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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.

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    $\begingroup$ this may be preliminary for answering your question: mathoverflow.net/questions/40666/… - which you may already know well. $\endgroup$ – wonderich May 1 at 3:48
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    $\begingroup$ In fact this completely answers the question. $\endgroup$ – abx May 1 at 4:05
  • $\begingroup$ what is the answer? I am asking the total Aut of $SU(N)$? $\endgroup$ – annie heart May 1 at 4:46
  • $\begingroup$ p.s. I am not asking Inn or Out of $SU(N)$. $\endgroup$ – annie heart May 1 at 4:51
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    $\begingroup$ 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. $\endgroup$ – abx May 1 at 5:58
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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: $$\mathrm{Aut}(\mathrm{SU}(n))\cong\left\{\begin{array}{ll}\mathrm{PSU}(n)\rtimes \mathbb{Z}/2\mathbb{Z},&\text{ if } n\geq 3\\ \mathrm{PU}(2),&\text{ if }n=2. \end{array}\right.$$

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  • $\begingroup$ PSU(2) = PU(2), yes? $\endgroup$ – annie heart May 6 at 19:27
  • $\begingroup$ 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: $$\mathrm{Aut}(\mathrm{PSU}(n))\cong\left\{\begin{array}{ll}\mathrm{PSU}(n)\rtimes \mathbb{Z}/2\mathbb{Z},&\text{ if } n\geq 3\\ \mathrm{PU}(2),&\text{ if }n=2. \end{array}\right. ?$$ $\endgroup$ – annie heart May 6 at 19:35
  • $\begingroup$ 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? $\endgroup$ – annie heart May 6 at 19:43
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    $\begingroup$ $PU(n)\cong PSU(n)$ 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. $\endgroup$ – Sean Lawton May 6 at 21:08
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    $\begingroup$ 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. $\Box$ $\endgroup$ – Sean Lawton May 6 at 21:17

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