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Given the compact symplectic Lie group $\mathrm{Sp}(n)$ of $2n \times 2n$ matrices and the unitary Lie group $\mathrm{U}(n)$ of $n \times n$ matrices, is the quotient ${\mathrm{Sp}(n) \over\mathrm{U}(n)}$ a Lie group as well?

Impossibear
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  • Yes, since that quotient has the same cardinal has $\Bbb R$. Actually, assuming the Axiom of Choice, every non-empty set has a group structure. – José Carlos Santos Sep 11 '20 at 10:59
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    Just to be clear, how are you viewing an element of $U(n)$ as an element of $Sp(n)$? That is, what is the inclusion map $U(n) \to Sp(n)$? – Michael Albanese Sep 11 '20 at 13:13
  • @Michael: $Sp(n)$ is the subgroup of $GL_n(\mathbb{H})$ preserving a Hermitian form on $\mathbb{H}^n$, so any embedding $\mathbb{C} \to \mathbb{H}$ produces a subgroup isomorphic to $U(n)$ by restricting attention to matrices with entries in that copy of $\mathbb{C}$, and all such embeddings are conjugate so we get the same subgroup up to conjugacy. – Qiaochu Yuan Sep 11 '20 at 21:34
  • @QiaochuYuan: I asked because the OP described $Sp(n)$ as $2n\times 2n$ matrices (as opposed to $n\times n$ matrices with quaternionic entries). – Michael Albanese Sep 11 '20 at 21:58
  • You can get a description in those terms by embedding $\mathbb{H}$ into $M_2(\mathbb{C})$. – Qiaochu Yuan Sep 11 '20 at 22:01

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Not in a way that has anything to do with the group structure of $Sp(n)$; generally $G/H$ inherits a group structure from $G$ iff $H$ is normal, and $U(n)$ is very far from being normal in $Sp(n)$ (as I mentioned in the comments, conjugates of $U(n)$ can be constructed by picking different embeddings $\mathbb{C} \to \mathbb{H}$). In fact $Sp(n)$ is simple and so has no nontrivial connected normal subgroups.

When $n = 1$ the quotient $Sp(1)/U(1)$ is the $2$-sphere $S^2$ so does not admit any topological group structures whatsoever.

Qiaochu Yuan
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  • I asked this question because, for the $n=1$ case, the topological charge corresponding to homotopy group $\pi_2 \left( S^2 \right)$ can be computed with a normalized vector $\hat{v}$ as an integral over the starting manifold, \begin{equation} S = {1 \over 4\pi} \int \hat{v} \cdot \left({\partial \hat{v} \over \partial x} \times {\partial \hat{v} \over \partial y} \right) dx dy \end{equation} where the vector $v$ does not change in magnitude over $x$ and $y$. This indicates $v$ is a conserved quantity, chosen, for instance, using isomorphism ${SU(2) \over U(1)} \cong S^2$ @QiaochuYuan – Impossibear Sep 16 '20 at 14:07
  • For general $n$, $v$ seems to instead be a vector that can change in magnitude for different values of $x$ and $y$ in the starting manifold. This suggests that $v$ is no longer a conserved quantity, which should be chosen based on the structure of $Sp(n)/U(n)$. I suspected this was because $Sp(n)/U(n)$ was not a Lie group itself and because of this should be computed with a vector that is not conserved in magnitude. I guess I don't understand whether this difference between the $n=1$ and general $n$ cases relates to your topological group structures comment. @QiaochuYuan – Impossibear Sep 16 '20 at 14:08
  • If you do have any idea of why there would be this difference in $v$ magnitude over $x$ and $y$ between the $n=1$ case and general $n$ case, I would really appreciate your comments. @QiaochuYuan. Thank you very much already for your thoughts. – Impossibear Sep 16 '20 at 14:10
  • I don't understand what $v$ is in your integral so I'm afraid I can't comment. (Is it a vector? A vector-valued function? A vector field?) – Qiaochu Yuan Sep 16 '20 at 17:13
  • $v$ is a vector field $v (x, y) = a (x,y) \hat{x} + b(x,y) \hat{y} + c(x,y) \hat{z}$, where $a (x,y)$, $b (x,y) $, $c(x,y) \in \mathbb{R}$. @QiaochuYuan – Impossibear Sep 17 '20 at 08:27
  • https://en.wikipedia.org/wiki/Magnetic_skyrmion might help in explaining. @QiaochuYuan – Impossibear Sep 17 '20 at 08:40