« first day (3267 days earlier)      last day (40 days later) » 

4:08 AM
@Ultradark Homeomorphism doesn't care about "pinches" so those are all homeomorphic
 
4:59 AM
Is Lebegue an alternative spelling of Lebesgue, or is it simply a typo/misspelling?
I have noticed Lebegue/Lebesgue is not in this list on meta.
 
5:16 AM
(But I asked mainly because I saw it in several posts on the main site.)
 
5:30 AM
Heh. Henri Leb\'egue (1856-1938) was a French palaeographer.
 
5:56 AM
@MartinSleziak it looks like Lebesgue always signed his letters to Borel with "H. Lebesgue"
 
6:51 AM
Thanks for the responses!
 
 
2 hours later…
8:43 AM
Hi. If $E[X \mid Y, Z]=E[X \mid Y]$ and $E[W \mid Y, Z]=E[W \mid Y]$, is it true that also $E[f(X,W) \mid Y, Z]=E[f(X,W) \mid Y]$ for measurable $f$?
 
@Julius does LOTUS help?
 
9:02 AM
Can you classify all possible Circle actions on spheres? Since $S^1$ is a Lie Group do all these actions generate global vector fields?
 
@LeakyNun, if it does, I don't see how. I don't think so, since those are conditional expectations. The result would be true for a single random variable since $\sigma(f(X))\subset \sigma(X)$. I think it's also true for $f(X,W)$ since, I think, each $A\in \sigma(f(X,W))$ can be written as a combination of unions and intersections of $B_i\in\sigma(X)$ and $C_i\in\sigma(W)$.
 
I mean they obviously do because you can always define the vector field as the one generated by the lie algebra of S^1 but how do these vector fields look like?
 
 
1 hour later…
10:34 AM
@RyanUnger Too many Homs
@Albas Linear actions correspond to $S^1$ subgroups of $SO(n)$. Classifying them is tantamount to understanding the maximal torus of $SO(n)$, which is the block diagonal matrix consisting of $2\times 2$ rotation matrix blocks (with a $1$ somewhere depending on the parity of $n$).
In that torus you can have many $S^1$-subgroups, by taking an appropriate rational slope curve.
For example, consider $S^1$ acting on $S^3$ by $\theta \cdot (z, w) = (e^{i\theta} z, e^{i\theta} w)$. This is such an action, it's the Hopf fibration explicitly.
But this corresponds to the $(1, 1)$ curve in the 2-dimensional maximal torus of $SO(4)$. You can take, I dunno, a $(p, q)$ curve. Explicitly, $\theta \cdot (z, w) = (e^{i\theta} z, e^{i\theta p/q} w)$, something like that.
I don't know about nonlinear actions.
You can probably do crazy shit. Take a homology $3$-sphere $X$ and look $S^2 X$. By the double suspension theorem, $S^2 X \cong S^5$. There's always a circle action on a suspension, you can rotate along the meridians. I bet this is a nonlinear $S^1$-action on $S^5$.
No, the above was nonsense. Scratch.
 
11:04 AM
@Mathein hab mich jetzt fuer die Studentenwohnheime beworben lol, mal schauen obs noch was gibt
 
@ÍgjøgnumMeg cool!
 
Die moderneren Studentenwohnheime in der Bahnstadt schauen mal fein aus
lol
 
If you take various complex line bundles on $S^2$, the total spaces of their unit circle bundles are linear quotients of $S^3$, I think, but I don't have an argument off the top of my head
For example $O(2)$ is the unit tangent bundle, and the total space of it's unit circle bundle is $\Bbb{RP}^3 = S^3/\Bbb Z_2$.
I think for $O(4)$ it's $S^3/Q_8$.
 
@ÍgjøgnumMeg Bahnstadt ist generell ein edles, modernes Viertel
 
Yeah alright. Circle bundles over $S^2$ are classified by the linking number of the fibers. So they are all quotients of the Hopf fibration $S^3 \to S^2$, which is $\Bbb Z_n$-equivariant, by $\Bbb Z_n$.
Wait
Yeah that seems right. So total space of unit bundle of $O(4)$ is... $S^3/\Bbb Z_4$? They are all lens space!
Alright, so this gives some picture of how some linear $S^1$-actions on $S^3$ behave. Look at $S^3 \to S^3/\Bbb Z_n \to S^2$, where the last map is the bundle projection of $O(n)$. Let $S^1$ act on $S^3/\Bbb Z_n$ by acting principally on the circle fibers of $S^3/\Bbb Z_n \to S^2$.
This lifts to an $S^1$-action on $S^3$, right? Use the $n$-fold covering map $S^1 \to S^1/\Bbb Z_n = S^1$ and a big commutative diagram.
Hm, so now to fix the nonlinear nonsense. I feel like I can cook up a nonlinear circle action on spheres someway, somehow.
 
11:28 AM
@Mathein schaut cool aus, ist wahrscheinlich besser als privates Leben
lol
 
 
1 hour later…
12:32 PM
A quick question: Does $\bigoplus_{n=1}^{\infty}\mathbb{Z}$ have uncountable cardinality?
I want to say yes because allow $f:\bigoplus_{n=1}^{\infty}\mathbb{Z}\to\mathbb{R}$ to be defined such that $f(x)=\text{lim}_{n\to\infty}\frac{x_{2n}}{x_{2n+1}}$
Where $x=(x_1,x_2,\cdots ,x_i,\cdots)$
Since this includes all rational numbers as well as the accumulation points of rational numbers, I believe the function is onto, but I feel like there might be some oversight on my part, because $\bigoplus_{n=1}^{\infty}\mathbb{Z}$ looks like it would be a countable union of countable sets.
(Actually, verified it by just doing a google search. It's the set of sequences of integers. Easy enough.)
 
12:50 PM
11
Q: Conjecture: "For every prime $k$ there will be at least one prime of the form $n! \pm k$" true?

MathphileUsing PARI/GP, I searched for primes of the form $n!\pm k$ where $k \ne 2$ is prime and $n\in \Bbb{N}$. With the help of user Peter, we covered a range of $k \le 17028457$ and couldn't find a prime $k$ for which $n!+k$ has no primes. Observations: $(1)$ When $n \ge k$, $n! \pm k$ cannot be ...

Any ideas, heuristics or guesses?
 
@Rithaniel For every $x\in\bigoplus\Bbb Z$ there is $N\in\Bbb N$ with $x_k=0$ for all $k\geq N$
So in particular $|\bigoplus_{n\in\Bbb N}\Bbb Z|=\sum_{n\in\Bbb N}|\Bbb Z|^n=\sum_{n\in\Bbb N}|\Bbb Z|=\aleph_0\cdot|\Bbb Z|=\aleph_0$
 
Is this a syntax error on my part? Because I know what you're referring to, and I meant to be specifying the group where infinitely many elements can be nonzero.
 
Then you're talking about $\prod_{n\in\Bbb N}\Bbb Z$, a very different object
 
Okay, then yeah, I was using the wrong syntax.
 
1:05 PM
That one is uncountable, in fact it has the same cardinality as the reals
 
I suspected that might be the case, and was trying to formulate a surjection in either direction to prove there existed a bijection, but the other direction wasn't coming to me.
 
That's $|\Bbb Z|^{|\Bbb N|}=2^{|\Bbb N|}=2^{\aleph_0}=|\Bbb R|$ just by cardinal arithmetic
 
I've not had any course which introduced me to cardinal arithmetic, unfortunately.
 
@Rithaniel Then you can inject $[0,1]$ into $\prod\Bbb Z$ by sending $0.a_1a_2a_3a_4\cdots$ to $(a_1,a_2,a_3,\cdots)$, it doesn't matter what you do with number with two decimal expansions, just pick of the them
Except with $1$ for which you want to use $0.\bar{9}$. Or inject $[0,1)$ instead
 
1:14 PM
But what if what I want is to surject $[0,1]$ or $[0,1)$ or $\mathbb{R}$ onto $\prod\mathbb{Z}$?
 
I think it's easier to inject $\prod\Bbb Z$ into $\Bbb R$
Actually the easiest is probably to inject $\prod\Bbb N$ into $\Bbb R$ with something like $a=(a_0,a_1,a_2,\cdots)$ goes to $0.a_0a_1a_2\cdots$ if $a$ is not eventually constantly $9$ and it goes to $3.a_0a_1a_2$ if $a$ is eventually constantly $9$
(and $\prod\Bbb Z$ and $\prod\Bbb N$ have the same cardinality because you can pick a bijection $\Bbb N\to\Bbb Z$ and use that on each coordinate to get a bijection)
 
Hm... I don't really follow the last one; if you meant decimal expansion then you cannot have arbitrarily large digits.
 
Oh lol right, there's stuff bigger than $9$ in $\Bbb N$
I guess it's time for continued fractions then
 
Perhaps something similar would work with continued fractions, but that's probably overcomplicating things. Something like that is here: Baire space homeomorphic to irrationals.
 
So, the theorem is that if there exists a surjection from A to B and a surjection from B to A, then there exists a bijection between then, but having an injection from A to B doesn't mean you have a surjection from B to A, right? Like, I can inject from $\mathbb{R}$ to $\mathbb{N}$ by taking every real number to it's first digit.
 
1:24 PM
Anyway, cannot we simply use that $\mathbb Z$ can be embedded into $\{0,1\}^{\mathbb N}$ and then use that $\{0,1\}^{\mathbb N}$ has the same cardinality as $(\{0,1\}^{\mathbb N})^{\mathbb N} \cong \{0,1\}^{\mathbb N\times\mathbb N}$?
 
The theorem is that if there is an injection $A\to B$ and an injection $B\to A$ then there is a bijection $A\to B$. With choice we can use surjections instead. If there is an injection $A\to B$ then there is a surjection $B\to A$. With choice if there is a surjection $A\to B$ then there is an injection $B\to A$
 
@Rithaniel Did you mean to write about surjection from $\mathbb R$ to $\mathbb N$ rather than injection?
 
Ah, right, that wouldn't be an injection, my mistake.
My brain for some reason had me thinking injection meant "surjects onto a subset."
Alright, yeah, thinking this stuff through with the proper definition in my head again, the world makes sense again.
 
1:40 PM
@BalarkaSen @Albas Fibered knots allow you to construct nonlinear circle actions on $S^3$ and $S^4$. I think these were first found in dimension 4 by Pao.
 
Are non linear circle actions not possible on $S^2$?
This brings me to another question whose answer I believe is in the negative but given a smooth manifold, is it true that every global vector field possible on the smooth manifold can be generated by the action of some Lie group?
 
2:26 PM
@BalarkaSen do you understand the point of these Lie theory papers
there's one on math.dg every day
I have no idea what their goal is
 
2:47 PM
Speaking of injections
My arm hurts
but at least I won't get chickenpox
By the way
If you're telling someone about Cantor stuff, and you say "There's more than one infinity"
 
3
Q: Does there exist some prime $k$ for which there will be exactly two primes of the form $n!+k$?

MathphileThis is a question related to my recent question Conjecture: “For every prime $k$ there will be at least one prime of the form $n!\pm k$” true? Using PARI/GP I searched for the number of primes of the form $n!+k$ possible for each prime $k\le 2300$. I observed that for some prime $k$, all number...

any ideas about this?
 
and they ask "How many infinities are there?"
How would you respond?
If you say "Infinitely many" the natural response, given what you've just told them, is "Which infinity"
In this imaginary conversation I'm imagining myself muttering something about Russell's paradox and losing them
In any case that's the vaccines done, now to go to the glasses place to get my eyes checked
 
3:04 PM
Alan Turing revealed as the new face of the £50 note!
 
@Albas No. You can prove this by hand by studying what the possible fixed point sets are
 
Hi Mike.
 
@Albas If the vector field has a flow for all time then flowing defines an action of R whose derivative at zero is the vector field. If you don't have a flow for all time then that's not possible.
Hi. I'm just responding to a question.
 
@AkivaWeinberger "many" would be my response.
But I think Russell's paradox is pretty accessible to the layman.
 
3:25 PM
@MikeMiller Ah, yes the one parameter way of talking about flows. Thanks.
 
Given a set $\Omega$, a ring of sets $\mathcal{R}$ on $\Omega$ and a pre-measure $\mu$ on $\mathcal{R}$, is there a good criterion to affirm the existence of non-$\mu$-measurable sets (assuming AC)? I'm particularly interested in the case $\Omega=\mathbb{R}$, $\mathcal{R}$ the ring of finite unions of half-open intervals and $\mu$ being absolutely continuous w.r.t. the Lebesgue measure.
 
3:45 PM
@MikeMiller Huh I see.
@Albas Any complete vector field can be thought as a homomorphism R -> Diffeo(M), like Mike said.
 
@BalarkaSen Yea, I saw it was written in my notes. Must have forgotten
Watcha upto @BalarkaSen?
 
Hm, if I let $S^1$ act on $S^2$ by rotation, and then I take suspension of this I get an $S^1$-action on $S^3$ which fixes a circle, right?
@Albas Nothing much, freewheeling.
Can I get $S^1$-actions on $S^3$ which fixes a knot?
Then I can suspend to get an $S^1$-action on $S^4$ which fixes a locally non-flat submanifold. That cannot be conjugate to a smooth action.
I dunno, weird questions maybe
 
4:08 PM
The following is the metric tensor for the $2$-sphere -----> $ds^2 = d\theta^2+\sin^2(\theta)d\phi^2$
 
There are no smooth $\Bbb Z/n$-actions on $S^3$ which fixes a knot.
In mathematics, the Smith conjecture states that if f is a diffeomorphism of the 3-sphere of finite order, then the fixed point set of f cannot be a nontrivial knot. Paul A. Smith (1939, remark after theorem 4) showed that a non-trivial orientation-preserving diffeomorphism of finite order with fixed points must have fixed point set equal to a circle, and asked in (Eilenberg 1949, Problem 36) if the fixed point set can be knotted. Friedhelm Waldhausen (1969) proved the Smith conjecture for the special case of diffeomorphisms of order 2 (and hence any even order). The proof of the general case...
This is too complicated for me lol
 
@BalarkaSen doesn't that say it can't fix only a knot
 
That's what I mean.
I'm not a precise person
I mean, otherwise the question is dumb lol.
Take the involution on $S^3$ which switches the two solid torii in it's genus 1 Heegaard decomposition. Take a $(p, q)$-torus knot on the fixed torus :P
 
4:30 PM
My construction must be wrong then
 
4:46 PM
Wait, this was your strategy?
 
@AkivaWeinberger at least 4
 
over 9000 ofc
 
@BalarkaSen I understand my failure now. I tried to get a circle action from a fibered knot. This is fine if the knot complement has a circle action giving the fibering.
But that's only going to be possible if the knot / fibering is trivial.
Apparently this strategy will actually work one dimension up.
 
Ah I see. So the only way to get an open book decomposition on the knot complement which fibers over S^1 is the trivial knot
Is that correct?
You wanted to act by S^1 on the total space by acting on the base of S^1 (it's like a family of Seifert surfaces parametrized over S^1)
 
5:02 PM
Any theorems out there for counting closed orbits on surface homeomorphic to $S^2$?
Clairut's relation seems appropriate here
 
@BalarkaSen Right. You would need the monodromy f to be a periodic diffeomorphism to cook up a circle action at all. But the monodromy is assumed the identity near the boundary and if it was periodic it would globally by the identity.
 
Aha, gotcha.
Very nice ideas
Hi @Ted
 
Hi, a @Balarka
 
There must be some way to relax this assumption on boundary monodromy
Don't know anything else
H
 
hi @ted
I applied for a physics instructor position in the fall at an undergraduate liberal arts college nearby yesterday
bit of a crapshoot but we'll see how it goes
 
5:18 PM
Hi @Semiclassic: It would probably be wiser to cast a wider net.
 
yeah
this is one which just happened to show up on my radar
(i.e. they need someone for a specific course in the fall)
 
Sounds like hardly any money.
 
5:32 PM
Ted do you have .a minute to be bothered by a differential geometry question?
 
We'll see what the question is. I'm half packing, half paying attention.
 
[More details here, but sadly not receiving much attention: math.stackexchange.com/questions/3290962/…
Here's the setup. We're on the n-sphere, and considering convex functions on the n-sphere
 
Oof too hard for me
 
What does this mean? It means for a map f: S^{n-1} -> R any geodesic gamam on the n-sphere must yield a convex map f \circ gamma.
*gamma
Wow. Let me try that again: for every geodesic $\gamma: [0, 1] \to S^{n-1}$, the composite map $f \circ \gamma$ should be convex.
 
5:35 PM
You might as well change coordinates to diagonalize $F$ in the first place.
 
Absolutely.
(my partial response does this)
in which case f looks like a weighted sum of squares.
 
Since geodesics are great circles, it should just suffice to study $F$ on a $2$-plane through the origin.
 
yes that's right. (that's what my answer does too)
So you can restrict to span{e_i, e_j} intersected with S^{n-1}
 
But if $F$ isn't positive-definite, can't you see it won't be convex on a $2$-plane with mixed sign (or even on which it's negative definite)?
 
Yielding a great circle, and e_i, e_j picked to be assocaited with eigen values which are distinct.
 
5:36 PM
@TedShifrin good luck with the packing!
 
Oh, I didn't see you'd answered your own question.
I hate it, but thanks, @Leaky.
 
Really? Does positive definiteness matter here? I think distinctness of eigenvalues is all one needs.
 
@TedShifrin Moving or going on holidays?
 
I doubt distinctness matters.
@Tobias: Moving 3 miles.
But that isn't so much simpler than moving 3000 miles.
 
Well what I think I showed is that if even two eigenvalues are distinct then you're hosed.
 
5:38 PM
well, at least it is not that far. But moving always sucks
 
But let me know if something is not convincing in my argument. Basically, we can restrict to the great circle in the plane spanned by the two eigenvectors associated to the two distinct eigenvalues.
 
Interesting. So $x^2+2y^2$ is not convex on the unit circle?
 
Parameterizing the function in this space, we get f(theta) = (lambda_1 - lambda_2) cos^2(theta) + constant.
 
That's $1+y^2$.
Hmm ...
 
Yes, that is true...
 
5:41 PM
Yeah, I guess you've convinced me. So you don't need me!
P.S. This isn't differential geometry :P
 
I'm confused, 1 +y^2 is convex, though
What am I misunderstanding.
 
You argued in your proof that $y^2$ is not convex. The $1$ is certainly irrelephant.
 
I never understood questions without a question mark
 
If we take the points $(0,\pm 1)$, it's pretty clear we violate convexity on the semicircle, isn't it?
LOL, @MikeM?
 
@MikeMiller Is that so
 
5:47 PM
@BalarkaSen hm?
 
lmao
 
@AkivaWeinberger I will say it this way: There are so many infinities such that the whole list is an infinity larger than any of them
(that's really what a proper class of cardinals is really...)
 
I agree with you, a @Balarka?
 
@MikeMiller Lol yeah. Your first reaction is to read it as a statement lol
 
Just to be concrete, you're saying we should essentially consider f(t) = sin^2(t) + 2cos^2(t) for t in [0, pi]
 
5:48 PM
Oh, wait, I got that precisely wrong, @Drew. Let's try the points $(\pm 1,0)$.
 
Nah, let's do $\sin^2 t$ on $[0,\pi]$.
hi, demonic @Alessandro
 
well yes its obvious that as a map [0, \pi] -> R it is not convex
Wait, I'm confused are you saying you agree with my sketch or that you don't agree with what I'd sketched @TedShifrin?
 
I'm thinking only about this discussion. So doesn't that function violate your definition of convexity on the circle?
 
Yes, this means that y^2 as map from the circle to R is not convex.
(when I said earlier that $1 + y^2$ is convex, I thought you meant as a function from $\Bbb R^n \to \Bbb R$ !!)
 
5:56 PM
So this says your argument must be right. I had never thought about this before.
 
Bizarre that in the tangent space this function is convex but not once on the manifold.
 
Why do musicians stick to one genre?
 
Anyways, thanks.
Btw, I agree this isn't serious differential geometry. It did reduce to polar coordinates after all ;)
 
@ultradark lol they don't
 
@Drew: I've never thought about this before, as I said, but it's the curvature of the circle/sphere that's causing problems.
 
5:59 PM
@BalarkaSen one could say they stick to music
 
good point
i am become defeat
 
I wonder if there's a famous artist (meaning musician) that's also a famous mathematician
 
little known fact: homer's last name was pythagoras
 
Raoul Bott was a somewhat accomplished pianist. Lots of mathematicians play instruments pretty seriously.
 
Huh I didn't know that
 
6:04 PM
Well, I have to know a few things you don't.
 
lol
 
6:17 PM
@TedShifrin im really good at being bad at the mandolin
 
hahaha
 
The linking number for two smooth manifolds $M,N$ is defined as the degree of the map $l(x,y) = \frac{x-y}{||x-y||}$. If $M$ is the boundary of an oriented manifold say $\Sigma$ disjoint from $N$, the map $l$ can be extended to a map $L : \Sigma \times N \to S^k$ which when restricted to $\partial \Sigma$ is just $l$. This is smooth and also well defined because $\Sigma$ is disjoint fro $N$. Using the boundary theorem, it implies that $\operatorname{deg}(l) = 0$. Is this right?
 
@Ultradark Tom Lehrer was probably more known for his music. But he was also a mathematician.
 
Yes, he even taught at MIT and Harvard.
I think he ended up in the Econ departments, though.
@Albas: You needed to say that $M,N\subset\Bbb R^{k+1}$, dimensions, etc. But yes.
 
Linking number is very good
You can actually say that if $\partial \Sigma = M$, then the linking number is the number of points of intersection between $\Sigma$ and $N$, once made transverse.
(Try to prove it @Albas)
 
6:28 PM
Cool. Will try to.
I heard it today from my project guide. Apparently they are used in TQFT's somehow.
 
each one counted with sign @Balarka @Albas
 
Yes ^ intersection number, rather.
 
** Star this if you are right handed **
** Star this if you are left handed **
 
@Ultradark I lost both my hands in a chemical accident.
 
Lefschetz actually did lose both his hands
 
6:42 PM
@RyanUnger before the accident were you right handed or left handed?
 
@s.harp It was a reference to him
 
6:55 PM
cool til @RyanUnger
 
7:14 PM
Does anyone else find it strange that the only numbers needed in analysis are the naturals and numbers of the form $\frac{1}{n}$ or $\frac{1}{2^n}$ for $n \in \Bbb{N}$?
 
@user193319 How would those numbers suffice for analysis?
 
At least those are the only numbers needed to prove certain "foundational" results in real analysis.
 
@user193319 having all the real numbers is kind of important
complex numbers too...
 
Well, e.g., whenever I do a "$\epsilon$-$\delta$ proof, I really only use natural numbers (don't need any irrational epsilons).
As another example, I only need naturals to describe what $\lim_{n \to \infty} f_n(x) = \infty$ means, where $f_n$ is, say, a sequence of continuous functions.
 
that is just because those always just require you to pick a small number
(or a large one)
 
7:18 PM
In fact, I needed the fact that to show that $\{x \in \Bbb{R} \mid \lim_{n \to \infty} f_n(x) = \infty \}$ is a countable intersection of $F_\sigma$ sets.
Just need the naturals--that's all.
I suppose that should be expected, though, since all other numbers can be constructed from them.
It's still strange though.
 
I think you've just discovered what an Archimedian field is
 
I guess so
So is anyone here a weeb?
 
Hi @Ryan
I have a functional analysis question which is either easy or a typo in my notes
 
@AlessandroCodenotti ok let's hear it
 
I have a densely defined operator $D$ on a $H$ separable Hilbert such that $D$ is self-adjoint and $(D\pm i)^{-1}$ are compact ($D$ comes from a spectral triple essentially)
Now I can show that the bounded transform of $D$, namely $D(1+D^2)^{-1/2}$, is Fredholm
I want to show that $D$ itself is Fredholm
 
7:33 PM
is $(D\pm i)^{-1}$ compact automatic or a hypothesis
 
Hi guys. How to represent conjugate of z and mod of z in mathJax? Thanks.
 
@RyanUnger hypothesis
In my notes I just wrote $D=D(1+D^2)^{-1/2}(1+D^2)^{1/2}$, where the first part is the bounded transform so it's Fredholm, while the second half is unitary, so $D$ is Fredholm
However I don't see why the second part is unitary now
 
you mean why is $(1+D^2)^{1/2}$ unitary?
 
yes
I don't see it
 
I don't buy it...look in finite dimensions and take D to be the identity
the compactness is automatic
the matrix is diagonal $\sqrt 2$ right
not unitary
 
7:39 PM
@RyanUnger Looks right to me
 
how many distinct ways can you torus fibrate a sphere
 
also unitary implies bounded and no way is that bounded right
 
waddup yo
 
Uhm wait so $1+D^2$ is invertible, right?
 
we decided that last time, right
how else do you define the bounded transform
 
7:43 PM
Yes
True
Doesn't have $0$ in its spectrum, I remember
 
@AlessandroCodenotti studierst du nun in Deutschland?
 
Ja, seit Oktober
 
so uhhh, idk. I haven't studied unbounded Fredholm operators
 
@AlessandroCodenotti aber studierst du auf Deutsch oder auf Englisch?
 
Auf Englisch
@RyanUnger I don't actually buy that $D$ is Fredholm now
 
7:48 PM
@Alessandro muss glaub ich in Heidelberg auf Deutsch studieren :(
 
Because we want to use the bounded transform to get a Fredholm module from a spectral triple, why not use $D$ directly if it is Fredholm?
Bonn is very international, so we have all classes in English
Almost 40% of the students in the maths master program are foreigners
 
I actually don't know what an unbounded Fredholm operator is
 
Fair, I think most of Heidelberg is German
 
Can't you use the same definition? finite dimensional kernel and cokernel?
 
maybe, but there's a very nice theorem in Conway that gives 7 equivalent definitions for bounded ones
idk how much of this carries over
 
7:54 PM
Oh ok so the idea is that you have a closed operator $D\colon\mathrm{dom}(D)\to H$, and it is bounded if you put the graph norm on its domain, so you can use the usual definition
(In this case $D$ is closed because it is self-adjoint)
 
Oh my god this guy is computing $[\nabla,\star]=0$ in coordinates
 
8:09 PM
Can $S^2$ be expressed as a 3D projection of a 4D shape?
 
 
2 hours later…
10:08 PM
I'm struggling to understand primitive roots modulo n -- is it true that for a primitive root r, and k in Z, r^k mod n is always coprime with n? Or does it only go "the other way", i.e. that all coprimes are generated, but not only them?
 
10:24 PM
@ÉricoMeloSilva did you sign up for healthcare
 

« first day (3267 days earlier)      last day (40 days later) »