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$SU(N)$ 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?

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    $\begingroup$ Pauli matrices generate the Lie algebra of $SU(2)$. The Lie group can be obtained from the Lie algebra by exponential mapping, $U=e^{i\vec{n}\cdot\vec{\sigma}}$. You do not have to explicitly include the identity since the exponential mapping takes care of that. For the counting, notice that there are $N(N-1)/2$ independent complex off-diagonal entries for a Hermitian matrix, which counts $N(N-1)$ since one complex number is really two real numbers. The diagonal entries of a Hermitian matrix are real, and there are $N-1$ of them (traceless), so in total we have $N(N-1)+N-1=N^2-1$. $\endgroup$ – Meng Cheng Apr 10 '15 at 13:13
  • $\begingroup$ Traceless is the easiest part: $1 = \det U = \det e^{iT} = e^{i tr T} $ so that $tr T=1$. $\endgroup$ – Valter Moretti May 15 '16 at 7:56
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Comments to the question (v2):

  1. 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 $SU(2)$.

  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 $$SU(N)~:=~\{U\in {\rm Mat}_{2\times 2}(\mathbb{C}) | U^{\dagger}U={\bf 1}_{N\times N}, ~\det(M)=1\} ~=~ e^{i~su(N)}.$$ The magic word in OP's quoted sentence is the word generate. One says that a Lie algebra generates a Lie group.

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

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