# Why is $Sp(2m)$ as regular set of $f(A)=A^tJA-J$, and, hence a Lie group.

I'm trying to prove that $Sp(2m)$ is a Lie group using this: defining a function $f(A)=A^tJA-J$ and trying to see that this is a submersion. But I've not realized yet what is the domains and the range of this function, neither why this function is actually a submersion. I can see that the tanget space at $I$ is actually the lie algebra of matrices $B$ whit the property $B^tJ+JB=0$; a simple calculation shows that the kernel of $df_I$ is exactly this set. Any help will be appreciated.

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I think you can define $f:GL(2m,\mathbb{R})\rightarrow \mathfrak{o}(2m)$, here $\mathfrak{o}(2m)$ denotes $2m\times 2m$ antisymmetric real matrices.
Claim: $f$ is a submersion.
For any $A\in GL(2m,\mathbb{R})$, we have $df_A(B)=B^tJA+A^tJB=A^tJB-(A^tJB)^t$. As $B$ ranges over all $2m\times 2m$ matrices, $A^tJB$ ranges over all $2m\times 2m$ matrices since $A^tJ$ is nonsingular. Hence $A^tJB-(A^tJB)^t$ ranges over all antisymmetric $2m\times 2m$ matrices, that is $df_A(M_n(\mathbb{R}))=\mathfrak{o}(n)$. Therefore, $df_A$ is surjective, i.e. $f$ is a submersion, and we prove the claim.
Since $Sp(2m)=f^{-1}(0)$ is the preimage of regular value $0$, $Sp(2m)$ is a submanifold of $GL(2m,\mathbb{R})$ hence a Lie group.