46

As indicated by Igor Rivin, the volume of the unitary group is given by
$vol(U(N))=(2\pi)^{(N^2+N)/2}/\prod_{k=1}^{N-1} k!$.
The denominator is the Barnes G-function, which is well-known :
http://en.wikipedia.org/wiki/Barnes_G-function
and in particular has a known Stirling-like asymptotic expansion for large $N$:
$\log(\prod_{k=1}^{N-1} k!) \sim$
$\...

42

A very good expository article (in Farsi) on recent work of Maryam Mirzakhani can be found here. (PDF)

42

If you add cuspidal curves, then $\overline{\mathcal{M}}_{1,1}$ will no longer be separated, which is the scheme/stack analogue of Hausdorff. Specifically, consider the families
$$y_1^2 = x_1^3 + t^6 \ \mbox{and}\ y_2^2 = x_2^3 + 1$$
(so the second family is a constant family with no $t$-dependence). For all nonzero $t$, they are isomorphic by the change of ...

answered Sep 25 '19 at 19:13

David E Speyer

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31

Riemann combines what is called Riemann-Roch and Riemann-Hurwitz nowadays.
He considers the dimension of the space of holomorphic maps of degree $d$ from the Riemann surface of genus $g$ to the sphere. He computes this dimension in two ways. By Riemann-Roch this dimension is
$2d-g+1$, for a fixed Riemann surface. (Indeed, Riemann-Roch says that
the dimension ...

answered Dec 1 '19 at 16:44

Alexandre Eremenko

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29

Suppose that $Br(X)$ is representable in the following sense: there exists a $k$-scheme $B_X$ such that for each $k$-scheme $S$ there is a natural bijection $B_X(S)=Br(X_S)$, or perhaps we should rigidify by asking for $B_X(S)=Br(X_S)/Br(S)$. In any case, since $B_X(k)\rightarrow B_X(l)$ is injective for all field extensions $k\rightarrow l$, we see that $Br(...

answered Oct 28 '14 at 23:08

Benjamin Antieau

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20

A formula for the number of isomorphism classes of curves over $\mathbf F_q$ is probably hopeless. As pointed out by Olivier Benoist and Qiaochu Yuan, the much more well behaved number is given by isomorphism classes weighted by their automorphism group, in other words, the groupoid cardinality of the groupoid $\mathcal M_g(\mathbf F_q)$:
$$ \# \mathcal M_g(\...

18

For standard universal properties, you need the scheme to behave well under base change, which in these cases would mean tensor products. Tensor product is right exact, so a quotient remain a quotient, not left exact, so a sub may not remain a sub.

17

Dan,
Although I'm no longer very active on MO, I thought I'd make a few comments, since your question is an interesting one (and you're not anonymous).
The paper of Looijenga referenced in Igor's answer would show that there are "algebro-geometric" representations of $\Gamma_g$ which don't factor through $Sp(2g,\mathbb{Z})$. In summary, he takes a finite ...

17

As Charles indicates, "the moduli stack of $G$-bundles on $E$" is not quite the right thing to consider, especially if you're not working over $\mathbf{C}$. This is for two (unrelated) reasons:
1) The geometric object $M_{G}$ that you associate to a group $G$ isn't something that you can access directly (at least by the construction I know): what you can ...

16

A bit of mastication of Katz-Mazur Theorem 13.7.6 and the surrounding text seems to yield the following description of the special fiber of $Y(p)$:
It is fundamentally $p+1$ copies of $\mathbb{P}^1$ (each with a nonempty finite set of punctures corresponding to cusps) all glued together at supersingular points.
The completed local ring at a $k$-rational ...

16

The Quillen-Drinfeld-Deligne-etc. philosopy should not be looked at as something too mysterious.
Namely, it reduces to the fact that if the set of objects one is interesting in the infinitesimal deformations of is not too wild, then it can be described in the form $f(v)+Q(v)=0$, where $f:V\to W$ is a linear function and $Q:V \to W$ is a quadratic function. ...

16

As the title to Mumford's famous paper "Toward an enumerative geometry..." suggests, knowing the cohomology / cycle theory of the moduli space of curves allows one to answer enumerative geometry questions for curves.
Here is an (very concrete) example that came up in real life for a student of mine. He had a family of genus 2 curves over $\mathbb{P}^2$ ...

15

It seems to me that this is not true and that a counterexample can be constructed as follows.
Take a double cover $\alpha \colon X \longrightarrow A$ of an abelian surface $A$, branched over a smooth divisor $B \in |2 L|$, with $L$ very ample. We have $$K_X=\alpha^* L, \quad \alpha_* \mathcal{\omega}_X = \mathcal{\omega}_A \oplus \omega_A (L),$$ hence $$K_X^...

answered May 12 '13 at 8:37

Francesco Polizzi

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15

I am answering your "later addon" only, although it seems actually to be a very different question than your original one.
This is perhaps one of the most misunderstood aspects of HoTT and particularly Univalence.
There is no squeezing of water.
Upon hearing that univalence "makes equivalent spaces equal to each other", it's natural to think that ...

answered Dec 7 '17 at 17:47

Mike Shulman

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15

The original paper of Riemann is his celebrated "Theorie der Abel'schen Functionen" in Crelle's Journal of 1854. This paper can be found online at https://www.maths.tcd.ie/pub/HistMath/People/Riemann/AbelFn/
There is an English translation of Riemann's Collected Papers (Kendrick Press, I believe).
Modern accounts can be found in several textbooks or ...

14

No. In fact more is true: the locus of all $n$-pointed curves of genus $g-1$ with a single elliptic tail $E$, such that $\mathrm{Aut}(E)=\mathbf Z/6$, has codimension two in $\overline M_{g,n}$ and consists of noncanonical singularities. This was famously determined by Harris and Mumford in their paper on the Kodaira dimension of the moduli space of curves (...

14

Here is an answer to your original question in the context of HoTT. An arbitrary map $f:X\to Y$ that isn't a fibration can't be viewed literally as a family of spaces varying continuously over $Y$, because the strict fibers don't vary continuously over $Y$; a path from $y_1$ to $y_2$ doesn't induce any sort of map from the fiber over $y_1$ to the fiber over ...

answered Dec 7 '17 at 17:54

Mike Shulman

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14

Hormander showed that there is a generic set of scalar linear PDE's that can be studied using general techniques, known as microlocal analysis. This can be linked to algebraic geometry as follows: Any scalar linear partial differential operator of order $k$ on an open set in $\mathbb{R}n$ can be written as
$$
Pu = \sum_{|\alpha|\le k} a^\alpha\partial_\...

14

First, when defining the stack you will have the issue that there are formal deformations of abelian varieties which do not extend to families of abelian varieties over any reduced scheme. These are the deformations that do not respect any polarization. (In the complex analytic world these correspond to deformations of complex tori) So unless you have some ...

13

This is a standard argument and there probably exists a reference but it's not hard once you rephrase it in terms of the period domain.
The moduli space of K3 surfaces is locally isomorphic to its period domain.
The period domain is an open subset of the vanishing locus of a quadratic polynomial in $\mathbb P^{21}(\mathbb C)$. See, for instance, Huybrechts' ...

12

As Francesco points out, the claim is false when $\dim X>1$. The question is discussed in
[B. Fantechi, R. Pardini, Automorphisms and moduli spaces of varieties with ample canonical class via deformations of abelian covers, Comm. Algebra 25 (1997), 1413-1441. math.AG/9410006] (see in particular Thm. 6.6).

12

It is certainly not true that $\mathcal X \to \mathcal X^H$ is a good moduli morphism, unless $H$ is linearly reductive, because when you push forward the cohomology of $H$ will come into play.
On the other hand $\mathcal X^H \to X$ is a good moduli space, because the pushforward $QCoh(\mathcal X^H) \to QCoh(X)$ can be factored as the pullback $QCoh(\...

12

Yes, Calabi-Yau manifolds have unobstructed deformations. This is due to Tian and Todorov; there is a nice algebraic proof in a paper by Kawamata, J. Algebraic Geom. 1 (1992), no. 2, 183–190.

11

Sure -- try the set of homomorphisms from $\pi_1^{\mathrm{et}}(E - O)$ to a fixed finite group G. This is a "non-abelian level structure" of the sort considered by de Jong and Pikaart
http://arxiv.org/abs/alg-geom/9501003
Rachel Davis, a 2013 Wisconsin Ph.D. working with Nigel Boston, wrote her thesis about this kind of stuff in the case of elliptic ...

11

If it was, $Y:=(X\times S)/(f\times g)$ would be isomorphic to $X\times F$. One way to see this is not the case is to look at 3-forms: $X\times F$ has no nonzero holomorphic 3-forms (because $H^0(F,\Omega ^2_F)=0$). On the other hand, let $\omega $ be the generator of $H^0(S,\Omega ^2_S)$; for any $\alpha \in H^0(X,\Omega ^1_X)$, the form $pr_1^*\alpha \...

11

As already said by Simon in his comment, this is a very vast topic.
Let us stick for simplicity to the case of smooth surfaces $S$ of general type: in this case it is well known that $h^0(S, \, T_S)=0$, hence $\textrm{Aut}(S)$ is a finite group and the moduli functor $\textrm{Def}_S$ is prorepresentable.
The existence of a quasiprojective coarse moduli ...

answered Dec 19 '14 at 9:18

Francesco Polizzi

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11

The volume of the unitary groups is standard (I recommend Lando, Zvonkin: Graphs on Surfaces and their applications, Cor. 3.5.2), it is:
$$\textrm{vol}~ U_n = \frac{(2\pi)^{(n^2+n)/2}}{\prod_{k=1}^{n-1} k!}.$$
The Euler characteristic of $\mathcal{M}_g$ is $$\chi_g = \frac{B_{2g}}{4g(g-1)},$$ as per Harer-Zagier.
Why the two sides of your formula should ...

answered Apr 25 '15 at 3:21

Igor Rivin

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11

Yes, this is a result of Michele Linch, 1974.

answered Oct 14 '16 at 19:59

Igor Rivin

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11

Question 1 (compare virtual fundamental cycles of different perfect obstruction theories on space underlying space): There is essentially no relation between $[X]_\varphi$ and $[X]_{\varphi'}$ for different perfect obstruction theories $\varphi:E^\bullet\to\mathbb L_X$ and $\varphi':E^{\prime\bullet}\to\mathbb L_X$. The "derived" structure on $X$ encoded ...

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