Jtdylktuy

The goal is to write down explicit expressions of inner / outer automorphism of SU($n$), for $ngeq 2$.

We know that SU(2) has an SO(3) ($supseteq mathbb{Z}_2$)-inner automorphism,

while SU(n) has a $mathbb{Z}_2$-outer automorphism. For simply connected simple Lie groups, the outer automorphisms come from the automorphisms of the Dynkin diagram. See also the discussion in MO.

• For SU(2), we can write the group element as
  $$ g_{text{SU(2)}} = expleft(thetasum_{k=1}^{3} i t_k frac{sigma_k}{2}right) $$
  where $(t_1,t_2,t_3)$ forms a unit vector [effectively pointing in some direction on a unit 2-sphere $S^2$], and $sigma_k$ are Pauli matrices:
  begin{align}
  sigma_1 &=
  begin{pmatrix}
  0&1\
  1&0
  end{pmatrix} \
  sigma_2 &=
  begin{pmatrix}
  0&-i\
  i&0
  end{pmatrix} \
  sigma_3 &=
  begin{pmatrix}
  1&0\
  0&-1
  end{pmatrix} ,.
  end{align}
  Notice that any group element on $SU(2)$ can be parametrized by some $theta$ and $(t_1,t_2,t_3)$. Also $theta$ has a periodicity $[0,4 pi)$.

The inner automorphism is given by,
$$
x g_{text{SU(2)}} x^{-1}=
expleft(thetasum_{k=1}^{3} (-i) t_k frac{sigma_k^T}{2}right)
expleft(thetasum_{k=1}^{3} (-i) t_k frac{sigma_k^*}{2}right)
=g_{text{SU(2)}}^*.
$$
where
$$x=e^{ifrac{pi }{2}sigma_2} = isigma_2= begin{pmatrix}
0&1\
-1&0
end{pmatrix} in text{SU(2)},$$

• For SU($n$), $n>2$,

Do we have a simple expression of $g_{text{SU(n)}}$?

(It looks like the answer given here in ME by Anon is negative. But the Refs here Ref 1, Ref 2, Ref 3 writing down suggestive expressions
$$ g_{text{SU(n)}} = expleft(thetasum_{k=1}^{n^2-1} i t_k frac{lambda_k}{2}right)??? $$

So the outer automorphism of SU(n) simply sends $g_{text{SU(n)}}$ to its complex conjugation
$$
g_{text{SU(n)}} to g_{text{SU(n)}}^*?
$$

What is the explicit $x$ such that, for $n=3,4,5, etc$?
$$
g_{text{SU(n)}} to g_{text{SU(n)}}^* = x g_{text{SU(n)}} x^{-1}?
$$

A complaint first about notation: I learned to use the notation $g^{ast}$ for the complex conjugate transpose; you seem to be using it for just complex conjugate. To avoid this issue, I'll write $overline{g}$ for the complex conjugate of the matrix $g$.

The outer automorphism of $SU(n)$ is, indeed, $g mapsto overline{g}$. By the defining property of unitary matrices, this is also $g mapsto (g^T)^{-1}$. If $H$ is a Hermitian matrix, then this automorphism sends $exp(iH)$ to $exp(-ioverline{H}) = exp(-i H^T)$. Of course, you can express this formula in terms of any basis for the Hermitian matrices you like.

Once $n$ is at least $3$, the matrices $g$ and $overline{g}$ are generically not conjugate. Let the eigenvalues of $g$ be $exp(i theta_1)$, $exp(i theta_2)$, .., $exp(i theta_n)$. Then the eigenvalues of $overline{g}$ will be $exp(-i theta_1)$, $exp(-i theta_2)$, .., $exp(-i theta_n)$. For generic $(theta_1, ldots, theta_n)$ with $sum theta_j=0$, the second list of eigenvalues will not be a permutation of the first, so $g$ and $overline{g}$ are not conjugate within $SU(n)$ (or even $GL_n$). Indeed, this is the easiest way to see that the automorphism is outer. So your request for a matrix $x$ with $overline{g} = x g x^{-1}$ doesn't make sense.

Editing to incorporate comments about the equation $overline{g} = x g x^{-1}$: Since $overline{g}$ and $g$ have different eigenvalues, they cannot be conjugate in $GL_n$. Is there some larger group where this could be true?

For any group $G$ at all, and any automorphism $sigma$ of $G$, we can embed $G$ into $H$ such that $g$ and $sigma(g)$ become conjugate. Namely, take $H = mathbb{Z} ltimes G$ with $k in mathbb{Z}$ acting on $G$ by $sigma^k$. Then $(1,e) cdot (0,g) cdot (1,e)^{-1} = (0, sigma(g))$, where $e$ is the identity of $G$.

When $G$ is a subgroup of $GL_n$ and $sigma$ has finite order $m$, then we can even embed $(mathbb{Z}/m) ltimes G$ into $GL_{mn}$. All of our matrices will consist of $m$ blocks, each of which is $n times n$. We send $(0,g)$ to the block diagonal matrix
$$begin{bmatrix}
g & 0 & 0 & cdots & 0 \
0 & sigma(g) & 0 &cdots & 0 \
0 & 0 & sigma(g) &cdots & 0 \
& & & ddots & \
0 &0 &0 & cdots & sigma^{m-1}(g) \
end{bmatrix}$$
and send $(1,e)$ to
$$begin{bmatrix}
0 & mathrm{Id} & 0 & cdots & 0 & 0 \
0 & 0& mathrm{Id} &cdots & 0 &0 \
0 & 0 & 0 &cdots & 0 &0 \
& & & ddots &ddots & \
0 &0 &0 & cdots & 0 &mathrm{Id} \
mathrm{Id} &0 &0 & cdots & 0 &0 \
end{bmatrix}.$$

Our in our particular example, let $H$ be the subgroup of matrices in $GL_{2n}$ of the block forms $left[ begin{smallmatrix} g&0 \ 0 & overline{g} \ end{smallmatrix} right]$ and $left[ begin{smallmatrix} 0&g \ overline{g}&0 \ end{smallmatrix} right]$, with $g in SU(n)$. The matrices of the former kind form a subgroup isomorphic to $SU(n)$. Conjugation by $left[ begin{smallmatrix} 0&mathrm{Id} \ mathrm{Id}&0 \ end{smallmatrix} right]$
takes $left[ begin{smallmatrix} g&0 \ 0 & overline{g} \ end{smallmatrix} right]$ to $left[ begin{smallmatrix} overline{g}&0 \ 0 & g \ end{smallmatrix} right]$, meaning that it acts on the subgroup $SU(n)$ by complex conjugation. As I tried to indicate in the previous paragraphs though, all of this is general nonsense about how to write any automorphism of a group as conjugation in some larger group and doesn't have much to do with the structure of $SU(n)$.