Timeline for If $A$ is the ring of continuous functions on a genus $g$ surface, can the genus of $X$ be seen by simple algebra in $A$?
Current License: CC BY-SA 3.0
26 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 31, 2016 at 19:01 | comment | added | Qiaochu Yuan | @Area: no, but they're related. For the correct statement look up the Hodge decomposition. On a compact Riemann surface $H^0(X, \mathcal{O}_X) \cong \mathbb{C}$ and $H^1(X, \mathcal{O}_X) \cong \mathbb{C}^g$ (not $2g$), and all higher cohomology vanishes. | |
| Jan 31, 2016 at 18:57 | comment | added | Elle Najt | @QiaochuYuan Thank you, that is a good point. Does sheaf of holomorphic functions on a projective Riemann surface have the same betti numbers as the Cech cohomology of the corresponding algebraic curve? I think I have heard something like this. So there is no good reason to suspect that the constant sheaf $\mathbb{C}$ is injective. | |
| Jan 31, 2016 at 18:44 | comment | added | Qiaochu Yuan | @Area: the sheaf of continuous functions to $\mathbb{C}$ is fine, so for reasonable $X$ that means it's acyclic (that is, all higher cohomology vanishes). Your argument doesn't seem to have enough resolution to distinguish between the sheaf of continuous functions and the sheaf of holomorphic functions. | |
| Jan 31, 2016 at 18:41 | comment | added | Elle Najt | What is unclear to me is why $H^1 (X, \mathbb{Z}) \to H^1 (X, \mathbb{C})$ is the zero map... $\mathbb{C}$ is divisible group, so an injective object in the category of abelian groups. Does it follow that the sheaf $\mathbb{C}$ is an injective object in the category of sheaves of abelian groups on $X$ (I'm concerned that non-uniqueness of the extension is an obstruction to gluing an extension) so it has no higher cohomology? | |
| Jan 31, 2016 at 18:41 | comment | added | Elle Najt | Thanks, very helpful! As for your exercise, sending $\lambda \to 0$ in $e^{\lambda f(z)}$ provides a nullhomotopy for exponentials, and conversely if $f(z)$ is in the connected component of the identity, then it induces trivial maps on $\pi_1$ and hence lifts to the universal cover of $\mathbb{C}^*$, i.e. is an exponent. | |
| Jan 31, 2016 at 17:59 | comment | added | Qiaochu Yuan | ...exact sequence in cohomology part of which goes $\dots \to H^0(X, \mathbb{C}) \to H^0(X, \mathbb{C}^{\times}) \to H^1(X, \mathbb{Z}) \to 0$. This shows that $H^1(X, \mathbb{Z})$ (once you believe that sheaf cohomology of the constant sheaf $\mathbb{Z}$ agrees with singular cohomology; let me assume $X$ is a reasonable space so this is true) is the quotient of the group of continuous functions $X \to \mathbb{C}^{\times}$ by the subgroup of exponentials of continuous functions $X \to \mathbb{C}$; this turns out to be the connected component of the identity (exercise). | |
| Jan 31, 2016 at 17:56 | comment | added | Qiaochu Yuan | @Area: on $C(X, \mathbb{C})$ the spectral radius and sup norm coincide. The group structure is induced from that of $S^1$. What I have in mind in general is that cohomology is represented by Eilenberg-MacLane spaces, and $S^1$ is the Eilenberg-MacLane space $B \mathbb{Z}$. In the particular case of $S^1$ you can also argue using the exponential sheaf sequence, as follows. There is a short exact sequence $0 \to \mathbb{Z} \to \mathbb{C} \to \mathbb{C}^{\times} \to 0$ of sheaves on $X$ (here each object stands for the sheaf of continuous functions into that object), which induces a long... | |
| Jan 31, 2016 at 9:20 | comment | added | Elle Najt | Is surjectivity easy? Where can I find a reference for this last fact? | |
| Jan 31, 2016 at 9:18 | comment | added | Elle Najt | The spectral radius topology induces the same topology on the group of units as the sup norm topology on $C(X,\mathbb{C}^*)$? So the connected components of the identity are those maps homotopic to a point. Is the group structure on the $[X,S^1]_+$ that induced from $S^1$? Some $f: X \to S^1$ induces a $\mathbb{Z} \to H^1(X)$, and $H^1(f) = 0$ implies $\pi_1(f) = 0$, so that via the contractible cover $f \cong 0$. $H^1(mult)(1) = (1,1)$ implies this is a homomorphism. It is not clear to me that this is a surjection in general. (For closed surfaces I see a cell structure that maybe shows it.) | |
| Jan 31, 2016 at 8:24 | comment | added | Qiaochu Yuan | @Area: the group of connected components of the group of units is the group of homotopy classes of maps $X \to \mathbb{C}^{\times}$. The latter space is homotopy equivalent to $S^1$, so equivalently we're talking about the group of homotopy classes of maps $X \to S^1$. And this is of course precisely $H^1(X, \mathbb{Z})$. | |
| Jan 31, 2016 at 8:21 | comment | added | Elle Najt | I just meant that the second edit doesn't distinguish between higher genus surfaces, because there are going to be maps from higher genus surfaces into $\mathbb{C}^*$ that don't lift via the cover $z \to z^2$ (no square root). The thought was that more could be seen from the continuous maps into the Lie group $(\mathbb{C}^*)^n$, by looking at covers like $(z_0, \ldots, z_n) \to (z_0^2, \ldots, z_n^2)$. But this doesn't seem to quite work, or at least the counting roots statement is more complicated. About your $K$-theory remarks: I don't know why the group of connected components is $H^1$. | |
| Jan 31, 2016 at 8:06 | comment | added | Qiaochu Yuan | @Area: I don't really understand your question; it seems to be at least two different questions at once, which seem worth separating. First, it sounds like you're worried that I haven't distinguished higher genus surfaces, but I have: this is covered in Edit #1. Second, there are lots of interesting things to say about, say, principal $G$-bundles and $G$-local systems on a surface but it's not obvious that this information is easily accessible starting from the ring of functions (although there are things to say here). | |
| Jan 31, 2016 at 8:04 | comment | added | Elle Najt | I guess the natural thing to do is to look at maps to $(\mathbb{C}^*)^n$. | |
| Jan 31, 2016 at 8:02 | comment | added | Elle Najt | Thanks for your answer! I will hopefully have a good excuse to some $K$ theory soon. I like the spirit of your second edit. Unfortunately, once we have a $\pi_1$ surjective map to $S^1$ from our surface I don't see how to get more information out of maps to $\mathbb{C}^*$ (in order to distinguish higher genus surfaces). Maybe a Lie group $G$ with higher rank $\pi_1$ and enough "algebraic" covers would provide access to finitely more information $\pi_1(S)$ via the group theory of $C(S,G)$? Or is the situation of the covers of $\mathbb{C}^*$ special? (In terms of their algebraic nature.) | |
| Jan 29, 2016 at 17:01 | comment | added | Qiaochu Yuan | Nice. I figured the argument would need to use compactness, which is used here to conclude that $w$ is bounded and so $|w|$ exists. | |
| Jan 29, 2016 at 16:56 | comment | added | David E Speyer | Choose $N$ large enough that $|w|/N < \pi$. Then $\exp(w/N)$ is disjoint from the negative real line, so $a \exp(w/N) + (1-a)$ is a unit for all $a \in [0,1]$. Thus $a v \exp((k+1) w/N) + (1-a) v \exp(k w/N)$ is a unit for all such $a$, and we can take a piecewise linear path through the points $v \exp(k w/N)$ for $0 \leq k \leq N$. | |
| Jan 29, 2016 at 16:50 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
deleted 26 characters in body
|
| Jan 29, 2016 at 16:42 | comment | added | David E Speyer | Oh, sorry. I am confused by what you mean by "straight line". You mean literally straight in the vector space. That sounds plausible but I need to think a bit about whether it is actually true. | |
| Jan 29, 2016 at 16:40 | comment | added | Qiaochu Yuan | @David: but not a straight line, right? Although maybe you're saying that path can easily be approximated by straight lines? (Which is the sort of argument I had in mind.) I wanted to avoid mentioning exponentials to keep things as algebraic as possible: talking about straight lines only requires a real vector space structure. | |
| Jan 29, 2016 at 16:38 | comment | added | David E Speyer | @QiaochuYuan Yes, you are right. Obviously, if two units are connected by a line of units, they induce the same map $H_1(X) \to H_1(\mathbb{C}^{\ast})$. Conversely, if $u$ and $v$ are maps $X \to \mathbb{C}^{\ast}$ inducing the same map on homology, then $u v^{-1}$ has a continuous logarithm $w$, and then $v \exp(a w)$ is a continuous path from one to the other, for $a \in [0,1]$. | |
| Jan 29, 2016 at 16:34 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
added 161 characters in body
|
| Jan 29, 2016 at 15:22 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
added 238 characters in body
|
| Jan 29, 2016 at 15:10 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
added 280 characters in body
|
| Jan 29, 2016 at 5:36 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
added 301 characters in body
|
| Jan 28, 2016 at 22:59 | comment | added | Qiaochu Yuan | @Arul: take the equivalence relation generated by the relation "two units are connected by a straight line consisting of units." I think this is the same as the same-component relation but I haven't thought about it in detail. | |
| Jan 28, 2016 at 22:48 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |