Given a complex variety, say $\mathbb{P}^1_\mathbb{C}$, I want to compute $$\text{Aut}_\mathbb{R}(\mathbb{P}^1_\mathbb{C}\vert_{\mathbb{R}}).$$ Using the definition of Weil restriction, one can show that for a complex variety $X$, $X\vert_\mathbb{R}\times_\mathbb{R}\mathbb{C}=X\times_\mathbb{C}X$, so that $$\text{Hom}_\mathbb{R}(\mathbb{P}^1_\mathbb{C}\vert_{\mathbb{R}},\mathbb{P}^1_\mathbb{C}\vert_{\mathbb{R}}) \ = \ \text{Hom}_\mathbb{C}(\mathbb{P}^1_\mathbb{C}\vert_{\mathbb{R}}\times_\mathbb{R}\mathbb{C},\mathbb{P}^1_\mathbb{C})\ = \ \text{Hom}_\mathbb{C}(\mathbb{P}^1_\mathbb{C}\times_\mathbb{C}\mathbb{P}^1_\mathbb{C},\mathbb{P}^1_\mathbb{C}).$$ However, I don't know which maps on the right correspond to automorphisms, and how to compute the group of them.

Edit: GetOffTheInternet has noted that the above formulas are not quite correct, but the questions below still stand:

  1. What is the answer in this case? Are they just the complex automorphisms composed with a Galois action?
  2. What about for other complex curves? For instance, are there $168\cdot 2$ automorphisms of the Klein quartic over $\mathbb{R}$?

$$\text{}$$ The reason I care about this is to get useful examples of Galois descent of schemes. Indeed, the varieties over $\mathbb{R}$ which go to to $X$ upon tensoring with $\mathbb{C}$ biject with $$H^1(\text{Gal}\mathbb{C}/\mathbb{R}, \text{Aut}_\mathbb{R}(X_0))$$ where $X_0$ is a particular one going to $X$. In the affine case $\text{Aut}(X_0)$ is too ugly to deal with, so I'm looking at the next simplest case of projective curves.

Edit: I had forgotten the actual result; it is $H^1(\text{Gal}(\mathbb{C}/\mathbb{R}, \text{Aut}_\mathbb{C} X)$, where $\text{Gal}$ acts on $\text{Aut}_\mathbb{C} X=\text{Aut}_\mathbb{C}(X_0\otimes_\mathbb{R}\mathbb{C})$ by conjugation.

  • @GetOffTheInternet Thanks so much for this. Could you maybe add what $\sigma \otimes 1$ looks like in the $\mathbb{C}/\mathbb{R}$ and $\mathbb{P}^1$ case (I think I can see what happens on affine pieces, but I'm not convinced that it will glue correctly)? – Meow Nov 8 at 23:08
  • @getofftheinternet Thanks for your thoughtful answer. – Meow Nov 10 at 11:07
up vote 2 down vote accepted
+50

I do not really understand the following, I am not sure what you really mean by:

$X \times_\mathbb{R} \mathbb{C} = X \times_\mathbb{C} (\mathbb{C} \times_\mathbb{R} \mathbb{C}) = X \times_\mathbb{C} X$ for a complex variety $X$$\ldots$

Don't we want to have a Weil restriction somewhere?

Also, if $X$ is a complex variety, well we can regard $X$ as an $\mathbb{R}$-scheme and then we can form $X \times_\mathbb{R} \mathbb{C}$, but:

First, $\ldots \times_\mathbb{R} \mathbb{C}$ means $\ldots \times_{\text{Spec}\,\mathbb{R}} \text{Spec}\,\mathbb{C}$, let us do this for $X = \text{Spec}\,\mathbb{C}$:$$X \times_\mathbb{R} \mathbb{C} = \text{Spec}(\mathbb{C} \otimes_\mathbb{R} \mathbb{C}) = \text{Spec}(\mathbb{C} \times \mathbb{C}) = \text{Spec}\,\mathbb{C} \cup \text{Spec}\,\mathbb{C},$$and similarly for $X$. So we get a union, not a product, and this is surely not what we want.

Given a complex variety, say $\mathbb{P}^1_\mathbb{C}$, I want to compute $$\text{Aut}_\mathbb{R}(\mathbb{P}^1_\mathbb{C}\vert_{\mathbb{R}}).$$

$\text{}$1. What is the answer in this case? Are they just the complex automorphisms composed with a Galois action?

Yes. Here is an analysis for a more general situation. Let $K/k$ be a Galois field extension. Let $X$ be a $k$-variety---it is important to start with a variety over $k$.

Question. What is $\text{Aut}_kX_K$---also denoted $\text{Aut}_k(X_K/k)$ or even $\text{Aut}(X_K/k)$?

Answer. Let first $\sigma$ in $\text{Aut}_KX_K$. Then $\sigma$ defines an automorphism $\sigma^*$ of $\kappa(X_K)/k$---function field. we have $\sigma^*(K) = K$ because $K/k$ is Galois. We therefore have a restriction homomorphism$$\text{Aut}_kX_K \to \text{Gal}(K/k), \quad \sigma \mapsto \sigma^*|_K.$$This gives a short exact sequence$$1 \to \text{Aut}_KX_K \to \text{Aut}_kX_K \to \text{Gal}(K/k) \to 1.$$A short remark---we also have an isomorphism$$\text{Aut}(\text{Spec}\,K \to \text{Spec}\,k) \to \text{Gal}(K/k), \quad \sigma \to \sigma^*.$$Let in the next step $\sigma^*$ be the element of $\text{Gal}(K/k)$ corresponding to a $\sigma$ as shown. This sequence has this splitting$$\text{Gal}(K/k) \to \text{Aut}_kX_K, \quad \sigma^* \mapsto \text{Id}_X \times_{\text{Spec}\,k} \sigma$$---the latter is often denoted by $\text{Id}_X \otimes_k \sigma$. So we have realized $\text{Aut}_kX_K$ in a canonical way as a semidirect product of $\text{Gal}(K/k)$ by $\text{Aut}_KX_K$.

Note also the following. The splitting defines a decomposition as a direct product if and only if $\text{Aut}_KX_K = \text{Aut}_kX$---more precisely, if the canonical inclusion $\text{Aut}_kX \to \text{Aut}_KX_K$ is an isomorphism. This is, however, not the case for $\mathbb{P}^1_\mathbb{R}$ and $\mathbb{C}/\mathbb{R}$.

Could you maybe add what $\sigma \otimes 1$ looks like in the $\mathbb{C}/\mathbb{R}$ and $\mathbb{P}^1$ case (I think I can see what happens on affine pieces, but I'm not convinced it will glue correctly)?

It is easy to describe algebraically:

On $\mathbb{A}^1_\mathbb{C} = \text{Spec}\,\mathbb{C}[x]$ it is given by$$\mathbb{C}[x] \to \mathbb{C}[x], \quad \sum_i a_i x^i \mapsto \sum_i \sigma(a_i) x^i.$$If we cover $\mathbb{P}^1$ by the usual two charts, we see that this works.

Note, however, that the automorphism on $\mathbb{A}^1_\mathbb{C}/\mathbb{R}$ does not have a "usual" geometric meaning. In particular, big warning, we have to be very careful if we try to apply this to a usual $\mathbb{C}$-valued point.

In fact, such a point $P$ is---by definition---a morphism$$\text{Spec}\,\mathbb{C} \to \mathbb{A}^1_\mathbb{C}$$over $\text{Spec}\,\mathbb{C}$, and to apply a morphism $\phi$ to it means to compose the two. So by definition$$(\text{Id} \times \sigma)(P) = (\text{Id} \times \sigma) \circ P.$$And conjugation $\circ$ $P$ is not even a $\mathbb{C}$-valued point!

However, we can still apply $\sigma$ to a point in a different way, namely by "conjugation"---has nothing to do with complex conjugation, it is just a coincidence that this is also around here.

Here we have by definition$$\sigma(P) = (\text{Id} \times \sigma) \circ P \circ \sigma^{-1}.$$This is again a $\mathbb{C}$-valued point. And it is the point we expect it to be if we "just apply $\sigma$ to the coordinates".

But, again warning, if we have any $\tau$ in $\text{Aut}_\mathbb{R}\mathbb{P}^1_\mathbb{C}$, we might not be able to apply it to a $\mathbb{C}$-valued point in any reasonable way.

Update: Let me try a second explanation of the original question. The general setting is that the real variety $X$ is the Weil restriction of a complex variety $Y$. Our wish seems to relate complex automorphisms of $Y$ and real automorphisms of $X$. This is a good start since by functoriality, a $\mathbb{C}$-automorphism of $Y$ gives rise to a real automorphism of $X$. The question is to see in which extent this provides everything.

The idea is to use Galois descent and to determine firstly $\mathbb{C}$-automorphisms of$$X_\mathbb{C} \cong Y \times_\mathbb{R} \overline{Y}$$and then to detect those arising from $\mathbb{R}$-automorphisms. The situation depends then much of the $\mathbb{C}$-variety $Y$.

For example if $Y$ is the affine line, then $X$ is isomorphic to the affine space and then $\text{Aut}\,X$ is much bigger than $\text{Aut}\,Y_\mathbb{C}$.

Our case is when $Y$ is the projective line. Then$$X_\mathbb{C} \cong Y \times Y$$and $\text{Aut}\,X_\mathbb{C}$ is the semidirect product of $\mathbb{Z}/2\mathbb{Z}$ by $\text{PGL}_2(\mathbb{C}) \times \text{PGL}_2(\mathbb{C})$, see the link below. With that we can conclude that$$\text{Aut}\,X= \text{PGL}_2(\mathbb{C}) \times \text{PGL}_2(\mathbb{C}) \rtimes \mathbb{Z}/2\mathbb{Z}$$where the action is the complex conjugation.

What is the automorphism group of $\mathbb P^1 \times \mathbb P^1$?

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