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

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39

From the point of view of physics, Fourier transforms are ubiquitous because they are expansions in eigenfunctions of the derivative operator - and the derivative operator is fundamental in many aspects. Just to give two examples: The derivative operator is the generator of translations (in space or time), and to learn about the natural world, it is crucial ...

38

Euclidean case
Using the formula for the tan of the half solid angle that Robin Houston quotes, and expressing everything in terms of edge lengths by using the cosine law to convert the dot products, I end up with the following linear relationship:
$$\frac{1}{P_i} = \alpha_E \cot\left(\frac{\Omega_i}{2}\right)+\beta_E$$
where:
$$\alpha_E = \frac{12 V}{...

30

This is not an answer to the question, but an experimental observation that suggests a sharper conjecture: it’s only written as an answer because I’d like to flesh it out a bit more than there’s room for in a comment.
The image of $e^{i\Omega}$ under the Möbius transformation $z\mapsto\frac{i-zi}{1+z}$ is $\tan(\frac\Omega2)$, so an equivalent statement of ...

29

Yes. I'll talk about why elliptic curve discrete log is harder than ordinary discrete log.
Suppose we have $g, h$ and want to find $n$ such that $g^n = h$.
The usual methods for solving the discrete logarithm problem in $\mathbb{Z} / p \mathbb{Z}$ (as opposed to on an elliptic curve) is to compute $g^k \mod p$ for lots of values $k$ and look for powers ...

answered Nov 22 '19 at 19:41

Adam P. Goucher

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29

Yes, ish. There is a community effort (https://github.com/ryankeleti/ega) to translate the EGA into English. I’m posting this now because we’ve just finished EGA I, and around 30% of the combined EGA 0.

28

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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26

Today (3 April 2020) his papers have been accepted for publication on RIMS journal.
https://www.nature.com/articles/d41586-020-00998-2

25

There's a distinction that I find striking but don't know how to formalize usefully or how to evaluate its importance: In algebraic geometry, moduli spaces get compactified, and this involves adding a relatively small set to the original space. Roughly speaking, the original space parametrizes some nice objects, and the compactification adds points "at ...

ag.algebraic-geometry dg.differential-geometry ct.category-theory gn.general-topology compactifications

answered Oct 14 '19 at 15:33

Andreas Blass

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25

Monstrous moonshine, the famous relationship between the
dimensions of the irreducible representations of the Monster group and the coefficients in the Fourier expansion of
the $j$-invariant. Borcherds' proof of monstrous moonshine uses the Goddard-Thorn
theorem, which comes out of string theory,
specifically the quantization of the bosonic string.

ag.algebraic-geometry dg.differential-geometry gr.group-theory rt.representation-theory string-theory

23

Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $\mathbb{C}^N$ as a closed complex submanifold.
For the "only if" direction, see Hörmander, An Introduction to Complex Analysis in Several variables, Theorem 5.3.9. For the converse, an argument is contained on pp 109-110 of Hörmander, ...

22

$G_1:=GL(2)(\cong SL(2)\rtimes \mathbb G_m)$ is isomorphic as a variety to $G_2:=SL(2) \times \mathbb G_m$ via the map
$$
A\mapsto \Big(\big(\begin{smallmatrix}\det(A)^{-1} & 0 \\ 0 & 1 \end{smallmatrix}\big)A , \det(A)\Big).
$$
We will show that $G_1$ and $G_2$ (which are both reductive) are not isomorphic as algebraic groups over fields of ...

answered Dec 23 '19 at 22:05

André Henriques

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22

As many people have said, the reason that people believe that factoring and discrete log in $\mathbb F_p^*$ are hard, and that discrete log in $E(\mathbb F_p)$ is even harder, is because a lot of smart people have worked on these problems, and we know the best algorithms they've come up with. ECDLP is delicate, there are certain elliptic curves on which ...

answered Nov 22 '19 at 23:16

Joe Silverman

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21

You don't need the Noetherianness hypothesis to talk about K-theory. But the definition you propose in your question is not suited for the most general case. From a notion of K-theory we want at least the following properties
K-theory of an affine scheme $\mathrm{Spec}\,R$ is given by the algebraic K-theory of projective $R$-modules in the sense of Quillen
...

20

Three vectors $v_1,v_2,v_3$ lie in a hyperplane $H:\alpha x+\beta y+\gamma z+\delta t=0$, in this plane we have $Q(v,v):=(\alpha x+\beta y)^2-(\gamma z+\delta t)^2=0,\forall v\in H$. Thus by polarization $Q(v,w)=\frac 14 (Q(v+w,v+w)-Q(v-w,v-w))=0$ for all $v,w\in H$ that yields a relation between columns of your matrix: if $v=(a,b,c,d), w=(A,B,C,D)$, then $Q(...

answered Oct 27 '19 at 6:45

Fedor Petrov

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20

For any list $p_i$ of primes, we can find such a curve over a number field that has reduction type $\Gamma_i$ at some prime lying over $p_i$.
We can iteratively blow up the strata of the stable graph stratification of $\overline{M}_g$ until they all have codimension $1$. Then by taking an intersection of general hyperplanes in the blow-up and applying ...

19

To add a representation theory perspective: if $G$ is a Lie group, and $f$ is a function (or more precisely a distribution) on $G$ then (under certain mild conditions on $f$ and $G$), the function $f$ is uniquely determined by its unitary matrix coefficients, i.e. the coefficients of the matrix $\rho(f)$ where $\rho:G\to GL_n$ goes over all isomorphism ...

19

In the general context, "regular" is a property of a scheme (or a ring, or local ring), and "smooth" is a property of a morphism of schemes.
"Regular" means exactly that at every point, the dimension of the (Zariski) tangent space is equal to the (Krull) dimension (of the local ring at that point).
A map $f: X \to Y$ is smooth if the fibers over geometric ...

18

I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $\mathrm{X}$ sheaf cohomology vanishes, $\mathrm{H}^n(\mathrm{X},-)=0$ for $n\geqslant 1$, and this property characterises Stein manifolds among complex manifolds. In the same way affine schemes are characterised among (nice) schemes by cohomology vanishing (this is a theorem of ...

18

I think the separated condition should be included, at least for purposes of sheaf cohomology. First let's consider the case where Y is a point. if X is a compact Hausdorff space, then sheaf cohomology over X commutes with filtered colimits. This is a nice property which fails if X is non-Hausdorff. For example, take X to be two copies of [0,1] glued ...

18

Let me give a solution for $n=2$, that can be easily generalized to all $n$.
Let us consider the morphism $$f \colon \mathbb{P}^2 \to \mathbb{P}^2, \quad f([x:y:z]) =[x^2:y^2:z^2].$$
This is a Galois cover, with Galois group isomorphic to the Klein group $\mathbb{Z}_2 \times \mathbb{Z}_2$ and ramified on the union of the three lines $x=0$, $y=0$, $z=0$, ...

answered Apr 17 '19 at 12:09

Francesco Polizzi

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18

It's inherently difficult to give a negative answer to a question like this, but here's a technical fact that pushes in that direction:
Let ZFC$_n$ be the subtheory of ZFC gotten by restricting Separation and Replacement to $\Sigma_n$ formulas. By the reflection principle,$^1$ for each $n$ the theory ZFC proves that there is an ordinal $\alpha_n$ such that $...

answered May 13 '19 at 17:26

Noah Schweber

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17

The work on analytic geometry is all joint with Dustin Clausen!
Your main question seems a little vague to me, but let me try to get at it by answering the subquestions. See also the discussion at the nCatCafe. Also, as David Corfield comments, much of this had been observed long before: https://nforum.ncatlab.org/discussion/5473/etale-site/?Focus=43431#...

17

A set of points that generate $E(\mathbb{Q})$ modulo torsion is given by
(1955516573881233507049678279 : -86467145649172260650105545143411861089140 : 1),
(49225691888888099223656060329/10201 : 67749663895993353685065159554645568700902610/1030301 : 1),
(61339810590192565389735634 : -440289331793622522908840423931186017125 : 1),
(...

17

Here is a locally Noetherian separated counterexample. I also give some motivation for this construction afterwards.
Definition. Let $Z$ be an infinite chain of affine lines: $Z = Z_1 \amalg_{p_1} Z_2 \amalg_{p_2} \ldots$, where $Z_i \cong \mathbf A^1_{\mathbf C}$ and $p_i$ is the point $1$ in $Z_i$ and the point $0$ in $Z_{i+1}$. Let $Y = \mathbf P^2_Z$, ...

answered Nov 13 '19 at 9:14

R. van Dobben de Bruyn

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17

One very important point, implicit in the comments by @zzy and @abx, is that valuative criteria allow to check a property (e.g. properness) of (say) an $S$-scheme $X$ directly in terms of the functor of points of $X$. This is of course especially convenient if $X$ is defined by this functor, classical examples being Picard, Quot or various moduli functors. ...

answered May 1 '19 at 7:48

Laurent Moret-Bailly

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17

Here is a counterexample. Fix a field $k$, and let $Y$ be built from two copies of the affine nodal curve $y^2=x^3+x^2$, glued together on the complement of the singular point. In other words $Y$ is a nodal curve with doubled singular point. Then $Y$ is not affine because it is not separated. However, there is a homeomorphism $\mathbb{A}^1\to Y$, which is ...

16

This is a twice updated version:
Thanks to @PiotrPstrągowski for pointing out some issues that require more care.
Some thoughts were in response to the original wording of the question. I left them here, because they might be interesting to some.
As @AndréHenriques points out this still doesn't work -- at least compared to the updated question. My initial ...

answered Dec 30 '19 at 19:44

Sándor Kovács

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16

Here goes the elementary proof of the claim by Robert Houston that the quadraples $(P_1^{-1},P_2^{-1},P_3^{-1},P_4^{-1})$ and $(\cot \frac{\Omega_1}2,\cot \frac{\Omega_2}2,\cot \frac{\Omega_3}2,\cot \frac{\Omega_4}2)$ are affinely equivalent. In the spherical case the first quadraple should be replaced to $\{\cot\frac{P_i}2\}$, in the hyperbolic case to ...

answered Aug 27 '19 at 21:53

Fedor Petrov

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16

I just take a quick opportunity to share what a prism is, and why it is called like that (as I learned from Lars Hesselholt). All the theory is developed relatively to a fixed prime $p \in \mathbb{N}$.
A prism is a couple $((A,\delta),I)$ where $A$ is a commutative ring with unity, $\delta \colon A \to A$ is a set theoretic map and $I \subseteq A$ is an ...

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