Mathematics

MatheinBoulomenos

12:01 AM

@Ted Since you're an expert on complex stuff, do you know a good reference to learn Hecke operators?

orbit-stabilizer

Question: Do you guys have personal websites?

Ted Shifrin

I don't know nothing about no Hecke operators. I mean, I've heard of 'em, but that's it.

They're more number theory :)

@orbit-stabilizer: I have my old one at UGA from being a professor for an eternity, but nothing personal.

MatheinBoulomenos

Well, all the complex analysis people at my uni are actually number theorists, so I guess I have skewed impression

orbit-stabilizer

@TedShifrin ah okay

Ted Shifrin

Yup, @Mathei, and I'm definitely not a complex analyst. But it's hard to do complex geometry/manifolds without knowing some single- and several complex variables.

MatheinBoulomenos

12:06 AM

I didn't mean to imply that you were, but I thought being a complex geometer still makes more of an expert on complex analysis than others

Actually, I'm happy that the exercises in complex analysis 2 don't match the lecture in that we're still doing some Riemann surfaces, although so far, it has been very basic

Ted Shifrin

Well, sure, than plenty of "others," but ... also way less than a lot of "others." :)

Leaky Nun

@TedShifrin for two days?

Ted Shifrin

Leaky, huh?

Leaky Nun

it's a rick and morty reference lol

Ted Shifrin

shrugs

MatheinBoulomenos

12:13 AM

We just globalized some local theorems from single complex variable like open mapping, Riemann's removable singularities, identity theorem and computed the meromorphic function on $\Bbb C \Bbb P^1$, so far

I guess it's better than nothing on Riemann surfaces

Ted Shifrin

LOL, but hardly anything interesting other than knowing what a chart is.

MatheinBoulomenos

yeah, I know :(

Simply Beautiful Art

Hi ted

MatheinBoulomenos

there are a lot of people in this course who have not seen abstract manifolds before

Simply Beautiful Art

Hi @AkivaWeinberger and @LeakyNun

Leaky Nun

12:14 AM

hi

@MatheinBoulomenos they have

everything is a manifold

Simply Beautiful Art

lol

Leaky Nun

you just have to repeat your definition manifold

(do people even know what "manifold" means in English lol)

it means twice, thrice, ...

MatheinBoulomenos

The ways in which this joke is unfunny are manifold

4

Ted Shifrin

I know, @Mathei. I always spent a day on that when i taught Riemann surfaces. I guess I probably had to spend a day on partitions of unity, too, because only people that had had graduate manifolds or undergraduate diff top knew those.

Argon

Is it correct to assert that there is one homomorphism $f: D_{9} \rightarrow \mathbb{Z}$ on the basis that $f(a^8)=8f(a)=0$ and $f(b^2) = 2f(b) = 0$?

Leaky Nun

12:18 AM

@Argon but $\Bbb Z$ has a non-trivial automorphism so if there is one then there must be another...

MatheinBoulomenos

@Argon yes, more generally, elements of finite order must be mapped to elements of finite order, of which $\Bbb Z$ has only one

Leaky Nun

never mind, it's the trivial one

what am I doing with group theory

Argon

Haha thanks guys

Leaky Nun

@MatheinBoulomenos if it was non-trivial must there be another?

There is another

►

MatheinBoulomenos

@LeakyNun yes, but that doesn't follow simply because of the fact that there is a nontrivial automorphism

Leaky Nun

12:20 AM

@MatheinBoulomenos hmm?

MatheinBoulomenos

it's the fact that there is a non-trivial automorphism which fixes only the identity element

otherwise you can't be sure that composing with the automorphism actually changes your embedding

Leaky Nun

oh, ok

is there a variation of this in nonsense?

MatheinBoulomenos

For example, consider the homomorphism from a cyclic group of order $2$ to the quaternion group $Q_8$. $Q_8$ has lots of automorphisms, but they all restrict to the identity on the unique subgroup of order 2

Semiclassical

@TedShifrin yet another neat thing about my surface, and this time it's an algebraic observation

MatheinBoulomenos

If you have two objects $A,B$ in a category $C$, then $\operatorname{Aut}(A)$ acts on $\operatorname{Hom}_C(A,B)$ from the left and $\operatorname{Aut}(B)$ acts on the right by composition (you can also let the whole endomorphism monoids act if you like monoid actions), moreover this two actions are compatible

but that's just a bunch of fancy words for saying that you can compose things

Leaky Nun

12:26 AM

alright

Semiclassical

$$\begin{vmatrix} 1 & z & x \\ z & 1 & y \\ x & y & 1 \end{vmatrix}=1+2xyz-x^2-y^2-z^2$$

Leaky Nun

@MatheinBoulomenos what is the f.g. of the horn torus?

Semiclassical

so the condition that $x^2+y^2+z^2=1+2xyz$ is equivalent to the vanishing of that determinant.

Leaky Nun

nvm it isn't a manifold

Semiclassical

moreover, if you take the surface to be $f=1+2xyz-x^2-y^2-z^2=0$

MatheinBoulomenos

12:27 AM

@LeakyNun fundamental groups make sense for general topological spaces

at least if they're path connected

Leaky Nun

what am I doing with my life if it isn't a manifold

then what is it?

@Semiclassical it looks like everytime I'm doing my torus you're doing your cone lol

Ted Shifrin

@Semiclassic: Well, that still doesn't settle things for the reason that I indicated. We really only care about the restriction of that quadratic form to the tangent space of your surface, not on all $\Bbb R^3$.

Semiclassical

then the hessian is $H=2\begin{pmatrix} -1 & z & x \\ z & -1 & y \\ y & x & -1\end{pmatrix}$

Alessandro Codenotti

What's a horn torus?

Ted Shifrin

Oh, wait ...

Leaky Nun

12:28 AM

@AlessandroCodenotti a torus with major radius = minor radius

i.e. innermost smallest "circle" becomes a point

Ted Shifrin

You're confusing me, @Semiclassic.

Semiclassical

so I can write that as $f=\det(2I+H/2)=0$ (if I'm not getting the algebra wrong)

Alessandro Codenotti

Its $\pi_1$ is $\Bbb Z$ then

Leaky Nun

hmm

Semiclassical

I'm not claiming anything re: convexity here, I"m just making an algebraic observation that I thought was cute

Ted Shifrin

12:30 AM

We started with $x^2+y^2+z^2-(1+2xyz)$?

Alessandro Codenotti

It's the same as a torus with a disk closing the hole of the donut

Semiclassical

yeah. though for these purposes I'm finding it better to flip the sign on that, i.e. $1+2xyz-x^2-y^2-z^2=0$

Leaky Nun

simple proof: in a torus, the (1,1)-knot and the Villarceau circle are homotopic

in the horn torus however, the Villarceau circle becomes a vertical circle

therefore we have ab=b, i.e. a=1

so it becomes $\Bbb Z$

Ted Shifrin

OK ... My comment still stands. We don't really want the whole determinant ... only the restriction to the tangent plane of the surface is relevant.

MatheinBoulomenos

I think it might actually be the free group on two generators

Semiclassical

12:31 AM

oh, sure

I'm just saying it's kinda neat

Leaky Nun

@MatheinBoulomenos which two?

the horizontal circle deforms to a point (the center)

MatheinBoulomenos

Oh yeah, nevermind

Alessandro Codenotti

I think it should be $\Bbb Z$, if I'm doing Seifert-Van Kampen right in my head

Ted Shifrin

There are all sorts of interesting happenstances with quadrics and cubics.

Eric Silva

@Semi have you showed it's convex yet?

Semiclassical

12:32 AM

not yet.

tbh I'm dragging my feet on that right now and I'm just playing around with equations

Eric Silva

its second fundamental form should be p easy to compute

Semiclassical

though there is one neat implication of the above

MatheinBoulomenos

Leaky Nun

deforms to a point in the center

Semiclassical

namely, the characteristic polynomial of $H$ at $\lambda=-4$ is $\det((-4)I-H)=\det(-2(2I+H/2))$. but I just noted above that the surface could be written as $f=\det(2I+H/2)=0$, so $\lambda=-4$ is an eigenvalue of $H$ for every point on $f=0$.

i dunno what I should conclude from that, but it's neat

Eric Silva

12:37 AM

something like convex iff the second fund. form is positive (negative) semi-definite is true

Semiclassical

kk

i'll have to check it out later

mostly I was just surprised to see $\lambda=-4$ as being an eigenvalue of $H$ on the entire surface.

back later

MatheinBoulomenos

@LeakyNun if you consider a standard torus, but with the hole in the middle filled by disk, then this deformation retracts to the horn torus. I find this filled torus easier for visualization/Seifert-Van-Kampen computation. You can set up Seifert-Van-Kampen with a small neighborhood contractible of the center of the disk as one open set and the torus + a punctured neighborhood of the center of the disk as you other open set.

The torus has fundamental group $\Bbb Z^2$, the intersection retracts to a circle which has fundamental group $\Bbb Z$, the contractible open set has trivial fundamental group, amalgamation in this case kills one generator of $\Bbb Z^2$, so we're left with $\Bbb Z$

Leaky Nun

ok

MatheinBoulomenos

I would draw pictures if I could

hope my description is clear enough

Alessandro Codenotti

@MatheinBoulomenos That's not actually needed, but if the space isn't path connected you need to specify at which point are the loops based

user76284

12:51 AM

Is there reason to believe that orthogonal matrix initialization (as opposed to say, He initialization) can improve performance for shallow (ReLU-activation) neural networks?

MatheinBoulomenos

@AlessandroCodenotti only if you're being pedantic

take any point in the intersection