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The group Sp(2m) consists of the 2m×2m matrices A with the property At (A transpose)JA = J, where J is the 2mX2m standard skew-symmetric matrix. How do I prove that the lie algebra associated with has the property that (B transpose) = 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.

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You should accept answers to the other questions you asked. –  martini Mar 22 '12 at 14:05

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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.

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I would like to add this: I've asked in another topic why $Sp(2m)$ is a Lie group and the answer is this: math.stackexchange.com/questions/133295/… –  matgaio Apr 18 '12 at 19:16

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