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.
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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 $Df_I:B\in M\mapsto B+\bar B^t\in M$.
Mariano Suárez-Álvarez
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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.
Name
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1The symplectic group is the kernel of a linear map? – Mariano Suárez-Álvarez Apr 26 '14 at 04:21
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1Ah, sorry. Let me rewrite. – Name Apr 26 '14 at 04:23
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1But actually, isn't Sp(n) the kernel of $\overline AA^T -I $ ? – Name Apr 26 '14 at 04:25
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1Well, that is not a linear map nor a group homomorphism. – Mariano Suárez-Álvarez Apr 26 '14 at 04:26
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1O.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. – Name Apr 26 '14 at 04:32
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1I don't know what an affine map is, but I would guess it is not an affine map, either... – Mariano Suárez-Álvarez Apr 26 '14 at 04:35
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1An affine map is a composition of a linear map and a translation. But we have a product, which screws up linearity. Never mind. – Name Apr 26 '14 at 04:38
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1But I'm still surprised at the fact that you can guess it is not an affine map when you don't have a working definition. – Name Apr 26 '14 at 04:46
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1Well, I know many thing which are affine, and none of them can be quadratic, and this map is quadratic. – Mariano Suárez-Álvarez Apr 26 '14 at 04:58
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1Actually, you're right that is not affine, because there is a multiplication involved, but I don't know if by quadratic you mean that there are terms that are squared: $(A+A')(B^T+B'^T)=AB^T+AB'^T +A'B^T+A'B'^T$ , so no terms is multiplied by itself. Where I was wrong is that multiplying is not linear. – Name Apr 26 '14 at 05:21
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1At any rate, my mistake, the map is neither linear nor affine. My bad. – Name Apr 26 '14 at 05:22
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1WHat I mean is that the entries of the matrix $A^tA-I$ are quadratic polynomials of the entries of the matrix $A$. – Mariano Suárez-Álvarez Apr 26 '14 at 05:29