I am trying to find that the real dimension of $SU(N)$ is $N^2-1$ but I make a mistake and I don't know where.
I would like to prove it directly on the group (I don't want to use the algebra).
What I did :
$$SU(N)=\{ A \in \mathcal{M}_N(\mathbb{C}) / A^{\dagger}A=I, det(A)=+1 \} $$
- I have $2N^2$ coefficient for $SU(N)$ (because complex).
- $M=A^{\dagger}A$ is an hermitian matrix. Thus by construction all the coefficients on the diagonal are reals : $N$ equation to determine them. I also have $2*(N-1+N-2+...+1)=N(N-1)$ extra diagonal coefficient to determine.
Finally I will have : $2N^2 - N - N(N-1)=N^2$ which should be the dimension of $SU(N)$... But it is wrong.
Where is my mistake ?
Is it because the determinant is $+1$ and I didn't used it ?
A unitary $N \times N$ matrix is equivalent to a orthonormal basis of $\mathbb{C}^N$ (for example, take its columns as the components of the vectors). We can choose an orthonormal basis as follows:
Take a unit vector $e_1$. The manifold $M_1$ of such vectors has $2N-1$ real dimensions.
Take a unit vector $e_2$ orthogonal to $e_1$. The manifold $M_2$ of such vectors is $(2N-3)$-dimensional over $\mathbb{R}$.
Take a unit vector orthogonal to $e_1$ and $e_2$. The corresponding manifold $M_3$ has $2N-5$ dimensions.
Continue taking unit vectors orthogonal to all the previous ones until $e_{N-1}$. The manifold $M_N$ has dimension $2N-(2N-3)=1$.
The dimension of the manifold $M_1\times M_2 \times \cdots \times M_N$ is, therefore: $$2N-1 + 2N - 3 + \cdots + 1 = N^2.$$ If we were to form a matrix with the components of this vectors we would get a unitary matrix. We can make it have unit determinant by taking the submanifold $M'_N$ of $M_N$ of the vectors with the suitable phase. $M'_N$ has dimension $0$. Thus $$ \operatorname{dim}(SU(N)) = \operatorname{dim}(M_1\times M_2 \times \cdots \times M_{N-1} \times M'_N) = N^2 - 1 $$