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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 ?

StarBucK
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  • The dimension of $U(n)$ is $n^2$; of course determinant maps $U(n)$ onto the circle group, which is one-dimensional. – Angina Seng Nov 11 '17 at 18:09
  • $U(N)=U(1)\otimes SU(N)$. The $\det=1$ condition fixes the $U(1)$ degree of freedom, leaving the $SU(N)$ group $N^2-1$ real dimensions. – Zhuoran He Nov 11 '17 at 18:12
  • @LordSharktheUnknown But $A^{\dagger}A=I$ implicates $|det(A)|=1$. Does the sign $+1$ for the det restrict more the dimension ? I thought it would only specify "a connex part" to take. – StarBucK Nov 11 '17 at 18:26
  • @StarBucK $|\det A|=1$ is not the same as $\det A=1$. – Angina Seng Nov 11 '17 at 18:30

1 Answers1

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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 $$ \operatorname{dim}(SU(N)) = \operatorname{dim}(M_1\times M_2 \times \cdots \times M_{N-1} \times M'_N) = N^2 - 1 $$

coconut
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