Automorphisms of Clifford Algebras 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.  
 A: 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.
A: $C_{n,0}$ 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


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*$\mathrm{Mat}_{k \times k}(\mathbb{R})$ 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 $\mathrm{Aut}(A \oplus A)$.)
