# Automorphisms of the Weil restriction

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.

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 $$\mathbb P^1 \times \mathbb P^1$$?