0

It's known that the set of meromorphic functions (functions to $\mathbb{C}\cup\{\infty\}$) on a complex variety $X$ forms a field, called the function field of $X$.

Edit: Thanks to the comment by @Qiaochu Yuan, I realize that $\mathbb{PC}^1(X)\neq$ the set of meromorphic functions, since we don't allow the constant function with constant value $\infty$ to be meromorphic.

So is it possible that, as a scheme over $\mathbb{C}$, the projective line $\mathbb{P}^1_\mathbb{C}$ is a group scheme?

What if $\mathbb{C}$ is replaced by a general field $K$ or even $\mathbb{Z}$?

(I have never seen an explicit projective group scheme of the form $\mathop{\mathrm{Proj}}S$ in any reference, i.e. stating the multiplication morphism. All explicit examples introducing the group schemes are the affine ones. Hence stating any projective group scheme of the form $\mathop{\mathrm{Proj}}S$ while stating the multiplication morphism in the comment or answer will be nice)

Z Wu
  • 1,721
  • 1
    What is the relevance of the first paragraph to your question? – Qiaochu Yuan Sep 14 '22 at 17:37
  • @QiaochuYuan It means that $\mathbb{CP}^1(X)$ is a field and in particular a group. Hence it is possible that $\mathbb{P}^1_\mathbb{C}(X)$ is a group when $X$ is considered as a scheme. – Z Wu Sep 14 '22 at 19:22
  • A field is not a group! It has a zero element. To form a field we also need to exclude the constant function with constant value $\infty$. – Qiaochu Yuan Sep 14 '22 at 19:23
  • @QiaochuYuan the addition part makes it a group. I just checked wiki again. – Z Wu Sep 14 '22 at 19:28
  • Then why mention that it's a field? In any case you still need to exclude the constant function with constant value $\infty$. So this is not a group-valued representable functor which I assume is what you had in mind. – Qiaochu Yuan Sep 14 '22 at 19:31
  • @QiaochuYuan Well, I didn't remember the part of excluding constant $\infty$ function, that's my bad. Now it looks like the first paragraph is indeed wrong. I will edit it now. – Z Wu Sep 14 '22 at 19:37
  • The argument had to be wrong, because if it were correct $\mathbb{CP}^1$ would have to be a group object! But at least now I understand the motivation you had in mind; to be clear I think this is a natural question to ask. – Qiaochu Yuan Sep 14 '22 at 19:50

1 Answers1

2

No. A stronger statement is true: $S^2 \cong \mathbb{CP}^1$ is not a topological group. There are several different ways to see this and you can check out e.g. this math.SE answer for details. Here it would suffice to prove the weaker claim that $S^2$ is not a Lie group, which can be done using the hairy ball theorem or by classifying $2$-dimensional Lie algebras. A general fact which implies this conclusion at-a-glance is that if a compact triangulable path-connected space admits a topological group structure then by the Lefschetz fixed point theorem it must have Euler characteristic $0$ (or be a point).

Edit: Actually for the desired statement it suffices to show that $\mathbb{CP}^1$ is not a complex Lie group which is even easier and can be done by classifying $1$-dimensional complex Lie algebras. There's only one, namely $\mathbb{C}$, so $\mathbb{C}$ is the unique simply connected $1$-dimensional complex Lie group.

Examples of projective group schemes are given by elliptic curves and more generally by abelian varieties. The group operation is a bit annoying to write down explicitly but very classical. Somewhat surprisingly they are all abelian, hence the name; see e.g. Lemma 39.9.5 in the Stacks Project.

In the complex analytic category elliptic curves are quotients $\mathbb{C}/\Gamma$ where $\Gamma$ is a lattice in $\mathbb{C}$, as they would have to be to be compatible with the above-mentioned argument about $1$-dimensional complex Lie groups and Lie algebras.

Qiaochu Yuan
  • 419,620
  • 1
    Thanks for the links. Maybe you want to add that the analytification functor respects terminal object and product hence group objects, so if $\mathbb{P}^1_{\mathbb{C}}$ is a group scheme, then its analytification $\mathbb{CP}^1=\mathbb{P}^1_{\mathbb{C}}(\mathbb{C})$ is a group object in the category of complex analytic spaces, and hence a topological group. Then it follows from your argument that it cann't be a topological group. I want to add that because I personally didn't know why it follows from your argument at first and so it would be nice to clear it out. – Z Wu Sep 14 '22 at 19:09
  • Yes, that's what I meant. Actually for this conclusion, as your argument suggests, it would suffice to show that $\mathbb{CP}^1$ can't be a complex Lie group; I've edited in details. – Qiaochu Yuan Sep 14 '22 at 19:14