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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

It helps to think of the columns of $A$ instead of the blocks of $A$.

A matrix $A$ in $SP_{2n}(\mathbb{R})$ has $2n$ columns, call them $e_1,\ldots,e_n,f_1,\ldots,f_n$. We want $A$ to satisfy $A^T\omega A = \omega$, so there are $(2n)^2$ constraints we can write down (one per entry of $\omega$):

  1. $e_i^T\omega e_j = 0$
  2. $f_i^T \omega f_j = 0$
  3. $e_i^T \omega f_j = \delta_{ij}$ and $f_i^T \omega e_j = -\delta_{ij}$.

By alternativity and antisymmetry of the form $\omega$, there are only $n(n-1)/2$ independent equations for each of #1 and #2.

By antisymmetry, there are only $n^2$ independent constraints in #3.

These are $2n^2 - n$ total independent constraints on the $(2n)^2$ possible $2n \times 2n$ matrices.

We are left with $2n^2 + n$ real degrees of freedom, so $SP_{2n}(\mathbb{R})$ has dimension $2n^2 + n$.

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