7

$SU(N)$ is the group of special unitary matrices of dimension $N$, i.e., the set of all unitary ($U^{\dagger}U=I$) $N\times N$ matrices with $\det(U)=1$.

For $N=2$, these matrices are spanned by the identity and the Pauli matrices, i.e. we can write

$$U_{2\times 2} = a_0 I_2+\vec{a}\cdot\vec{\sigma}$$

so I would say that the basis for $SU(2)$ is $\{I_2,\sigma_1,\sigma_2,\sigma_3\}$.

  1. However, I've read in several lecture notes (and wikipedia seems to agree) that "$SU(N)$ is generated by traceless Hermitian matrices and so has (real) dimension $N^2-1$." Can someone explain this?

  2. Why traceless? This would immediately rule out the identity which, in my view, is needed for $SU(2)$.

  3. How is it $N^2-1$ and not $N^2-2$, as we have 2 conditions (traceless and hermitian) on the basis elements?

Qmechanic
  • 220,844
TMG
  • 81

1 Answers1

5

  • Note that the Lie group $SU(2)$ is not a vector space; only a manifold. But it is a subset of the vector space ${\rm Mat}_{2\times 2}(\mathbb{C})$. So $\{{\rm 1}_{2\times 2},\sigma_1,\sigma_2,\sigma_3\}$ is formally speaking a (complex) basis for ${\rm Mat}_{2\times 2}(\mathbb{C})$; not $SU(2)$.

  • The lecture notes refer to the Lie algebra $$su(N)~:=~\{M\in {\rm Mat}_{2\times 2}(\mathbb{C}) | M^{\dagger}=M,~ {\rm tr}(M)=0\} $$ rather than the Lie group $$\begin{align}SU(N)~:=~&\{U\in {\rm Mat}_{2\times 2}(\mathbb{C}) | U^{\dagger}U={\bf 1}_{N\times N}, ~\det(M)=1\}\cr ~=~& e^{i~su(N)}.\end{align}$$ An important word in OP's quoted sentence is the word generate. One says that a Lie algebra generates a Lie group.

  • Note that a Lie algebra is a vector space, and hence has a basis. It is straightforward to calculate the dimension $$\dim_{\mathbb{R}} su(N)=N^2-1.$$

Qmechanic
  • 220,844