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).

  • 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. $\endgroup$ May 3, 2017 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, 2017 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, 2017 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, 2017 at 8:23

1 Answer 1


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.

  • $\begingroup$ 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, 2017 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. $\endgroup$ May 5, 2017 at 23:11
  • $\begingroup$ I have to work through this, but I think your argument seems to be correct. $\endgroup$
    – John Klein
    May 5, 2017 at 23:43
  • $\begingroup$ I added something about the general case to the answer. $\endgroup$ May 6, 2017 at 12:41

Your Answer

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