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Looking at this and this question about the Lagrangian Grassmannian, and its linked Wikipedia description as the quotient of $Sp(N)$ by the unitary group $U(n)$, I wondering what is the explicit embedding $U(N) \hookrightarrow Sp(2n)$.

As far I understand, we have (also by Wikipedia) that $Sp(n)$ is the group of quaternionic unitary matrices of order $n$, and so, a subset of $M(2n,\mathbb{C})$. By analogy with the case of the usual Grassmannian, I would embed $U(n)$ in the bottom right hand corner, and put $1$'s on the remaining diagonal entries. However, it is not clear to me that this is actually contained in $Sp(n)$. Moreover, will it make a difference if I instead embed it in the top left hand corner. Because of less Dynkin diagram symmetries in the $C$-series case, I feel this should not be isomorphic.

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  • $\begingroup$ Wikipedia says $Sp(n)$, not $SP(2n)$. Just consider the subgroup $U(n)\subset Sp(n)$ containing all matrices with entries in $\mathbb C$. It fixes the Lagrangian subspace $\mathbb C^n\subset\mathbb H^n$. $\endgroup$
    – Sebastian Goette
    Commented Feb 19, 2017 at 0:08
  • $\begingroup$ ok, so my question now matches Wikip - thanks for pointing out the error $\endgroup$
    – Alesandro Levi
    Commented Feb 19, 2017 at 0:13
  • $\begingroup$ But if we embed $U(n,\mathbb{H})$ into $M(n,\mathbb{C})$, what does $U(n,\mathbb{C})$ then look like? $\endgroup$
    – Alesandro Levi
    Commented Feb 19, 2017 at 0:17
  • $\begingroup$ The group $U(n)$ preserves a hermitian form $h$ on $\mathbb{C}^n$. If you view $\mathbb{C}^n$ as $\mathbb{R}^{2n}$, and replace the Hermitian form by it imaginary part $\Omega$ , you see that $U(n)$ preserves $\Omega$ as well. This is the embedding $U(h)\subset Sp_{\Omega}$. $\endgroup$
    – Venkataramana
    Commented Feb 19, 2017 at 2:37
  • $\begingroup$ For $(\mathbb R^{2n},\omega_0)$ where $\omega_0(u,v)=u^T [ {\begin{array}{cc} 0 & I \\ -I & 0 \\ \end{array} } ] v$ the Lagrangian Grassmannian $\Lambda (n)=U(n)/O(n)\cong Sp(n)/GL(n)$ $\endgroup$
    – user21574
    Commented Oct 15, 2017 at 12:37

1 Answer 1

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Like any normal person, I'm going to use the symplectic form where $\langle e_i,e_{j}\rangle =\pm \delta_{j,2n-i+1}$ with $1$ if $i\leq n$ and $-1$ if $i>n$. The compact symplectic group you have in mind are the unitary matrices preserving this form. In these coordinates, you have to embed $U(n)$ as $A\mapsto \big[\begin{smallmatrix}A & 0\\ 0&A^{-1}\end{smallmatrix}\big]$; this is the subgroup of the compact symplectic group which preserves the Lagrangian subspace spanned by the vectors $e_1,\dots, e_n$.

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