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

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 $$ g_{\text{SU(2)}} = \exp\left(\theta\sum_{k=1}^{3} i t_k \frac{\sigma_k}{2}\right) $$ 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, $$ x g_{\text{SU(2)}} x^{-1}= \exp\left(\theta\sum_{k=1}^{3} (-i) t_k \frac{\sigma_k^T}{2}\right) \exp\left(\theta\sum_{k=1}^{3} (-i) t_k \frac{\sigma_k^*}{2}\right) =g_{\text{SU(2)}}^*. $$ where $$x=e^{i\frac{\pi }{2}\sigma_2} = i\sigma_2= \begin{pmatrix} 0&1\\ -1&0 \end{pmatrix} \in \text{SU(2)},$$

  • 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 $$ g_{\text{SU(n)}} = \exp\left(\theta\sum_{k=1}^{n^2-1} i t_k \frac{\lambda_k}{2}\right)??? $$

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 $n=3,4,5, etc$? $$ g_{\text{SU(n)}} \to g_{\text{SU(n)}}^* = x g_{\text{SU(n)}} x^{-1}? $$

  • $\begingroup$ An inner automorphism, by definition, is conjugation by an element of the group. So to find an inner automorphism of order $2$ just find some order $2$ elements of the group. $\endgroup$ Aug 17, 2018 at 4:39
  • $\begingroup$ The element I used for conjugation is $$x=e^{i\frac{\pi }{2}\sigma_2} = i\sigma_2= \begin{pmatrix} 0&1\\ -1&0 \end{pmatrix} \in \text{SU(2)},$$ which is in the order 2 ($\mathbb{Z}_4$) rather than the order 4 ($\mathbb{Z}_2$), because $x^4=1$. But it works. Any more comments? Thanks! $\endgroup$
    – wonderich
    Aug 17, 2018 at 14:18

1 Answer 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 $\exp(-i\overline{H}) = \exp(-i H^T)$. 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 $SU(n)$.

  • 1
    $\begingroup$ I dont see why "request for a matrix $x$ with $\overline{g} = x g x^{-1}$ doesn't make sense. " ---- the $x$ is not in the SU(n), but such an $x$ may still be possible in a larger group? $\endgroup$ Jan 8, 2019 at 18:00
  • $\begingroup$ Such $x$ isn't in $GL_n$ either. Of course, it is possible in some group: If $G$ is any group and $\alpha$ is an automorphism, then $\alpha$ becomes inner if we embed $G$ into $G \rtimes \mathbb{Z}$ where the generator of $\mathbb{Z}$ acts by $\alpha$. $\endgroup$ Jan 8, 2019 at 18:04
  • $\begingroup$ In this case, we could embed $SU(n)$ into $SU(n) \times SU(n)$ by $g \mapsto \left( \begin{smallmatrix} g & 0 \\ 0 & \overline{g} \end{smallmatrix} \right)$ and then take $x = \left( \begin{smallmatrix} 0 & \mathrm{Id}_n \\ \mathrm{Id}_n & 0 \\ \end{smallmatrix} \right)$. But I assume that isn't what is being asked for. $\endgroup$ Jan 8, 2019 at 18:06
  • $\begingroup$ I am interested in knowing that to offer the bounty, if there are more details -- I am all ear! Thank you! (I will check myself too) $\endgroup$ Jan 9, 2019 at 19:58
  • $\begingroup$ What does it mean by "Our in our particular example" in your sentence? -- thanks... $\endgroup$
    – wonderich
    Feb 16, 2019 at 19:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .