Motivation behind definition of complex sympletic group

Question

One definition of complex sympletic group I have encountered is (sourced from Wikipedia): $$Sp(2n,F)=\{M\in M_{2n\times 2n}(F):M^{\mathrm {T} }\Omega M=\Omega \}$$

What is the motivation for imposing the condition $M^{\mathrm {T} }\Omega M=\Omega$ instead of others such as $M^{-1}\Omega M=\Omega$?

Answer 1

In general, if $V$ is a $n$-dimensional $F$-vector space equipped with a bilinear form $b\colon V \times V \to F$ and $T\colon V \to V$ is an endomorphism such that $b(Tx,Ty) = b(x,y)$ for all $x,y \in V$, one may take a basis $\mathcal{B} = (e_1,\ldots, e_n)$ for $V$, let $b_{ij} = b(e_i,e_j)$ and $Te_j = \sum_{i=1}^n T^i_{~j}e_i$, and compute $$b_{ij} = b(e_i,e_j) = b(Te_i,Te_j) = b\left(\sum_{k=1}^n T^k_{~i}e_k, \sum_{\ell=1}^r T^\ell_{~j}e_j\right) = \sum_{k,\ell=1}^n T^k_{~i}T^\ell_{~j}b_{k\ell}.$$If $[T]_{\mathcal{B}} = (T^i_{~j})_{i,j=1}^n$ and $[b]_{\mathcal{B}} =(b_{ij})_{i,j=1}^n$, the above identity then reads $$[b]_{\mathcal{B}} = [T]_{\mathcal{B}}^\top [b]_{\mathcal{B}}[T]_{\mathcal{B}}.$$So, once a basis $\mathcal{B}$ (and hence the matrix $B = [b]_{\mathcal{B}}$) is fixed, the isomorphism $T\mapsto [T]_{\mathcal{B}}$ between ${\rm End}(V)$ and ${\rm Mat}(n,F)$ restricts to an isomorphism $$\{T \in {\rm End}(V) \mid b(Tx,Ty) = b(x,y) \mbox{ for all }x,y \in V \}\cong \{ M \in {\rm Mat}(n,F) \mid M^\top BM = B \}.$$The fact that the dimension of the space $V$ is even and that $\Omega$ is symplectic is irrelevant, this is a general mechanism regarding the relation between the matrix of a bilinear map and the matrix of its pull-back under a linear map.

Answer 2

Such issues become clearer when we do not commit to a basis of a vector space, and do not depend on matrices to express either linear maps or quadratic forms or...

So, given a finite-dimensional $k$-vectorspace $V$, with $k$ a field of characteristic not $2$, we can consider non-degenerate skew-symmetric bilinear forms $J$ on $V$. Two such ($J_1$ and $J_2$) might be said "equivalent" when there is an automorphism $A$ of $V$ such that $J_1(Ax,Ay)=J_2(x,y)$ for all $x,y\in V$.

If/when we describe the bilinear forms $J_i$ as $J_i(x,y)=y^\top S_i x$ for skew-symmetric matrices $S_i$, we obtain the formulaic condition you mention.