Difficulty understanding a proposition in Fulton and Harris regarding $\mathrm{Sp}_{2n}(\mathbb{C})$

Question

In Fulton & Harris, Proposition 23.1 (computing the fundamental groups of classical complex Lie groups) and the Exercises after it, specifically in dealing with $\mathrm{Sp}_{2n}(\mathbb{C})$,

the book seems to suggest that the following two submanifolds are diffeomorphic to each other: $$ M = \left\{ ((x_1,x_2),(y_1,y_2)) \in \mathbb{C}^{2n}\times\mathbb{C}^{2n} : x_1^Ty_2 - x_2^Ty_1 = 1 \right\} , \\ M' = \left\{ ((x_1,x_2),(y_1,y_2)) \in \mathbb{C}^{2n}\times\mathbb{C}^{2n} : x_1^Tx_1 + x_2^Tx_2 + y_1^Ty_1 + y_2^Ty_2 = 1 \right\} . $$ In the book these are written as $$ M = \{ (v,w) \in \mathbb{C}^{2n}\times \mathbb{C}^{2n} : Q(v,w) = 1\}, \\ M' = \{ z \in \mathbb{C}^{4n} : z^T z = 1\} , $$ where $Q$ is an alternating non-degenerate quadratic form on $\mathbb{C}^{2n} \times \mathbb{C}^{2n}$, which I've put in "standard form" by picking a symplectic basis.

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My guess is, $M$ and $M'$ are shown diffeomorphic by showing there is a change-of-basis on $\mathbb{C}^{4n}$ which takes the $M$ expression to the $M'$ one.

However, I'm not sure how to find such a change-of-basis; how do I complete the argument here that $M$ and $M'$ are diffeomorphic?

Perhaps I'm missing something? Any help would be much appreciated!

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Edit: as Stephen has pointed out in the comments, $(v,w) \mapsto v^Tv + w^Tw$ is not at all a bilinear form (as I'd incorrectly written earlier).

Answer 1

Isn't this just, you say that $Q$ is a symplectic form on $\mathbb{C}^{2n}$, and use that there exists a basis (sometimes called (standard) symplectic basis) in which the symplectic form has the matrix representation $$ \begin{pmatrix}0_n&-1_n\\1_n&0_n\end{pmatrix}? $$ So I don't see any problem with your argument.

Answer 2

Some of the quadratic equations you wrote are (in various coordinates) the equations for vectors being "isotropic" for the symplectic form $Q$, that is, $Q(v)=0$. EDIT2: others are $\{v:Q(v)=c\}$ for some constant $c$.

It is a not-completely-trivial theorem that, up to isomorphism (change-of-basis) there is a unique non-degenerate alternating form on a vectorspace of a given (even) dimension. The relevant change-of-basis is linear over whatever groundfield (maybe we need characteristic not 2), $v\to Av$, so $v\to v^\top Qv$ becomes $v\to (Av)^\top Q(Av)$, which means that the matrix of the symplectic form transforms by $Q\to A^\top QA$.

EDIT: And, if we want $\{v:Q(v)=c\}$ for any constant $c$, Witt's theorem assures that the symplectic group is transitive on such vectors.

Answer 3

So after some initial confusion, I've managed to work out the details of the argument.

We have $$ M = \{ z \in \mathbb{C}^{4n} : B(z,z) = 1\} $$ where $B : \mathbb{C}^{4n} \times \mathbb{C}^{4n} \rightarrow \mathbb{C}$ is the symmetric non-degenerate complex bilinear form given w.r.t. the standard basis by the matrix $$ W = \frac{1}{2}\begin{bmatrix} & & & I\\ & & -I & \\ & -I & & \\ I & & & \end{bmatrix} \,. $$ Now, because the matrix $W$ is real and symmetric (and nonsingular), in particular (by first applying the spectral theorem) it is possible to show $A^TWA = I$ for some change-of-basis matrix $A$.