What are the automorphisms of real Clifford algebras $Cl_{n,0}$? Of course, I'm interested in the case where they are not central simple.
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$\begingroup$ A detailed discussion that might be helpful can be found in these notes by A. Trautman. $\endgroup$– Igor KhavkineCommented Mar 10, 2015 at 0:05
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$\begingroup$ The notes referenced by @IgorKhavkine: Trautman - Clifford algebras and their representations. $\endgroup$– LSpiceCommented Jul 8, 2020 at 0:09
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2 Answers
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$C_{n,0}$ is either a full matrix algebra over $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$, or the direct sum of two such algebras that are isomorphic. The exact description depends on the residue class of $n$ (mod $8$) and can be found in textbooks or on wikipedia http://en.wikipedia.org/wiki/Classification_of_Clifford_algebras
$\mathrm{Mat}_{k \times k}(\mathbb{R})$ and $\mathrm{Mat}_{k \times k}(\mathbb{H})$ are central simple, so their automorphisms are all inner. (So the automorphism group is $\mathrm{PGL}_k(\mathbb{R})$ or $\mathrm{PGL}_k(\mathbb{H})$, respectively.)
$\mathrm{Mat}_{k \times k}(\mathbb{C})$ is not central simple over $\mathbb{R}$, so, in addition to the inner automorphisms, there is also complex conjugation. (So the group $\mathrm{PGL}_k(\mathbb{C})$ of inner automorphisms is a subgroup of index $2$ in $\mathrm{Aut} \bigl( \mathrm{Mat}_{k \times k}(\mathbb{C}) \bigr)$.)
An automorphism of $A \oplus A$ can act independently on the two summands. Since $A$ is simple, the only additional automorphism interchanges the two summands. (So $\mathrm{Aut}(A) \times \mathrm{Aut}(A)$ is a subgroup of index $2$ in $\mathrm{Aut}(A \oplus A)$.)
answered Mar 10, 2015 at 6:33
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$\begingroup$ That was exactly my thought. Do you know any reference of the second point, I mean that there are no more. $\endgroup$ Commented Mar 11, 2015 at 0:50
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$\begingroup$ This is a straightforward exercise. Let $\alpha \in \mathrm{Aut} \bigl( \mathrm{Mat}_{k \times k}(\mathbb{C}) \bigr)$. The center $\mathbb{C}$ must be invariant under $\alpha$, so the restriction of $\alpha$ to $\mathbb{C}$ is an $\mathbb{R}$-linear automorphism of $\mathbb{C}$. So it is either the identity or complex conjugation. If the restriction is the identity, then $\alpha$ is $\mathbb{C}$-linear, so it must be an inner automorphism. If the restriction is complex conjugation, then composing $\alpha$ with complex conjugation yields a $\mathbb{C}$-linear automorphism, which must be inner. $\endgroup$ Commented Mar 11, 2015 at 1:59
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$\begingroup$ Thanks. In general an automorphism of $A\oplus A$ does not act independently on the two summands. This happens since A is simple, I guess that's what you meant. $\endgroup$ Commented Mar 11, 2015 at 18:03
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$\begingroup$ Since $A$ is simple, the only idempotents in the center of $A \oplus A$ are $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$. Any automorphism $\alpha$ must either fix both $(0,1)$ and $(1,0)$, or interchange them. If $\alpha$ fixes the two nontrivial idempotents, then it fixes both $A \oplus \{0\}$ and $\{0\} \oplus A$ (setwise), so it is in $\mathrm{Aut}(A) \times \mathrm{Aut}(A)$. If $\alpha$ interchanges the two nontrivial idempotents, then the composition with the automorphism $\sigma(a,b) = (b,a)$ fixes the idempotents, and therefore the composition is in $\mathrm{Aut}(A) \times \mathrm{Aut}(A)$. $\endgroup$ Commented Mar 11, 2015 at 18:21
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$\begingroup$ I understood what you said, I was just pointing that maybe you could put "since A is simple" before. Another way to see that is that if $A\oplus A=A_1\oplus A_2$ then $A\oplus A=\alpha(A_1)\cap A_1\oplus\alpha(A_1)\cap A_2\oplus \alpha(A_2)\cap A_1\oplus \alpha(A_2)\cap A_2$. Anyway, Thanks a lot. $\endgroup$ Commented Mar 11, 2015 at 19:01
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I don't think that, strictly speaking, the question by the OP has been properly addressed. Although it is true that real Clifford algebras are isomorphic (through a non-canonical isomorphism of unital associative algebras) to some matrix algebra or direct sum thereof, Clifford algebras are more than just matrix algebras.
Let $(V,g)$ be a real regular quadratic space and let $Cl(V,g)$ be its associated Clifford algebra. There exists a canonical injection $i: V\hookrightarrow Cl(V,g)$ and Clifford algebras should be understood indeed as unital associative algebras $Cl(V,g)$ with a preferred choice of linear subspace $V\subset Cl(V,g)$. As such, automorphisms of $Cl(V,g)$ should be understood as automorphisms of the unital associative algebra $Cl(V,h)$ that preserve $V\subset Cl(V,h)$. With this provisos in mind, we would arrive to the following definition of automorphism group $Aut(Cl(V,g))$ of $Cl(V,g)$:
$Aut(Cl(V,g)) := \left\{ g\in AlgAut(Cl(V,g)) \,\, |\,\, g(V) = V\right\}$
where $AlgAut(Cl(V,h))$ denotes the automorphisms of $Cl(V,g)$ as a unital associative algebra. Note that since $V$ generates $Cl(V,g)$ elements of $Aut(Cl(V,g))$ preserve the $\mathbb{Z}_{2}$-grading of $Cl(V,g)$. Now, computing $Aut(Cl(M,g))$ is a much harder task, but it is the relevant object when one considers Clifford algebras in the context of spin geometry beyond spin groups.
