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How do you find the dimension of the symplectic group $Sp(2n,\mathbb{R})$?

$Sp(2n,\mathbb{R})\subset Gl(2n,\mathbb{R})$ is the group of invertible matrices $A$ such that $\omega = A^T\omega A$, where $$\omega = \bigg(\array{0 & Id_n\\ -Id_n & 0}\bigg)$$

I have tried to find the dimension of this group by dividing a matrix $A$ in four blocks, $$A = \bigg(\array{X & Y\\ Z & W}\bigg)$$ and use the defining property to put conditions on the blocks. I find the equations $X^TZ = Z^TX$, $Y^TW = W^TY$ and $Y^TZ - W^TX = Id_n$. The first two are obviously independent, and I think they put $2n^2$ restrictions on the group, but how should I go about the third one?

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Do you know the Lie algebra of the symplectic group? –  23rd Apr 11 '13 at 18:36
    
@richard: Yes, and it should have dimension $2n^2-n$, and so also the group would have the same dimension. But I want to find it without passing from the Lie algebra, if possible. –  Daniel Robert-Nicoud Apr 11 '13 at 18:41
    
I have no idea how to read the dimension from your three equations. By the way, the dimension should be $2n^2+n$. –  23rd Apr 11 '13 at 18:49

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