2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Lord Shark the Unknown Aug 17 '18 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 '18 at 14:18
4
+50
$\begingroup$

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 $SU(n)$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.