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I’m trying to visualize or understand intuitively what does the holomorphic endomorphism $[z] \mapsto [cz]$ from a complex torus $X=\mathbb{C}/\Lambda$ to itself look like (here we assume that $\Lambda=\mathbb{Z}+\tau \mathbb{Z}$ with $\operatorname{Im}(\tau)>0$, $c\Lambda \subseteq \Lambda$, and $[z]$ designs the equivalence class of $z$ in $X$).

Thanks to Riemann-Hurwitz, I know this map doesn’t have any ramification points, so it must be a topological covering map. Thanks to this answer to my previous question, I now understand why the degree of that map must be equal to $|c|^2$, but I’m still trying to wrap my head around how to visualize it.

For instance, if $c=1+i$ and $\Lambda=\mathbb{Z}[i]$, then the map has to have degree $2$, meaning the anti-image of any given point must consist of two points. But what would these two points be? I’m still thinking of that map as some kind of dilation and I don’t see how two points could be mapped to a single one in that case. Sorry if this is kind of a trivial question.

Gokimo
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Remember that two complex numbers represent the same point if they differ by an element of the lattice. Also, these maps have the happy advantage of being group homomorphisms, so you really only need to understand the pre-image of the origin to understand the pre-image of anything.

In your example with $c = 1+i$, I assume your lattice is the Gaussian integers themselves. As for the origin, note that $(1-i)/2$ is not $0$ (it is not a Gaussian integer), but that when you multiply it by $(1+i)$ you get $1$ which is in the lattice. So the origin itself and this point are the two pre-images of the origin. For any other point, its pre-images are given by taking one-preimage and adding $(1-i)/2$ to the other.

hunter
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  • Thank you, this is helpful! Is there also a way to visualize why the degree must be equal to $|c|^2$, or is that asking too much? I’m still trying to “convince” myself of that fact in an intuitive way. – Gokimo Apr 30 '23 at 13:57
  • Also yes, I was meaning $\Lambda =\mathbb{Z}[i]$ in that case. – Gokimo Apr 30 '23 at 13:57
  • @Gokimo if $c$ is an integer, this is straightforward to visualize-- take the fundamental domain, and mark the $c^2$ evenly spaced points of the form $a/c + ib/c$.

    If instead $c$ has some non-trivial imaginary component, there is some rotation in addition to stretching. I don't know a great visual proof for $|c^2|$ in this case although it's quick to write down the kernel of the map and verify how many points are in it. In your example, the point in the middle of the square is going to get rotated to the axis and then stretched until it hits the lattice.

     – hunter Apr 30 '23 at 14:34
  • @Gokimo here is a simple argument if $c$ is not an integer. It's sort of clear that $c$ and $\overline{c}$ should have the same size of kernel, since they only differ by a reflection that has no kernel. But $c\overline{c}$ is an integer, which has a kernel of clear size $|c|^4$, so the kernel of $c$ must have size $|c|^2$. – hunter Apr 30 '23 at 14:59
  • Thank you, that last argument was very useful. – Gokimo Apr 30 '23 at 15:43