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comment Which bundles does the character variety parameterize?
You don't need to be careful about punctures unless you care about complex-analytical 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 dg-scheme
Apr
18
comment Finite group action on quasi-projective varieties
You don't need quasi-projectiveness. Being of finite type is enough: given any (irreducible) finite-type $X$ 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 sub-extensions.
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 $S^{n-1}$ 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?