I am trying to show that the symplectic group $Sp(n) =\{A\in GL(n,\mathbb{H})\mid \overline{A}^TA=I\}$ is a regular submanifold of $GL(n,\mathbb{H})$ but I am stuck. Any help would be appreciated.
2 Answers
Check that the map $f:M\to M$ from the space $M$ of square matrices of size $n$ to itself such that $f(A)=\overline A^tA$ has I as a regular value. To do this,, you need to compute the rank of its differential at all $A$ in the sympectic group.
Let us compute the differential of $f$ at a matrix $A$, which is a real linear map $Df_A:M\to M$ such that \begin{align} Df_A(B)&=\lim_{h\to0}\frac{f(A+hB)-f(A)}{h}\\ &=\lim_{h\to0}\frac{(\bar A+h\bar B)^t(A+hB)-\bar A^tA}{h}\\ &=\lim_{h\to0}\frac{h\bar A^tB+h\bar B^tA+h^2\bar B^tB}{h}\\ &=\bar A^tB+\bar B^tA \end{align} Notice that for all $A$ and $B$ we have $Df_A(B)=Df_I(\bar A^tB)$. It follows that the differential $Df_A:M\to M$ is the composition of the linear map $X\in\mapsto \bar A^tB\in M$, which is an ismorphism if $A$ is in the symplctic group, and $Df_I$. In particular, the rank of the linear map $Df_A$ is equal to the rank of $Df_I$ whenever $A$ is in the symplectic group.
You should now be able to compute the rank of the map $Df_I:B\in M\mapsto B+\bar B^t\in M$.
Sorry, I don't have 50 points, so I cannot make a comment. The usual strategy is to show that the symplectic group is the preimage of a regular value, then use the regular value theorem.
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1$\begingroup$ The symplectic group is the kernel of a linear map? $\endgroup$ Apr 26, 2014 at 4:21
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1$\begingroup$ But actually, isn't Sp(n) the kernel of $\overline AA^T -I $ ? $\endgroup$– NameApr 26, 2014 at 4:25
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1$\begingroup$ Well, that is not a linear map nor a group homomorphism. $\endgroup$ Apr 26, 2014 at 4:26
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1$\begingroup$ O.K, it is an affine map, and so Sp(n) is just the zero set. I guess that doesn't buy much towards showing it is a submanifold. $\endgroup$– NameApr 26, 2014 at 4:32