n-1}{\ell-1}\binom{k}{\ell}-\binom{n-\ell-1}{\ell-1}\binom{k}{\ell}$$ $$=\binom{n+k-1}{n}-\sum_{0\leq \ell\leq k}\binom{n-\ell-1}{\ell-1}\binom{k}{\ell}$$

[Edit: I now see that this argument was already given by Vladimir Dotsenko. There seems to be some disagreement about his answer though, so I will leave this here as independent confirmation.]

Answer by Tom Church for How to calculate symmetric tensor products of SO(10) representations?

2010-06-05T22:25:03Z

I'm not familiar with your notation, but is $V$ the standard representation of $\text{SO}_{10}\mathbb{C}$ ? If so, $\text{Sym}^k V$ is the representation of $\text{GL}_{10}\mathbb{C}$ corresponding to the partition $\lambda=(k,0,...,0)$. Then you can apply Littlewood's formula to see how this irreducible decomposes when you restrict from $\text{GL}_{10}\mathbb{C}$ to $\text{SO}_{10}\mathbb{C}$ : the multiplicity of the irreducible representation $V_{[\mu]}$ corresponding to the partition $\mu$ will be

$\sum_\eta C_{\eta\mu}^\lambda$

where $C_{\cdot\cdot}^\cdot$ is the Littlewood-Richardson coefficient, and the sum is over all partitions $\eta=(\eta_1,\eta_2,\eta_3,\eta_4,\eta_5)$ with each $\eta_i$ even. (Fulton-Harris "Representation theory", Equation 25.37, p. 427) Hopefully you can check whether this agrees with your expected answer.

Answer by Tom Church for Given an object in a Lie groupoid, does there exist a subgroupoid for which the object has no automorphisms but retains its equivalence class?

2010-06-01T06:36:49Z

Consider the case of a group $K$ acting on $X$. Restricting to the orbit of the point $x$, by the orbit-stabilizer theorem $X$ is identified with $K/\text{Stab}(x)$. Your question seems to amount to asking whether the projection $K\times K/\text{Stab}(x)\to K/\text{Stab}(x)\times K/\text{Stab}(x)$ given by $(\gamma,p)\mapsto (\gamma\cdot p,p)$ admits an immersed section. Restricting such a section to a fiber $K/\text{Stab}(x)\times \text{pt}$ would give an section of the map $K\to K/\text{Stab}(x)$, which will not exist in general. (Edit: rewritten. Thanks to David Carchedi for pointing out the mistake in my notation.)

A concrete counterexample: consider the action of $K=\mathbb{R}$ on $X=S^1$ by translation. Then $G$ is diffeomorphic to $\mathbb{R}\times S^1$, and $H$ would have to be diffeomorphic to $S^1\times S^1$, but $S^1\times S^1$ does not immerse in $\mathbb{R}\times S^1$. Moreover the restriction to a fiber $S^1\times \text{pt}$ would be an immersion of $S^1$ into $\mathbb{R}$, which also cannot exist.

But this is not a phenomenon just of fundamental group or of equi-dimensional immersions. For example, consider $K=\text{Isom}(\mathbb{H}^2)=\text{PSL}_2\mathbb{R}$ acting on a hyperbolic surface $\Sigma$. A section of $\text{PSL}_2\mathbb{R}\times \Sigma\to \Sigma\times \Sigma$ restricts to an immersed section of $\text{PSL}_2\mathbb{R}\to \Sigma$. Of course $\Sigma$ does immerse, in fact embed, in $\text{PSL}_2\mathbb{R}$ (it's a 3-manifold). But if we had a continuous section $\varphi:\Sigma\to \text{PSL}_2\mathbb{R}$, we could translate a nonzero vector $v$ to each point $p$ by the isometry $\varphi(p)$. This would yield a nonzero vector field on $\Sigma$, contradicting the Gauss-Bonnet theorem. The same works for $\text{SO}(3)$ acting on $S^2$ or $\mathbb{R}P^2$, where the fundamental group is not the issue.

Answer by Tom Church for Free splittings of one-relator groups

2010-06-01T05:49:38Z

I'm not sure whether it helps here, but your question reminds me of the Freiheitssatz. As you probably know, this is the theorem of Magnus that says that if $G=\langle x_1,\ldots,x_n|r\rangle$ and $r$ involves the generator $x_n$, then the elements $x_1,\ldots,x_{n-1}$ generate a free group of rank $n-1$ inside $G$. Certainly your assumption implies this hypothesis with respect to any basis $x_1,\ldots,x_n$.

Also, I feel like we don't want to count HNN extensions as free products -- can't we just exclude the example of $\mathbb{Z}=\langle a,b|b\rangle$ by fiat? There aren't any other such counterexamples, right?

Answer by Tom Church for Automorphisms of $\pi_1$ induced by pseudo-Anosov maps

2010-05-08T22:49:43Z

No. Let $\Gamma_i$ be the lower central series defined by $\Gamma_1=\pi_1(X,x)$, $\Gamma_{i+1}=[\Gamma_1,\Gamma_i]$. The Johnson filtration $\text{Mod}_g(k)$ is the descending filtration of the mapping class group relative to $x$ defined by:

$f\in \text{Mod}_g(k)\iff f$ acts trivially on $\Gamma_1/\Gamma_k$

The first term $\text{Mod}_g(2)$ is the Torelli group, consisting of diffeomorphisms acting trivially on homology. The next term $\text{Mod}_g(3)$ is the Johnson kernel. By a beautiful theorem of Johnson, this is the subgroup generated by Dehn twists around separating curves.

By residual nilpotence of surface groups, we have $\bigcap \text{Mod}_g(k)={1}$, but every individual term in the filtration is nontrivial. It is not hard to see that every term of the Johnson filtration contains pseudo-Anosovs. Indeed every normal subgroup of the mapping class group contains pseudo-Anosovs (see Lemma 2.5 of Long, "A note on the normal subgroups of mapping class groups") from which Long concluded that any two normal subgroups intersect nontrivially!

Thus since $\text{Mod}_g(k)$ is normal, if we take $f\in\text{Mod}_g(k)$ we have $\gamma^{-1}\cdot g^{-1}fg(\gamma)\in \Gamma_k$ for all $g$ and all $\gamma$.

Answer by Tom Church for Are extensions of linear groups linear?

2010-04-28T08:16:57Z

The universal central extension $\widetilde{\text{Sp}_{2n}}\mathbb{Z}$ is the preimage of $\text{Sp}_{2n}\mathbb{Z}$ in the universal cover of $\text{Sp}_{2n}\mathbb{R}$ , and fits into the sequence

$1\to \mathbb{Z}\to \widetilde{\text{Sp}_{2n}}\mathbb{Z}\to \text{Sp}_{2n}\mathbb{Z}\to 1$ .

Deligne proved that $\widetilde{\text{Sp}_{2n}}\mathbb{Z}$ is not residually finite; the intersection of all finite-index subgroups of is $2\mathbb{Z}<\widetilde{\text{Sp}_{2n}}\mathbb{Z}$ . In particular, this implies that $\widetilde{\text{Sp}_{2n}}\mathbb{Z}$ is not linear. But certainly $\mathbb{Z}$ and $\text{Sp}_{2n}\mathbb{Z}$ are. If you want an arithmetic group, you can take the corresponding $\mathbb{Z}/k\mathbb{Z}$-extension of $\text{Sp}_{2n}\mathbb{Z}$ , which will not be linear as long as $k\neq 2$.

I learned the proof of this theorem from Dave Witte Morris, who has written up his fairly-accessible notes as "A lattice with no torsion-free subgroup of finite index (after P. Deligne)" (PDF link).

Answer by Tom Church for Knot complement diffeomorphism groups and embedding spaces

2010-04-18T16:40:51Z

In dimension $k=1$, i.e. for embeddings $\sqcup_k S^1\hookrightarrow S^3$, and using a totally unlinked embedding as your basepoint $e$, this is what's called the ring group or the loop group. It is closely related to the braid group and has been studied a ton, but two places with references you could follow are Brendle-Hatcher "Configuration spaces of rings and wickets" and Brownstein-Lee "Cohomology of the group of motions of n strings in 3-space".

The fundamental group $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)$ can been identified with McCool's "symmetric automorphism group". This is all the automorphisms of a free group $\langle x_1,\ldots,x_k\rangle$ which take each generator $x_i$ to a conjugate of some generator $x_j$. (A loop around one component of the link has to go to a loop around some component of the link.)

This is the image of $\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ in $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$, but since $S^3\setminus\sqcup_k S^1$ is not aspherical, this doesn't give us that $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)=\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ yet. I would be glad to see an argument that a diffeomorphism acting trivially on $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$ must be isotopic (or even homotopic) to the identity.

Answer by Tom Church for Does IP = PSPACE work over other rings?

2010-04-16T18:05:12Z

This does not directly answer your question, but it is related to issues of relativizing IP and PSPACE.

For a random oracle A, IP^A ≠ PSPACE^A with probability 1.

The problems coincide relative to a random oracle with probability either 0 or 1 by Komolgorov's zero-one law. That they are distinct for almost all oracles was proved by Chang-Chor-Goldreich-Hartmanis-Håstad-Ranjan-Rohatgi in "The random oracle hypothesis is false." J. Comput. System Sci. 49 (1994) 1 24--39. Note however that if you pass from IP to IPP, where the error probability is only ≤ 1/2, you do have that IPP^A = PSPACE^A for a random oracle A, so these are subtle issues.

Answer by Tom Church for Is the pure braid group on three strands generated as a normal subgroup of the braid group by the six-crossing braid?

2010-04-13T00:14:55Z

No.

Let $P_n$ be the pure braid group on $n$ strands. Forgetting the first strand gives a projection $P_3\to P_2$. The kernel is the free group $F_2=\pi_1(\mathbb{C}\setminus \{2,3\},1)$ given by moving the first strand around the other two while they are held in place. Since $P_2$ is just $\mathbb{Z}$, we have a short exact sequence

$1\to F_2\to P_3\to \mathbb{Z}\to 1$

Your element $b$ is contained in the kernel of this sequence (see the picture below) so its normal closure is contained in $F_2\lt P_3$. For an appropriate basis $\langle x,y\rangle=F_2$, the element $b$ represents $xyx^{-1}y^{-1}$.

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In fact, removing any of the three strands results in a trivial 2-strand braid (as can be seen from the picture). Closing up the ends of the braid thus results in a 3-component link so that if any one component is removed, the other two are unlinked. This is the well-known Borromean rings. There are three different surjections $P_3\to \mathbb{Z}$, given by forgetting each strand separately, and the group of Brunnian braids is exactly the intersection of their kernels. Since $xyx^{-1}y^{-1}$ normally generates the commutator subgroup of $F_2$, the normal closure $N$ of $b$ will be exactly the group of Brunnian braids. In particular, $N$ is an infinite-rank free group.

Answer by Tom Church for Are fundamental groups of aspherical manifolds Hopfian?

2010-04-09T12:29:36Z

The Baumslag-Solitar group $B(2,3)=\langle a,b\vert ba^2b^{-1}=a^3\rangle$ is not Hopfian. But it has a natural $K(\pi,1)$ given by the double mapping cylinder of $S^1 \rightrightarrows S^1$ where the maps are $z\mapsto z^2$ and $z\mapsto z^3$. This is a finite CW complex.

Edit: The double mapping cylinder can be constructed like this. Take a circle $S^1$ and a cylinder $S^1\times I$. Glue one end of the cylinder to the circle by the degree 2 map $z\mapsto z^2$, and glue the other end of the cylinder to the circle by the degree 3 map $z\mapsto z^3$. The $a$ in the presentation above is the loop around the circle, while the $b$ is the loop that goes along the cylinder (whose ends have been brought together, forming a loop).

To see that this is a $K(\pi,1)$, you can check that its universal cover is the product $T_5\times \mathbb{R}$ of the infinite $5$-regular tree $T_5$ with the line $\mathbb{R}$. (Think about what this CW complex looks like locally: away from the circle where we glued everything together, it's locally a $2$-manifold. At the circle, we have $2+3=5$ half-planes meeting along their edges.) There is a picture of this universal cover on page 3 of Farb-Mosher, "A rigidity theorem for the solvable Baumslag-Solitar groups", Inventiones 131 2 (1998), 419-451.

Answer by Tom Church for Presentation of the monoid of surfaces

2010-04-05T23:55:54Z

I assume you want your surfaces-with-boundary to be compact? Anyway, this cannot be generated by the $P[k]$ and $T[k]$, since you are leaving out the genus 0 surfaces (spheres with holes). Since connect-sum-with-a-disk is the same as removing an open disk, I would work instead with the generators $P$, $T$, and the disk $D$; the $D$ won't interact with the other generators. Since every surface-with-boundary is a surface-minus-some-disks, it seems a presentation is $\langle P,T,D\vert P^3=PT\rangle$.

In answer to your second question: it can be very worthwhile to know the proofs of these classical facts, but that doesn't mean you need to learn the classical proofs. For the classification of surfaces, I have seen Benson Farb give a very nice proof (fitting with the "modern" perspective on mapping class groups etc.) hinging upon the fact that the sphere has the maximal Euler characteristic among surfaces. If I can find any notes of that lecture or a written version, I'll update with a link.

Answer by Tom Church for Given a ramified cover of a Riemann surface, is there a good choice of basis for H_1 of the source?

2010-04-03T11:03:49Z

You might check out "Representations of Galois Groups on the Homology of Surfaces", by Koberda-Silberstein (arxiv/0905.3002). If your branched cover $S\to T$ is Galois (i.e. regular) the Lefschetz fixed point theorem tells you how $H_1(S)$ decomposes as a representation of the deck transformation group $G=\text{Gal}(S/T)$.

For example, for an unbranched cover of closed surfaces, every nontrivial element $g\in G$ acts freely on $S$, so Lefschetz gives that the trace $\text{tr}\big(g_\ast\colon H_1(S)\to H_1(S)\big)$ is 2; of course the trace of the identity is $2g(S)$. It follows that $H_1(S)$ decomposes as two copies of the trivial representation, plus $\big(2g(S)-2\big)/|G|=2g(T)-2$ copies of the regular representation of $G$. Extensions of this argument work for branched covers as well.

I've heard that this paper addresses the question of finding canonical generators for these subrepresentations, although I can't confirm that myself. It might be too high-powered than you need for your application, though.

Answer by Tom Church for Is an invertible biset necessarily a bitorsor?

2010-03-26T20:54:36Z

A torsor is a faithful transitive $G$-set. If the left $G$-action on $X$ is not faithful, the left $G$-action on $X\times_G Y$ will not be faithful. If the left $G$-action on $Y$ is not transitive, the left $G$-action on $X\times_G Y$ will not be transitive. By symmetry, it follows that a $G$-biset with a left and right inverse is a $G$-bitorsor.

Answer by Tom Church for Are there results about the group of homeomorphisms of $(T^2-\{,\})$ up to isotopy?

2010-03-26T18:36:45Z

There are many results in this field, and such groups, called mapping class groups, are well-studied. In the case of a torus the situation is totally understood; the mapping class group of the torus is $\text{SL}_2(\mathbb{Z})$. The only problem is that I am not sure what $\{*,*\}$ means. Note: the group $\text{Homeo}(T^2\setminus \{p\})/\sim$ is called the extended mapping class group, denoted $\text{Mod}^{\pm}(T^2\setminus\{p\})$ , while the subgroup of orientation-preserving homeomorphisms is the mapping class group $\text{Mod}(T^2\setminus\{p\}):=\text{Homeo}^+(T^2\setminus \{p\})/\sim$ .

If you mean a one-element set, something like $\{(0,0)\}$ : in the case of a torus, the missing point turns out not to matter: $\text{Mod}(T^2\setminus\{p\})=\text{Mod}(T^2)$ . This group, of orientation-preserving homeomorphisms up to isotopy, is isomorphic to $\text{SL}_2(\mathbb{Z})$. Your group is then an extension of $\mathbb{Z}/2\mathbb{Z}$ by this group, corresponding to the action on the orientation. \[1\to \text{Mod}(T^2\setminus\{p\})\to \text{Mod}^{\pm}(T^2\setminus\{p\})\to \mathbb{Z}/2\mathbb{Z}\to 1\] which can be written as \[1\to \text{SL}_2(\mathbb{Z})\to \text{Homeo}(T^2\setminus \{p\})/\sim\to \mathbb{Z}/2\mathbb{Z}\to 1\]

If you mean a two-element set, then first consider the subgroup $\text{PMod}(T^2\setminus\{p,q\})$ of homeomorphisms that don't "switch" the two punctures. The map given by "filling in the puncture $q$" gives an extension \[1\to\pi_1(T^2\setminus\{p\},q) \to\text{PMod}(T^2\setminus\{p,q\})\to \text{Mod}(T^2\setminus\{p\})\to 1\] which can also be written \[1\to F_2\to\text{PMod}(T^2\setminus\{p,q\})\to \text{SL}_2(\mathbb{Z})\to 1\] since $\pi_1(T^2\setminus\{p\},q)$ is a free group of rank two. The mapping class group is an extension of $\mathbb{Z}/2\mathbb{Z}$ by this group, corresponding to whether the punctures are switched: \[1\to \text{PMod}(T^2\setminus\{p,q\})\to \text{Mod}(T^2\setminus\{p,q\})\to \mathbb{Z}/2\mathbb{Z}\to 1\]

A good reference for all these things is Farb-Margalit's "A Primer on Mapping Class Groups". In particular, the useful fact that there is no difference between homotopy and isotopy in dimension 2, or between considering homeomorphisms and diffeomorphisms, is covered in Chapter 1. The mapping class group of the torus is described in Chapter 2, starting with Theorem 2.15 on page 70.

Answer by Tom Church for Fundamental group of a compact space form.

2010-03-26T01:46:38Z

For the record, no compact 3-manifold can ever admit two different geometries; this isn't just for space forms. You should be able to show this using quasi-isometric techniques, as in Sergei Ivanov's answer above (although I haven't thought about this in a while). One nice corollary is that you can determine the geometry just from the fundamental group without a lot of technical conditions (this is copied from Wikipedia with minimal editing, since it is laid out so nicely there).

Assume $M$ is a compact 3-manifold which admits one of the 8 geometries. Then:

If $\pi_1(M)$ is finite then the geometry on $M$ is spherical.
If $\pi_1(M)$ is virtually cyclic but not finite then the geometry on $M$ is $S^2×\mathbb{R}$.
If $\pi_1(M)$ is virtually abelian but not virtually cyclic then the geometry on $M$ is Euclidean.
If $\pi_1(M)$ is virtually nilpotent but not virtually abelian then the geometry on $M$ is Nil geometry.
If $\pi_1(M)$ is virtually solvable but not virtually nilpotent then the geometry on $M$ is Sol geometry.
If $\pi_1(M)$ virtually splits as a semidirect product with $\mathbb{Z}$ but is not virtually solvable then the geometry on $M$ is the universal cover of $SL_2(\mathbb{R})$.
If $\pi_1(M)$ has an infinite normal cyclic subgroup but not of the above form and is not virtually solvable then the geometry on $M$ is $\mathbb{H}^2\times\mathbb{R}$.
If $\pi_1(M)$ has no infinite normal cyclic subgroup and is not virtually solvable then the geometry on $M$ is hyperbolic.

This is false for finite-volume non-compact manifolds, for example the complement of a trefoil knot, which admits both $\widetilde{SL_2(\mathbb{R})}$ and $\mathbb{H}^2\times\mathbb{R}$ geometries.

Answer by Tom Church for Can all G-connections on a Riemann surface X be induced by maps from X to G

2010-03-20T23:38:49Z

José Figueroa-O'Farrill has already pointed out one necessary condition, namely that your connection must be flat. The remaining condition is that the monodromy should be trivial. In what follows $X$ is any connected smooth manifold, not necessarily a surface, and $G$ is any Lie group.

Let's first consider the analogous situation when $G$ is replaced by $\mathbb{R}$. You can think of a one-form $\omega\in \Omega^1(X;\mathfrak{g})$ as potentially being the derivative of a map $X\to G$, just as a one-form $\eta\in \Omega^1(X;\mathbb{R})$ is potentially the derivative of a map $X\to \mathbb{R}$. We want to know when these really are the derivative of some map, i.e. when we can integrate these forms. (You mentioned the exponential map, but I think integration is the right metaphor here.)

There is a local obstruction, namely that if $\eta$ is to be integrable (meaning $\eta=df$ for some $f$) it must be closed, meaning $d\eta=0$; the Poincaré lemma tells us this is a sufficient condition for $\eta$ to be locally integrable. Then there is also a global condition, that the integral of $\eta$ around every closed loop must be 0 (unlike $d\theta$ on the circle, which has integral $2\pi$); Stokes' theorem tells us this is a necessary condition for $\eta$ to be globally integrable. If we have these conditions, recovering the map $f$ from $\eta$ is easy; just write $f(p)=\int_\ast^p \eta$, which is well-defined by the above two conditions.

Now let's try to do the same for $\mathfrak{g}$-valued one-forms. Start with a connection on the trivial $G$-bundle $X\times G$ with connection form $\omega\in \Omega^1(X;\mathfrak{g})$. We've talked about the connection being flat, which means that $d\omega+\frac{1}{2}[\omega,\omega]=0$; but what does that have to do with flatness or integrability? Well, you can show that $d\omega+\frac{1}{2}[\omega,\omega]$ measures the Lie bracket of two horizontal vector fields, or rather measures the vertical part of the Lie bracket. Thus if this vanishes, the bracket of two horizontal vector fields is horizontal. By the Frobenius integrability theorem, this implies that the horizontal distribution of the connection is integrable; another way to say this is that parallel transport is locally well-defined. Now pick a basepoint $\ast$ and restrict your attention to a small neighborhood $U$ of $\ast$. Since parallel transport is well-defined on $U$, we get a function $T\colon U\to G$ by saying that the parallel transport from $\ast$ to $u$ (along any path) is multiplication by $T(u)$.

Key point: if you pull back the tautological form on $G$ by this "parallel transport" map $T$, the form you get is the same as your original $\omega$!

What this tells us is that if a flat connection on $X\times G$ comes from a map $f:X\to G$, then you can recover $f$ by looking at the parallel transport of the connection. (The analogue is that if $\eta=f^\ast(dx)$ for some $f\colon X\to \mathbb{R}$, you can recover $f$ by integrating $\eta$, also known as the fundamental theorem of calculus.) Thus flatness, in the form of the Maurer-Cartan equation, is the local obstruction to integrability; here the Frobenius integrability theorem plays the role that the Poincaré lemma does in the real case. To prove the key point is really just a matter of definitions: think about the correspondence between a connection, its connection form, and its parallel transport.

In particular, this tells us that parallel transport must be not just locally well-defined, but globally well-defined (meaning independent of the path), since transport along any path from $\ast$ to $p$ is always multiplication by $f(p)\in G$. The monodromy of a flat connection is the map $\pi_1(X,\ast)\to G$ which sends a loop to the parallel transport around that loop, and so another way to say "parallel transport is globally well-defined" is that the monodromy is trivial.

This can all be summed up by saying that if $X$ is simply connected, we have an on-the-nose bijection $C^\infty(X,G)\longleftrightarrow \{\omega\in \Omega^1(X;\mathfrak{g})\vert d\omega+\frac{1}{2}[\omega,\omega]=0\}$ . (Here on the left we assume the maps take the basepoint $\ast\in X$ to $1\in G$.) If $X$ has fundamental group, we need to add on the right side the additional condition that the monodromy of $\omega$ be 0. This is hard to write down just in terms of $\omega$, but for the corresponding connection it is just that parallel transport is totally path-independent.

Answer by Tom Church for Homotopy type of set of self homotopy-equivalences of a surface

2010-03-19T02:53:57Z

Hi Andy,

Here is a proof for the case with marked points (see below for some ideas for the case of closed surfaces).

Proof: straight-line homotopy.

Less tersely: let $HE_0(\Sigma,\ast)$ be the identity component of the monoid of self-homotopy equivalences of $\Sigma$ fixing the basepoint; in particular, each $f\in HE_0(\Sigma,\ast)$ is homotopic rel $\ast$ to the identity. Fix a hyperbolic metric on $\Sigma$ and thus an identification of the universal cover $\widetilde{\Sigma}$ with the hyperbolic plane $\mathbb{H}^2$, and a basepoint $\ast$ in $\mathbb{H}^2$.

Each $f\colon \Sigma\to \Sigma$ has a unique lift to $\mathbb{H}^2$ fixing the basepoint. because $f$ acts trivially on $\pi_1(\Sigma)$, $f$ commutes with the deck transformations. Thus we may take the straight-line homotopy $f_t(x)=tx-(1-t)f(x)$, where by this convex combination I mean to move with unit speed along the geodesic from $f(x)$ to $x$. Since the deck transformations act by isometries on $\mathbb{H}^2$, this homotopy descends to $\Sigma$; each $f_t$ is still a homotopy equivalence. We can perform this straight-line homotopy for all $f$ simultaneously; since the lifts of $f$ are uniformly continuous, this homotopy is continuous on $HE_0(\Sigma,\ast)$ and gives a contraction to the identity.

There must be some work needed to get from this to the case for closed surfaces, because this proof works for a genus 1 surface with marked point, and of course $HE_0(T^2)$ is homotopy equivalent to $T^2$ itself. But it seems to me like a reduction should be possible; I think the important thing is that $\pi_1(\Sigma)$ is centerless.

Acknowledgement: I learned this idea from Rita Jimenez Rolland, based on conversations she had with Mladen Bestvina about the related case of $\text{Aut}(F_n)$.

Answer by Tom Church for What are the worst notations, in your opinion ?

2010-03-19T02:30:18Z

I get very frustrated when an author or speaker writes "Let $X\colon= A\sqcup B$..." to mean:

$A$ and $B$ are disjoint sets (in whatever the appropriate universe is),
and let $X\colon= A\cup B$.

If they just meant "form the disjoint union of $A$ and $B$" this would be fine. But I've seen speakers later use the fact that $A$ and $B$ are disjoint, which was never stated anywhere except as above. You should never hide an assumption implicitly in your notation.

Answer by Tom Church for A slick proof of the Bruhat Decomposition for GL_n(k)?

2010-02-17T02:50:58Z

Here is some standard machinery generalizing this result, which is more combinatorial than the standard reductive-groups proofs (but gives far less understanding than e.g. Matt Emerton's explanation above).

Assume $G$ acts strongly transitively on a thick building $\Delta$. Let $B$ be the stabilizer of the fundamental chamber, $N$ the stabilizer of the fundamental apartment, $T$ the subgroup fixing the fundamental apartment, and $W$ the quotient $N/T$. Then $(G,B,N)$ is a $BN$-pair, also called a Tits system, for $G$. In particular, you have the Bruhat decomposition $G=\coprod_{w\in W}BwB$, plus lots more.

In this case, take $G$ to be $GL(V)$, and take $\Delta$ to be the flag complex of subspaces of $V$. The stabilizer $B$ of the fundamental chamber is the upper-triangular matrices. The stabilizer $N$ of the fundamental apartment is the monomial matrices; the subgroup $T$ fixing the fundamental apartment is the diagonal matrices; and the Weyl group $W$ is the quotient $N/T$, which can be identified with the permutation matrices.

You can find all the above, including proofs, in Chapter V.2 of "Buildings" by Brown. For the special case of $GL(V)$, you could also look at Exercises 7 and 8 in Chapter 2.4 of "Groups and Representations" by Alperin-Bell.

Answer by Tom Church for Size of the smallest group not satisfying an identity.

2010-02-12T09:49:25Z

The asymptotic version of this question raised by Bjorn Poonen has been studied by Khalid Bou-Rabee for general groups, not just free groups. That is, given G a residually finite group, for each g we can ask: how large is the smallest finite group F which detects g, meaning there exists f: G -> F so that f(g) is nontrivial? Now fix a word metric on G, and ask how the maximum of this "detection number" grows as you consider words of length at most n.

See "Quantifying residual finiteness" and "Asymptotic growth and least common multiples in groups" (with Ben McReynolds) for his results. For example, as long as G is linear, the growth function is polylog, meaning asymptotically less than (log n)^k for some k, if and only if G is virtually nilpotent.

To answer your question, by considering congruence quotients of SL₂Z, Bou-Rabee concludes that for every word of length n in the free group F₂, there is a finite group of order O(n³) where the word is not an identity. The same bound can be obtained uniformly as follows. Ury Hadad gives a lower bound in "On the shortest identity in finite simple groups of Lie type" which implies that the shortest identity in PSL₂(F_q) has length at least (q-1)/3. Since the size of PSL₂(F_q) is order q³, this implies that every word of length at most n fails to be an identity in one single group PSL₂(F_q) of order O(n³)!

I learned this argument from Martin Kassabov and Francesco Matucci's paper "Bounding the residual finiteness of free groups". In it, they use a nice analysis of finite groups with elements of "large order" to construct a word of length n in the free group F₂ which is trivial in every finite group of order at most O(n^2/3). This improved on the lower bound of n^1/3 due to Bou-Rabee and McReynolds; I believe this is now the best lower bound known.

Answer by Tom Church for Is there a finitely complete category with terminal object but NO subobject classifier?

2010-02-09T23:00:21Z

You should basically never expect anything to have a subobject classifier -- it's a ridiculously strong condition. Some examples showing how quickly it fails to exist:

The category of groups is complete. If this category had a subobject classifier Ω, every subgroup would be the kernel of a map to Ω. But not every subgroup is normal.
The category of Hausdorff spaces is complete. If it had a subobject classifier Ω, every subspace would be the preimage of a point in Ω. But not every subspace is closed.
The category of rings with identity is complete. The terminal object is the zero ring (with 0 = 1). But the zero ring has no maps to any nonzero ring.
The category of compact Hausdorff spaces is complete. As above every closed subspace should be the preimage of a point $\ast$ in Ω under a unique map to Ω. For the subspace {1} in {1,2} to be classified by a unique map, $\Omega\setminus \ast$ must be a single point, so Ω is discrete on two points. But then there is only one map from $S^1$ to Ω, while it has uncountably many subobjects.

Edit: the opposite of any cocomplete category will (with probability 1) give examples as well.

The category of groups is cocomplete. A subobject classifier in the opposite category would be a group Ω so that every surjection G -> H is a pushout of Ω -> 1 and Ω -> G. This is the amalgamated free product of G with the trivial group over Ω, that is the quotient of G by the image of Ω. But if Ω is a group surjecting to the kernel of every map, that surjection cannot be unique (just consider Z -> 1).
The category of sets (also Hausdorff spaces) is cocomplete. In the opposite category Ω would be a set (space) so that every surjection X -> Y was the quotient of X by the image of a map Ω -> X; this would imply that at most one fiber of X -> Y is not just a single point.
The category of commutative rings with identity is cocomplete. A subobject classifier for the opposite category (the category of affine schemes) would be an affine scheme Ω with a $\mathbb{Z}$-point Spec $\mathbb{Z}$ -> Ω so that any monomorphism Y -> X is the base change to $\mathbb{Z}$ of a unique map X -> Ω. This is ridiculous; note that Spec $\mathbb{Q}$ -> Spec $\mathbb{Z}$ is injective, and you'll never get Spec $\mathbb{Q}$ as the fiber product of two $\mathbb{Z}$-points.
The partial order category of ordinals is cocomplete. The initial object is 0; but no nonzero ordinal maps to (is ≤) 0. Thus Ω would have to be 0, but then we would have for any α ≥ β that α was the coproduct (supremum) of β with 0, which is false whenever α > β.

Answer by Tom Church for Describing $SU(n,C)$

2010-02-04T18:11:21Z

Use the fact that matrices act on vectors. $SU(n)$ acts transitively on the space of unit-length vectors; the stabilizer of a point is $SU(n-1)$ by Thorny's argument. For example, for the vector $(1,0,...,0)$ the stabilizer is the subgroup $\left(\begin{matrix}1&0\\ 0&A\end{matrix}\right)\approx SU(n-1)$. Now by the orbit-stabilizer theorem, the space of unit-length vectors is identified with $SU(n)/SU(n-1)$. Fixing one vector $(1,0,...,0)$ fixes this identification, and then each other vector corresponds to a coset $gSU(n-1)$ which is the family you describe.

This isn't really using much about matrices or geometry; I referred to this as the "orbit-stabilizer theorem" above, but it is really just the basic structural feature of group actions. It's certainly something you can understand by yourself; if it's not immediately obvious, you can think about some simpler examples. In the group of permutations $S_n$, consider the family of permutations that map $1\mapsto 3$ -- how do elements of this family differ from each other? You could also try extending your argument to understand the first column of matrices in $GL(n,\mathbb{R})$; you will of course find a similar answer, but the details are interestingly different. Another fun example is to consider linear functions from $\mathbb{R}\to\mathbb{R}$, and look at the family of linear functions taking $2\mapsto 7$.

Answer by Tom Church for Stokes' theorem etc., for non-Hausdorff manifolds

2010-01-26T23:44:14Z

The existence of flows in the direction of a vector field seems to require Hausdorff; indeed, consider the vector field $\frac{\partial}{\partial x}$ on the line-with-two-origins. We have no global existence of a flow for any positive t, even if we make our space compact (that is, considering the circle-with-one-point-doubled). If the nonexistence of the flow is not visibly clear, consider instead the real line with the interval [0,1] doubled.

Also, partitions of unity do not exist; for example, in the line with two origins, take the open cover by "the line plus the first origin" and "the line plus the second origin". There is no partition of unity subordinate to this cover (the values at each origin would have to be 1).

For me, a basic example of the beauty of this function-theoretic approach is the definition of a vector field as a derivation $D\colon C^\infty(M)\to C^\infty(M)$. The proof that such a derivation defines a vector field hinges upon the fact that $Df$ near a point p only depends on $f$ near the point p. To prove this fact you use the fineness of your sheaf $\mathcal{O}_X$, i.e. the existence of partitions of unity. (It is true though that the failure of fineness in the non-Hausdorff case is of a different sort and might not break this particular theorem.) I feel that the existence of partitions of unity, and the implications thereof, is one of the basic fundamentals of approaching smooth manifolds through their functions; more importantly, a good handle on how partitions of unity are used is important to understand the differences that arise when the same approach is extended to more rigid functions (holomorphic, algebraic, etc.).

Now that the question has been edited to ask specifically about Stokes' theorem, let me say a bit more. Stokes' theorem will be false for non-Hausdorff manifolds, because you can (loosely speaking) quotient out by part of your manifold, and thus part of its homology, without killing all of it.

For the simplest example, consider dimension 1, where Stokes' theorem is the fundamental theorem of calculus. Let $X$ be the forked line, the 1-dimensional (non-Hausdorff) manifold which is the real line with the half-ray $[0,\infty)$ doubled. For nonnegative $x$, denote the two copies of $x$ by $x^\bullet$ and $x_\bullet$, and consider the submanifold $M$ consisting of $[-1,0) \cup [0^\bullet,1^\bullet] \cup [0_\bullet,1_\bullet]$. The boundary of $M$ consists of the three points $[-1]$ (with negative orientation), $[1^\bullet]$ (with positive orientation), and $[1_\bullet]$ (with positive orientation); to see this, just note that every other point is a manifold point.

Consider the real-valued function on $X$ given by "$f(x)=x$" (by which I mean $f(x^\bullet)=f(x_\bullet)=x$). Its differential is the 1-form which we would naturally call $dx$. Now consider $\int_M dx$; it seems clear that this integral is 3, but I don't actually need this. Stokes' theorem would say that

$\int_M dx=\int_M df = \int_{\partial M}f=f(1^\bullet)+f(1_\bullet)-f(-1)=1+1-(-1)=3$.

This is all fine so far, but now consider the function given by $g(x)=x+10$. Since $dg=dx$, we should have

$\int_M dx=\int_M dg=\int_{\partial M}g=g(1^\bullet)+g(1_\bullet)-g(-1)=11+11-9=13$. Contradiction.

It's possible to explain this by the nonexistence of flows (instead of $df$, consider the flux of the flow by $\nabla f$). But also note that Stokes' theorem, i.e. homology theory, is founded on a well-defined boundary operation. However, without the Hausdorff condition, open submanifolds do not have unique boundaries, as for example $[-1,0)$ inside $X$, and so we can't break up our manifolds into smaller pieces. We can pass to the Hausdorff-ization as Andrew suggests by identifying $0^\bullet$ with $0_\bullet$, but now we lose additivity. Recall that $M$ was the disjoint union of $A=[-1,0)$ and $B=[0^\bullet,1^\bullet] \cup [0_\bullet,1_\bullet]$. So in the quotient $\partial [A] = [0]-[-1]$ and $\partial [B] = [1^\bullet]-[0]+[1_\bullet]-[0]=[1^\bullet]+[1_\bullet]-2[0]$, which shows that $\partial [M]\neq \partial [A]+\partial [B]$. This is inconsistent with any sort of Stokes formalism.

Finally, I'd like to point out that Stokes' theorem aside, even rather nice non-Hausdorff manifolds can be significantly more complicated than we might want to deal with. One nice example is the leaf-space of the foliation of the punctured plane by the level sets of the function $f(x,y)=xy$. The leaf-space looks like the union of the lines $y=x$ and $y=-x$, except that the intersection has been blown up to four points, each of which is dense in this subset. In general, any finite graph can be modeled as a non-Hausdorff 1-manifold by blowing up the vertices, and in higher dimensions the situation is even more confusing. So for any introductory explanation, I would strongly recommend requiring Hausdorff until the students have a lot more intuition about manifolds.

Answer by Tom Church for Weil's theorem about maps from a discrete group to a Lie group.

2010-01-22T23:24:39Z

Some tangential comments:

$\textrm{PSL}_{2} \mathbb{R}$ is 3-dimensional; you get a Riemann surface by taking a lattice $\Gamma$ in $\textrm{PSL}_2\mathbb{R}$ and taking the double quotient $\Gamma \backslash \textrm{PSL}_2\mathbb{R} / \textrm{SO}(2)$. (That is, it's $\Gamma \backslash \mathbb{H}^2$ where $\mathbb{H}^2$ is the hyperbolic plane.) But this is compact iff $\Gamma \backslash$ $\textrm{PSL}_2\mathbb{R}$ is, since $\textrm{SO}(2)$ is compact.

If your Riemann surface is non-compact but of finite type, then when you uniformize, the complete hyperbolic metric will have infinite cusps at the punctures. This means that your representation $\Gamma\to$ $\textrm{PSL}_2\mathbb{R}$ must send the elements $\gamma\in\Gamma$ corresponding to loops around the punctures to parabolic elements of $\textrm{PSL}_2\mathbb{R}$ . Putting this restriction on a representation will force the quotient to have finite volume, and then you have the same theorem that the discrete faithful representations are open in the representation variety.

You might also look at Section 4 of Peter Shalen's paper "Representations of 3-manifold groups". For representations of hyperbolic 3-manifold groups into $\textrm{PSL}_2\mathbb{C}$ we have Mostow rigidity, which says that any two discrete faithful representations are conjugate; thus the appropriate subpace of $\textrm{Hom}(\Gamma,\textrm{PSL}_2\mathbb{C})/\textrm{PSL}_2\mathbb{C}$ is just a point (in stark contrast to the case for surface groups which you attribute to Goldman). But you still have algebraic deformations in the character variety in the case of manifolds with cusps, and these were analyzed by Thurston. In particular, Shalen says that Thurston generalized Weil's results to finite-volume cusped hyperbolic 3-manifolds, by imposing the condition mentioned above that cusp subgroups map to parabolic subgroups of $\textrm{PSL}_2\mathbb{C}$ .

Answer by Tom Church for Why are abelian groups amenable?

2010-01-19T05:11:13Z

Here is (I believe) a simpler argument, combining 1--6 into one step. (I assume here the group is countable; I can't tell if you're interested in uncountable discrete groups, but I have no idea what issues, if any, arise there.)

Let $G$ be a countable abelian group generated by $x_1,x_2,\ldots$. Then a Følner sequence is given by taking $S_n$ to be the pyramid consisting of elements which can be written as

$a_1x_2+a_2x_2+\cdots+a_nx_n$ with $|a_1|\leq n,|a_2|\leq n-1,\ldots,|a_n|\leq 1$.

The invariant probability measure is then defined by $\mu(A)=\underset{\omega}{\lim}|A\cap S_n| / |S_n|$ as usual.

A more natural way to phrase this argument is:

The countable group $\mathbb{Z}^\infty$ is amenable.
All countable abelian groups are amenable, because amenability descends to quotients.

But I would like to emphasize that there is really only one step here, because the proof for $\mathbb{Z}^\infty$ automatically applies to any countable abelian group. This two-step approach is easier to remember, though. (The ideas here are the same as in my other answer, but I think this formulation is much cleaner.)

Answer by Tom Church for Why are abelian groups amenable?

2010-01-18T19:36:02Z

The simplest argument for me to keep in my head is (none of this is original, even to this thread):

$\mathbb{Z}^n$ is amenable (because the cube of radius $k$ gives a Følner sequence).
Thus finitely generated abelian groups are amenable (because amenability descends to quotients).
Thus (discrete) abelian groups are amenable (because locally amenable groups are amenable).

Each of these properties is most easily proved using a different characterization of amenability, so this might not be pedagogically optimal, but since we already know the characterizations are equivalent we don't have to worry about that. Each of the parentheticals above is a property that everyone should have in mind when dealing with amenable groups, and so this argument also serves to remind me what's true in case I forget. Of course 1 and 2 could be combined (the ball of radius $k$ gives a Følner sequence in any f.g. abelian group, no need to be free) or deduced differently (you might prefer to justify 2 instead by "virtually amenable groups are amenable" or "amenable-by-amenable groups are amenable"), but I prefer this approach because it doesn't depend on the classification of f.g. abelian groups.

[Edit: upon rereading, I see that more of this than I realized was already present in Yemon's answer and the comments, so I'm making this community wiki.]

Second edit: in comments, Tom Leinster asked about seeing directly that f.g. abelian groups are amenable. Here are my thoughts: if you assume the classification and write $G=\mathbb{Z}^n\times T$ for $T$ finite, this is easy. Geometrically the most natural Følner sequence is the cube-of-radius-$k\times T$, but anything you try will work. If you don't want to use the structure theorem, I would note that if $G$ is abelian and has rank $n$, then it has polynomial growth of rank at most $n$; it follows that the ball of radius $k$ gives a Følner sequence. (Above I implicitly criticized this approach; that was because to bound the growth rate without knowing the structure of the group, you dominate it by the growth of the free abelian group. Thus this seemed to just be a hidden appeal to my step 2 above. It's perfectly valid though.)

Both approaches give a Følner sequence (because that's the definition of amenable I understand best). If you prefer the invariant mean definition, that's fine, but there is still value in sticking with Følner sequences as long as you can. For the simplest possible proof without relying on any equivalences, I would do the following:

Finitely generated abelian groups admit Følner sequences (e.g. the ball of radius $k$).
If $G$ is countable and every f.g. subgroup of $G$ admits a Følner sequence, then $G$ is amenable (admits an invariant f.a. probability measure).

To see 2, choose an increasing sequence of f.g. subgroups $G_i < G$ which exhausts $G$, and let $S_i^n$ be a Følner sequence for $G_i$; we can consider the $S_i^n$ as subsets of $G$. Then the measure we get is the "asymptotic density"

$\mu(A) = \underset{i\to \omega}{\lim}\underset{n\to\omega}{\lim}\ \ \vert A\cap S_i^n\vert / \vert S_i^n\vert$

where $\omega$ is a non-principal ultrafilter so that the limits exist. This is clearly a finitely additive probability measure, and to see that it's invariant note that every $\gamma\in G$ lies in all $G_N$ for $N$ sufficiently large.

Answer by Tom Church for Cohomology of fibrations over the circle: how to compute the ring structure?

2010-01-15T05:43:16Z

This is a continuation of Ryan's answer above, but it has become too large for a comment. I wanted to work out the details of Ryan's example explicitly, so that we can see explicitly where your conditions fail to determine the cohomology; perhaps this can help you to pin down precisely what conditions you want. It doesn't seem that we actually need Kitano here, just Johnson's classic results.

Let $S_g\to M^3\to S^1$ be a mapping torus of an element of the Torelli group, i.e. a diffeomorphism $S_g\to S_g$ acting trivially on homology. Such a bundle admits cohomology classes satisfying the Leray-Hirsch condition [this is a fun exercise], implying that as $H^{\ast}(S^1)$-modules, $H^\ast(M^3) = H^\ast(S_g) \otimes H^\ast(S^1)$. Thus the following do not depend on the monodromy:

$Q = H^\ast(S_g)$,
$I$, which is $Q$ with grading shifted by 1 (if $H^\ast(S^1) = \mathbb{Z}[t]/t^2$, this is $tQ$)
the action of $Q$ on $I$ (just the action of $Q$ on $tQ$),
and the Massey products on $Q = H^\ast(S_g)$ [although perhaps I misunderstand what you mean here].

However, Johnson's work implies that your 3-manifold has the same cohomology ring as the product $S_g \times S^1$ iff the monodromy lies in the kernel of a certain homomorphism called the Johnson homomorphism; in particular, the ring $H^\ast(E)$ depends on the monodromy. It seems this shows that the answers to 1) and 2) are both "No".

Now we can compare this with your conditions to see exactly what information we're missing; it turns out to be exactly the "Johnson homomorphism". The exact sequence above $0\to I\to H^\ast(E)\to Q\to 0$ has a splitting as abelian groups $H^\ast(E) = Q\oplus tQ$ coming from the Leray-Hirsch theorem as above. The only information we don't know automatically is the cup product on $Q$ in this splitting with itself. We know when we project back to the $Q$ factor we recover the cup product there, which means that the missing information is the projection onto the $tQ$ factor. Letting e.g. $Q(1)$ denote the degree 1 part, the cup product is a map $Q(1) \wedge Q(1) \to H^2(E)$. Projecting onto the $tQ$ factor, we have $Q(1) \wedge Q(1) \to tQ(2)$. But both $Q(1)$ and $tQ(2)$ are isomorphic to $H^1(S_g)$, so this projection of cup product is a map $\bigwedge^2 H^1(S_g) \to H^1(S_g)$. This exactly encodes the data that is not determined by your conditions; Johnson's beautiful result is that this map is exactly the Johnson homomorphism, originally defined from the algebraic properties of the monodromy. In particular he showed that this missing data could be zero or nonzero, and in fact can be anything in the subspace $\bigwedge^3 H^1<\textrm{Hom}(\bigwedge^2 H^1,H^1)$.

This was first laid out in Johnson's survey "A survey of the Torelli group" (MR0718141), and the details are worked out carefully in Hain, "Torelli groups and geometry of moduli spaces of curves" (MR1397061). What Kitano is doing is different, or rather a generalization of this: showing that just as the cup product on $H^\ast(E)$ detects the Johnson homomorphism, the higher Massey products on $H^\ast(E)$ detect "higher Johnson homomorphisms" measuring deeper algebraic invariants. (If any of this is useful, please consider it a partial repayment for your beautiful summary of Hodge theory in this answer.)

Answer by Tom Church for The De Rham Cohomology of $\mathbb{R}^n - \mathbb{S}^k$

2009-11-24T00:57:59Z

To apply the Mayer-Vietoris sequence, you need subspaces whose interiors cover your space (see e.g. Wikipedia, or Hatcher, p. 149). This is not true in your example, because a k-disk in Rⁿ has empty interior for k<n.

You might also enjoy deriving this result from Alexander duality.

Edit: Carsten makes an excellent point, which is that the hard part is to show that the homology of the complement is independent of the embedding. You did say that this is proved in your book, but I wanted to point out that this is quite difficult and arguably surprising.

1) One embedding that satisfies the conditions of the theorem is the Alexander horned sphere, a "wild" embedding of S² into S³ (this animation is quite nice too). While it's true that the outer component of the complement has the homology of a point, it is very far from being simply connected -- in fact its fundamental group is not finitely generated. (You can find an explicit description of its fundamental group in Hatcher, p. 170-172.)

2) Every knot is an embedding of S¹ into S³. The fundamental group of the complement is a strong knot invariant, and is usually much more complicated than just Z. Since H₁ of the knot complement is the abelianization of the knot group, the result you are using implies that all knot groups have infinite cyclic abelianization. This is true (it can be seen nicely from the Wirtinger presentation), but it's not obvious.

3) It is important that the ambient space is a sphere (or equivalently Rⁿ). For a simple example where the theorem breaks down, consider embeddings of S¹ into a surface Σ_g of genus g≥2. Taking g=2 for simplicity, we see that there are three topologically inequivalent ways of embedding a circle into Σ₂: A) a tiny loop enclosing a disk; B) a loop encircling the waist of the surface and separating it into two components, each of genus 1; and C) a loop going through one of the handles, which does not separate the surface at all.

The homology groups of Σ₂ are H₀=Z, H₁=Z⁴, and H₂=Z. For both A) and B), the complement of S¹ has homology groups H₀=Z² because the curve separates, and H₁=Z⁴. However, for C) we have H₀=Z because the complement is connected, and H₁=Z³ because we have "interrupted" one of the elements [you can see where it went by looking at the Mayer-Vietoris sequence]. Thus we see that the homology of the complement depends essentially on the embedding into the surface Σ_g, in contrast with the classical case of embedding a circle into the sphere S².

Answer by Tom Church for Does $Aut(Aut(...Aut(G)...))$ stabilize?

2009-11-15T21:17:30Z

I don't know about non-stabilizing, but rigidity provides many examples that stabilize quickly.

1) Let π be the fundamental group of a finite volume hyperbolic manifold M of dimension ≥ 3 with no symmetries (that is, no nontrivial self-isometries). Negative curvature implies that π is centerless, so the map π -> Aut(π) is injective. Mostow-Prasad rigidity says that Out(π) = Isom(M), so the lack of isometries implies that Out(π) is trivial and Aut(π) = π. [This works verbatim for lattices in higher-rank semi-simple Lie groups subject to appropriate conditions.]

2) Let π=F_d be a free group of rank 2≤d<∞. Then Aut(F_n) is a much larger group; however, Dyer-Formanek showed that Out(Aut(F_n)) is trivial. Thus since Aut(F_n) is clearly centerless, we have Aut(Aut(F_n)) = Aut(F_n).

3) Interpolating between these two examples, if π=π₁(S_g) is the fundamental group of a surface of genus g≥2, then Aut(π) is the so-called "punctured mapping class group" Mod_g,*, which is much bigger than π. Ivanov proved that Out(Mod_g,*) is trivial, and since Mod_g,* is again centerless, we have Aut(Aut(π₁(S_g))) = Aut(π₁(S_g)).

In each of these cases, rigidity in fact gives stronger statements: Let H and H' be finite index subgroups of G = Aut(F_n) or Mod_g,*. (This class of groups can be widened enormously, these are just some examples.) Then any isomorphism from H to H' comes from conjugation by an element of G, by Farb-Handel and Ivanov respectively. In particular, Aut(H) is the normalizer of H in G. Rigidity gives the same conclusion for H = π₁(M) as in the first example and G = Isom(Hⁿ) [which is roughly SO(n,1)]. It seems that by carefully controlling the normalizers, you could use this to construct examples that stabilize only after n steps, for arbitrary large n.

Edit: I find the examples of D₈ and D_∞ unsatisfying because even though Inn(D) is a proper subgroup of Aut(D), we still have Aut(D) isomorphic to D. Here is a general recipe for building similarly liminal examples. Let G be an infinite group with no 2-torsion so that Aut(G) = G and H¹(G;Z/2Z) = Z/2Z. (Edited: For example, by rigidity, any hyperbolic knot complement with no isometries has these properties; by Thurston, most knot complements are hyperbolic.) The condition on the 2-torsion implies that for any automorphism G x Z/2Z -> G x Z/2Z, the composition

G -> G x Z/2Z -> G x Z/2Z -> G

is an isomorphism. From this we see that Aut(G x Z/2Z) / G = H¹(G;Z/2Z) = Z/2Z. By examination the extension is trivial, and thus Aut(G x Z/2Z) = G x Z/2Z. However, the image Inn(G x Z/2Z) is the proper subgroup G.

Comments: looking back, this feels very close to your original example of R x Z/2Z. Interesting that it's (seemingly) much harder to find group-theoretic conditions to force the behavior the way you want, while topologically it's easy.

Also, if you instead take G with H¹(G;Z/2Z) having larger dimension, say H¹(G;Z/2Z) = (Z/2Z)², this blows up quickly. You get Aut(G x Z/2Z) = G x (Z/2Z)², but then Aut(Aut(G x Z/2Z)) is the semidirect product of H¹(G;Z/2Z²) = (Z/2Z)⁴ with Aut(G) x Aut(Z/2Z²) = G x GL(2,2). Already the next step seems very hard to figure out. However, if you had enough control over the finite quotients of G, perhaps you could show that the linear parts of these groups don't get "entangled" with the rest, so that the automorphism groups would act like a product of G x (Z/2Z)ⁿ with something else, with n going to infinity. If so, this could yield an example where the isomorphism types of the groups never stabilize.

Comment by Tom Church

2010-07-22T00:11:36Z

@Ben: A presentation for this group has to make disjoint transpositions commute; in particular, it has to make transpositions at distance $n$ commute for all $n$ (except possibly some small numbers, of course). This should imply that $H_2(G)$ has infinite rank and so the group is not finitely presented. For a simpler example along the same lines, consider the lamplighter group $\mathbb{Z}\wr \mathbb{Z}=\langle t,s_i| ts_it^{-1}=s_{i+1}, [s_i,s_j]=0\rangle$. Hopf's formula shows that $H_2(\mathbb{Z}\wr \mathbb{Z})=\mathbb{Z}^\infty$, with basis e.g. $\{[s_7,s_{7+k}]|k\in\mathbb{Z}\}$ .

Comment by Tom Church

2010-07-20T07:42:35Z

Here is a link that should work: en.wikipedia.org/wiki/Caribou_%28musician%29

Comment by Tom Church

2010-07-20T04:40:23Z

You could record electronic music, win the Polaris Music Prize, and break the US Billboard 100: en.wikipedia.org/wiki/Caribou_(musician)

Comment by Tom Church

2010-07-19T06:17:37Z

I have restored the author's original tags to the question. I see no reason to have removed these (eminently applicable) tags. If the remover feels that they clearly deserve to be deleted, then he should have no trouble waiting for someone else to do so.

Comment by Tom Church

2010-07-18T19:11:41Z

I have taken the liberty of changing "/" to "\" (setminus) above, to avoid confusion with the quotient space.

Comment by Tom Church

2010-07-15T03:05:35Z

"Ordinary Differential Equations" is the only one I have read, but it is among the best math books I've come across. I wouldn't have called it difficult, though -- Arnold's style makes it a comfortable read for most graduate students, I'd say. (It actually makes quite nice "pleasure reading" when you want a break from more strenuous stuff.)

Comment by Tom Church

2010-07-13T17:16:11Z

These examples are great! Do you know any examples of non-homeomorphic closed manifolds whose tangent bundles are *non*trivial and diffeomorphic?

Comment by Tom Church

2010-07-08T19:40:51Z

The point I was missing is that an isomorphism between the original rings would give an isomorphism over $\mathbb{C}$ between $\text{SL}_n\mathbb{C}$ and $\mathbb{A}^{n^2-1}$, which would certainly imply that their $\mathbb{C}$--points are homeomorphic. My confusion was that I thought you were arguing that $\mathbb{C}[x_1,\ldots,x_{n^2-1}]$ and $\mathbb{C}[y_1,\ldots,y_{n^2}]/(\text{det}=1)$ are non-isomorphic as rings, which as far as I can tell doesn't follow from your argument, rather than as $\mathbb{C}$-algebras. (Of course this isn't what was asked.)

Comment by Tom Church

2010-07-08T15:39:16Z

Your argument seems to also prove that $\mathbb{R}$ and $\mathbb{R}^2$ are not isomorphic as groups (they're not homeomorphic!). And there are varieties which are conjugate but whose $\mathbb{C}$-points are not homeomorphic (I know Serre constructed a non-singular projective surface which is an example, not sure about affine examples). Can you explain how to get the reduction to the classical topology (without going through étale cohomology or something like that)?

Comment by Tom Church

2010-07-07T01:46:29Z

The benefit of Arturo's suggested notation is that $[G_i,G_j]\subset G_{i+j}$.

Comment by Tom Church

2010-06-25T23:07:08Z

You seem to be confusing "bijection" with "function", perhaps?

Comment by Tom Church

2010-05-31T18:57:41Z

Another way to state this condition is that every subgroup $H<G$ has commensurator $\text{Comm}_G(H)$ equal to all of $G$. The commensurator in $G$ of a subgroup $H$ consists of all those $g$ for which $H\cap gHg^{-1}$ has finite index in $H$ and in $gHg^{-1}$.) (Note that in your question you need to intersect $H$ with $gHg^{-1}$ to ensure you get a subgroup of $H$.) Do you have examples of not-virtually-solvable groups for which this condition holds?

Comment by Tom Church

2010-05-29T03:59:04Z

I mean this not as criticism, but as a request for clarification. Can you give another example of a "top-down" description of mathematics? I can't think of what it could be, except the old "mathematics is what's done by mathematicians".

Comment by Tom Church

2010-05-29T03:56:22Z

How is category theory useful in saying what mathematics is? Of course there are entire fields of mathematics (underrepresented here) with which category theory has had no interaction. But even if we set those aside, category theory describes not only actual mathematics (e.g. semigroups) but also meaningless things which are certainly not mathematics (e.g. sets with a binary operation satisfying a(bc) = b(b((bc)(ac)))). To distinguish them you need to already know the difference. The English language describes everything I consider to be important, but that doesn't make it a top-down theory.

Comment by Tom Church

2010-05-16T17:24:46Z

1. The point is that $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ acts on the collection of all curves defined over $\mathbb{Q}$. The connection is Belyi's theorem, that curves defined over $\mathbb{Q}$ are exactly those with a map to $P^1$ ramified only at $\{0,1,\infty\}$. 3. Grothendieck suggested that by decomposing curves into "low-complexity" curves (small dimensional moduli), it suffices to understand $M_{0,4}$, $M_{1,1}$, $M_{0,5}$, and $M_{1,2}$. 4. If your points are ordered, then $\pi_1(\mathcal{M}_{0,n})$ is the pure braid group. If unordered, then the braid group.

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