The tutoring thing isn't happening :/

I was only gonna be available for a month, and they found someone who could be there long term

REPLACED

free market strikes again

Probably a wise choice on their part, DogAteMy.

Let $X,Y$ be two random variables and let $f$ be continuous and differentiable then does there always exist a random variable $C$ such that

$$f(X) - f(Y) = f'(C) (X-Y)$$

(a mean value theorem for random variables)

Adriana opens a savings account with an initial deposit of \$1000. The annual rate is 6%, compounded continuously. Adriana pledges that each year, her annual deposit (continuous) will exceed that of the previous year by $500. How much will be in the account at the end of the tenth year?

I did $D' = D + 500, D(0) = 1000$ to find the annual deposit $D = 1500e^t - 500$, then $P' = \frac{3}{50}P + 1500e^t - 500, P(0) = 1000$, but then I get $P(10) \approx \$35\ million$. This seems too high, what am I doing wrong?

Is there a closed form for the Maclaurin series of $\log \zeta(s)(s-1)$?

For example $\gcd(5,25)=5$ are all in the magma, but $5^2\neq5$

*all elements of $\gcd(5,25)=5$ are in the magma

$5^2=\gcd(5,5)=5$

oh right, the binary algebraic operation is $\gcd$!

r/unexpectedfactorial

mornin' y'all

morning

Morning

Mornin' dawg

wuz goood

nothin' much. Wuss poppin'

Does anyone know a good source to learn proofs of Fermat's last theorem

For n very small, not the general case lol.

There's Kummer's proof for regular primes

n=4 is elementary, it's possible to do that case by the theory of Pythagoreen triples

I'm looking at Kummer's proof. They mention Galois group but I'm guessing it's not necessary to know what that is.

@J.Doe Murty/Esmond - Problems in Algebraic Number Theory has n=3 and n=4 at least

Learning some algebraic number theory would be very helpful for understanding Kummer's proof. Kummer's proof basically spawned ANT

@J. Doe You might find this interesting: pearl.plymouth.ac.uk/bitstream/handle/10026.1/14185/…

;)

shameless plug

Thanks guys. I'll check those out.

@MatheinBoulomenos ... and the full proof?

scienzamedia.uniroma2.it/~eal/Wiles-Fermat.pdf

@ÍgjøgnumMeg If I learn what's contained in that paper, I'm gonna start charging people just for talking to me. :]

lol

@MatheinBoulomenos so "fourth power" in Latin is squared-squared

indeed

Anyone member of OEIS ?

I consider joining and it asks for some personal information, but there's members who just has their name. Did they join before this requirement or is that enough?

Does a primitive $n$-th root of unity always exist? More precisely, given a field $F$ and the polynomial $x^n - 1$ over $F$, we know there exists a field extension $E$ in which $x^n-1$ splits (i.e., $E$ is the splitting field). So, there exist $x_1,...,x_n \in E$ (not necessarily distinct) such that $x_i^n - 1 = 0$ for $i=1,...,n$. My question is, is it true that one of the $x_i$ is a primitive $n$-th root of unity?

@user193319 assuming the characteristic of $F$ doesn't divide $n$, yes

If $F$ has characteristic $p$ then every $np$-root of unity is also a $n$-root of unity

Just because $x^{np}-1=(x^n-1)^p$ in characteristic $p$

the reason is that every finite subgroup of the multiplicative group of a field is cyclic, so the group of $n$th roots of unity in $E$ is a cyclic group and a generator of that group will be a primitive $n$th root of unity

@J.Doe if you square that congruence, you get that $a^{2^n} \equiv 1 \pmod{2^n+1}$, so the order of $a$ is a divisor of $2^n$. Since all proper divisors are of the form $2^k$ for $k<n$ and hence divide $2^{n-1}$, we only need to check that $a^{2^{n-1}} \neq 1 \pmod{2^n+1}$ which is clear since it is $-1$

@MatheinBoulomenos So $G = \langle x_1,...,x_n \rangle$ is a finite subgroup of $E^\times$?

@user193319 yes. What happens if you multiply two $n$-th roots of unity?

@J.Doe the result you quoted works, too

@MatheinBoulomenos Oh, very nice! Thanks!

you don't need to take the group generated by $x_1, \dots, x_n$. The set of all $n$th roots unity itself is a subgroup

Ah, you get another one, and since there can be at most $n$, the product equals $x_i$ for some $i=1,...,n$.

The result I quoted I've been trying to prove it, so using it was cheating. Actually your solution gives me some insight as to how to prove it I think.

@MatheinBoulomenos How does a normal person define tangent bundle to a smooth scheme? I explicitly want the tangent bundle to be a scheme as well, not the dual of the sheaf $\mathcal{I}/\mathcal{I}^2$ (pulled back to the $X$) where $\mathcal{I}$ is the ideal sheaf to the diagonal $\Delta \subset X \times X$. I also don't want functor of points business, $\text{Hom}_{Sch}(\mathrm{Spec} k[\varepsilon]/(\varepsilon^2), X)$ stuff.

@BalarkaSen what's wrong with dual of I/I^2?

It's not a scheme, it's a sheaf.

I remember asking this question to my AG professor

@BalarkaSen wait is the tangent bundle a scheme?

yeah it should be

I think you modulo details can take the relative Spec of that sheaf

@LeakyNun Tangent bundle to a variety is a variety, so it better fucking be

wait I know that the cotangent bundle is a scheme

I really don't remember any AG :/

also there's another construction apart from I/I^2

@Alessandro Hm I see, what's the relative Spec

you can define it on the affine opens, and use the Kahler differentials

@LeakyNun Sure, but I am sure Grothendieck wouldn't like your definition ;)

I want some construction which is non-local, but like an actual scheme.

what's the tangent bundle of $V(X^2-Y^3) \subseteq \Bbb A^2$?

That's a singular dude. I said smooth scheme.

oh

It's tangent bundle is meaningful, but in a stratified sense.

The relative spec takes a quasi-coherent sheaf of $\mathcal O_X$-algebras and spits out a scheme over $X$ with some nice universal property iirc

The Zariski tangent space at $(0, 0)$ is $2$ dimensional

so it's like a blow-up?

@Alessandro Oh shit, that might be what I need then.

@MatheinBoulomenos I don't quite see why we have to assume the characteristic doesn't divide $n$ in order for the statement to hold.

@LeakyNun No. Just take $T\Bbb A^2$ and look at the union of the Zariski tangent spaces to each point in $x^2 - y^3 = 0$.

The subspace you get is the Zariski tangent bundle to that variety.

and is it a variety?

It gives an equivalence of categories between schemes over X and quasi coherent $\mathcal O_x$-algebras I think, it's called relative Spec because that's similar to what the usual Spec does

@LeakyNun Yes.

if you have a quasi-coherent $\mathcal{O}_X$-algebra $A$, then you can glue the spectra $\mathrm{Spec}(\Gamma(U,A))$ where $U$ ranges over affine-opens of $X$

that's the relative spec

and the relative spec comes with a morphism $\mathrm{Spec}_S(A) \to S$

mmm $\operatorname{Spec}k$

@MatheinBoulomenos That's pretty cool. Do you know a reference where I can read about it?

It's in Vakil most likely

Gotcha, I'll have a look.

I think smoothness of $X$ over $S$ should make $\Omega_{X/S}$ locally free, so that you actually get a bundle

@LeakyNun It's all pairs $(x, v) \in T\Bbb A^2 = \Bbb A^2 \times \Bbb A^2$ such that $x \in X = V(f)$ and $df_x(v) = 0$, isn't it? Those are the algebraic equations that cuts out the affine Zariski tangent bundle, irregardless of whether or not $X$ is smooth.

$\Omega_{X/S}$ is always quasi-coherent, but not locally free if you don't assume smoothness I think

so maybe you want schemes of finite type

Ah I see. That's very relevant, @Mathein, thank you.

@LeakyNun I concur it's sort of like a blowup. The projection $TX \to X$ has fiber over any point away from $(0, 0)$ one-dimensional, and at that point it's two dimensional.

I'd just call it a "stratified vector bundle" :)

Question: for which $n \times n$ matrices $A$ is the map $x \mapsto x^T A x$ convex as a function of $x \in M$, where $M = S^{n-1} \subset \mathbf{R}^n$?

Here we consider a different notion of convexity, not the usual one.

Let $M$ be a Riemannian manifold. A map $f : M \to \mathbf{R}$ is called convex (w.r.t. $M$), provided that $f(\gamma(t)) \leq t f(\gamma(0)) + (1 - t) f(\gamma(1))$ for every geodesic $\gamma: [0, 1] \to M$.

That escalated quickly!

(Indeed if $M = \mathbf{R}^n$ with the usual flat metric, then this reduces to the usual definition.)

In case $M = \mathbf{R}^n$, the answer is easy, the answer is that the matrix $A$ needs to be symmetric positive nonnegative definite.

But I wonder if there is a nice way to answer this question on the sphere as I explained above in the geodesic definition.

I'm not super comfortable with Riemannian geometry, so I thought I'd ask here :)

Given a (non-compact) manifold, can I always find a Riemannian metric with a non-compact isometry group?

generally I can make it so that the local isometries at a certain point are non-compact, but I won't be able to extend them to the entire manifold

If so I would expect to cook it up as follows: by construction (use a compact exhaustion) every noncompact manifold admits a nonvanishing complete vector field; you can choose it so that not all orbits are circles. Then try to prove that for any complete nonvanishing vector field, there is some complete metric which has it as a Killing field.

The last line is the only thing I'm unsure of

Why is the "smoothness" of a function $f$ correlated with the convexity of $\ln(f)$? In fact, I just read that the gamma function is considered extremely smooth? But then wouldn't a regular infinitely differentiable function be considered smooth? And yes not all of them are logarithmically convex...

However it seems difficult: if you didn't already know that the irrational straight line vector field on the torus was a killing field, how would you prove it?

@Mike to get rid of examples like that one could try to make it so that the flow lines of the vector field are closed, which could work for non-compact manifolds

one then gets a foliation of the manifold by the $1$d flow lines, if you define a metric smoothly on a local section it will get transported smoothly along the lines, then I should ask why I can this a globally well defined way

I don't have any idea how you would make the flowlines closed.

If you could then this would be relatively easy. You can take a global section of the quotient M -> M/R.

Ah, I had thought about making the flow leave every bounded set, but my idea for that doesnt even work in 2d, it was to ask that $\|X\|>$ constant (wrt an auxilliary metric), but that doesnt imply that the flow leaves any disk

Why is that the fundamental vector field generated by the action of a lie group always symplectic?

I have to prove that $d(\iota _ {X} \omega) = 0$ but I don't see how I would prove that.

Yes, well I mean quite seriously I don't know any way to resolve that issue. And you probably want to get rid of circles too.

Anyway you soon get topological obstructions.

@Albas Sorry, what is the setup?

Umm so you have a Lie group $G$ which is acting on $M$, $G \times M \to M$ where $(g,m) \to \phi_g(m)$ where $\phi _g$'s are symplectomorphisms. Now you can define a vector field $X_M = \frac{d}{dt} \phi_{\operatorname{exp}tX}(m)$(evaluated at $t=0$) where $X$ $\in$ Lie algebra of $G$. I have to show that $d(\iota _{X_M} \omega) = 0$, that is $X_M$ is a symplectic vector field @MikeMiller

Where $(M,\omega)$ is the symplectic manifold

I don't know how to prove that, and explicitly calculating it is a massive mess

Never mind these $\phi_g's$ are symplectomorphisms

That would imply that $L_{X_M} \omega = 0$

Since the flow of $X_M$ is determined by $\phi_g$ for a given $g$

I should just type things out loud. You sometimes miss the details.

Problem: If $A \in M_n(\Bbb{R})$ with $A^T A = AA^T$, then $A$ is diagonalizable...Regarded as a complex matrix, $A$ is normal, so the spectral theorem says there exists a unitary $U \in M_n(\Bbb{C})$ such that $U^* A U$ is diagonal...

How do I argue that $U$ is actually a real matrix?

How can I prove that the image of normal space under continuous and closed map is also normal?

Sorry for vanishing @Albas. I'm glad you worked it out, I got busy suddenly

That's fine @MikeMiller :)

In the Gelfand-Naimark equivalence of categories, how do we reconstruct a compact Hausdorff space given a commutative unital C*-algebra?

Oh, use the Gelfand transform to get an isomorphism with the continuous functions on the spectrum of the C*-algebra

the emotional zigzag between "my code works" and "my could does not work" has been real today

I finally got it to compute for next case of interest...but the result doesn't have the correct symmetry and so can't be correct

I've been staring at this for the past hour and I can't see why $\alpha$ starts from $1$... I know it's something silly but I just don't get it, if you ask me it should start from $p$...

anyone here familiar w/ geodesics on the n-sphere

..yes I know ask, dont ask to ask...but i asked a while ago to crickets, so.

Welcome to the club

Except, my question is irrelevant

The following question seems like it should be obvious but I'm a bit unsure now

Suppose I have a plane containing three lines each 120 degree angles. If I have a geometric figure which has reflection symmetry with respect to two of the lines, must it also have reflection symmetry w/r/t the third?

I think this just comes down to the fact that any two of the transpositions in S3 will generate the rest of the group.

this is a question regarding mathjax

so which one of the following is correct

a) {eq} x^2 {/eq}

b) $$x^2$$

[{MathJax fullWidth='false' x^2 }]

The first one doesn't register at all if I start up chatjax, though that's not quite the same thing

{[MathJax fullWidth='false'

x^2}]

not a typo its an option

weird

}]

all options are these

@Semiclassical where can I check MathJax syntax

dunno. my knowledge of it begins/ends with "it does latex"

ha :) ok

@MikeMiller just a remark about one of the things you said: look at the field $y\partial_x + x\partial_y$ in $\Bbb R^2$. The flow of this field is $(x,y)\mapsto (\cosh(t) x + \sinh(t)y, \sinh(t)x+\cosh(t)y)$. There can be no metric having this field as a Killing field as its flow acting on $T_0\Bbb R^2$ generates a subgroup of $GL(T_0\Bbb R^2)$ that does not have compact closure.

@FuzzyPixelz, could I ask what's $D_\alpha$? I know not helping - but I'm just curious.

@AlessandroCodenotti remember that the equivalence is for proper maps as morphisms for the locally compact Hausdorff category and proper $C^*$-morphisms for the commutative $C^*$-algebra category

Hmm I'm pretty sure you can just use continuous maps and unit preserving *-homomorphisms

Hi @Ted

Hi demonic @Alessandro

Oh, wait, I'm thinking about the easy case, compact spaces and unital commutative $C^\ast$-algebras

@DrewBrady Assuming you mean the round $n$-sphere, they're all intersections of the sphere with $2$-dimensional subspaces of $\Bbb R^{n+1}$.

Hey Ted and Alessandro!

hi nerdly Demonark

@Semiclassic: What's the composition of those two reflections? Isn't it a rotation that gives you the third axis? However, I don't know where you're getting $S_3$.

Hi @Dami

yep

Wouldn't it be S3? That's generated by (12) and (23), for instance

Anyhow, the composition of symmetries is another symmetry. Now play with the composition of the rotation and one of the reflections.

So you're arguing as I am, I guess, that the group of symmetries must contain that $S_3$.

Right

OK

For context, I was trying to decide which of my two methods of counting something was right

one which assumed symmetry but was otherwise brute-force, and one which didn't assume symmetry but was supposed to be systematic

Had the advising appointment today, turns out the reason they do it this early is that they're trying to handle TA schedules and want to do it around the classes/seminars you plan to take

What's the difference between "brute force" and "systematic"?

yeah, that's a bad way to put it

Demonark, I figured as much, but you might change your mind before September.

I guess we didn't have that problem for typical first-year students at UGA because they were very limited in what teaching they could do. But I ended up scheduling those students after advising the week before classes started, pretty much.

It's sorta interesting so I'l explain. The problem goes like this. Suppose I've got a high-dimensional polytope P determined as an intersection of finitely-many half-spaces.

@Daminark wot, you're already in wisc

I have a linear map T from P down to a low-dimensional space (3D in this case)

Nah it was over the phone

oh

Then the image T(P) will be a 3D polytope, and I want to figure out its vertices/facets

This is true, I guess we'll have to see how it goes. A couple contingencies were referenced so I guess we'll see how things go. It seems the plan now is to do AG first year, and it depends on how I do on the analysis qual. If I'm comfortable with measure theory then chances are I'll do analytic NT and rep theory first semester, and functional analysis + possibly smooth manifolds? unless something better comes up, second semester (I don't have the course offerings for then yet)

One obvious way to do that: The image of a convex hull of some set is the convex hull of the image of that set.

So compute the vertices of the high-dimensional polytope P, map them via P, and take their convex hull.

Problem is that the first step gets bad fast.

If it feels like maybe my measure theory isn't fully up to snuff then I'll swap out either analytic NT or rep theory for that

The smarter approach: Since T(P) is a polytope, I can view it as the feasible set for a linear program. So if I pick an objective function in the projection space and solve the resulting LP using the simplex method, I'm sure to get one of the vertices of T(P)

Sounds like cool courses, I'm also considering doing analytic NT in the next semester!

jesus dami thats a lot of classes

Moreover, you can trace that back from an LP on T(P) to an LP on P without any issue

3 each semester, I think that's standard there

oh I thought that was all in the first sem

So you can determine vertices of the projected polytope by doing LP's on the original polytope, which is waaaaaay easier than enumerating all the vertices of the original polytope

The trouble is: It's not at all obvious how many such LP's you have to do, or how to choose them

Planning two semesters ahead? Not even the Germans are that organized

I have literally no idea what is being offered in 2 semesters

LOL @Alessandro

Oh no, it'd be fall semester AG1 + 2 of {analytic NT, rep theory, measure theory} and spring semester AG2 + functional analysis + smooth manifolds or something else maybe

I mean the course schedule isn't up for spring but functional analysis, AG2, and smooth manifolds are offered every spring

@TedShifrin So my first method (which I called brute-force but which I guess I should call probabilistic) was to randomly choose a bunch of different objective functions and solve the resulting LP's

@RyanUnger Same, they publish the courses for the next semester at the end of the previous one here

in addition, I know that my projected polytope has to have certain symmetries

I would probably only take FA depending on who is teaching it

so I enforce those symmetries at the same time

Like I might take Riem geo because Marques is

Where are the set theory and logic courses? :P

I'm also keeping an eye on physics

@AlessandroCodenotti That's big brain math, not offered at Princeton

Actually there is one logic course

It was offered in the spring

In that way, I get a method which works fast and seems reliable.

That's kinda sad

However, I don't really like using a probalistic algorithm

Lmao, Madison's actually quite the place for complexity theory which honestly feels less cool to me than something like model theory but prob better than set theory

The department is small so there's only a few topics that seem to be well covered

So I wanted to trade that out for a deterministic one, and I came across a good one

I might agree that model theory is the best tbh

Trouble is, I'm having to implement it myself in mathematica

and despite having worked on it for a while, I'm pretty sure at this point that what I've got just doesn't work like it should

What's the FA course about?

That's not a problem with the algorithm, but my implementation probably just isn't sound

Which is frustrating...

@Semiclassic: I don't think about this stuff, as you know, but I don't see why you expect the projected thing to have any (nontrivial) symmetries at all.

it's not obvious from my description, no

So the division of topics isn't exactly specified from measure theory

e.g. maybe first semester there's time to do interpolation of L^p spaces or something

There's the dreaded Princeton Jackson EM course

But it's built into the nature of 1) the constraints that I've put onto the original polytope P, and 2) the particular linear map I'm using

Might be cool to sit in on

So I do have good reason to say that those symmetries are there

But yeah basically it should cover through distributions, Fourier analysis, Sobolev spaces, spectral theory

See @AlessandroCodenotti

Fourier analysis

On the note of the other conversation: I've gotten interested into the history of probability theory lately

and I'm realizing that my ignorance of measure theory is a handicap

@RyanUnger Pfff

Not too fond of it Alessandro?

So yeah, I may have to learn some measure theory eventually if I'm to redress that :P

@Daminark I never learned it

And Ryan was surprised by this fact a few days ago

Ah, yeah kinda same honestly, I've been saying I should learn it forever (just like covering space) but I've never gotten around to it, aside from the "$e^{inx}$ is an ONB of $L^2$" mantra

Hey guys, what do you guys think would be a better buy, Harvard's advanced calculus book or Rudin's principles? (learning for the sake of learning, not a math student)

Main thing I'm curious is when in history the phrase "Hilbert space of random variables" shows up and when it becomes common

I'll start doing noncommutative fourier analysis just to annoy you @Ryan

Hilbert space of random variables

That's interesting

I mean, once you've gotten measure theory as a foundation of probability theory, then it's natural enough to connect that with Hilbert space

Oh I meant more historically

ah, yeah

The best precedent for it that I know of is: Suppose you're taking observations of quantities X,Y,Z

Then you can form observation vectors x=(x1,x2,...xn) and so forth

And you end up in $L^2$ because of finite variance or something ...

Well, you're doing finite observations so you really can't get infinite variance in this setting

But yeah, once you move from vectors of finitely-many observations to the random variables itself

then you definitely need it

The question is when people started thinking of it like that. There's a paper of Fisher from 1915 where he uses that idea (viewing observation vectors in terms of n-dimensional space)

And iirc he goes as far as to connect the angle between such observation vectors with their correlation coefficient

But it's not something he draws much attention to

First place where I've seen somebody talk about an inner product space of random variables is a paper of de Finetti from 1937

There's an earlier paper of Frechet where he considers Lp spaces of random variables, but only as metric spaces

I can't help but think quantum mechanics is partly (mostly?) responsible.

(There's also a book by Frechet which addresses this, but I haven't tracked it down yet to see)

@TedShifrin Actually, not so much that I've seen. de Finetti's paper is more in the realm of what he got from Frechet

but it is an interesting question about where von Neumann fits into this story (if at all)

Quantum mechanics can motivate (and in many instances, did motivate) large parts of functional analysis

Yeah

A common story is that the name "Spectrum" of an operator was defined in analogy to the spectrum of light going through a prism, or spectral lines of elements

it's more of a historical accident, isns't it?

then it turned out that the spectral lines of elements turned out to be the spectrum of their hamiltonians

I always told my students the former ... didn't know the latter.

I also like this story:

> In the 1960s Friedrichs met Heisenberg and used the occasion to express to him the deep gratitude of mathematicians for having created quantum mechanics, which gave birth to the beautiful theory of operators on Hilbert space.

> Heisenberg allowed that this was so; Friedrichs then added that the mathematicians have, in some measure, returned the favor. Heisenberg looked noncommittal, so Friedrichs pointed out that it was a mathematician, von Neumann, who clarified the difference between a self-adjoint operator and one that is merely symmetric. "What's the difference," said Heisenberg.

(more precisely, the energy of the spectral lines corresponds to the distances between eigenvalues of the Hamiltonian operator)

@Semiclassical oh yes, the transition from one eigenstate to another can emit a photon of that energy

right

I guess I've never heard of "symmetric" in the general hermitian context.

it is fair to say, though, that the spectrum of the Hamiltonian directly determines the spectral lines of the elements

@TedShifrin If an operator is unbounded (like the multiplication with $x$ on $L^2(\Bbb R)$) then there are problems with the domain. But there still is a good notion of self-adjoint, which is stronger than just the condition $ \langle v, x w\rangle = \langle xv, w\rangle$ for all $v,w$ in the domain.

If $D$ is a densely defined unbounded operator on an Hilbert space then $D$ is symmetric means $(D\phi,\psi)=(\phi,D\psi)$ for all $\phi,\psi\in\mathrm{dom}(D)$, and it is self-adjoint if it symmetric and $\mathrm{dom}D=\mathrm{dom}D^\ast$

Oh no Alessandro

I thought I recalled densely defined operators being called self-adjoint. I haven't thought about this in many decades.

Oh, I didn't recall that distinction. Thanks @Alessandro

@RyanUnger What?

Looked worst without Tex

The intuition is that symmetric means $D\subseteq D^\ast$ and self-adjoint means $D=D^\ast$

Yes

essentially self-adjoint or go home

So do you know the functional calculus now

Essentially self adjoint means that $\overline{D}=D^\ast$ to finish the intuitive picture

@RyanUnger I just know how it's defined now, but I never really worked with it

You need to look into heat kernels

Or semigroups in general

I'll try to take a course called "semigroups and the evolution equation" next term

Yeah but as we've discussed, that course doesn't really make sense without having a PDE in the back of your mind

"try" because I don't know whether I can manage with my (absence of) analysis background

I have one, it's really faaaaaaaar back into my mind

I like Pazy's book on semigroups

It's unapologetically about PDEs

There might also be a followup NCG course so I might do that to get the analysis credits I need

can you fit a curve to this desmos.com/calculator/dw20a5du2y ?

@AlessandroCodenotti I was told this paper is beautiful ma.utexas.edu/mp_arc/c/08/08-191.pdf

if you want to work wiht spectral theory and essentially self-adjoint operators

That's a cool paper

I think the Reed and Simon books are incredible

I agree that they are great

@RyanUnger I'm looking at the index and skimming the first chapter and I must say it doesn't look like my cup of tea, but maybe I can learn something different once in a while

the first chapter is very....pure

chapters 7 and 8 are where it gets good

I was carefully avoiding PDEs heavy chapters :P

people have managed to exploit these ideas for Ricci flow and have proved some very strong results

projecteuclid.org/euclid.ajm/1333976883

You won't convince me to become an analyst

😭

(even though my master's thesis is likely to be at least somewhat analytical at this point, but we'll see how that ends up)

Yeah but that’s soft analysis

Semigroups stuff heads in the direction of hard analysis

Only you and the Russians enjoy hard analysis

🤔

I find Russian math pretty incomprehensible

arxiv.org/abs/1707.05940 that's the kind of semigroups I can deal with in "analysis"

There's a lot of hard analysis in geometric analysis. There's plenty of non-Russians working in that, silly @Alessandro.

hue hue hue, if you want to do C* algebras I can wholey recommend the book by Bratteli & Robinson

at one point they introduce 7 topologies on a von Neumann algebra, it blew my mind to pieces

s.harp what is it you do anyway

@Ryan I am doing geometry

@s.harp Ok now I'm interested

What kind of geometry

@s.harp why should it have compact closure? The ambient manifold is not compact.

I figured the geometry part out by myself...

@AlessandroCodenotti (its not actually that complicated, but the book is really nice)

This does work as an obstruction for non Seifert fibered 3-manifolds.

I'll check it out

@MikeMiller the isometries preserving a point should act as a subgroup of $O(n)$ on $T_xM$

so should have compact closure in $GL(T_xM)$, even if the manifold is not compact

@MikeM: I was puzzling over that, too, for a bit. But it's cuz of Riemannian.

@RyanUnger well differential geometry, more concretely Lorentz geometry and Cartan geometries and the question of compact space with noncompact automorphisms, but I'm just a student atm

Undergrad? Grad?

masters student, which makes me grad student i believe

Oh ur in Germany

yes

Noch ein Deutscher?

jawohl

Es gibt so viele in diesem Zimmer :P

K.

Wir sind überall

An amusing and useless criterion is that a vector field on a closed manifold is Killing for some metric iff its closure in the diffeomorphism group is compact.

I have never really thought about Killing vector fields ... a definite lacuna in my background.

@RyanUnger Ich sehe einen Amerikanischen Spion

@MikeMiller the metric being given by averaging over the group?

The original question was: "does every open manifold admit a metric with noncompact isometry group?"

Hey I was born in Germany

@s.harp yeah

Oh, sorry, I didn't know!

@TedShifrin yeah, they're all paths along great circles essentially right

but I guess I had a specific question regarding composite maps w/ geodesics

Looks like I missed ya tho