The group $\mathrm{Sp}(2m)$ consists of the $2m\times 2m$ matrices $A$ with the property $A^tJA = J$, where $J$ is the $2m\times 2m$ standard skew-symmetric matrix. How do I prove that the lie algebra associated with has the property that $B^t = JBJ$ where $B$ is the matrix belonging to the lie algebra. I am trying to use $\exp(B)=1+B=A$. But after using it with the definition of the symplectic group I am not getting the required result.
[Math] Lie algebra associated with the symplectic group $\mathrm{Sp}(2m)$
group-theorylie-algebraslie-groups
Best Answer
Multiplying by $J$ in both side of the carachterizing property you gave, we see that the associated algebra is characterized by $B^tJ+JB=0$. The function $f(A)=A^tJA-J$ is differentiable and its derivative at the identidy $I\in Sp(2m)$ can be calculated by its action on tangent vectors, i.e. matrices, by newton quotient: let $B$ be a matrix, so
$df_I(B)=lim_{t\rightarrow 0}\frac{f(I+tB)-f(I)}{t}=B^tJ+JB$
So, as long as the tangent space of $Sp(2m)$ at $I$ is the kernel of $df_I$, we have the result. But here I have something I have not solved yet: $f$ is supposed to be a submersion and, in fact, $Sp(2m)$ is the inverse image of (the regular value) $0$. I coudn't see yet why is $0$ a regular value and neither what is the domain and the image of f. If anyone can help, I'd appreciate.