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Martin Sleziak
Martin Sleziak
01:47
I'll repost this here in the main chatroom, maybe somebody will be able to help with that problem:
in Calculus and analysis, 15 hours ago, by johnny09
If we have a $1$st order ODE $\dot{u}=f(u)$ where $f(u)=u^2-4$ how can we do the linearisation about a fixed point?
@BalarkaSen When I read in his profile that he described himself as a failed mathematician, then I do know what I am.
user276387
user276387
02:22
Is there a faster method than the Euclidean algorithm for finding the $\gcd$?
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
10 hours ago, by Partly Putrid Pile of Pus
Is the transcendence base of $E/F$ just the subset of $E$, consisting of all transcendentals of $E$ over $F$?
Martin Sleziak
Martin Sleziak
02:38
@PartlyPutridPileofPus I do not not know anything about this topic. But just by having look at what Wikipedia says about this, it seems that the basis definitely does not contain all transcendentals. Elements of the basis have to be algebraically independent.
They list $\pi$ as a basis of $\mathbb Q(\sqrt2,\pi)$ over $\mathbb Q$.
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
And algebraic independence of a set of $n$ elements, $\{a_1,\cdots,a_n\}$ means that there exists no $f(x_1,\cdots,x_n)\in F[x_1,\cdots,x_n]$ such that $f(a_1,\cdots,a_n)= 0$
And the set of all transcendental elements should satisfy this property, but I could be wrong. Surely there are no algebraic elements in the set
Oh yes I see. There are many transcendentals generated by $\pi$...
Martin Sleziak
Martin Sleziak
No non-trivial $f$. (I.e. non-zero?)
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
Yes, forgot to write that
Martin Sleziak
Martin Sleziak
Like $\pi$, $2\pi$, $\pi+1$. Again, Wikipedia seems to say something along those lines: en.wikipedia.org/wiki/Algebraic_independence
L33ter
L33ter
02:54
I would like to discuss something related to CW complex
I am trying to wrap my head around the way we do the identification
@MartinSleziak familiar with algebraic topology ?
Martin Sleziak
Martin Sleziak
No. But a few people who know about this area frequent this chat. There even is a chat room called Algebraic Topology & Homological Algebra.
L33ter
L33ter
oh oke
 
3 hours later…
Balarka Sen
Balarka Sen
06:14
@L33ter What is le question?
Julian Rachman
Julian Rachman
I asked this in the Homotopy Theory chat room and recieved no answer so i guess I will ask it here.
Given the example,
Let $[n]$ be the finite chain poset $[0\to1\to\codts\to n]$. Then define an infinite sequence
$F:\text{Seq}\to C$
in $C$ to be essentially finite if $F$ factors through some $[n]$, i.e., as the composite
$\text{Seq}\to[n]\to C$
What does it mean to "factor through?"
Balarka Sen
Balarka Sen
A map $f : X \to Y$ is said to factor through $Z$ if there are maps $p : X \to Z$ and $q : Z \to Y$ such that $f = q \circ p$.
Julian Rachman
Julian Rachman
Ok. So I was given this this other day:
Categorically, you can represent an infinite sequence in C as a functor F: Seq-->C, where Seq is the posetal category [0-->1-->2-->...]. Would you agree that F is equivalent to an infinite decreasing sequence iff it does not factor through a finite chain category [0-->1-->...-->n]?
Does this mean that we can show properties or characteristics from one category to another using this method?
Balarka Sen
Balarka Sen
There's a lot of undefined words in there, but I am no category theorist. You'd have to ask someone else.
@MartinSleziak :) I believe both you and him are good mathematicians.
Julian Rachman
Julian Rachman
06:30
Ok. So even if I defined everything you still would not be able to provide me with an answer?
Balarka Sen
Balarka Sen
I don't know. I am not reading the question carefully, mostly because I do not want to.
@L33ter I answered.
Julian Rachman
Julian Rachman
Well... Ok.
Know of anyone that I could ask?
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
Why don't you ask it on the main site?
Balarka Sen
Balarka Sen
Sorry, I don't. We don't have too many category theorists here: most of them belong in the homotopy theory chat. If they're not answering, probably they are not interested either.
@PartlyPutridPileofPus's suggestion is good.
Julian Rachman
Julian Rachman
I got downvoted a lot for my first attempt and ended up deleting the question
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
06:39
Perhaps write out like so:

* Question

* Context

* Definitions

* Thoughts

Or some rearrangement.
Julian Rachman
Julian Rachman
Ok. I'll form a question out of that format.
Mike Miller
Mike Miller
I give up on this question. Maybe someone can figure out how to finish my answer.
Julian Rachman
Julian Rachman
But actually, my problem does not require that much as you saw above.
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
@JulianRachman Regardless, I doubt you will be downvoted for that format. Think of it as covering your backside more than anything else.
Balarka Sen
Balarka Sen
@MikeMiller Interesting idea.
Julian Rachman
Julian Rachman
06:57
@PartlyPutridPileofPus Got it. :)
Julian Rachman
Julian Rachman
07:32
@PartlyPutridPileofPus Done: math.stackexchange.com/questions/1632880/…. Thanks again. Let's see what the turnout is going to look like.
 
3 hours later…
Idle001
Idle001
10:41
cant understand the proof about the naturality of binomials
Huy
Huy
11:34
@BalarkaSen: are you still alive?
Martin Sleziak
Martin Sleziak
12:20
When I saw this question, it reminded me that I have heard or read somewhere that some mathematician re-invented differential and integral calculus on his own as a kid. He had his own symbols for derivatives and integrals, but he was able to get at least some basic results. Anybody heard the same story? Do you remember the name? (I think he was Russian.)
Huy
Huy
@DanielFischer: "Let $p: \tilde{S} \to S$ be any covering space. By a lift of a closed curve $\alpha$ to $\tilde{S}$ we will mean the image of a lift $\mathbb{R} \to \tilde{S}$ of the map $\alpha \circ \pi$, where $\pi: \mathbb{R} \to S^1$ is the usual covering map. Note that a lift is different from a path lift, which is typically a proper subset of a lift." - what's a path lift?
is that what I get if I take a path in $S$ with $p$ on the path and then pick some point "above" $p$ in $\tilde{S}$ and then get a unique path in $\tilde{S}$ through that point above $p$?
Idle001
Idle001
i think that comes togather along with huge time/sacrifice/vain effort, people can reach same deductions in maths without inter-contacting that proves the uniqueness of maths
user189740
12:39
Anyone with a moment or two for a group theory question?
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
@byteofthat People pop in and out from my observation. If you ask, people may or may not answer depending on how busy or interested they are. Ask away.
Daniel Fischer
Daniel Fischer
12:59
@Huy I would guess that they mean a (continuous, of course) map $\tilde{\beta}\colon [0,1] \to \tilde{S}$ with $p\circ \tilde{\beta} = \beta$, where $\beta \colon [0,1]\to S$ is a path.
Huy
Huy
ok, so what I meant
@DanielFischer: do you understand their notion of a lift? it seems later on that these lifts are not unique, but I don't exactly understand how to get different lifts
Martin Sleziak
Martin Sleziak
We have moved with byteofthat into the algebra room. And his question is basically the proof that reduced residue system is a group with multiplication.
Just in case somebody will have some comments on that. (BTW I am surprised I did not find this question on the main. Probably I have to learn to search better; since this is rather common question.)
user189740
@MartinSleziak thanks for notifying everyone here
Martin Sleziak
Martin Sleziak
That was more an effort to advertise that room, since it is not used that much.
Daniel Fischer
Daniel Fischer
@Huy If you have two coverings $p_1 \colon E_1 \to B_1$ and $p_2 \colon E_2 \to B_2$, and a (continuous) map $f \colon B_1 \to B_2$, then, under suitable hypotheses, there is a continuous map $\tilde{f}\colon E_1 \to E_2$ with $p_2 \circ \tilde{f} = f\circ p_1$. That sort of lift they mean. And of course you can compose such a thing with deck transformations to get a different lift.
Huy
Huy
13:08
@DanielFischer: Are you sure? They never talk about two coverings, but maybe they implicitly mean it. For more context, later on they write "suppose $\tilde{S}$ is the universal cover and $\alpha$ a simple closed curve in $S$ that is not a nontrivial multiple of another closed curve. In this case, the lifts of $\alpha$ to $\tilde{S}$ are in natural bijection with the cosets in $\pi_1(S)$ of the infinite cyclic subgroup $\langle \alpha \rangle$."
Daniel Fischer
Daniel Fischer
@Huy They explicitly mention the covering $\pi \colon \mathbb{R}\to S^1$. I'm not sure I guess right, but going by what they say about lifts, and that they distinguish them from path lifts, that would make sense.
Huy
Huy
13:22
ok, it makes sense now
I had to draw the diagram, composing three maps is too much for my brain
Daniel Fischer
Daniel Fischer
$$\require{AMScd}\begin{CD} E_1 @>\tilde{f}>> E_2 \\ @Vp_1VV @VVp_2V \\ B_1 @>>f> B_2\end{CD}$$
Huy
Huy
:D
thanks
is the fact that the deck transformation group of a universal cover is isomorphic to the fundamental group trivial?
(it is just stated here and I don't immediately see why)
Daniel Fischer
Daniel Fischer
13:41
Trivial would be an exaggeration. But take a deck transformation $t$, and a path from $x_0$ to $t(x_0)$. Pushing that path down gives you a loop in $p(x_0)$, and that loop is nullhomotopic if and only if its lift - the original path - is closed, so if and only if $t = \operatorname{id}$. Verify that this map is a group homomorphism. Then you already have an embedding $\operatorname{Deck}(p) \hookrightarrow \pi_1(B,p(x_0))$.
For surjectivity, take $[\alpha] \in \pi_1(B,p(x_0))$ and lift $\alpha$ to the cover, starting at $x_0$. Let $x_1 = \tilde{\alpha}(1)$ be the end point of that lift. Show there is a deck transformation $t_{\alpha}$ with $t_{\alpha}(x_0) = x_1$ - you can skip that if you already know that the deck transformation group acts transitively on the fibres. Of course that's only true if the spaces are nice enough (sufficiently connected etc.).
iwriteonbananas
iwriteonbananas
13:55
Hey @DanielFischer. The operator $$T:L^2(\Omega)\to L^2(\Omega),\quad f\mapsto \int_{\Omega} k(-,y)f(y)d\mu(y)$$ is compact, right? Here $(\Omega,\mathcal A,\mu)$ is some measure space and $k\in L^2(\Omega\times \Omega)$.
Daniel Fischer
Daniel Fischer
@iwriteonbananas I'm not entirely sure whether we must assume $\mu$ nice enough, or if it holds without restrictions on $\mu$. But at least for decent $\mu$ it holds. If $(\psi_i)$ is an ONB of $L^2(\Omega,\mu)$, then - at least for decent $\mu$ - $(\psi_i \otimes \psi_j)$ is an ONB for $L^2(\Omega\times \Omega,\mu\otimes\mu)$, where $(\psi_i\otimes \psi_j)(x,y) = \psi_i(x)\cdot \psi_j(y)$. And then you have an easy approximation by finite-rank operators.
iwriteonbananas
iwriteonbananas
@DanielFischer Hm, ok what will the approximation by finite-rank operators be in that case?
Balarka Sen
Balarka Sen
@Huy barely
Daniel Fischer
Daniel Fischer
@iwriteonbananas Write $k = \sum_{i,j} c_{ij} \psi_i \otimes \psi_j$. Take a suitable finite subset of the $(i,j)$.
Balarka Sen
Balarka Sen
@Huy the deck transform. group acts on the fiber. the fundamental group acts on the fiber. you use the two.
iwriteonbananas
iwriteonbananas
14:12
@DanielFischer I see, thanks for the help.
user3165667
user3165667
14:27
Hi to everyone! What is the definition of a trivial intersection ? I googled but havent really found anything definitive. My guess, based of search results, that a trivial intersection it is an intersection whose result is one element, is it correct ?
Huy
Huy
context?
user3165667
user3165667
general topology
Huy
Huy
that's almost as specific as "mathematics"
user3165667
user3165667
the proof of Tychonoff ’s Theorem for the general case
topologywithouttears.net here is the book the proof is on page 257
Yavor Paunov
Yavor Paunov
Hi sweets, can anyone provide some help in R?
s.harp
s.harp
14:43
@user3165667 I cannot open your webpage, but I would say the intersection is trivial if it is the empty set, ie the sets are disjoint
user3165667
user3165667
Thank you!
s.harp
s.harp
If you are talking about subgroups or subvector spaces though then "trivial intersection" means that the intersection is the unit or the 0 element respectively
Balarka Sen
Balarka Sen
15:03
Hi @JulianRachman.
Julian Rachman
Julian Rachman
@Balarka Hello
s.harp
s.harp
Does anybody know of an example of a proper ideal that is not contained in a proper maximal ideal? (Preferably in the setting of normed algebras)
Mike Miller
Mike Miller
@s.harp: If you believe Zorn's lemma, that's not possible.
Huy
Huy
morning @Mike
Mike Miller
Mike Miller
Morning.
Huy
Huy
15:11
@Mike you said once that you like the primer on mcgs from farb & margalit
have you read all of it?
Mike Miller
Mike Miller
No. I thought the parts I read were well-written and I liked the table of contents.
Huy
Huy
ok
Balarka Sen
Balarka Sen
@Huy What are you studying?
Huy
Huy
@BalarkaSen: said book
Mike Miller
Mike Miller
what's up?
Balarka Sen
Balarka Sen
15:13
ah, ok.
Huy
Huy
but my topology knowledge is ... lacking
not sure if I can already tackle it
Balarka Sen
Balarka Sen
I think you can pick up the needed topology.
s.harp
s.harp
@MikeMiller In the beginning of the book by Murphy on C* Algebras he notes that zorns lemma gives that every proper modular ideal is contained in a maximal ideal. Since he specifically mentions that it holds for modular ideals I would have thought that there must be counterexamples for non-modular ideals
Mike Miller
Mike Miller
You'll have to remind me what a modular ideal is.
s.harp
s.harp
an ideal $I$ in an algebra $A$ is modular if there exists an element $u$ in $A$ so that for all $a$ in $A$ $a -a u$ lies in $I$
sorry both $a-au$ and $a-ua$ must lie in $I$
Mike Miller
Mike Miller
15:17
@s.harp: I think I see my error; my claim above that ideals are contained in a maximal ideal only works if the algebra is unital, but I'm guessing your operator algebra needn't be.
Huy
Huy
@Balarkasen: where do I go to revise basics on covering spaces, deck transformations and the relationship to fundamental groups? just munkres or do you know of something more suited?
Balarka Sen
Balarka Sen
@Huy Hatcher :)
Huy
Huy
his algtop book on his website?
Balarka Sen
Balarka Sen
Nod.
Huy
Huy
I wish I had taken algtop at some point
Balarka Sen
Balarka Sen
15:18
It can be a bit verbose, but just read the theorems and proofs if you don't want to read those.
Hatcher gives nice intuition though.
s.harp
s.harp
@MikeMiller sorry to bother you, but do you know where I can find a proof of how the maximal containing ideal exists (in unital algebra or modular ideal case)? I think that would help me find counter examples
Balarka Sen
Balarka Sen
off-topic: i can't understand any of the recent questions of this full-twist guy.
sounds like he got some really twisty terminologies.
Mike Miller
Mike Miller
@s.harp: Given an increasing chain of ideals $I_n$, their union is clearly an ideal. It's a proper ideal because the union does not contain 1 since none of the ideals $I_n$ contain 1. Apply Zorn.
Huy
Huy
@BalarkaSen: please never become a comedian
Balarka Sen
Balarka Sen
i thought that was a better pun than half of Ted's ones?
Mike Miller
Mike Miller
15:22
Are operaror ideals supposed to be norm-closed?
s.harp
s.harp
@MikeMiller in my definitions not, since the algebras themselves also need not be complete
Huy
Huy
@BalarkaSen: if half of his puns are terrible and half are incredible, then yours could still be terrible
Mike Miller
Mike Miller
Gotcha.
Balarka Sen
Balarka Sen
"I wish half of his puns would be half as terrible as half of mine, but then i don't know half of his good puns half as well as mine"
Mike Miller
Mike Miller
15:44
@s.harp OK, I don't have a counterexample but I've convinced myself they exist. I'm gonna stop thinking about this.
s.harp
s.harp
@MikeMiller thank you for your time, I think I have gained more understanding for the topic ^ I will ask the question on the website also since it doesn't look trivial anymore
Mike Miller
Mike Miller
15:58
@s.harp I have an example.
Take $R$ to be the ring of "polynomials in $\Bbb Q^+$". Explicitly, elements are finite sums $\sum a_i T^{\lambda_i}$, where $\lambda_i$ is positive and rational. The sum and product are what you expect.
Let $I_a$ be the ideal consisting of those elements such that $\lambda_i > a$. Another way of putting this is that it's the ideal generated by $T^a$.
No proper ideal contains all of the $I_a$.
So Zorn fails here. My money is that there are no maximal ideals. You should verify this.
s.harp
s.harp
How much money are we talking about?
But thank you I will look into it
Mike Miller
Mike Miller
One dollar and thirty three cents.
morphic
morphic
16:18
Grad school life bank account balance
Andrew Thompson
Andrew Thompson
16:30
Hellu @MikeMiller.
iwriteonbananas
iwriteonbananas
@DanielFischer By the way, do you know an example of an integral operator as above which is compact despite $k\notin L^2(\Omega\times \Omega)$?
Mike Miller
Mike Miller
Hi @AndrewT.
iwriteonbananas
iwriteonbananas
Hey Mike
Mike Miller
Mike Miller
morning
I'm an artist
I'm an artist
16:45
Are you done with this one?
user image
Don't miss it ...
Daniel Fischer
Daniel Fischer
@iwriteonbananas No. It's already not easy to pick a $k \notin L^2(\Omega\times \Omega)$ such that $$x \mapsto \int_{\Omega} k(x,y)f(y)\,d\mu(y)$$ defines an operator $L^2(\Omega) \to L^2(\Omega)$. If that is possible at all.
iwriteonbananas
iwriteonbananas
@DanielFischer Ah, good point
I'm an artist
I'm an artist
By the way, I sent the above integral to more students and professors and no one did it so far. The interesting thing is that the integral is not hard. What is the missing picture in the puzzle then?
@AndrewThompson can you tell me a single useful step for the integral above? Not a full solution.
Andrew Thompson
Andrew Thompson
17:01
@I'manartist No.
Mike Miller
Mike Miller
@AndrewT: How are your courses going?
What are you taking again?
I'm an artist
I'm an artist
@AndrewThompson OK
Andrew Thompson
Andrew Thompson
@MikeMiller Pretty well so far, although the undergraduate project is the only one I've really started. Lie Theory, K-theory and undergradthesis.
Mike Miller
Mike Miller
What are you working on there? You said something about homological algebra.
Andrew Thompson
Andrew Thompson
Yes, more precisely hochschild homology. I want to see if some results from the Auslander-Reiten world can simplify computations in some particular cases.
At the moment I'm learning about Morita equivalences and functors between module categories in general.
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
17:06
@I'manartist Why would one calculate this?
Mike Miller
Mike Miller
@AndrewThompson: So this is some sort of representation theory, I guess?
I'm an artist
I'm an artist
@PartlyPutridPileofPus For at least the same reasons one would do other math stuff as crazy as it seems at first sight mine, no matter it's about different fields.
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
@I'manartist Sorry, I didn't understand this?
Andrew Thompson
Andrew Thompson
@MikeMiller I do know reptheory can be useful in this setting, although the relation is as of yet not entirely clear to me.
I'm an artist
I'm an artist
What I post here is perfectly as fine as any other math others post here.
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
17:09
Oh, I don't know what you are responding to. I meant, what is the use of solving it?
I'm an artist
I'm an artist
@PartlyPutridPileofPus What would you expect to find after solving a math problem? What is the use you have in mind for the math you do?
@PartlyPutridPileofPus My math is an art to me, so I do art.
Partly Putrid Pile of Pus
Partly Putrid Pile of Pus
Okay, so you are doing recreational mathematics, that was the (type of) answer I was looking for.
I was just curious about the 'without pen and paper'. (I have no idea how that would be possible)
Andrew Thompson
Andrew Thompson
@MikeMiller What I want to learn on the homological algebra is definitely closely linked to representation theory, though. (More precisely I want to work in the stable module category, Auslander-Reiten's Representation Theory of Artin Algebras is probably the canonical reference.) The Hochschild homology part of things: not so sure.
I'm an artist
I'm an artist
@PartlyPutridPileofPus What is the use you have in mind for the math you do?
Mike Miller
Mike Miller
@AndrewThompson I think the representation theorists are into Hochschild stuff.
But I don't understand, algebra, so : o )
I'm an artist
I'm an artist
17:14
Can you name a great mathematician for which mathematics wasn't considered an art to him/her? I ony want to know out of curiosity.
The question is available to all users here.
Just name the mathematicians and I immediately write their names down. To recollect once in a while that in their case wasn't about art.
BBL (I have to finish some stuff)
Balarka Sen
Balarka Sen
@AndrewThompson Want to tell me what Hochschild homology is about?
Andrew Thompson
Andrew Thompson
@BalarkaSen I can tell you that Wikipedia is inaccurate :)
Jokes aside,
I know the definitions and can compute some simple cases. I know its important to (our) topologists because it acts as an invariant in K-theory in some precise sense.
Balarka Sen
Balarka Sen
@AndrewThompson Heh.
@AndrewThompson Ah, ok. I was hoping for a small expository, though. E.g., what is it about and why is it relevant to algebra.
If you're not busy that is.
Andrew Thompson
Andrew Thompson
Alright, gotta do some stuff, back in five minutes. Then I can tell you what little I know (so far, in a few months the plan is I will know significantly more.)
Balarka Sen
Balarka Sen
Cool!
Thanks in advance.
Andrew Thompson
Andrew Thompson
17:26
No problem. On second thought: I could just send you my project outline which gives an exposition on what I want to do. Do you have an e-mail?
got it.
Hello @TedShifrin!
morphic
morphic
hi @TedShifrin
Ted Shifrin
Ted Shifrin
Hi @AndrewT, mr eyeglasses, Balarka
Balarka Sen
Balarka Sen
Hi @TedShifrin!
I am ill again :(
Ted Shifrin
Ted Shifrin
stop that!
morphic
morphic
@BalarkaSen you always get sick so often
Ted Shifrin
Ted Shifrin
17:41
Math makes him sick — no sleep.
Haven't seen you in ages, mr eyeglasses
Huy
Huy
it's not math, it's his decision to not follow a somewhat normal sleeping pattern
morphic
morphic
Was just a little depressed for a while
Balarka Sen
Balarka Sen
@TedShifrin Trying :(
Ted Shifrin
Ted Shifrin
There's correlation, perhaps not causation ....
Sorry to hear that, mr eyeglasses
Balarka Sen
Balarka Sen
@morphic Hi :)
What's up?
@TedShifrin: It's because of Huy.
Scroll to 9:50.
Mike Miller
Mike Miller
18:21
Morning @Ted.
Anthony
Anthony
Is anyone here into foundations?
Mike Miller
Mike Miller
I don't use much makeup, no.
5
No, but really, @Anthony, you should just go ahead and ask.
Semiclassical
Semiclassical
18:56
afternoon chat
Mike Miller
Mike Miller
19:18
Start with a reasonably nice topological space with a reasonably nice involution $i$ on it. (This includes eg smooth manifolds with smooth involutions.) Then if the space is path-connected, so is the fixed point set. I wonder if there's an elementary proof of that.
s.harp
s.harp
@MikeMiller what about $S^1$ with $i(z)=1/z$? Fixpoints are $1,-1$. The map is smooth, and $S^1$ is a very nice space that also is path connected
Mike Miller
Mike Miller
Huh. I'm confused.
Ah, ok, I see my error. Whoops! Thanks.
What I said is true if the space is contractible (or just has trivial $\Bbb Z/2$-homology). But the need for that assumption makes me feel it's unlikely to have a simple proof.
Balarka Sen
Balarka Sen
I am worried if contractibility is sufficient. What if I take cone on a Hawaiian earring and reflect along the obvious hyperplane?
Ted Shifrin
Ted Shifrin
Goodnight @MikeM ... and hi @Anthony
Balarka Sen
Balarka Sen
You'll get some sort of a broom space - which is not path-connected.
Mike Miller
Mike Miller
19:32
But I guess this still gives the nice statement "A smooth involution on $\Bbb R^n$ has path-connected fixed point set". ($\Bbb Z/2$-acyclic, even.)
Ted Shifrin
Ted Shifrin
I think the fixed point set is connected in that case, @Balarka.
Balarka Sen
Balarka Sen
Connected, yep. But not path-connected.
Ted Shifrin
Ted Shifrin
And path-connected, unless your cone on Hawaii is different from mine. Everything connects at the vertex point.
Oh, are you doing a double cone?
Nah, even that works, I think.
Balarka Sen
Balarka Sen
I am not sure what you mean. By cone I mean the unreduced cone: $H \times I$ with the top $H \times 1$ pinched up. If you slice along the 2-plane of symmetry, you'll get the broom space, no?
Ted Shifrin
Ted Shifrin
But a broom is path connected.
Mike Miller
Mike Miller
19:35
Nobody should be talking about Hawaiian earrings even before I define sufficiently nice because that's obviously nice nice by virtually any definition. Sufficiently nice means there's a CW complex structure such that the involution is cellular.
Semiclassical
Semiclassical
is there a simple statement/example of what K-theory is about? I feel like I should know that much at least.
Ted Shifrin
Ted Shifrin
To me, @Semiclassic, it's about vector bundles :) But I don't think about it.
Semiclassical
Semiclassical
gotcha
Ted Shifrin
Ted Shifrin
@MikeM: If you can put a metric on the space that's involution-invariant, then the fixed point set will be a submanifold, right?
Mike Miller
Mike Miller
@Semiclassical: Consider vector bundles over a topological space. You can take the direct sum of vector bundles, so you might want to turn it into a group by adding "negative rank vector bundles" so that you have inverses. You just defined K-theory.
Ted Shifrin
Ted Shifrin
19:38
Well, you need an equivalence relation or two ...
Semiclassical
Semiclassical
'negative rank vector bundles'?
Mike Miller
Mike Miller
@Ted: What kind of metric?
Ted Shifrin
Ted Shifrin
Riemannian, @MikeM.
Mike Miller
Mike Miller
Yes, fixed point sets are submanifolds (and you can of course pick an invariant metric).
Ted Shifrin
Ted Shifrin
Oh, I see, so that only answers the question locally, not globally. Sorry.
Mike Miller
Mike Miller
19:39
@Semiclassical: Direct sum adds the rank of vector bundles. If I wanted to turn the set of vector bundles into a group, what would the inverse of the trivial rank 1 vector bundle be?
Balarka Sen
Balarka Sen
@TedShifrin Ok, I am thinking about collection of rays from origin in $[0, 1]^2$ of slope $1/n$ plus $(0, 1/2]$. If you look at all of $[0, 1]$, of course it'll be path connected. Right. But I can look at the modified $H \times I/H \times 1$ minus $x_0 \times [0, 1/2)$ where $x_0$ is the interesting point.
It also has a natural 2-plane of symmetry, reflecting along which gives me the non-path connected broom space.
@MikeMiller lol
Ted Shifrin
Ted Shifrin
What if we use stable equivalence, @MikeM?
Semiclassical
Semiclassical
not a clue. evidently its direct sum with said trivial rank 1 vector bundle would be rank zero, but beyond that
shrug
Mike Miller
Mike Miller
@Semiclassical: Have some imagination!
Semiclassical
Semiclassical
psh
Mike Miller
Mike Miller
19:41
If direct sum adds rank, it should have "rank -1".
Ted Shifrin
Ted Shifrin
@Semiclassic: Mike wants you to draw us a $-1$-dimensional vector space.
Balarka Sen
Balarka Sen
it's just standard group completion, as far as I know.
add in the inverses.
Semiclassical
Semiclassical
i don't have intuition when it comes to vector bundles, to be honest
Mike Miller
Mike Miller
@Semiclassical: Then work with vector bundles over the point.
Balarka Sen
Balarka Sen
nobody can visualize a rank -1 bundle rolleyes
Semiclassical
Semiclassical
19:42
let me rephrase that. I don't have intuition when it comes to bundles
not formally, anyways. i probably have something if I can turn it into a physics example
Mike Miller
Mike Miller
@Semiclassical: You have intuition about vector bundles over the 1-point space.
Ted Shifrin
Ted Shifrin
OK, I'm off to meet friends for lunch. Y'all have fun drawing $-1$-dimensional vector spaces.
Balarka Sen
Balarka Sen
Bubye.
Semiclassical
Semiclassical
when you say "over the 1-point space" that's the base space?
Mike Miller
Mike Miller
Yes.
Semiclassical
Semiclassical
19:43
then wouldn't a vector bundle over that just be any vector space?
Mike Miller
Mike Miller
Yes. Hence my claim that you understand them.
Semiclassical
Semiclassical
pffff
Balarka Sen
Balarka Sen
(...)
Mike Miller
Mike Miller
Think about what I'm saying for vector spaces up to isomorphism. With direct sum that's the monoid $\Bbb N$. If you wanted to make that a group, you'd add negative numbers.
Semiclassical
Semiclassical
sure. but how?
Balarka Sen
Balarka Sen
19:45
@Semiclassical A fiber bundle is a map $p : E \to B$ such that fiber over any point is $F$, i.e., $p^{-1}(x) = F$ for any $x \in B$ and $p$ locally looks like a projection map $F \times U \to U$. But you can ignore this if you know it already.
Mike Miller
Mike Miller
Of course these are not literal objects. You make K-theory by taking the so-called monoid of vector bundles and passing to the "Grothendieck group" (its just a formal construction, it's the biggest thing you can get if your monoid suddenly had inverses.)
Semiclassical
Semiclassical
yeah, that sounds sensible @balarka
Mike Miller
Mike Miller
But the "intuition" is that suddenly I've created some object called a negative rank vector bundle.
Semiclassical
Semiclassical
i can see the goal, but I guess I have to trust that said construction makes sense
Karl Kronenfeld
Karl Kronenfeld
Bleh, mathematicians.. Creating things out of thin air, as usual.
Mike Miller
Mike Miller
19:47
What Ted was saying was that you can avoid this if you're willing to let the trivial rank n vector bundle be zero in your group.
Semiclassical
Semiclassical
Which remark was that?
i couldn't tell which part of that conversation was re: K-theory and what was for involutions
Balarka Sen
Balarka Sen
Hey @KarlKronenfeld.
Karl Kronenfeld
Karl Kronenfeld
Sup.
Mike Miller
Mike Miller
You equate vector bundles that are isomorphic upon summing with a large trivial bundle. Then this thing, under direct sum, is an actual group.
Mambo
Mambo
@Semiclassical May I know about your experience with representations of locally compact groups?
Semiclassical
Semiclassical
19:49
less than i'd like
not enough to answer questions, tbh
@MikeMiller hrm
Balarka Sen
Balarka Sen
@KarlKronenfeld How's life?
Frank Science
Frank Science
Locally compact abelian groups?
Mambo
Mambo
Not non abelian also
Semiclassical
Semiclassical
what i'm confused about is that the phrase i'd expected to hear about is 'clifford algebra'
Frank Science
Frank Science
But I heard that the theory of representations for noncompact groups isn't complete.
Mambo
Mambo
19:51
I have issue with Peter Weyl theorem for compact groups
Frank Science
Frank Science
There's a theory about $SL_n$, say, in Lang's book.
Balarka Sen
Balarka Sen
@Semiclassical i was confused upon hearing that phrase in a physics conference discussing k-theory, on the other hand.
Semiclassical
Semiclassical
heh
Mambo
Mambo
For a compact group every unitary representation is on finite dimensional hilbert space
Balarka Sen
Balarka Sen
you guys are weird
Mike Miller
Mike Miller
19:53
@Semiclassical: There's a way of talking about K-theory in terms of Clifford algebras. It is a later development, not the origin of the idea of K-theory. Topological K-theort just comes from wanting to understand bundles.
Semiclassical
Semiclassical
i wondered if that was the issue
Frank Science
Frank Science
Mostly about the decomposition of $L^2(G)$ as a Hilbert direct sum for compact groups.
Mambo
Mambo
Exactly
Semiclassical
Semiclassical
applications of K-theory rather than K-theory itself, I guess?
Frank Science
Frank Science
When $G$ is Abelian, it's Fourier transformation (to the dual group).
Mike Miller
Mike Miller
19:54
No, it's not about applications. It's just a different viewpoint.
Semiclassical
Semiclassical
hmm
Mambo
Mambo
Why does $L^2(G)$ have exactly corresponding dimension of unitary repns copy
Karl Kronenfeld
Karl Kronenfeld
@BalarkaSen There are way too many ways to answer that one. Perhaps "good" will suffice?
Semiclassical
Semiclassical
is there a typical name for that perspective? I see adjectives like 'topological', 'algebraic' and 'operator' applied to K-theory
Frank Science
Frank Science
Maybe via character theory.
Balarka Sen
Balarka Sen
19:56
@KarlKronenfeld Truth be told, logic has turned you into a mutant.
Mambo
Mambo
No character theory
Karl Kronenfeld
Karl Kronenfeld
@BalarkaSen I cannot deny that.
Mambo
Mambo
When characters are considered we should consider lower class
namely central functions
Semiclassical
Semiclassical
which I suspect means "K-theory according to Kitaev"
Karl Kronenfeld
Karl Kronenfeld
@BalarkaSen I promise, I'm not so bad "in person," though.
(Not focused on logic btw)
Balarka Sen
Balarka Sen
19:57
@KarlKronenfeld I presume that means you haven't agreed with that either, mr. constructivist.
Mike Miller
Mike Miller
@Semiclassical: Those aren't reslly approaches, those are apparitions of the same idea in different fields (topology, algebra, operator algebras).
Semiclassical
Semiclassical
hmm
Balarka Sen
Balarka Sen
@KarlKronenfeld I believe you :)
Mike Miller
Mike Miller
I would probably just say "K-theory via Clifford algebras"
Semiclassical
Semiclassical
i had the impression is that "operator K-theory" would be the closest
Mambo
Mambo
20:00
@FrankScience When commutative groups are under consideration, we can definitely think about characters
Frank Science
Frank Science
OK, maybe we cannot study characters for infinite groups.
Semiclassical
Semiclassical
but anyways. is there a good place to read up on "K-theory via Clifford algebras"?
Mambo
Mambo
No we can study characters for infinite groups
Mike Miller
Mike Miller
Probably googling it will give you plenty of sources. I don't have one for you.
Semiclassical
Semiclassical
fair enough
Mambo
Mambo
20:03
For infinite compact groups, every unitary representation is finite dimensional
Define the character as trace of representation
Here characters are central functions in the sense $\chi(xyx^{-1}) = \chi(y) ; x,y \in G$
I'm an artist
I'm an artist
@Semiclassical no one succeeded with this one so far
user image
Semiclassical
Semiclassical
not surprised, it looks beastly
Frank Science
Frank Science
I have no idea on abstract harmonic analysis.
Mambo
Mambo
And these characters form basis for subspace of central functions
Frank Science
Frank Science
But is Peter-Weyl theorem easy to prove?
Semiclassical
Semiclassical
20:06
is there a reason the first and third terms in the numerator are the same?
Mambo
Mambo
yes it is easy to prove
I'm an artist
I'm an artist
@Semiclassical It's just a matter of arranging the integrand.
Frank Science
Frank Science
OK, if I have enough time, I'll try it.
Semiclassical
Semiclassical
alright
Frank Science
Frank Science
How much functional analysis is needed?
Mambo
Mambo
20:08
If you assume every unitary representation of compact group is finite dimensional every thing else will be easy
It uses spectral theorem for compact operators
Sorry it contains a finite dimensional subrepresentation
So every irreducible unitary representation is finite dimensional
Semiclassical
Semiclassical
@I'manartist I'm actually a bit confused by how Ei is being used here. Typically, one defines it on a cut plane (along the negative real axis) which makes the values for $x<0$ ambiguous in terms of their imaginary part
and if one ignores that imaginary part, then one has $-Ei(-x)=Ei(x)$. so i'm perplexed as to the purpose of those minus signs
I'm an artist
I'm an artist
20:25
Exponential integral
In mathematics, the exponential integral Ei is a special function on the complex plane. It is defined as one particular definite integral of the ratio between an exponential function and its argument. == DefinitionsEdit == For real non zero values of x, the exponential integral Ei(x) is defined as The Risch algorithm shows that Ei is not an elementary function. The definition above can be used for positive values of x, but the integral has to be understood in terms of the Cauchy principal value due to the singularity of the integrand at zero. For complex values of the argument, the definition...
@Semiclassical See convergent series
Semiclassical
Semiclassical
the $x\neq 0$ one?
I'm an artist
I'm an artist
@Semiclassical Yeap
Semiclassical
Semiclassical
hmm.
Frank Science
Frank Science
@Mambo But the multiplicities of irreducible representations are determined in the Peter-Weyl theorem
Mike Miller
Mike Miller
20:50
The OP managed to figure out how to fix my argument here. Conclusion: No matter how terrible a surface in $\Bbb R^3$ is, there's still a way to go from one side to the other while only crossing the surface once.
A modified version of this should apply to any embedded, connected polyhedron.
Mike Miller
Mike Miller
21:07
@Danu: I don't know if this will ping you.
Guess not.
Mambo
Mambo
21:20
@FrankScience Basically I want to investigate why not more than the multiplicity
Vader
Vader
21:57
user image
what does it mean to "express" something?
it just looks like the 1 was swapped for a 2
wolframalpha steps are hard to read
user174558
22:10
@Vader It means write both sides in base 2.
Vader
Vader
oh i see
ofc
user174558
Are you a native English speaker?
user174558
Hey @I'manartist.
user174558
Surprisingly few books prove that the elementary functions are transcendental.
I'm an artist
I'm an artist
22:35
@Jasper JASPER!!!
L33ter
L33ter
hi @BalarkaSen
thank you for your answer
I understand now many things
I'm an artist
I'm an artist
@Jasper JASPER!!! How are you doing?
L33ter
L33ter
I have computed the homology of Torus,Project plane, Klein bottle but there is some technicalities I don't completely understand
so today I am just gonna cover chapter 0 neatly then go back to chapter 2 re read simplicial homology again and if I have any questions I will ask you or ask on forums.
I want to go over singular homology this week or next but I don't want to do everything fast in order to have a good grasp of what is going on
I will be back in few hours when I have some questions
again thanks again @BalarkaSen
btw simplicial homology made the computations of $S^1$ sooo easy
I'm an artist
I'm an artist
There are some issues with the internet connection again.
I'm an artist
I'm an artist
23:26
I clap hands looking at these answers by Cleo
17
A: Two integral involving logarithm and polylogarithm function

Cleo$$\begin{align*}{\large\int}_0^1\frac{\ln(1-x)\,\operatorname{Li}_3\left(\frac{1+x}2\right)}xdx&=\frac{29\,\zeta(5)}{16}-\frac{19\pi^2}{96}\zeta(3)+\frac{5\,\zeta(3)}{16}\ln^22+\frac{\ln^52}{40}\\&-\frac{5\pi^2}{72}\ln^32+\frac{11\pi^4}{1440}\ln2-3\operatorname{Li}_5\left(\tfrac12\right).\\ \\ {\...

Don't understand me wrongly, the good work must be appreciated, but I would like people realize that this stuff can be done by many with enough training.
What is hard in my understanding is far away from these integrals. The more we learn the more the meaning of difficulty changes.
How a good start would look like in my view? Beat Ramanujan! Or try to do that.
Let me be clearer, maybe my English is not good enough: both problems above are easy (say, at most of average difficulty, in the worst scenario - I'm done with both).
L33ter
L33ter
23:49
@BalarkaSen ping me when you come back.

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