Finding the dimension of the symplectic group How do you find the dimension of the symplectic group $\operatorname{Sp}(2n,\mathbb{R})$?
$\operatorname{Sp}(2n,\mathbb{R})\subset \operatorname{Gl}(2n,\mathbb{R})$ is the group of invertible matrices $A$ such that $\omega = A^T\omega A$, where $$\omega = \bigg(\array{0 & \mathit{Id}_n\\ -\mathit{Id}_n & 0}\bigg)$$
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 $Y^TZ - W^TX = \mathit{Id}_n$. 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?
 A: You could directly deduce the dimension from your own equations:


*

*$X^T Z = Z^T X \implies X^T Z = (X^T Z)^T$,

*$Y^T W = W^T Y \implies Y^T W = (Y^T W)^T$,

*$Y^T Z -W^T X = \mathbb{I}_n$.


The first two are independent constraints of the form $P = P^T$, so each gives $(n^2-n)/2$ constraints. The third is also an independent constraint, of the form $P-Q = \mathbb{I}_n$, so this gives $n^2$ constraints. This leaves $4n^2 - n^2 - (n^2-n) = 2n^2+n$ degrees of freedom.
To explicitly see that the constraints decouple, from their rather suggestive form, define the complex matrices $P=X + i Z, Q= Y + i W$. Then constraints (1) and (2) are $Im(P^\dagger P ) = 0, Im(Q^\dagger Q) =0$ respectively. Constraint (3) however becomes $Im(Q^\dagger P) = \mathbb{I}_n$.
A: 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 $\omega$):


*

*$e_i^T\omega e_j = 0$

*$f_i^T \omega f_j = 0$

*$e_i^T \omega f_j = \delta_{ij}$ 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 $2n^2 + n$.
A: The dimension of the Lie group $G = Sp_{2n}(\mathbb{R})$ is the same as that of its tangent space $\mathfrak{g} = \mathfrak{sp}_{2n}(\mathbb{R})$ at the origin. Linearizing the defining relation for $G$ (in your notation above, $g^t \omega g = \omega)$ gives 
$$\mathfrak{g} = \{X\in\mathfrak{sl}_n(\mathbb{R}):\, \omega X = -g^t X\}$$
Writing $X$ in terms of $n\times n$ blocks as you described gives $\dim \mathfrak{g}$.
