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I know that inner forms of $GL(2)$ are quaternion algebras. However, I cannot find the proof myself.

First, since quaternion algebras are forms of $GL(2)$ by the usual embedding in matrices, they are automatically inner by Skolem-Noether theorem.

But how can I prove the converse, that is to say if I have a group "inner isomorphic" (i.e. by conjugation) to $GL(2)$ over an algebraic closure, then it is a quaternion algebra?

Thanks for any clue or reference!

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    $\begingroup$ $GL(2)$ is a multiplicative group, not an algebra. Maybe you mean the algebra of matrices instead of $GL(2)$. $\endgroup$
    – YCor
    Nov 16, 2017 at 9:05
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    $\begingroup$ More likely the OP means unit groups of quaternion algebras: any nontrivial inner form of $GL_2$ is $GL_1(B)$ for some quaternion algebra $B$. $\endgroup$
    – David Loeffler
    Nov 16, 2017 at 10:47
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    $\begingroup$ See Ch. III, section 1.4 of Serre's book Galois Cohomology (and one should speak of using a separable closure rather than an algebraic closure in case $k$ may not be perfect). Also, "inner form" entails using the action of $k_s$-points of the algebraic group quotient $G/Z_G =: G^{\rm{ad}}$ modulo the schematic center, so beyond the case when $Z_G$ is a split torus (as holds for ${\rm{GL}}_2$ but not ${\rm{SL}}_2$, for example) the action by $G^{\rm{ad}}(k_s)$ might not arise from the action of $G(k_s)$ by conjugation. Hence, the word "inner" shouldn't be taken too literally in general. $\endgroup$
    – nfdc23
    Nov 16, 2017 at 13:31

1 Answer 1

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If you have a group $G$ over $k$ such that $G_{\overline{k}}$ is isomorphic to $GL_2$ and, and each element of $\operatorname{Gal}(\overline{k}|k)$ acts on this $GL_2$ by the standard action times conjugation by some (unique up to scalars element), then we can use the same cocycle to form a twisting of the algebra $M_2(k)$, because $PGL_2$ acts by conjugation on $M_2$.

By Hilbert 90, the twisted form of $M_2(K)$ remains a $4$-dimensional vector space, with algebra structure. Any two-sided ideal must remain a two-sided ideal over $\overline{k}$, and similarly with central elements, so it is a (rank 4) central simple algebra, i.e. a quaternion algebra.

Because the group of units of this algebra is $GL_2$, and this restriction is compatible with the action of $PGL_2$ and Galois, so the multiplicative group of this algebra is a twisted form of $GL_2$.

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  • $\begingroup$ Isn't Hilbert 90 saying that $H^1(k,GL_n)=0$, i.e. there are no forms of $GL(2)$? I thought the correspondence you describe relates forms of $PGL_2$ and quaternion algebras. $\endgroup$
    – Matthias Wendt
    Nov 16, 2017 at 9:41
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    $\begingroup$ @MatthiasWendt The forms of $G$ are parametrised by $H^1(k, \operatorname{Aut} G)$, not by $H^1(k, G)$. So the fact that $H^1(k, GL_n) = \{1\}$ vanishes implies that vector spaces have no twisted forms. This is being used for $n = 4$ to deduce that any twist of the algebra $M_2(k)$ is isomorphic as a $k$-vector space to $k^4$ -- the twist only affects the algebra structure. $\endgroup$
    – David Loeffler
    Nov 16, 2017 at 10:45
  • $\begingroup$ Of course, sorry, I was being silly $\endgroup$
    – Matthias Wendt
    Nov 16, 2017 at 10:48

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