Mathematics - 2025-01-19

one potato two potato

09:10

Is a covering of a parallelizable manifold parallelizable?

X4J

10:32

Is N a normal subgroup of G precisely when the set G/N is invariant under conjugation of elements in G?

Balarka Sen

11:26

@onepotatotwopotato Yes.

@Thorgott I do not think this is trivial. Even the following is famously open: does simply connected closed smooth $4$-manifold $M$ have a handle decomposition without $1$-handles?

handan_toddler

Welcome back, pal.

Sine of the Time

sup toddler

handan_toddler

chillin' u?

Sine of the Time

same

one potato two potato

Hi Balarka I saw you on arxiv

Balarka Sen

11:36

Hi potato. We uploaded a few things, yeah (and revisions are underway).

one potato two potato

@BalarkaSen Because I suddenly wonder if every oriented 3-manifold with boundary is also paralleliable (probably not).

It seems parallelizability is important in some cases. I recently studied Chern-Simons in 3-manifold

Balarka Sen

@onepotatotwopotato It sounds right to me.

Just double the 3-manifold along the boundary and you have a closed 3-manifold. The tangent bundle of the entire thing restricts to one piece.

You know the tangent bundle of the entire thing is trivial. So...

one potato two potato

Ah I didn't think about doubling!

Balarka Sen

@onepotatotwopotato I know very little about gauge theory in general. What interesting things did you learn?

X4J

12:05

I’m trying to find a group $G$ and a subgroup $H$ of G s.t there exists $g \in G$ with $gcd(o(g), [G:H])=1$ whereas $g \notin H$.

I know that for a normal subgroup $H$, one cannot find such an element

nickbros123

12:45

im trying to prove AC $\implies$ Zorns, so I take this Poset P and consider the set of all elements that are comparable with every other element. Basically i look at $S:=\{y \in P: x \leq y \text{ or } y \leq x \ \forall x \in P\}$. Is this the correct direction to think on? or is this consideration somewhat too narrow

ok it might be useless to look at this set

PM 2Ring

13:07

Hello. I've been sick chat.stackoverflow.com/transcript/message/57854012#57854012 so I haven't been posting or even lurking for a month or so.

Here's a lovely animation of the Poincaré disk model of the hyperbolic plane, created by Greg Egan, from Escher's Circle Limit III mathstodon.xyz/@gregeganSF/113842854457093299

Soumik Mukherjee

@PM2Ring I hope you recover soon.

PM 2Ring

@SoumikMukherjee Thanks

Circle Limit III on Dini's surface: mathstodon.xyz/@gregeganSF/113804496169577405

Sine of the Time

13:25

@PM2Ring get well soon :'(

Thorgott

@BalarkaSen interesting, though I would expect that to be a much harder problem since it's about actual spaces and not just homotopy types

PM 2Ring

14:10

Here's a silly Quadratic approximation to pi. The smaller solution of $x^2 - 18800692x + 59064106 = 0$ is $\pi$ to ~16 sig figs. 3.141592653589793449

Thorgott

@Ben @Balarka Just got this really cool argument from an MSE answer by Lee Mosher, much sharper than the usual CW-approximation. I'm surprised I never considered it before:
*Proposition:* If $X$ is an $n$-dimensional $k$-connected CW-complex and $k+2\le n$, then $X\simeq X^{\prime}$ for an $n$-dimensional CW-complex $X^{\prime}$ with trivial $k$-skeleton.
*Proof:* Induction over $k$ (with $n$ varying). The case $k=0$ is proven by collapsing a maximal spanning tree in the $1$-skeleton. In the inductive step, let $X$ be a $(k+1)$-connected $n$-dimensional CW-complex, $n\ge k+3$. The inductive…

The conclusion is also true for $k+2>n$, but for different reasons. If $k\ge n$, then Hurewicz implies $X$ is contractible, so the claim is trivial.
If $k=n-1$, assume $n\ge 2$ (the case $n=1$ is again just collapsing a maximal spanning tree). Then, there is a map $\bigvee_{i\in I}S_i^n\rightarrow X$ inducing an isomorphism on $H_n$ (since the Hurewicz map is an iso in degree $n$ for both spaces) and both spaces have vanishing homology in all other degrees by assumption and are simply connected, so this is a homotopy equivalence by Whitehead.

Ben Steffan

huh, that's neat

one potato two potato

14:37

@BalarkaSen not much, especially from a gauge theory point of view. I just learned the very definition of it. On a closed hyperbolic 3-manifold, CS invariant is a topological invariant but in case of infinite volume hyperbolic 3-manifold, hyperbolic structures can be deformed so CS invariant becomes a functional on the associated deformation space (of hyperbolic structures). It seems this CS invariant is closely related to so-called renormalized volume of hyperbolic 3-manifolds.

I don't know in detail (or even actually true in fact) but I was told that exp(renormalized volume functional + i CS invariant functional) becomes a holomorphic function.

As an independent interest, I recently learned (in gauge theory) that there are various types of connections. Ehresmann connection is really about ''connecting'' two incident tangent spaces. I always wondered why connection is called ''connection''.

yesterday, by one potato two potato

what would be a geometric meaning of the integral of the mean curvature over a codimension 1 (Riemannian) submanifold?

All I know is that the integrated mean curvature appears in the first variation formula of area functional.

Shaun

Does SL(2,q) in GL(2,q), such as it is, mean GL(2,q) is one-headed? I think so but I can't quite prove it. Every example I check seems to work, such as I understand them by looking on Group Names.

Ryder Rude

15:10

4

Q: Topos Theory as a link between logic and geometry

I was reading about topos theory and in many ways some people said that TT can be used to unify logic and geometry. What does that mean? I have an OK background in category theory (at least up to adjunction theorems) and topology (general, differential and a little bit of algebraic topology) so f...

can someone explain this idea in simple terms (like non category theory terms)?

Xander Henderson

It is topos theory. I don't think that non-categorical explanations exist...

Ryder Rude

oh

do you personally find this idea profound? like one of your favorite connections in math

Xander Henderson

I didn't read the question.

Ryder Rude

it is supposed to link logic and geometry

Xander Henderson

Topos theory is not something I know much about, nor is it something I care to learn.

Ryder Rude

15:13

oh

Ben Steffan

asking Xander whether he thinks a hardcore category theory topic is profound lol

Ryder Rude

i understand... thanks @XanderHenderson

i feel extremely hyped when I read i read stuff like "unifying logic and geometry"

there was a paper I came across which was formulating quantum mechanics using topos theory

it uses this connection between logic and topos

Xander Henderson

My feeling is that people who make big claims like "This will unify [x] and [y]!" tend to overpromise and underdeliver.

Ryder Rude

maybe... i don't have many examples of such instances except the one from Descartes about uniting algebra and geometry. that one really over-delivered :P

Edward Frenkel talks about something called the Langland's program which treats different math fields like continents and they want to connect everything

Ben Steffan

it's very easy for the langlands experts to make all sorts of miraculous statements about the program since nobody else seems to have even a remote chance of following what's going on

Xander Henderson

15:21

@BenSteffan I lol'd.

Ben Steffan

I'm being serious, mostly

Ryder Rude

xD

Xander Henderson

@RyderRude Lots of folk have found links between different fields. Descates put things together in a somewhat novel way (though I think that he was kind of a right place / right time dude---someone would have done what Descartes did), but I don't think that he made grandiose claims about what he was doing mathematically.

Ben Steffan

I've talked to postdocs in here working in adjacent fields and they say that you cannot keep up with the program as a side interest: the sheer amount of theory you need to have a decent understanding of is just too much

Xander Henderson

People who came after noted the value of his contributions (ditto other famous names, e.g. Gauss, Hilbert, Euler, etc.).

Ben Steffan

15:23

you go to talks and people pull out obscure facts about, just, whatever and you're expected to understand these

as for impact, well, you can find lost of papers making use of langlands in algebra, algebraic geometry, and things of the sort

Ryder Rude

@XanderHenderson yeah.. Newton's work also relied on this idea

Ben Steffan

other areas? well I for one haven't really seen much

Xander Henderson

@RyderRude And Newton didn't really do anything all that novel, either. :D

Ryder Rude

@XanderHenderson Descartes probably made metaphysical claims instead :P

Xander Henderson

He built so much on the work of others---the time was just ripe for calculus (as evidenced by its independent invention by Leibniz).

Ryder Rude

15:26

yeah.. Newton's ideas also built upon previous peoples' ideas

no one invents ideas in a vacuum. Apparently, Descartes had some version of momentum conservation, Hooke had a force law and maybe even the inverse r squared law

Newton just built upon that by working on other complex problems

i recently loved the connection between group theory and manifold theory

like, homology groups etc

@BenSteffan they're like the string theorists of math :P

Ben Steffan

perhaps that's a fair comparison

Xander Henderson

@BenSteffan I feel like category theorists are like the string theorists of math. Langlands is something else entirely.

Ben Steffan

@RyderRude calling homology a "connection between group theory and manifold theory" is a could way to not get taken seriously

there's little group theory going on with that

Ryder Rude

does the langland crowd not use category theory too much

Xander Henderson

Like, Langlands adherents seem to overpromise and underdeliver, but there are at least some concrete mathematical results which they are building towards. There are genuine conjectures running around.

Ryder Rude

15:33

@BenSteffan oh

Ben Steffan

@XanderHenderson not really. category theory is a well established working tool. string theory is a unproven and perhaps unfalsifiable theory known for being bespoke

Xander Henderson

I don't think that category theory has really gotten there, yet. It might. But right now, it feels like a lot of abstractification for the sake of abstractification, with no real down-in-the-weeds results of the kind that seem to be promised.

Ryder Rude

@BenSteffan yes. i think the idea is more about studying the manifolds themselves rather than group theory

Ben Steffan

@XanderHenderson well I for one use category theory quite a lot in my day to day mathematical life and I disagree

although you do have a point about category theory for its own sake

Ryder Rude

i would also identify string theory with tons of conjectures :P

the entire castle is built on conjectures

Ben Steffan

15:35

You see, category theory was born to solve specific abstraction problems, and that it does well, when applied in the right degree of moderation

Xander Henderson

@BenSteffan I am not claiming that people don't use category theory. My claim is mostly that the people who do use it are all category theorists (and some algebraic geometry / algebraic topology people, as those areas seem to be where category theory was born).

Ben Steffan

I find that borderline insulting :P
I'm not a category theorist, and I don't wish to be associated with the term very much

Xander Henderson

Category theorists often make big claims about being able to abstractly describe structures and unite all of mathematics, but they don't seem to have actually produced a lot of results which genuinely push other fields forward.

It seems very in-looking (there is another term I would like to use, but it is kind of rude. It involves circles, and a word which rhymes with "perk").

Note that I am not saying that category theory is a waste of time---maybe it will eventually do some of the things that a lot of category theorists claim that it is going to do. I just haven't seen it, yet.

@BenSteffan Nope, you are a category theorist, now!

:P

Ben Steffan

@XanderHenderson I'm not sure what claims you have in mind. For my purposes category theory delivers on its basic claims very well.

Xander Henderson

@BenSteffan Note that I said category theorists, not category theory.

Ben Steffan

15:42

I also suspect that the "true" category theorists are over-represented. There's not that many of them. Most people do a mix of category theory and things like algebraic topology, and these people mostly hold back on their claims. But the "true" category theorists are, well

I don't think I can find a good word to say about them

but they seem to be very present, especially online

Xander Henderson

No true Scotsman

No true Scotsman or appeal to purity is an informal fallacy in which one modifies a prior claim in response to a counterexample by asserting the counterexample is excluded by definition. Rather than admitting error or providing evidence to disprove the counterexample, the original claim is changed by using a non-substantive modifier such as "true", "pure", "genuine", "authentic", "real", or other similar terms. Philosophy professor Bradley Dowden explains the fallacy as an "ad hoc rescue" of a refuted generalization attempt. The following is a simplified rendition of the fallacy: == Occurrence... ==

Ben Steffan

@XanderHenderson That's a nice fallacy, but it doesn't really apply.

Well, anyways

Xander Henderson

@BenSteffan You are making a distinction between "true" category theorists, who don't make the kinds of claims I am suggesting get made by a lot of category theorists, and all other category theorists.

Note that John Baez was on my phd committee. He is currently one of the moderately large fish in the category theory pond. He doesn't seem to make a lot of the big claims that a lot of category theorists make, but his students certainly do.

Ben Steffan

@XanderHenderson Well, no. I'm making a distinction between "true" category theorists, who do make the kinds of claims you are suggesting get made by a lot of category theorists, and other category theorists. But I'm making the distinction not based on this behavior, but what their mathematical interests are.

I would very much not like to speak about John Baez.

Xander Henderson

@BenSteffan No? He is a lovely man...

Ben Steffan

15:48

I know people who disagree.

Xander Henderson

Heh.

Ben Steffan

Anyways, let's drop this topic before I step on too many people's toes :))

Thorgott

3

Ben Steffan

lol

back to learning about 3-manifolds :/

psie

I have a basic question. A linear map from a finite dimensional normed space to another normed space is continuous. Now, I'm just curious, are there linear spaces which are not normed linear spaces? I guess one can not speak of continuity in such "nonnormed" linear spaces, or?

Ben Steffan

15:55

@psie being normed is structure, not property

in other words, every linear space is "non-normed" until you put a norm on it

you'll have no trouble convincing yourself that you can always put a norm on a finite-dimensional vector space

psie

ah ok 👍

Xander Henderson

16:10

This is my nightmare:

Thorgott

16:44

@BenSteffan I wish I knew more $3$-manifold theory, to be honest

Ben Steffan

cool, do you wanna take this exam for me then? :)

I find it hard to care

Thorgott

why are you taking an exam on $3$-manifolds if you don't care?

Ben Steffan

ah, for the oldest reason in academia

credit points

also because there's a voice in me that says I should learn this material

something something broaden your background something

the course is not very hard

Ryder Rude

@BenSteffan do you use category theory as just another tool, or do you also think that ideas like "all mathematics should be founded on category theory instead of set theory" hold water?

Thorgott

@BenSteffan fair, I suppose

I think knot theory is really cool, and closely related to $3$-manifold topology

Ben Steffan

16:56

@Thorgott slowly backs away

Thorgott

wish I could've taken a course like that for credits, if you look at my academic résumé you'd think I'm an algebraist (perhaps I am)

@BenSteffan deriving the Wirtinger presentation is the single coolest application of Seifert-van Kampen there is

Ben Steffan

@RyderRude I couldn't care less about foundations

Thorgott

foundations are boring

Ben Steffan

and about the category theory and foundations thing, well, suffice it to say that HoTT is a bit of a running joke around here

it's a bit of a "2025 will be the year of linux on the desktop" theory, at best

Ryder Rude

@BenSteffan oh

is it true that there is a "category theory perspective" of looking at mathematical entities, just like "set theory perspective"? e.g. i think of arithmetic as a set of natural numbers, and a bunch of binary operators defined on it. this is the set theory perspective

Ben Steffan

17:09

ask yourself whether the last question you asked would be meaningful at all if no such perspective existed

Thorgott

any monoid is in particular a category, but that's hardly a meaningful observation

foundations are pretty orthogonal to the act of actually doing arithmetic (or most of any mathematics)

Ryder Rude

@BenSteffan yeah... So the perspective does exist

Ben Steffan

evidently

Ryder Rude

i currently think of spacetime as a set, with some structure on it (like topology and smooth structure and some functions defined on the set). how would one switch to the category theory perspective

and is it true that the cat theory perspective makes morphisms the primary objects. while in set theory perspective, the set elements are primary and functions are defined in terms of them

Ben Steffan

@Thorgott "no, but, you see, this observation immediately allows to generalize. for instance you can directly say what an $\infty$-monoid should be."

:^)

Ryder Rude

17:16

i am not a Platonist. but how i pragmatically imagine the situation is that we have some fixed mathematical universe (like a Platonic model). and we have two different perspectives of thinking about it. would this be a good way to imagine the situation

And that the set perspective prioritises set elements while the category perspective prioritises morphisms

Thorgott

@BenSteffan if only it was that easy to actually understand what an $\infty$-monoid is, though

@RyderRude I don't think this is a meaningful question, at the very least not a priori

Ryder Rude

@Thorgott i understand...

i think i am asking bad questions

Ryder Rude

17:32

one last question... regardless of whether we take the category view or the set view, we would be using something like a formal axiomatic system to define our system, right?

and we would use the usual classical logic deduction to prove theorems

e.g. one can define natural numbers as a set or as a category. the use of classical logic would be common in both approaches

or one can also use neither sets nor category to define arithmetic. e.g. the Peano axioms

leslie townes

@psie continuity is most frequently thought of as a topological notion. the notion of continuity that you see in the normed vector space context is a special case of the topological notion of continuity, where the topology in this setting is definable via a norm. it is very possible to define topologies on vector spaces that provably do not arise from norms. there is still some 'rigidity' in the finite dimensional setting though, this is primarily an infinite dimensional phenomenon

Balarka Sen

17:56

@Thorgott Ah, I didn't realize you were considering homotopy types.

Thorgott

yeah, the original question (in case you didn't see) was whether a simply connected $4$-manifold can be an Eilenberg-MacLane space

Balarka Sen

I saw that briefly, but did not have anything useful to say.

Thorgott

it's a surprisingly annoying problem

Balarka Sen

Usually, the way to remove $1$-handles in simply connected manifolds in higher dimensions is to trade them for $3$-handles. This is possible in dimension $\geq 5$, but I suppose upto homotopy type it is possible in all dimensions.

Attach a $2$-cell to the $1$-cell, and then a $3$-cell cancelling the newly attached $2$-cell.

Thorgott

cause there are no obvious homological or homotopical obstructions that stop $K(\mathbb{Z}[p^{\infty}],2)$ for $p$ an odd prime from being realized by an open $4$-manifold

Balarka Sen

17:59

Then cancel the $1$- and $2$-cell

Thorgott

@BalarkaSen yup, that's effectively what the proof I wrote up earlier does

Balarka Sen

Gotcha.

Thorgott

except it's an elementary collapse instead of a handle cancellation

Balarka Sen

Right.

I was thinking about a related question a while back. Suppose $X$ is a CW-complex, and suppose $Y = (X \vee (\vee_{i = 1}^k S^2)) \cup_{j = 1}^l e^3_j$. Suppose the inclusion $X \to Y$ is an isomorphism in $\pi_2$. When can you perform Whitehead moves to $Y$ so as to obtain a subcomplex $X \subset Z \subset Y$ such that $(Y, Z)$ has only $3$-cells and $X \to Z$ is a homotopy equivalence?

It appears to be possible if every stably free $\Bbb Z[\pi_1(X)]$-module is free. It also appears to be always possible if I am allowed to replace $Y$ by $Y \vee (\vee^n S^3)$. It also appears to be always possible if I am allowed to have $4$-cells in $(Y, Z)$, too.

Ultimately the reason I wanted this became redundant but nevertheless, an interesting question.

Correction: in all of the above, I want some kind of Whitehead torsion to vanish. So also need the hypothesis $Wh(\pi_1) = 0$

Thorgott

"Whitehead moves" being the elementary collapses/expansions?

Balarka Sen

18:13

Right. I also include what I call "cell-slides": these are simple homotopy equivalences between $(X \cup_f e^n) \cup_g e^n$ and $(X \cup_f e^n) \cup_{f + g} e^n$ where $f + g$ is the addition of the homotopy classes of $f$ and $g$ in $\pi_{n-1}(X)$, $n \geq 3$.

But of course, that's strictly speaking redundant because they can be written as an expansion followed by a collapse.

(I suppose there's no issue with $n = 2$ either, I was being overly careful.)

The argument is an observation you made above: the $3$-complex gives a two-term free resolution of $\Bbb Z[\pi_1]$. Namely, we have the cellular chain map $\partial : C_3(\widetilde{Y}, \widetilde{X}) \to C_2(\widetilde{Y}, \widetilde{X})$ which is a morphism between two free $\Bbb Z[\pi_1]$-modules. This is surjective, because $H_2(\widetilde{Y}, \widetilde{X}) = 0$ by the $\pi_2$ hypothesis.

If the kernel was free, then we could choose a basis of the kernel, and a basis of a section of $\partial$ (which always exists for epimorphisms between free modules), and get a basis for $C_3(\widetilde{Y}, \widetilde{X})$. Then consider the basechange matrix from this basis to the standard basis of 3-cells of $(\widetilde{Y}, \widetilde{X})$. This is a matrix in $GL_n(\Bbb Z[\pi_1])$, for some $n$.

If this matrix represented the zero element in $Wh(\Bbb Z[\pi_1])$, then we are done. We can add more cancelling pairs of $(2, 3)$-cells by elementary expansions, so that the matrix can be stabilized, and row reduced to the identity matrix. Then apply Whitehead moves on $Y$ according to the stabilization and the row reduction process (cell-slide here).

This gives some definite collection of $3$-cells which algebraically cancel the $2$-cells, and then some more. Collect all the $2$-cells as well as the $3$-cells of the first kind, and call $Z$ that.

$(Y, Z)$ then clearly consists of $3$-cells, and $X \to Z$ is an isomorphism in $\pi_1$, and $\widetilde{X} \to \widetilde{Z}$ is an isomorphism on $H_2$ and $H_3$. Hence, $X \to Z$ is a homotopy equivalence by the homological Whitehead theorem.

Thorgott

18:42

wow, that's a rather involved argument

I never really learned the 'main theorem' of simple homotopy theory

Balarka Sen

yeah I was rather surprised that this question has not been addressed in detail before.

what's the main theorem?

Thorgott

that $A\hookrightarrow X$ is simple if and only if the torsion vanishes

or that the geometrically defined Whitehead group is isomorphic to the algebraically defined one

which is basically the same thing

basically the concrete principle that translates geometric moves into algebra

Balarka Sen

Ah, I think the main idea is the "cell-slide" move I described above. It is not difficult to check that the effect of cell-slide is an elementary transformation of the third kind to the matrix of $\partial : C_{even} \to C_{odd}$.

Thorgott

there's a nice brief chapter on this in Rosenberg's "Algebraic $K$-Theory and Its Applications" (I happen to own a copy) that I should read again in more detail, but it only provides the sketch

psie

@leslietownes ok 👍

Thorgott

18:55

@BalarkaSen yeah, I just never sat down and did the computation/drew the picture myself to really grok it

Balarka Sen

To be fair, it is a pretty smart observation that you can do topology guided by linear algebra

All of this is due to Whitehead, originally, right?

Thorgott

@BalarkaSen I believe it is, yeah

Balarka Sen

Smart chap

Thorgott

it's crazy how much he did back in the day

have I told you the story about his love for pigs before?

Balarka Sen

yes, i remember lol

I am not familiar with Rosenberg's book, but I would recommend reading Milnor's article on the Whitehead torsion

Thorgott

19:14

yeah, I've had it downloaded for years lol

hbghlyj

20:32

I am reading a proof of Dirichlet's theorem on Fourier series in the PDF https://people.math.harvard.edu/~knill/teaching/math22b2019/handouts/lecture30.pdf
I have trouble understanding the sentence "Even so $f$ is only piecewise continuous, the
function $y → F_x(y)$ is continuous and $2π$-periodic."

In $F_x(y)=\frac{f(x+y)-f(x)}{2 \sin (y / 2)}$ the numerator is only piecewise continuous, so how can the function $y → F_x(y)$ be continuous?

leslie townes

20:50

@hbghlyj at first glance i don't see how it would be. is the hypothesis actually used anywhere? if so, how?

hbghlyj

It is used in the following Part VI (Riemann-Lebesgue lemma) for continuous functions g, h

leslie townes

its pretty common for fourier analysis books to make hypotheses that are way more restrictive than necessary just so they don't need to use subtler arguments. i wonder if in the background either (a) it's never used at all, or (b) it's used only in the sense that for some relatively simple argument, it would suffice to assume that f is in fact continuous

hbghlyj

Ok

leslie townes

okay, well the riemann lebesgue lemma would certainly also hold for piecewise continuous functions.

hbghlyj

Yes.

leslie townes

20:55

without criticizing the substance of this, it seems kind of sloppily written and dashed off. i wrote a lot of stuff like this when i taught. i wouldn't think too much of the details maybe not making literal sense here and there. it's probably written with the understanding that whoever is most immediately reading it can ask questions and hear "oh yeah, of course i meant x."

yeah, there is no reason why F would be continuous at a jump discontinuity of f.

it's certainly continuous wherever f is differentiable, which would be most of the time because of the "piecewise" part of piecewise smooth.

smoothness also is way more restrictive than necessary for a lot of this, but again, the proofs get harder if you start looking for minimal hypotheses and eventually there aren't any

Silly Goose

is there a deep connection between flat connections on $U(1)$-bundles and harmonic analysis?

leslie townes

to this day, i do not think there is any 'simpler' (i.e. other than a rephrasing of exactly this property) characterization of "the fourier series of f converges pointwise a.e. to f." there are tons of natural hypotheses that imply that, but the condition itself is i think provably weaker than all of them.

i've said before, it's a cruel joke on humanity that pointwise convergence was basically the starting point of focus of the theory of fourier series, when it is the almost impossibly hardest part of the subject. or it's a blessing on humanity because it required us to invent the rest of analysis to understand it

Silly Goose

to my knowledge, flat connections are classified by $\rho: \pi_1(X) \to U(1)$, but these look just like characters of $\pi_1(X)$

or i guess the bundle need not be $U(1)$

psie

21:48

I'm reading the theorem that every positive operator on $V$ has a unique positive square root (in LADR by Axler).

Suppose $T\in\mathcal L(V)$ is positive. Let $v\in V$ be an eigenvector of $T$. So there exists a real number $\lambda\geq0$ such that $Tv=\lambda v$. Now let $R$ be a positive square root of $T$. Axler proves $Rv=\sqrt{\lambda}v$, because

> "This will imply that the behavior of $R$ on the eigenvectors of $T$ is uniquely determined. Because there is a basis of $V$ consisting of eigenvectors of $T$ [...] this will imply that $R$ is uniquely determined."

I struggle with understanding these two sentences. What does it mean for the behavior of $R$ on the eigenvectors of $T$ to be uniquely determined?

psie

21:59

I've looked in a couple of places for a proof of the uniqueness of the square root of a positive semi-definite matrix, but the proofs I think are quite difficult.

Ben Steffan

@Thorgott Do you know of an argument that $H^{\dim M}(M) \cong \mathbb{Z} / 2$ when $M$ is a non-orientable compact manifold that isn't Bredon's?

It's not hard to show it's either $\mathbb{Z} / 2$ or $\mathbb{Z} / 4$, but I'm struggling to rule out the second case

Ben Steffan

22:40

actually I cannot rule out $\mathbb{Z} / 2^k$ I think

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