# Automorphism group of the projective unitary group PU(N) and SO(N)

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.

• 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. May 7 '19 at 18:12
• 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. May 7 '19 at 19:00
• 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 $\big(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix})$. May 9 '19 at 1:01
• @arctic tern, bounty added May 9 '19 at 18:18

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.

• Thanks +1, but I hope to get a very precise answer for the precise PU(N) though. Out(𝑃𝑈(𝑁)) and Aut(𝑃𝑈(𝑁)). May 10 '19 at 15:37
• I hope we get to the bottom May 10 '19 at 16:45
• 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.
– jdc
May 10 '19 at 22:38
• You need to include: Out(𝑃𝑈(𝑁)) = ? and Aut(𝑃𝑈(𝑁)) = ? and fill the question marks please. thanks May 10 '19 at 23:09
• I don't understand; I think the above is a complete answer. Do you disagree with some aspect of the reasoning?
– jdc
May 12 '19 at 0:42

@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: $$\text{Inn}(SO(N))=SO(N).$$

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

There is an exception that $$\text{Out}(SO(8))=S_3.$$

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

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

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

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

I hope this completely solve your puzzle.

• Despite the edit this answer still seems incorrect, as $\mathrm{Out}(PU(N))=\mathbb Z_2$ for all $N\geqslant 3$ regardless of parity. See Bourbaki, Lie IX, §4, Exercise 3b which says: for every almost simple connected compact group $\mathrm G$ of type $\mathrm A_n$ $(n\geqslant2)$ the group $\mathrm{Aut(G)/Int(G)}$ is cyclic of order $2$. May 30 '20 at 11:16