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It remains to prove the claim. Let $l=n-k$. Let $e_1, \ldots, e_k, f_1, \ldots, f_l$ be a basis of eigenvectors for $g$. We may assume that the first $k$ vectors are a basis for the subspace ${\mathbb C}^k$ fixed by $g$, and the last $l$ vectors are a basis for the orthogonal complement of ${\mathbb C}^k$. Let $\mu_1, \ldots, \mu_k, \nu_1, \ldots, \nu_l$ be the corresponding eigenvalues. By assumption, each $\mu_i$ is equal to $1$, and each $\nu_j$ is a unit complex number different from $1$.

A point in a small open neighborhood of $g$ in $U(n)$ is determined by the following data: an ordered $n$-tuple of pairwise orthogonal lines that are close to the lines spanned by the eigenvectors, together with corresponding eigenvalues, that are close enough to the corresponding eigenvalues of $g$. In particular we always assume that the last $l$ eigenvalues are all different from $1$. A point in the neighborhood belongs to $\mathcal D$ if and only if at least one of the first $k$ eigenvalues equals $1$. Thinking of $g$ as an element of $U(k)\times \{g_l\}$, where $g_l$ is the restriction of $g$ to the orthogonal complement of ${\mathbb C}^k$, it is clear that $g$ has a product neighborhood of the form $U\times V$, where $U$ is an open neighborhood of the identity in $U(k)$, and $V\cong {\mathbb R}^{n^2-k^2}$ is determined by the normal bundle of $U(k)\times\{g_l\}$ in $U(n)$. Whether a point in the neighborhood belongs to $\mathcal D$ is determined by the first $k$ eigenvalues, so it is a condition that only depends on the first factor in the product $U\times V$. From here, I think it is easy to see that the intersection $U\times V\cap {\mathcal D}\cong (U\cap {\mathcal D})\times V$, which is what we wanted to know.

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.

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 $U_k\times {\mathbb C}^{n^2-k^2}$ 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 $l=n-k$. Let $e_1, \ldots, e_k, f_1, \ldots, f_l$ be a basis of eigenvectors for $g$. We may assume that the first $k$ vectors are a basis for the subspace ${\mathbb C}^k$ fixed by $g$, and the last $l$ vectors are a basis for the orthogonal complement of ${\mathbb C}^k$. Let $\mu_1, \ldots, \mu_k, \nu_1, \ldots, \nu_l$ be the corresponding eigenvalues. By assumption, each $\mu_i$ is equal to $1$, and each $\nu_j$ is a unit complex number different from $1$.

A point in a small open neighborhood of $g$ in $U(n)$ is determined by the following data: an ordered $n$-tuple of pairwise orthogonal lines that are close to the lines spanned by the eigenvectors, together with corresponding eigenvalues, that are close enough to the corresponding eigenvalues of $g$. In particular we always assume that the last $l$ eigenvalues are all different from $1$. A point in the neighborhood belongs to $\mathcal D$ if and only if at least one of the first $k$ eigenvalues equals $1$. Thinking of $g$ as an element of $U(k)\times \{g_l\}$, where $g_l$ is the restriction of $g$ to the orthogonal complement of ${\mathbb C}^k$, it is clear that $g$ has a product neighborhood of the form $U\times V$, where $U$ is an open neighborhood of the identity in $U(k)$, and $V\cong {\mathbb C}^{n^2-k^2}$ is determined by the normal bundle of $U(k)\times\{g_l\}$ in $U(n)$. Whether a point in the neighborhood belongs to $\mathcal D$ is determined by the first $k$ eigenvalues, so it is a condition that only depends on the first factor in the product $U\times V$. From here, I think it is easy to see that the intersection $U\times V\cap {\mathcal D}\cong (U\cap {\mathcal D})\times V$, which is what we wanted to know.

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 $l=n-k$. Let $e_1, \ldots, e_k, f_1, \ldots, f_l$ be a basis of eigenvectors for $g$. We may assume that the first $k$ vectors are a basis for the subspace ${\mathbb C}^k$ fixed by $g$, and the last $l$ vectors are a basis for the orthogonal complement of ${\mathbb C}^k$. Let $\mu_1, \ldots, \mu_k, \nu_1, \ldots, \nu_l$ be the corresponding eigenvalues. By assumption, each $\mu_i$ is equal to $1$, and each $\nu_j$ is a unit complex number different from $1$.

A point in a small open neighborhood of $g$ in $U(n)$ is determined by the following data: an ordered $n$-tuple of pairwise orthogonal lines that are close to the lines spanned by the eigenvectors, together with corresponding eigenvalues, that are close enough to the corresponding eigenvalues of $g$. In particular we always assume that the last $l$ eigenvalues are all different from $1$. A point in the neighborhood belongs to $\mathcal D$ if and only if at least one of the first $k$ eigenvalues equals $1$. Thinking of $g$ as an element of $U(k)\times \{g_l\}$, where $g_l$ is the restriction of $g$ to the orthogonal complement of ${\mathbb C}^k$, it is clear that $g$ has a product neighborhood of the form $U\times V$, where $U$ is an open neighborhood of the identity in $U(k)$, and $V\cong {\mathbb R}^{n^2-k^2}$ is determined by the normal bundle of $U(k)\times\{g_l\}$ in $U(n)$. Whether a point in the neighborhood belongs to $\mathcal D$ is determined by the first $k$ eigenvalues, so it is a condition that only depends on the first factor in the product $U\times V$. From here, I think it is easy to see that the intersection $U\times V\cap {\mathcal D}\cong (U\cap {\mathcal D})\times V$, which is what we wanted to know.

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 $U_k\times {\mathbb C}^{n^2-k^2}$ 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 $l=n-k$. Let $e_1, \ldots, e_k, f_1, \ldots, f_l$ be a basis of eigenvectors for $g$. We may assume that the first $k$ vectors are a basis for the subspace ${\mathbb C}^k$ fixed by $g$, and the last $l$ vectors are a basis for the orthogonal complement of ${\mathbb C}^k$. Let $\mu_1, \ldots, \mu_k, \nu_1, \ldots, \nu_l$ be the corresponding eigenvalues. By assumption, each $\mu_i$ is equal to $1$, and each $\nu_j$ is a unit complex number different from $1$.

A point in a small open neighborhood of $g$ in $U(n)$ is determined by the following data: an ordered $n$-tuple of pairwise orthogonal lines that are close to the lines spanned by the eigenvectors, together with corresponding eigenvalues, that are close enough to the corresponding eigenvalues of $g$. In particular we always assume that the last $l$ eigenvalues are all different from $1$. A point in the neighborhood belongs to $\mathcal D$ if and only if at least one of the first $k$ eigenvalues equals $1$. Thinking of $g$ as an element of $U(k)\times \{g_l\}$, where $g_l$ is the restriction of $g$ to the orthogonal complement of ${\mathbb C}^k$, it is clear that $g$ has a product neighborhood of the form $U\times V$, where $U$ is an open neighborhood of the identity in $U(k)$, and $V\cong {\mathbb C}^{n^2-k^2}$ is determined by the normal bundle of $U(k)$ in $U(n)$. Whether a point in the neighborhood belongs to $\mathcal D$ is determined by the first $k$ eigenvalues, so it is a condition that only depends on the first factor in the product $U\times V$. From here, I think it is easy to see that the intersection $U\times V\cap {\mathcal D}\cong (U\cap {\mathcal D})\times V$, which is what we wanted to know.

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 $U_k\times {\mathbb C}^{n^2-k^2}$ 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 $l=n-k$. Let $e_1, \ldots, e_k, f_1, \ldots, f_l$ be a basis of eigenvectors for $g$. We may assume that the first $k$ vectors are a basis for the subspace ${\mathbb C}^k$ fixed by $g$, and the last $l$ vectors are a basis for the orthogonal complement of ${\mathbb C}^k$. Let $\mu_1, \ldots, \mu_k, \nu_1, \ldots, \nu_l$ be the corresponding eigenvalues. By assumption, each $\mu_i$ is equal to $1$, and each $\nu_j$ is a unit complex number different from $1$.

A point in a small open neighborhood of $g$ in $U(n)$ is determined by the following data: an ordered $n$-tuple of pairwise orthogonal lines that are close to the lines spanned by the eigenvectors, together with corresponding eigenvalues, that are close enough to the corresponding eigenvalues of $g$. In particular we always assume that the last $l$ eigenvalues are all different from $1$. A point in the neighborhood belongs to $\mathcal D$ if and only if at least one of the first $k$ eigenvalues equals $1$. Thinking of $g$ as an element of $U(k)\times \{g_l\}$, where $g_l$ is the restriction of $g$ to the orthogonal complement of ${\mathbb C}^k$, it is clear that $g$ has a product neighborhood of the form $U\times V$, where $U$ is an open neighborhood of the identity in $U(k)$, and $V\cong {\mathbb C}^{n^2-k^2}$ is determined by the normal bundle of $U(k)\times\{g_l\}$ in $U(n)$. Whether a point in the neighborhood belongs to $\mathcal D$ is determined by the first $k$ eigenvalues, so it is a condition that only depends on the first factor in the product $U\times V$. From here, I think it is easy to see that the intersection $U\times V\cap {\mathcal D}\cong (U\cap {\mathcal D})\times V$, which is what we wanted to know.