18
$\begingroup$

Let $U(n)$ denote the unitary group (this is a manifold of dimension $n^2$). Let $$ {\cal D} \subset U(n) $$ denote the subspace of those matrices having a non-trivial $(+1)$-eigenspace.

Background: It is known that $\cal D$ has vanishing homology in dimension $n^2$. It is also not difficult to show that $H_{n^2-1}({\cal D}) \cong \Bbb Z$. Furthermore, it is known that ${\cal D}$ has the structure of a finite CW complex of dimension $n^2-1$.

For $g\in {\cal D}$, define the local homology by $$ H_{n^2-1}({\cal D}\, |\, g;\Bbb Q) := H_{n^2-1}({\cal D},{\cal D} \setminus g;\Bbb Q)\, . $$

Question: For arbitrary $g\in {\cal D}$ is the rank of this group known?

If yes, can anyone provide a reference?

Remark: Based on a direct computation when $n=2$, it seems reasonable to conjecture that the rank of $H_{n^2-1}({\cal D}\, |\, g;\Bbb Q)$ is equal to the dimension of the $(+1)$-eigenspace of $g$ (i.e., the multiplicity of the eigenvalue +1).

$\endgroup$
  • 5
    $\begingroup$ You can filter $U(n)$ by the dimension of the relevant eigenspace. The $k$'th filtration quotient is (essentially by the Cayley transform) the Thom space of the bundle of hermitian endomorphisms of the tautological bundle over the Grassmannian $G_k(\mathbb{C}^n)$. This circle of ideas also shows that $\mathcal{D}^c$ is just $\mathbb{R}^{n^2}$. The filtration is stably split by a theorem of Haynes Miller. This does not obviously answer your question, but it seems like relevant context. $\endgroup$ – Neil Strickland May 3 '17 at 15:15
  • 1
    $\begingroup$ This is pure speculation, but can you use the map $U(n)\to SP^n(S^1)$ which takes a unitary matrix to its spectrum of eigenvalues? The symmetric power is not a manifold, but maybe its local homology is easier to relate to multiplicities of eigenvalues than that of $\mathcal{D}$. $\endgroup$ – Mark Grant May 5 '17 at 15:30
  • $\begingroup$ @MarkGrant I am aware of that map. It can be used to equip $U(n)$ with the structure of a Whitney stratified space (I learned this from David Ayala). But how can you use the map to compute the local homology of $\cal D$? $\endgroup$ – John Klein May 5 '17 at 16:05
  • $\begingroup$ @JohnKlein: I don't know yet. Two unitary matrices are unitarily similar iff they have the same eigenvalues, so that map is actually the orbit map of the conjugation action, and $\mathcal{D}$ is the preimage of the inclusion $SP^{n-1}(S^1)\subset SP^n(S^1)$ obtained by adding a 1. Surely its not a bundle though. Greg's answer seems to be more a propos. $\endgroup$ – Mark Grant May 6 '17 at 8:23
10
+50
$\begingroup$

Let us first consider the case when $g=e$ is the identity matrix. Let $U$ be an open neighbourhood of the identity in $\mathcal D$. We want to calculate the local homology of $U$ at $e$.

We may assume that $U$ is mapped homeomorphically by the (inverse of) the exponential map onto its image in the tangent space of $U(n)$. The tangent space can be identified with the space of skew-Hermitian $n\times n$ matrices.

Elements of $\mathcal D$ that are close to the identity correspond under the exponential map to non-invertible skew-Hermitian matrices. So you are asking about the local homology in degree ${n^2-1}$ of the space of non-invertible skew-Hermitian matrices. By Alexander duality, this is isomorphic to the reduced homology in degree $0$ of the space of invertible skew-Hermitian matrices. So we need to count the path components of this space.

Such a matrix will have non-zero purely imaginary eigenvalues, of the form $ir$. The path component of a matrix is determined by the number of eigenvalues for which $r>0$. It follows that there are $n+1$ components, so the reduced homology has rank $n$, which confirms your conjecture.

Added later: For the general case, suppose $g$ is a unitary matrix that fixes a subspace ${\mathbb C}^k\subset {\mathbb C}^n$. Let ${\mathcal D}_k\subset U(k)$ be the subspace of matrices that fix a non-zero subspace of ${\mathbb C}^k$. I claim that $g$ has an open neighborhood $U\subset\mathcal D$ that is homeomorphic to $U_k\times {\mathbb R}^{n^2-k^2}$ where $U_k$ is an open neigborhood of the identity in ${\mathcal D}_k$, by a homeomorphism that takes $g$ to $e\times 0$. It follows easily that $H_*(U, U\setminus\{g\})\cong H_{*-(n^2-k^2)}(U_k,U_k\setminus\{e\})$, so the general case follows from the special case $g=e$.

It remains to prove the claim. Let $l=n-k$ and let ${\mathbb C}^l$ be the orthogonal complement of ${\mathbb C}^k$ in $\mathbb C^n$. Consider the eigenspace decomposition of $g$. One eigenspace is ${\mathbb C}^k$, with associated eigenvalue $1$. The remaining eigenspaces form an orthogonal decomposition of ${\mathbb C}^l$, and their eigenvalues are unit complex numbers different from $1$.

We may identify $g$ with the element $(e_k,g_l)\in U(k)\times \{g_l\}$ where $e_k$ is the identity of $U(k)$ and $g_l$ is the restriction of $g$ to ${\mathbb C}^l$. Since it is a submanifold, $g$ has a product neighborhood of the form $V_1\times V_2$ where $V_1$ is a sufficiently small open neighborhood of the identity in $U(k)$ and $V_2\cong{\mathbb R}^{n^2-k^2}$ is a small tubular neighborhood of $V_1$ in $U(n)$. We want to understand the intersection of $V_1\times V_2$ with $\mathcal D$. Consider the eigenspace decomposition of an element of $V_1\times V_2$. It will have eigenspaces of two types: some are very close to ${\mathbb C}^k$ and some are very close to $\mathbb C^l$. The eigevalues of first type are unit complex numbers close to $1$, and eigenvalues of second type are unit complex numbers distinct from $1$. The element belongs to $\mathcal D$ if an only if at least one of the eigenvalues of first type equals $1$. I think it is easy to see from here that an element of $V_1\times V_2$ belongs to $\mathcal D$ if an only if its $V_1$ complonent belongs to $\mathcal D_k$. It follows that $(V_1\times V_2)\cap {\mathcal D}= (V_1\cap {\mathcal D}_k)\times V_2\cong U_k\times {\mathbb R}^{n^2-k^2}$, which is what we wanted to know.

$\endgroup$
  • $\begingroup$ Greg, I should have said that knew how to do that case already. I am pretty certain that that the argument for $g \ne I$ is going to be more complicated because the stratification is different at different points of $\cal D$. What idea do you you have in mind to reduce the general case to the one in your post? I think that may be the crux of the matter! $\endgroup$ – John Klein May 5 '17 at 22:31
  • 1
    $\begingroup$ Suppose $g$ fixes a subspace of dimension $k$ in ${\mathbb C}^n$. Then a small open neighborhood of $g$ in $\mathcal D$ is homeomorphic to a vector bundle of dimension $n^2-k^2$ over an open neighborhood of the identity in ${\mathcal D}_k\subset U(k)$. I think it follows that the local homology of $\mathcal D$ at $g$ is isomorphic to the local homology of ${\mathcal D}_k$ at the identity shifted up by $n^2-k^2$, essentially by a Thom isomorphism. $\endgroup$ – Gregory Arone May 5 '17 at 23:11
  • $\begingroup$ I have to work through this, but I think your argument seems to be correct. $\endgroup$ – John Klein May 5 '17 at 23:43
  • $\begingroup$ I added something about the general case to the answer. $\endgroup$ – Gregory Arone May 6 '17 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.