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It is well known that $U(n)/U(n-k) \cong V_k(\mathbb{C}^n)$ where $U(n)$ is the unitary group, and $V_k(\mathbb{C}^n)$ is the appropriate Stielfel manifold.

I further understand that $V_k(\mathbb{C}^n)=\{ U \in \mathbb{C}^{n\times k} \ | \ U^{\dagger}U=I \}$ can be thought of as a set of non square matrices and given the structure of a manifold and that it inherits a canonical metric $g_{C}(Δ,Δ)=Tr(Δ^{∗}(I-(1/2)UU^{∗})Δ)$.

How can I construct this isomorphism in the cases $U(n)/U(1)$ and $U(n)/U(2)$? Given a $U(1) = \{e^{\beta b} \ | \ b \in \mathfrak{u}(n), \beta \in \mathbb{R} \}$ subgroup of $U(n)$, I seek a map $\phi: U(n) \rightarrow V_k(\mathbb{C}^n)$ which takes a unitary matrix and a gives me the non square matrix in $V_k(\mathbb{C}^n)$ corresponding to the correct coset.

Furthermore, are the geodesics on these spaces known and is a formula for the Riemannian logarithm on Steifel manifolds known in the complex case?

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    $\begingroup$ Did you study the answers at: mathoverflow.net/questions/84955/… already? $\endgroup$ – Suvrit Apr 5 '17 at 21:42
  • $\begingroup$ Yes, it doesn't actually answer either thing. Particularly, it didn't help me find the isomorphism. $\endgroup$ – Benjamin Apr 5 '17 at 21:44
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The map $U(n)\to V_k(\mathbb C^n)$ simply maps a matrix $U$ to its first $k$ columns, which are $k$ orthonormal vectors in $\mathbb C^n$. Otherwise put, $U$ is mapped to $(Ue_1,\dots,Ue_k)$ and the preimage of $(e_1,\dots,e_k)$ under this map is evidently isomorphic to $U(n-k)$ (via acting on $e_{k+1},\dots,e_n$). (So this concerns a specific subgroup of $U(n)$ - or its conjugates - and not any subgroup of $U(n)$ which is isomorphic to $U(n-k)$.)

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  • $\begingroup$ Does this map really not depend on $b$? I think it should as this determines the cosets comprising the quotient space. The action of every different $U(1)$ subgroup certainly doesn't just fix the same first column. $\endgroup$ – Benjamin Apr 6 '17 at 17:29
  • $\begingroup$ You get the Stiefel manifold only for the standard $U(1)\subseteq U(n)$ and its conjugates. Other 1-dimensional subgroups yield different spaces. $\endgroup$ – Friedrich Knop Apr 6 '17 at 19:57
  • $\begingroup$ What do you mean by the standard $U(1)$? $\endgroup$ – Benjamin Apr 6 '17 at 21:27
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    $\begingroup$ As I said in the reply, the statement is not about any subgroup of $U(n)$ isomorphic to $U(n-k)$ but about a specific subgroup (or more precisely a conjugacy class of subgroups). The "standard $U(n-k)$" in this sense is the stabilizer of the linear subspace $\mathbb C^k\subset \mathbb C^n$ and the corresponding conjugacy class consists of stabilizers of $k$-dimensional subspaces. It is only for those subgroups that you really get a Stiefel-manifold. $\endgroup$ – Andreas Cap Apr 7 '17 at 7:24

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