What is the fundamental group of the multiplicative group of the complex numbers $\mathbb{G}_m(\mathbb{C})$ with respect to the Zariski topology. More precisely, what are the homotopy classes of continuous loops $f:[0,1]\rightarrow \mathbb{G}_m(\mathbb{C})$ with a fixed base point?
-
What is $G_{m}(\mathbb{C})$? You mean $C^{*}$ or something different? – Bombyx mori Jan 02 '13 at 21:33
-
1That is not the correct definition of the fundamental group even in the topological sense (you want based homotopy and you haven't chosen a basepoint). $[0, 1]$ is contractible, so unbased homotopy classes of maps from $[0, 1]$ correspond to path components. – Qiaochu Yuan Jan 02 '13 at 22:39
-
Good point - corrected. – Bob G Jan 02 '13 at 23:59
1 Answers
The Zariski topology on $\mathbb{C}^{\ast}$ is the cofinite topology, so a map into $\mathbb{C}^{\ast}$ is continuous if and only if the preimage of every point is closed. (This is an extremely general class of maps.) If $b \in \mathbb{C}^{\ast}$ is a basepoint and $f, g : [0, 1] \to \mathbb{C}^{\ast}$ are two maps based at $b$, consider the homotopy
$$H(x, t) = \begin{cases} f(x) & \text{ if } t = 0 \\ g(x) & \text{ if } t = 1 \\ b & \text{ if } x = 0, 1 \\ \text{arb}(x, t) & \text{ otherwise} \end{cases}$$
where $\text{arb}(x, t)$ is an arbitrary bijection from whatever part of the domain hasn't already been covered to $\mathbb{C}^{\ast}$. Then the preimage $H^{-1}(y)$ of any point which is not $b$ is the union of a point and two closed sets, which is closed, and the preimage of $b$ is the union of four closed sets, so $H$ is continuous. ($H$ is not much of a homotopy, but then, the cofinite topology is not much of a topology.)
In other words, $\pi_1(\mathbb{C}^{\ast})$ is trivial.
But this is almost certainly not what you want. Taking the topological fundamental group with respect to the Zariski topology is not a good notion of fundamental group for varieties. You should be either taking the topological fundamental group with respect to the analytic topology or taking the étale fundamental group.
- 419,620
-
@Qianchu Yuan: I suspect he is talking about etale fundamental group. – Bombyx mori Jan 02 '13 at 23:21
-
@user: I don't think so. Look at the second sentence of the question. – Qiaochu Yuan Jan 02 '13 at 23:22
-
1@user32240 Also, it would be very strange to say "in Zariski topology" and then mean "but use the etale site." – Matt Jan 02 '13 at 23:23
-
@Matt: That's true. I was misled by my impression of Milne's book. – Bombyx mori Jan 03 '13 at 00:25