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Which bundles does the character variety parameterize?
You don't need to be careful about punctures unless you care about complexanalytical type data at the punctures. But (perhaps this is nitpicking), you do need to be careful about "instability". When $C$ is anything but a sphere with no punctures, the character variety parametrizes based maps $C\to BG$, i.e. principal $G$bundles with chosen trivialization at the base point. (For $GL_n$, this is equivalent to vector bundles: in general, there is a standard dictionary, see en.wikipedia.org/wiki/Frame_bundle). When $C$ has $\pi_2 \neq 0$ with $G$ not discrete, the two may not be the same 
Apr
22 
answered  Cotangent complex of certain dgscheme 
Apr
18 
comment 
Finite group action on quasiprojective varieties
You don't need quasiprojectiveness. Being of finite type is enough: given any (irreducible) finitetype $X$ with $G$ action, let $Y\subset X$ be a closed divisor such that the complement $X\setminus Y$ is affine. Then the complement $X\setminus G\cdot Y$ is an affine scheme, and Sean's answer applies. 
Apr
8 
accepted  What word can I use for a poset with equivalences 
Apr
8 
asked  What word can I use for a poset with equivalences 
Apr
8 
awarded  Nice Answer 
Apr
8 
comment 
Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
Let us continue this discussion in chat. 
Apr
8 
comment 
Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
Right. (I removed my earlier comment about algebraicity and edited the answer instead). Re your second question, if I remember correctly, the category of extensions of an algebraically closed field $k$ of transcendence degree 1 is equivalent to the category of closed curves with finite morphisms, and you can use that there are no nonconstant morphisms from $\mathbb{P}^1$ to a curve of higher genus. 
Apr
8 
revised 
Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
minor edit 
Apr
8 
revised 
Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
minor typo 
Apr
8 
comment 
Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
Answer edited to fix this. 
Apr
8 
revised 
Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
Fixed characterization of finite extensions 
Apr
8 
comment 
Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
Yeah, by "minimal" I mean in the to say that we impose the relation $K<L$ if there exists a map $K\to L$. (In which case $k(x) \cong k(x^2)$.) But you're absolutely right about there being more extensions with finite automorphism group. Instead, I think you can characterize finite extensions by the property that they have finitely many subextensions. 
Apr
8 
revised 
Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
fixed typo 
Apr
8 
answered  Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes? 
Apr
8 
comment 
Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
I'm not sure what you mean by your part 1. I think you can say something along these lines, but the statement that there is a map $K\to L$ if $L$ has higher transcendence degree is in general false. (There are no maps from an extension of $\mathbb{F}_p$ to the function field of $\mathbb{P}^1_{\mathbb{F}_p}.$) 
Mar
10 
comment 
Proof of the Belyi's theorem: where is the hypothesis really used?
note that this hypothesis is essential: a morphism ramified at 0, 1, and $\infty$ is determined by combinatorial data, and can be shown to be defined over $\overline{\mathbb{Q}}$ 
Feb
25 
awarded  Organizer 
Feb
22 
comment 
Existence of partitions of $S^{n1}$ with hypercubes
"isometries of R^n" isn't very clear. If you want isometric maps from the hypercube to the sphere, there are none because a sphere has nonzero curvature and the hypercube is flat. 
Feb
19 
answered  Is there a standard name for the following type of linear operator? 