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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.