Unitary Matrices: $SU(n)$

Now for a different example. $U(n)$ is the group of unitary $n\times n$ matrices, i.e., complex matrices satifying $A^*A={\bf 1}$. An easy computation shows that $\vert\det(A)\vert=1$. $SU(n)$ is the subgroup for which the determinant is 1 (unimodular matrices). Unlike the situation with $O(n)$ and $SO(n)$, the dimensions of $U(n)$ and $SU(n)$ (as manifolds) differ by 1.

The Lie algebras of $U(n)$ and $SU(n)$ are denoted $u(n)$ and $su(n)$, respectively. Differentiating $A^*A={\bf 1}$ we conclude that $u(n)$ consists of anti-Hermitian matrices: $B^*=-B$. Note that $B$ is anti-Hermitian if and only if $iB$ is Hermitian.

Fact: if $A(0)={\bf 1}$, then $\left.\frac{d\det A(t)}{dt}\right\vert _{t=0}={\rm tr\ }A'(0)$ (where tr = trace). (Expanding by minors does the trick.) This makes one half of the following fact obvious: the Lie algebra for the Lie group of unimodular matrices consists of all the traceless matrices.

For the special case $SU(2)$, things work out very nicely. Since $\det(A)=1$, one can write down the components for $A^{-1}$ easily, and equating them to $A^*$, one concludes that $SU(2)$ consists of all matrices of the form:

$$\left[ \begin{array}{cc} a+id & c+ib \\ -c+ib & a-id ... ...ft[ \begin{array}{cc} i & 0 \\ 0 & -i \end{array}\right]$$

$$= a{\bf 1}+b{\bf i}+c{\bf j}+d{\bf k}, \quad a^2+b^2+c^2+d^2=1$$

defining i,j,kas the given $2\times 2$ matrices in $SU(2)$. Exercise: these four elements satisfy the multiplication table of the quaternions, so $SU(2)$ is isomorphic to the group of quaternions of norm 1. (The somewhat peculiar arrangement of $a,b,c,d$ in the displayed element of $SU(2)$ is dictated by convention.)

Next, an arbitary anti-Hermitian matrix looks like:

$$\left[ \begin{array}{cc} i(a+d) & c+ib \\ -c+ib & i(a-d... ...ft[ \begin{array}{cc} i & 0 \\ 0 & -i \end{array}\right]$$

$$= ai{\bf 1}+b{\bf i}+c{\bf j}+d{\bf k}$$

This is traceless if and only if $a=0$. So we have a canonical 1-1 correspondence between $su(2)$ and ${\bf R}^3$, and so also with $so(3)$: $b{\bf i}+c{\bf j}+d{\bf k} \leftrightarrow (b,c,d) \leftrightarrow b\mbox{$\bf\hat{x}$}+c\mbox{$\bf\hat{y}$}+d\mbox{$\bf\hat{z}$}$.

It turns out that this correspondence is a Lie algebra isomorphism. $SU(2)$ and $SO(3)$ are locally isomorphic, but not isomorphic-- as we will see next.

$SU(2)$ acts on $su(2)$ via the adjoint representation. But we have a 1-1 correspondence between $su(2)$ and ${\bf R}^3$, so we have a representation of $SU(2)$ in the group of real $3\times 3$ matrices. Let $A$ be an element of $SU(2)$ and $v=b{\bf i}+c{\bf j}+d{\bf k}$ be an element of $su(2)$. Note that $\det v = b^2+c^2+d^2$. Since the map $v\mapsto AvA^{-1}$ preserves determinants, it preserves norms when considered as acting on ${\bf R}^3$. So the adjoint representation maps $SU(2)$ into $O(3)$. Fact: it maps $SU(2)$ onto $SO(3)$.

Incidentally, you can see directly that $v\mapsto AvA^{-1}$ preserves anti-Hermiticity by writing it $v\mapsto AvA^*$.

The kernel of the adjoint representation for $SU(2)$ is its center, which clearly contains $\pm{\bf 1}$-- and in fact, consists of just those two elements. So we have a 2-1 mapping $SU(2) \rightarrow SO(3)$. Our double cover! I'll look at the topology of this in a moment.

Physicists prefer to work with the Pauli spin matrices instead of the quaternions. The Pauli matrices are just the Hermitian counterparts to i, j, and k:

${\bf i}=i\sigma_x$,     ${\bf j}=i\sigma_y$,      ${\bf k}=i\sigma_z$

They form a basis (with 1) for the vector space of Hermitian $2\times 2$ matrices:

$$\left[ \begin{array}{cc} a+d & b-ic \\ b+ic & a-d \end... ...ft[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]$$

$$= a{\bf 1}+b\sigma_x+c\sigma_y+d\sigma_z$$

$SU(2)$ acts on the space of traceless Hermitian $2\times 2$ matrices in the same way as on $su(2)$: $h\mapsto ghg^{-1}$.

© 2001 Michael Weiss

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