I would like to determine the automorphism group of the projective unitary group $G=PU(N)=PSU(N)$ and $G=SO(N)$. We also knew that $$ 0 \to \text{Inn}(G) \to \text{Aut}(G) \to \text{Out}(G) \to 0. $$

For $G=PU(2)=PSU(2)$, we have:

  • $\text{Inn}(PU(2)) = PU(2)$,
  • $\text{Out}(PU(2)) = 0$,
  • And so $\text{Aut}(PU(2))=PU(2)$.

For $N > 2$, we have:

  • the center $\text{Z}(PU(N)) =0$,
  • $\text{Inn}(PU(N)) = PU(N)$,
  • I am not quite sure that $\text{Out}(PU(N)) =0 $, $\mathbb Z_2$, or others? $\text{Out}(PU(N))=?$

  • I am not quite sure that $\text{Aut}(PU(N))= PU(N) $, or others ? $\text{Aut}(PU(N))= ?$

For example, $PU(4)=PSU(4)=SU(4)/\mathbb{Z}_4=Spin(6)/\mathbb{Z}_4=SO(6)/\mathbb{Z}_2,$ what will be $\text{Out}(PU(4))=?$ and $\text{Aut}(PU(4))=?$

  • I think that $\text{Out}(SO(N))=\left\{\begin{array}{l} 0, \text{ if $N$ is odd} \\ \mathbb{Z}_2, \text{ if $N$ is even} \end{array}\right. ?$

  • I suspect that $\text{Aut}(SO(N))= \left\{\begin{array}{l} SO(N), \text{ if $N$ is odd} \\ O(N), \text{ if $N$ is even} \end{array}\right. $ ?

Would you be able to answer these? Thank you.

Explicit answers need to be included in order to get accepted and get the bounty.

  • $\begingroup$ Outer automorphisms of Lie groups correspond to outer automorphisms of lie algebras, and outer automorphisms of simple lie algebras are the symmetries of the Dynkin diagram. $\endgroup$ – arctic tern May 7 '19 at 18:12
  • $\begingroup$ That I knew -- but do you mean that $\text{Out}(PU(2)) =0 $, and $\text{Out}(PU(N)) =\mathbb Z_2$ for $N>2$? (which is the case for $SU(N)$ for sure. $\endgroup$ – annie heart May 7 '19 at 19:00
  • $\begingroup$ I don't know how the outer automorphisms of a real lie algebra relates to its complexification. Assuming ${\rm Out}({\frak g})\to{\rm Out}(\mathfrak{g}_{\Bbb C})$ is onto, I assume this means ${\rm Out} PU(n)=\Bbb Z_2$ for all $n>2$. The outer automorphism is $\tau(A)=(A^{-1})^T$. When $n=2$, this is equivalent to conjugation by $\big(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix})$. $\endgroup$ – arctic tern May 9 '19 at 1:01
  • $\begingroup$ @arctic tern, bounty added $\endgroup$ – annie heart May 9 '19 at 18:18

@annie, since you like to get to the bottom of the answer, simply use the relation $$ 0 \to \text{Inn}(G) \to \text{Aut}(G) \to \text{Out}(G) \to 0. $$ and the fact about jdc share, and also: https://math.stackexchange.com/a/59919/79069

We can get: $\text{Inn}(SO(N))=SO(N). $

$\text{Out}(SO(N))=\left\{\begin{array}{l} 0, \text{ if $N$ is odd} \\ \mathbb{Z}_2, \text{ if $N$ is even} \end{array}\right. $

$\text{Aut}(SO(N))= \left\{\begin{array}{l} SO(N), \text{ if $N$ is odd} \\ O(N), \text{ if $N$ is even} \end{array}\right. $

$\text{Inn}(PU(N))=PU(N). $

$\text{Out}(PU(N))=\left\{\begin{array}{l} 0, \text{ if $N$ is odd} \\ \mathbb{Z}_2, \text{ if $N$ is even} \end{array}\right. $

$\text{Aut}(PU(N))= \left\{\begin{array}{l} PU(N), \text{ if $N$ is odd} \\ PU(N) \rtimes \mathbb{Z}_2, \text{ if $N$ is even} \end{array}\right. $

I hope this completely solve your puzzle.


I think most of the answer is in your MathOverflow post, which you should link.

Any automorphism of a Lie group $G$ induces a Lie algebra automorphism on the Lie algebra $\mathfrak g$. Moreover, a compact Lie group $G$ is generated by the image of a small neighborhood of the identity in $\mathfrak g$ under the exponential map, so that the tangent map $$\mathrm{Aut }\, G \to \mathrm{Aut}\, \mathfrak g$$ is injective. In fact, if $G$ is simply-connected, then this map is bijective, as mentioned in the comments to your MO question. I believe that you already know the automorphisms of the Lie algebra, so if $\pi\colon\tilde G \to G$ is the universal cover, you know $\mathrm{Aut}\,\tilde G$ as well.

From the injectivity of the tangent map, you thus expect to be able to identify $\mathrm{Aut }\, G$ with a subgroup of $\mathrm{Aut}\, \tilde G$ that has the same tangent information (and so, particularly, looks the same near the identity, $\pi$ being a finite-sheeted covering. If we write $K = \ker \pi$, so that $G = \tilde G/K$, then a (continuous) automorphism $\tilde \phi$ of $\tilde G$ will induce a well-defined automorphism $\phi$ of $G$ by $\phi(gK) := \tilde\phi(g)K$ if and only if $\tilde\phi(K) = K$. (Moreover, since $\pi$ is a local bijection near $1 \in \tilde G$, this is the only option.)

So assuming you understand $\mathrm{Aut}\, \mathfrak g \cong \mathrm{Aut}\, \tilde G$, what remains is to identify $\ker(\tilde G \to G)$ and to determine which automorphisms preserve it.

In the case of the special unitary group, the kernel is the center of $\mathrm{SU}(N)$ (which is the subgroup of diagonal matrices $\zeta \cdot I$ for $\zeta$ an $N^{\mathrm{th}}$ root of unity). But the center of a group is preserved by any automorphism.

The corresponding question for $\mathrm{SO}(2N)$ may be more interesting because the kernel of the covering map $\mathrm{Spin}(2N) \to \mathrm{SO}(2N)$ is not the whole center and so there is something left to check.

  • $\begingroup$ Thanks +1, but I hope to get a very precise answer for the precise PU(N) though. Out(𝑃𝑈(𝑁)) and Aut(𝑃𝑈(𝑁)). $\endgroup$ – annie heart May 10 '19 at 15:37
  • $\begingroup$ I hope we get to the bottom $\endgroup$ – annie heart May 10 '19 at 16:45
  • $\begingroup$ Annie, I could be mistaken, but I believe my penultimate paragraph gives an answer. To descend to an automorphism of PSU($N$), an automorphism of SU($N$) should need only to preserve the center, but all automorphisms must preserve the center, so I believe all should descend. $\endgroup$ – jdc May 10 '19 at 22:38
  • $\begingroup$ You need to include: Out(𝑃𝑈(𝑁)) = ? and Aut(𝑃𝑈(𝑁)) = ? and fill the question marks please. thanks $\endgroup$ – annie heart May 10 '19 at 23:09
  • $\begingroup$ I don't understand; I think the above is a complete answer. Do you disagree with some aspect of the reasoning? $\endgroup$ – jdc May 12 '19 at 0:42

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