Explicit expressions of inner / outer automorphism of special unitary group SU(n)

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

• 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. – Lord Shark the Unknown Aug 17 '18 at 4:39
• 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! – wonderich Aug 17 '18 at 14:18

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 $$SU(n)$$.

• 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? – annie heart Jan 8 at 18:00
• 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 $\alpha$. – David E Speyer Jan 8 at 18:04
• 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. – David E Speyer Jan 8 at 18:06
• 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) – annie heart Jan 9 at 19:58
• What does it mean by "Our in our particular example" in your sentence? -- thanks... – wonderich Feb 16 at 19:19