Local homology of a space of unitary matrices 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).
 A: 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.
