$\DeclareMathOperator\Inn{Inn}\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Out{Out}\DeclareMathOperator\Pin{Pin}\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SO{SO}\DeclareMathOperator\PSO{PSO}$If I understand correctly, $\Aut\Spin(𝑛)$ automorphism of Spin groups splits as a semidirect product $$\Aut\Spin(𝑛) = \Out\Spin(𝑛) \ltimes \Inn\Spin(𝑛).$$ (This is not in general true for automorphism groups, although I am not sure about the situation for Lie groups specifically).

  • The inner automorphism group is given by $\Spin(𝑛)/𝑍(\Spin(𝑛))$, which equals the projective special orthogonal group $\PSO(𝑛):= \SO(𝑛)/Z(\SO(𝑛))$

  • The outer automorphism group is $\Out\Spin(𝑛)≅\mathbb{Z}_2$ when $n \geq 4$. When $n=3$, $\Out\Spin(𝑛)≅0$. When $n=8$, then the $\Out\Spin(𝑛)≅S_3$.

The definition of Pin group is given in https://en.wikipedia.org/wiki/Pin_group.

My question is that what is

  • the outer and inner automorphism group of $\Pin^{+}(n)$ and $\Pin^{-}(n)$ groups?

  • the outer and inner automorphism group of $\Pin^{+}(n,1)$ and $\Pin^{-}(n,1)$ groups?

  • $\begingroup$ Do you have a reference for the outer automorphism groups of Spin? $\endgroup$ Sep 18, 2021 at 12:49
  • $\begingroup$ I only know this ref: oxford.universitypressscholarship.com/view/10.1093/acprof:oso/… $\endgroup$ Sep 18, 2021 at 16:43
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    $\begingroup$ @KonradWaldorf: For a connected reductive group $G$ over an algebraically closed field $k$ (of char. 0?), the exact sequence $$1\to {\rm Inn\,} G\to {\rm Aut\,}G\to {\rm Out\,}\to 1$$ admits a splitting. Namely, any pinning of $G$ gives a splitting. See, e.g., Brian Conrad, Proposition 1.5.5. $\endgroup$ Sep 21, 2021 at 9:06
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    $\begingroup$ @KonradWaldorf: I think that the same holds also for connected compact groups over ${\Bbb R}$. For a proof in the special case of a simply connected simple compact $\Bbb R$-group, see Borovoi and Evenor, Lemma 4.1. $\endgroup$ Sep 21, 2021 at 9:19
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    $\begingroup$ @KonradWaldorf: The exact sequence in my comment above should read $$1\to{\rm Inn\,}G\to{\rm Aut\,}G\to{\rm Out\,}G\to 1.$$ $\endgroup$ Sep 21, 2021 at 9:27


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