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I thought I understood something, but now I don't think I do. Let $E/F$ be a separable quadratic extension with nontrivial automorphism $\tau$. Let $\Gamma_E = \operatorname{Gal}(E/F)$. Define a one-cocycle $c:\Gamma_E \rightarrow \operatorname{Aut}_E(\GL_{n,E})$ by $c(1_E) = 1$, and $c(\tau)$ the automorphism defined on $E$-points by $x \mapsto J^tx^{-1}J$, where

$$J = \textrm{antidiag}(1,...1)$$

$\Gamma_E$ acts on the scheme $\GL_{n,E}$ in the usual way (pointwise action). General theory tells us there must exist an algebraic group $G$ over $F$ and an isomorphism $G_E \rightarrow \GL_{n,E}$ which corresponds to this one-cocycle. If we identify $G(E) = G_E(E)$ with $\GL_{n,E}(E) = \GL_n(E)$, then $\Gamma$ acts on $G(E)$ by $\tau.x = J^t\overline{x}^{-1}J$, and

$$G(F) = \{ x \in \GL_n(E) : J^t\overline{x}^{-1}J = x \}$$

where the bar denotes the pointwise action of $\tau$. This $G$ is called the unitary group.

Now, I can base change $G$ to $\overline{F}$ and get an action of $\Gamma = \operatorname{Gal}(\overline{F}/F)$ on $G_{\overline{F}}$. Identifying $G(\overline{F})$ with $\GL_n(\overline{F})$, what exactly is this action? (and if possible, can I define $G$ from the very beginning from a one cocycle on $\operatorname{Gal}(\overline{F}/F)$?

I thought it would be $\sigma.x = x$ if $\sigma|_E =1$, and otherwise $\sigma.x = J^t\overline{x}^{-1}J$, where bar denotes pointwise action. But this doesn't seem to be correct, since then a map like

$$f:\mathbb G_a \rightarrow G_{\overline{F}}, x \mapsto \begin{pmatrix} 1 & x \\ & 1 \\ & & \ddots \\ & & & 1 \end{pmatrix}$$

is not defined over $E$ (clearly does not satisfy $\sigma \circ f \circ \sigma^{-1}$ for all $\sigma \in \Gamma$). But such a map should be defined over $E$ ($G$ is a split algebraic group over $E$). What am I doing wrong here?

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We have

$$G(\overline{F}) = \{ x \in GL_n(E \otimes \overline{F}) : \tau x = x \}$$

where the tensor product is over $F$ and the conjugation in the definition of $\tau$ acts on the left factor $E$ only and $\text{Gal}(\overline{F}/F)$ acts on the right factor $\overline{F}$ only. Since $E$ is a Galois extension we have

$$E \otimes \overline{F} \cong \prod_{g \in \Gamma_E} \overline{F}$$

where $\Gamma_E$ acts on the RHS by swapping the two factors and $\text{Gal}(\overline{F}/F)$ acts on the RHS as follows. Write $\varphi : \text{Gal}(\overline{F}/F) \to \Gamma_E$ for the natural quotient map. Then $\sigma$ sends $(x, y) \in \overline{F} \times \overline{F}$ to $(\sigma x, \sigma y)$ if $\varphi(\sigma) = 1$ and to $(\sigma y, \sigma x)$ otherwise. (This is the key step in the computation. Note that the fixed points of this action consist of pairs $(x, \tau x), x \in E$.)

Hence we can identify $GL_n(E \otimes \overline{F})$ with

$$GL_n(\overline{F} \times \overline{F}) \cong GL_n(\overline{F}) \times GL_n(\overline{F})$$

which gives

$$G(\overline{F}) = \{ (x, y) \in GL_n(\overline{F}) \times GL_n(\overline{F}) : y = J^t x^{-1} J \}$$

which can be identified with $GL_n(\overline{F})$ by projecting to either $x$ or $y$. Picking the projection to $x$, the action of $\text{Gal}(\overline{F}/F)$ is the following: $x \mapsto \sigma x$ if $\varphi(\sigma) = 1$, and $x \mapsto J^t (\sigma x)^{-1} J$ otherwise. (Note that your action has the wrong fixed points: the fixed points of the action should be $G(F)$.)

You can define $G$ from the very beginning with a $1$-cocycle on $\text{Gal}(\overline{F}/F)$ by pulling back along $\varphi$, and when you do you find that the Galois action above is given precisely by modifying the usual Galois action using this $1$-cocycle; this is one way to describe where $1$-cocycles come from.

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