45

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 $N$:
$\log(\prod_{k=1}^{N-1} k!) \sim$
$\...

41

First of all, let me recommend a book: J. Hubbard, Teichmüller theory, vol. 1.
Let me try to list briefly Teichmüller's own contribution to Teichmüller theory.
Bers's papers of 1960-s are good primary sources. The few papers of Teichmüller himself
that I read are also exciting, but my poor knowledge of German does not allow me to
read all of them.
Perhaps ...

answered Nov 28 '12 at 17:40

Alexandre Eremenko

60.5k66 gold badges169169 silver badges287287 bronze badges

41

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

40

The easiest way (I know) to see that there are no nonconstant holomorphic maps from a complete elliptic curve $E$ to the stack $M_g$ is to observe that such a map $f$ would lift to a holomorphic map of the universal covers $\tilde{f}: {\mathbb C} \to T_g$, where $T_g$ is the Teichmuller space. The latter is a bounded domain in ${\mathbb C}^{3g-3}$, so ...

40

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

122k1010 gold badges308308 silver badges585585 bronze badges

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

4,52811 gold badge2020 silver badges3434 bronze badges

27

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

60.5k66 gold badges169169 silver badges287287 bronze badges

22

There does not exist a map of a smooth complete genus 2 curve to $M_3$.
Such a map would give rise to a surface $S$ (of general type) which violates the Bogomolov-Miyaoka-Yau inequality $c_1(S)^2 \leq 3c_2(S)$. This inequality is equivalent to $3\sigma (S) \leq e(S)$ where $\sigma$ and $e$ are the signature and topological euler characteristic of the ...

20

First consider the case when $M_g$ is the stack:
Over $\mathbb C$ this is a consequence of the Torelli theorem: A map from a rational or elliptic curve to $M_g$ is the same as a smooth family over that curve. Then considering the period map gives a map from the same curve to the parameter space of Hodge structures. However, that is hyperbolic, so any ...

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(\...

19

I will give an answer, but first I would like to clarify the question. It seems to me that most commenters have misinterpreted the question. The question is not how people managed to construct different examples of moduli spaces before they had the tool of the language of functors. The question is the following: A moduli space is supposed to be a space whose ...

18

If $C$ is a smooth curve of genus $g$ and $f:C\to M_g$ is a non-constant morphism to the stack, then the total space of the induced family of curves $S\to C$ is a smooth surface and the underlying oriented 4-manifold has non-trivial signature: it is given by a multiple of the pullback of the first Chern class of the Hodge bundle on $M_g$, which is an ample ...

18

Stable sheaves are simple, i.e., $\textrm{End}E\simeq \mathbb{C}$. One thing that you want to avoid is the jumping of the automorphism group in a family.
A classical example is to consider a hyperelliptic curve $X$, and $[L]\in\textrm{Pic}^{g-1}X$.
If $\pi:X\to \mathbb{P}^1$ is the $g^1_2$, then Grothendieck-Riemann-Roch plus Riemann-Hurwitz tell you that
$...

17

All these deductions on non-existence of an elliptic curve as in the question from the fact that $M_g$ is hyperbolic in various senses were very interesting. Here is a preposterous variant. If such an elliptic curve exists, then passing to a finitely generated field and specializing any transcendental (if necessary) we can assume that the elliptic curve and ...

answered Jul 31 '12 at 13:13

Felipe Voloch

28.4k55 gold badges6969 silver badges134134 bronze badges

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

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.

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

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 ...

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

52.9k33 gold badges141141 silver badges226226 bronze badges

15

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. ...

15

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

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

Here is how the function field version of Shafarevich's conjecture (=Arakelov-Parshin Theorem) implies that there are no elliptic curves or (at most) twice punctured rational curves in $M_g$:
(See Noam's comment to Felipe's answer)
Suppose there exists a smooth non-isotrivial family $f:X\to C$ of curves of genus $g$ for some fixed $g>1$ parametrized by a ...

answered Jul 31 '12 at 23:58

Sándor Kovács

38.6k22 gold badges8989 silver badges131131 bronze badges

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

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_\...

13

"Il arrive que..." means "sometimes". So the paragraph says that sometimes the minimal model of $E_\eta$ over $\mathbf{C}[[t]]$ has bad reduction, which is true.
You're starting with an elliptic curve over $\mathbf{C}((t))$, not over $\mathbf{C}$. There's no reason that $E_\eta$ admits a level $n$ structure over $\mathbf{C}((t))$ (as opposed to over some ...

13

Your $F_N$ is the functor people would usually mean when they talk about the functor classifying elliptic curves with full level N structure (though it's a bit nicer if you replace $(Z/nZ )^2$ with $\mu_n \times Z/nZ$, so that the determinant takes values in $\mu_n$ on both sides.)
Wikipedia is not wrong. (Wikipedia is surprisingly seldom wrong!) Your F_2 ...

13

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

48.4k55 gold badges102102 silver badges258258 bronze badges

12

Yes, in the following sense. Pick a trivalent graph $G$ with $v$ vertices and regard it as the dual complex of the stable curve $E$ consisting of one copy of $\mathbb P^1$ for each vertex of $G$ and one node for each edge. The genus $g$ of $E$ is given by $2g-2=v$. The stack of stable curves of genus $g$ is irreducible, so any smooth curve $C$ of genus $g$ ...

Only top voted, non community-wiki answers of a minimum length are eligible

#### Related Tags

moduli-spaces × 673ag.algebraic-geometry × 532

algebraic-curves × 93

reference-request × 61

riemann-surfaces × 47

complex-geometry × 42

stacks × 41

vector-bundles × 40

dg.differential-geometry × 37

deformation-theory × 35

nt.number-theory × 34

coarse-moduli-spaces × 33

arithmetic-geometry × 32

elliptic-curves × 32

sg.symplectic-geometry × 30

abelian-varieties × 30

at.algebraic-topology × 26

modular-forms × 24

gromov-witten-theory × 23

hilbert-schemes × 19

birational-geometry × 17

teichmuller-theory × 17

gt.geometric-topology × 16

hyperbolic-geometry × 15

algebraic-stacks × 13