# Automorphism group of the special unitary group $SU(N)$

Let us consider the automorphism group of the special unitary group $$G=SU(N)$$.

We know there is an exact sequence: $$0 \to \text{Inn}(G) \to \text{Aut}(G) \to \text{Out}(G) \to 0.$$

For $$G=SU(2)$$, we have:

• $$\text{Z}(SU(2)) =\mathbb Z_2$$,
• $$\text{Inn}(SU(2)) = SO(3)$$,
• $$\text{Out}(SU(2)) = 0$$,

And so $$\text{Aut}(SU(2))=SO(3)$$.

For $$N > 2$$, we have:

• $$\text{Z}(SU(N)) =\mathbb Z_{N}$$,
• $$\text{Inn}(SU(N)) = PSU(N)$$,
• $$\text{Out}(SU(N)) = \mathbb Z_2$$.

My question is:

Does $$\text{Aut}(SU(N))=PSU(N) \times \mathbb Z_2$$? If not, does this answer depend on whether $$N$$ is odd or even?

It looks to me that there is a nontrivial fibration depending on something like $$H^2(B\mathbb Z_2,PSU(N))$$ due to $$B\text{Inn}(G) \to B\text{Aut}(G) \to B\text{Out}(G) \to B^2\text{Inn}(G) \to$$ and thus $$BPSU(N) \to B\text{Aut}(G) \to B\mathbb Z_2 \to B^2PSU(N) \to$$

But I do not know how to define $$H^2(B\mathbb Z_2,PSU(N))$$, if this is a correct thing to ponder.

• this may be preliminary for answering your question: mathoverflow.net/questions/40666/… - which you may already know well. – wonderich May 1 at 3:48
• In fact this completely answers the question. – abx May 1 at 4:05
• what is the answer? I am asking the total Aut of $SU(N)$? – annie heart May 1 at 4:46
• p.s. I am not asking Inn or Out of $SU(N)$. – annie heart May 1 at 4:51
• The answer is that it is a semi-direct product $\operatorname{PSU} (N)\rtimes \mathbb{Z}/2$, with $\mathbb{Z}/2$ acting on $\operatorname{PSU}(N)$ by conjugation. – abx May 1 at 5:58

## 1 Answer

Let $$G$$ be a compact simple simply connected Lie group. Then any automorphism of $$G$$ determines an automorphism of its Lie algebra $$\mathfrak{g}$$ and visa versa. So $$\mathrm{Aut}(G)$$ is naturally isomorphic to the linear group $$\mathrm{Aut}(\mathfrak{g})$$.

The sequence $$1\to \mathrm{Inn}(\mathfrak{g})\to \mathrm{Aut}(\mathfrak{g})\to \mathrm{Out}(\mathfrak{g})\to 1$$ is split.

Moreover, $$\mathrm{Out}(G)\cong\mathrm{Out}(\mathfrak{g})\cong \mathrm{Aut}(D_\mathfrak{g})$$ where $$D_\mathfrak{g}$$ is the Dynkin diagram of $$\mathfrak{g}$$.

The upshot is:

$$\mathrm{Aut}(G)\cong \mathrm{Aut}(\mathfrak{g})\cong \mathrm{Inn}(\mathfrak{g})\rtimes \mathrm{Out}(G)\cong \mathrm{Inn}(\mathfrak{g})\rtimes \mathrm{Aut}(D_\mathfrak{g}).$$

For types $$A_1, B_n, C_n, G_2, F_4, E_7, E_8$$ there are no symmetries of the Dynkin diagram. For $$A_n$$ ($$n>1$$), $$D_n$$ ($$n\not=4$$), and $$E_6$$, we have $$\mathrm{Aut}(D_\mathfrak{g})\cong \mathbb{Z}/2\mathbb{Z}$$. And in the final case of $$D_4$$, the symmetry group is the symmetric group on three letters.

In particular, as stated in the comments: $$\mathrm{Aut}(\mathrm{SU}(n))\cong\left\{\begin{array}{ll}\mathrm{PSU}(n)\rtimes \mathbb{Z}/2\mathbb{Z},&\text{ if } n\geq 3\\ \mathrm{PU}(2),&\text{ if }n=2. \end{array}\right.$$

• PSU(2) = PU(2), yes? – annie heart May 6 at 19:27
• Does it mean that the $\mathrm{Aut}(\mathrm{PU}(n))=\mathrm{Aut}(\mathrm{PSU}(n))=\mathrm{Aut}(\mathrm{SU}(n))$, since they have the same Lie algebra: $$\mathrm{Aut}(\mathrm{PSU}(n))\cong\left\{\begin{array}{ll}\mathrm{PSU}(n)\rtimes \mathbb{Z}/2\mathbb{Z},&\text{ if } n\geq 3\\ \mathrm{PU}(2),&\text{ if }n=2. \end{array}\right. ?$$ – annie heart May 6 at 19:35
• Do we have $$\mathrm{Out}(G)\cong\mathrm{Out}(\mathfrak{g})?$$ $$\mathrm{Inn}(G)\cong\mathrm{Inn}(\mathfrak{g})?$$ $$\mathrm{Aut}(G)\cong\mathrm{Aut}(\mathfrak{g})?$$ in general? – annie heart May 6 at 19:43
• $PU(n)\cong PSU(n)$ in general. For simply-connected Lie groups, $\mathrm{Aut}(G)\cong \mathrm{Aut}(\mathfrak{g})$ and $\mathrm{Out}(G)\cong \mathrm{Out}(\mathfrak{g})$.The same argument does not work for $\mathrm{Aut}(PG)$ since $PG$ is not generally simply-connected. The inner automorphisms of the Lie algebra are the image of $G$ via the adjoint. I think you need to generally figure out the kernel. – Sean Lawton May 6 at 21:08
• I am not sure. I think it is true for simple, simply-connected compact Lie groups. "Proof:" Since $G$ is simple, the Lie algebra of $\mathrm{Inn}(\mathfrak{g})$ is $\mathfrak{g}$ itself. Thus, since $G$ is simply-connected and compact, the abstract Lie group structure on $\mathrm{Inn}(\mathfrak{g})$ is a finite central quotient of $G$. Since the center is in the kernel, it must be $PG$. And $\mathrm{Inn}(G)\cong PG$ always. $\Box$ – Sean Lawton May 6 at 21:17