How to show the symplectic group is a submanifold of $GL(n,\mathbb{H})$? 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.
 A: 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 $Df_I:B\in M\mapsto B+\bar B^t\in M$. 
A: 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.
