« first day (2204 days earlier)      last day (3113 days later) » 
00:00 - 19:0020:00 - 00:00

00:01
(expanding the comment to bolbteppa) for instance, if n=1, S^1 is a nbhd of itself but S^1 is not homeo to an open subset of R^1
Good point, damn $n = 0$ cases
00:16
The question is meaningless without specifying the topology
A 0D manifold must be discrete, and the rationals with the standard subspace topology are not.
@bolbteppa actually
I've been thinking ahead to my senior thesis
I think de Rham cohomology and spectral sequences would not be bad at all
@0celo7 sure, would really love to know wtf a spectral sequence really is, especially using the language of obstructions
"Mayer Vietoris says that to work out the obstructions in AuB, that's like taking disjoint copies of A and B and looking at their defects, and then counting the ones you counted twice from their intersection that you ignored: it is the inclusion exclusion principle. The snake lemma tells you that if you were looking at n-dim holes, then you next need to take into account the n-1 dim holes - or the boundaries of the n-dim holes."
https://www.physicsforums.com/threads/exact-sequences.164213/
God damn algebra
01:01
Wow just reading this chat makes me feel dumb lol
01:49
@bolbteppa That might be the most incoherent thing I've ever read on basic algebraic topology. That whole thread is embarrassing.
@PVAL Incoherent?
@PVAL are you sure you understand the subject if it's that confusing for you?
02:04
lol
@bolbteppa why do you always accuse people of not understanding things
I think it's a threat to people who spend time learning math in a complicated fashion to find out it was all children's blocks and pictures
because it's not
Children's blocks and pictures are not legitimate proofs
Something you don't understand
You find this obstruction stuff all over the place in algebraic topology, even Hatcher whips it out i.sstatic.net/9pKqq.png
Notice the multiple viewpoints perspective, just because you took one doesn't invalidate the other
A picture is a heuristic, nothing more.
A painting is a bunch of atoms, nothing more
02:11
What?
"There are three different ways in which mathematicians have reached the subject now called algebraic topology:
Classification, properties and invariants. One looks for ways of distinguishing topological spaces.
Obstructions. What are the difficulties in getting certain types of maps between spaces.
Categorical. What is the structure of the category of topological spaces or some subcategories of it?
*As one realises during the study of algebraic topology all these are different ways of stating the same question--and the answer is to apply the methods of algebraic topology.*"
@bolbteppa Using transversality one can show that there exists a vector field on a compact manifold with only finitely many zeros.
02:27
@PVAL did you even know this? The people you're mocking in that thread are professors not students waving their hands...
03:10
@0celo7 its very difficult understand based on the way its written and has some sentences that say nothing meaningful (maybe most poorly exposited would be better). Anyway I won't be baited by personal insults because I am criticizing someones internet posts.
Apology accepted ;)
Apology?
03:41
GG Bott & Tu, no 12.4 on the page...
 
1 hour later…
04:44
I've got a question I'm thinking of asking, and am seeking some feedback before posting (or possibly avoiding asking something too trivial)
is anyone awake in this room?
maybe
so . . . I was messing around with some customized vectors in this project i'm working with . . . and made a function I'm calling Vector.excite and I refuse to believe there isn't already a name for this function
the function takes the summation of values in the vector and than multiplies the normalized vector by this sum
I'd be surprised, that's not a "good" function
it's a great function!
It's basis dependent
04:50
the summation of a vector is equal to the magnitude of it's excited form
excited form?
what's that
vector.excite => the summation of values in the vector and than multiplies the normalized vector by this sum
I'm calling it excite because I can't find an existing term for it
uhhh
04:52
I was wondering if there was already a name for this
that's still basis-dependent
why do you need this?
what do you man basis-dependent?
mean*
it depends on your coordinate axes
as all things vector
No, definitely not
04:55
well, it seems to work with vectors of any length
why wouldn't it work with any vector?
I'm saying that you get a completely different vector if you rotate your coordinates a bit
completely different length, that is
right, but the new vector, once rotated, follows the same principle V.excite.sum == V.magnitude
no, you don't rotate the vector
you rotate the coordinates
but rotating a vector does not change its magnidtude
you're not rotating a vector
05:01
it's just an interesting relationship, I thought
what relationship?
between the sum of a vector's elements and the magnitude of a vector
I don't see any relationship
V = Vector.
V.sum = summation of elements.
V.norm = normalized vector
V.mag = magnitude of vector

the relation i find interesting is this:
(V.norm*V.sum).mag = V.sum
and the excite function being = (V.norm*V.sum)
Why is that interesting?
V.norm gives you something of length one
And that's wrong
It should be (V.norm*V.sum).mag = abs(V.sum)
And that's trivial. If v is a unit vector, then cv has mag |c|.
05:11
I know it's somewhat basic, which is why I am sure there exists already a name for the excite function
Proof: mag(cv)=sqrt(<cv,cv>)=sqrt(c^2<v,v>)=sqrt(c^2)=|c|.
@punkerplunk No, because I'm not convinced it's useful. (It's coordinate system dependent, which is never good.)
Good night.
use is application. I find it useful- because it implies every vector is the 'excited' form of another vector, and there fore each vector belongs to a set of recursive uses of the excite function.
05:37
@punkerplunk (v.norm*v.sum).mag=abs(v.sum) (which is totally not how math people would notate it btw), is completely trivial
applying something to itself does not make it useful
application means solving a problem that could have been encountered independently, or deriving an interesting result that can be understood independently
I know it's trivial. I'm just wondering if there is a name for the function of (v.norm*V.sum)
don't see why it would have a name
if you want to notate it, it would be [the dot product of v and (1,...,1)] times v / magnitude(v)
(avoiding math notation)
I'm not exactly sure it'll ever be useful, but it does yield an series when applied recursively to a vector, and maybe one day someone will find it useful? idk, I suppose i just ask my question for posterity's sake?
it's also (the projection of (1,...,1) onto v) times the magnitude of v
06:23
Hello, is there someone who understand Portuguese?
i need translation of this
 
2 hours later…
user227867
08:28
@Vrouvrou Try translate.google.com
08:47
@bolbteppa I agree with PVAL that this particular paragraph (not the whole thread; I think the defect explanation is standard but computationally and conceptually useless) is close to being incoherent. I think anyone familiar with these kind of topology would agree.
I don't think it's "confusing"; the OP's trying to say that inclusion-exclusion of Euler char is a result of Mayer-Vietoris but that provides no insight on what M-V is.
And he's trying to fit it in the context in a weird way using "defects" - those aren't a thing. Topological spaces in itself don't have "defects".
I should say that I am also surprised that you're accusing a topologist about not understanding topology, even though you're not one.
09:43
Hello.
Can anyone help clarify vector projections a bit for me?
I've watched an example where the projection of vector x onto l is defined as:
$Proj_{L}(x) = c\cdot v = (\frac{x\cdot v}{v \cdot v}) \cdot v$
But doesn't vector orientation matter here? The example dots:
2 . 2
3   1
But isn't this invalid? A (2x1) . (2x1) operation isn't allowed?
Well this is bad, with my reasoning even the length of a vector is invalid because any vector $u$ cannot be multiplied by itself since it doesn't fit matrix multiplication requirements.
Okay. I suppose that there is a difference between the row/column vector of a matrix and a vector in the sense that they are using them.
10:09
Because the only way that I can reconcile the dot product of a vector with itself is to write it as: $v^{2} = v \cdot v^{T}$ if v is a horizontal vector. And $v^{T} \cdot v$ if it's a vertical one.
@Owatch o/
What's $c$ and $v$ in $Proj_{L}(x) = c\cdot v = (\frac{x\cdot v}{v \cdot v}) \cdot v$ ?
v is apparently the unit vector of L.
and c is the thing in the parentheses.
I'm not sure why you need to use a unit vector here, to me the reason why you have a $\frac{}{v\cdot v}$ is to normalize it
Anyhow it's not wrong
Oh I see where your confusion arises
My problem isn't so much with this particular problem.
The scalar product of two vectors $u,v$ is defined as $u\cdot v=tr(^t u\times v)$ where $\times$ is the usual matricial multiplication
10:15
It's with rules regarding dot product, and how it seems to differ when concerning things like dotting a vector with itself. Something not allowed when I consider the dimensions.
Since it's a transpose the dimensions are fine
What does that definition mean?
$tr$ is the trace
Lets say I have: u = [1, 0, 2]
Yes?
What is it you don't understand in the definition I put above ? What part ?
10:17
Well I'm not sure what trace is.
It's the sum of the diagonal terms
In the case of $n\times1$ vectors the multiplication leads to a $1\times1$ matrix so it's the matrix's sole coefficient
For instance say $u=\begin{bmatrix}1\\0\\2\end{bmatrix},v=\begin{bmatrix}3\\4\\5\end{bmatrix}$
Then $u\cdot v=tr(\begin{bmatrix}1&&0&2\end{bmatrix}\times \begin{bmatrix}3\\4\\5\end{bmatrix} )=tr([3+10])=13$
Okay. I see, it makes sense. I get the same answer no matter what the orientation.
I wish I'd learnt that earlier.
I've been oblivious to this until now.
Also I have another question.
Are Euclidean vectors always supposed to be "vertical" ?
Never mind
Well I can reconcile this with frobenius norm then. It's defined as the square root of the sum of all elements squared in a matrix. For a vector, it works the same way. The square root of the sum of all elements squared.
Or the two-norm.
10:35
@Owatch Both vertical and horizontal vectors are used. It's a matter of convention.
Fix a convention, work with it. I use vertical vectors.
Okay, thanks! I will use vertical too.
Its what I've always seen used anyways.
I have a small normal distribution question. If I wanted to find the probability at a point, like x = 50, would the integral look like this? $$\lim\limits_{\substack{a\rightarrow 50 \\ b\rightarrow 50}} \int^{b}_{a} p(x)dx$$
11:03
@TheBro21 Yeah, and that equals 0
11:38
Nice
Well, I can just try a short interval, or discrete probability instead, most probably
@BalarkaSen I'm not surprised, was expecting this from you, I've given you professors giving explanations of this and articles by professors explaining there are 3 ways to look at algebraic topology, and all that is thrown out the window by you guys for no reason with emotional claims, that should say it all
That particular paragraph you linked was written by a professor? I doubt.
11:53
Well your doubts are the same doubts calling that thread incoherent despite the fact they illustrate one of three ways of viewing the subject, we can go with the students or the experts on this one
I didn't call the thread incoherent: just that particular paragraph.
But I do believe that although thinking of homology as measuring defect is standard, it's not helpful. There are more geometrically transparent (and computationally useful) ways of thinking of it.
Ok, incoherent and conceptually useless
Ok what is the snake lemma saying then
I highly doubt any expert wrote that paragraph.
Well again, those doubts stem from the same place doubting this interpretation or viewing it as useless
@bolbteppa Suppose $X$ is a manifold, $A$ a subspace inside. The snake map $H_n(X, A) \to H_{n-1}(A)$ is essentially sending an embedded manifold with boundary (call it $M$) inside $X$ with boundary $\partial M \subset A$ to it's boundary $\partial M$ in $A$.
It's essentially taking a manifold w/ boundary and sending it to it's boundary, if I had to say it simply.
Homology classes are best visualized as embedded manifolds.
12:13
@bolbteppa I don't see any actual interpretation in that paragraph. They say, "inclusion exclusion mumble mumble n-dimensional holes mumble".
That sounds pretty cool, I don't see it yet but it sounds cool
@BalarkaSen you call it mumble mumble yet professors in that thread, and people in a thread titled "Mathematically mature way to think about Mayer–Vietoris" mathoverflow.net/q/23175/38721 are fine saying inclusion exclusion, I'm just saying...
I did not object to the use of "inclusion-exclusion" :P
I objected to the random mumbling.
@BalarkaSen Hey
Hi @0celo7
12:28
Did you ever figure out how to link the homological Euler char with GP's Euler char?
No, but I am going to think about it soon. I figured out the correlation between homological orientation and smooth orientation yesterday.
I haven't thrashed out the details, but the idea is pretty clear to me now.
@BalarkaSen Well, I'll try to work on a proof that GP's Euler char gives the Betti number one.
Does GP not explain that later on? They start from the crazy diagonal thing then draw triangulations later on right?
They don't prove anything
G&P is not an algebraic topology book.
12:34
@BalarkaSen Unfortunately, I think the answer will be fairly involved. I can sort-of see a way forward from what I've read in Bott & Tu. But it will involve spectral sequences.
lol oh noes
I still say it would be just the Lefschetz fixed point formula.
Yes, that's indeed it
@0celo7 the Lefschetz GP keep referring to who proved all these things is the guy you called a crank btw ;)
Wait, what was I trying to prove?
Lefschetz is a crank? Huh?
12:35
How does GP define the thing?
Self intersection no. of diagonal in $X\times X$.
@BalarkaSen Lefschetz never wrote down a correct proof in his life
@BalarkaSen Yeah but in terms of Lefschetz?
Oh, that may be true.
That stance is completely against what I find beautiful about mathematics
And precisely why I despise physics
12:37
@0celo7 Alternating sum of traces of the maps in homology.
The paradox is that he invented a lot of it, hmm, haha
Hi @SemiC
So I need to show that $$I(\Delta,\Delta)=\sum_q (-1)^qH^q(\mathrm{id})?$$
(I'll use cohomology btw)
ok, that's my project for today.
12:39
$H^q$ with rational coefficients, btw.
@BalarkaSen how about de Rham
Wait a second, I can't make sense of that. You should say trace(H^q(id)).
I am reading a book on advanced de Rham theory, after all
Aka, dim H^q
oh crap, typo.
I meant the trace, yeah
Which is the dimension
Oh, I see a proof!
It uses Poincare duality.
and characterstic classes
I'm probably waaaaay overcomplicating this
12:42
Yes, I think so too.
The snake lemma can't be that simple and obvious can it
The right hand side is the # of fixed points of a the identity map after homotoping it slightly, say.
We know from GP that the Lefschetz number is a sum of signs of determinants of $Df_x-1$
That should be the same as the left side because we're making the diagonal intersect with the graph.
The proof should go something like that, methinks.
where here $f$ is...the identity?
12:44
@bolbteppa Simple and obvious are both relative.
And true only after you understand it.
Aha, let $\eta$ be the Poincare dual of $\Gamma(f)$ in $M\times M$. Then $$L(f)=\int_\Delta \eta$$
that's the hard part to prove
But I bet one can show $$\int_\Delta\eta=\int_Me(TM)=\chi(M)=\sum(-1)^ib^i$$
if you let $f=\mathrm{id}$
ok, I'll work on that
toodles
@0celo7 Here's my suggestion. $f : X \to X$ be a result of homotoping the identity map slightly. $I(\Gamma_f, \Delta) = I(\Delta, \Delta)$ should hold, agree? Then $I(\Gamma_f, \Delta)$ is the # of fixed points of $f$: by Lefschetz that's $\sum_i (-1)^i \text{tr} H^i(f; \Bbb Q)$.
what's wrong with the $\Gamma(f)$ notation?
Now $H^i(f; \Bbb Q) = H^i(\text{id}; \Bbb Q)$.
Trace of that is dimension of $H^i(X; \Bbb Q)$.
Aka, you get back $\chi(X)$
It's not the number of fixed points, it's the number of fixed points with orientation taken into account
and I don't see how "by Lefschetz" that's $\sum_i(-1)^i\mathrm{tr} H^i$
I think that "by Lefschetz" is the major nontrivial thing here
unless you've already proved that
12:51
It's called the Lefschetz-Hopf theorem.
@0celo7 Sure, we're counting upto multiplicities.
Lefschetz Hopf, according to wiki, is that if the Lefschetz number is not zero, it has a fixed point.
No, that's the Lefschetz fixed point theorem.
It's that sum of multiplicities of each fixed point is the Lefschetz number.
how are they defining multiplicity?
It's not in Hatcher
(lefschetz-Hopf, that is)
Degree of $(x - f(x))/\|x - f(x)\|$ near $x$.
Ok I like my wildly more complicated proof
for which I have no details available yet
12:55
whatever floats your boat
@BalarkaSen I'll try to figure out the details
@0celo7 Isn't it in Guillemin-Pollack though?
@BalarkaSen Certainly not using (co)homology.
unless I missed it...
Sure, but I think I saw a version of it.
I haven't studied that much yet.
maybe in the exercises?
Well, I want to see if my idea works first
12:58
OK, good luck.
I need to get back to work.
13:57
@DanielFischer Are you around?
14:16
@BalarkaSen Depends. If you have a question about physics, then no. Otherwise maybe.
Hi, someone can explain what a holomorphic square root is intuitively?
I know that a holomorphic function is a complex function which is differentiable
@HirotoTakahashi A holomorphic square root of $f$ (which needs to be a holomorphic function) is a holomorphic function $g$ that satisfies $g(z)^2 = f(z)$ on the domain.
@DanielFischer Oh, thanks!
14:31
@DanielFischer Hehe, no, it's not physics.
I am looking at the following generalization of Phragmen-Lindelof theorem:
$S$ be a sector in $\Bbb C$ with a vertex at $0$, forming an angle of $\pi/\beta$. $F$ be a continuous function on $\text{cl} S$ which is holomorphic on the interior, and $|F(z)| \leq 1$ on the boundary. Suppose moreover $|F(z)| \leq C e^{c |z|^\alpha}$ for some constant $c, C > 0$, where $0 < \alpha < \beta$
@Hippalectryon
The claim is that $|F(z)| \leq 1$ throughout $S$.
14:34
@BalarkaSen whereas if it was physics, that's where I'd come in :P
@Semiclassical Indeed, I'd have asked it to you instead.
where are you in physics right now, btw?
@BalarkaSen Looks good. What's the question?
@DanielF OK, so this is not too hard to prove: essentially an application of the maximum modulus theorem. However, I am thinking of the following conclusion.
$f$ be a holomorphic function on all of $\Bbb C$ with $|f| \leq 1$ on the positive real axis and $|f(z)| \leq Ce^{c|z|^\alpha}$ where $\alpha < 1/2$. Then $f$ satisfies the hypothesis of the theorem with $\beta = 1/2$, so $|f| \leq 1$ uniformly on all of $\Bbb C$. Hence it's constant.
Is there a more direct way to see this?
It strikes strange to me because being of growth rate at most like $e^{|z|^{1/2}}$ sounds too weak for $f$ to be constant.
@Semiclassical Work/power/energy.
@BalarkaSen Right, consider $\cos \sqrt{z}$.
We get non-constant entire functions with growth rate $\exp \bigl(\lvert z\rvert^{1/k}\bigr)$ via $$\sum_{n = 0}^{\infty} \frac{z^n}{(kn)!}.$$
In P-L, we need that the growth rate has an exponent strictly smaller than $\beta$.
14:45
@Hippalectryon try this from one single shot ...
If $1>a>0$, show that
$$\int_0^1 x^{-(1-a)}\log(x)\left(\frac{1}{1-x}-x^{-a^2}\frac{1-x^a+a(1-2a(1-x^a))}{(1-x^a)^2}\right) \textrm{d}x=\frac{1}{a}-\frac{\pi^2}{\sin^2(a\pi)}
$$
@DanielFischer Right, but $\cos(z^{1/3})$ e.g. has growth rate with exponent strictly smaller than $1/2$. It's absolute value is also bounded by $1$ on the positive real axis. So what did I do wrong?
Is the conclusion I have drawn for $\beta = 1/2$ wrong?
@BalarkaSen Well, $\cos \bigl(z^{1/3}\bigr)$ isn't an entire function.
Oh, you're right, whoops.
@user1618033 I'll be here in 25 mins approx
@Hippalectryon OK. However I have to leave soon and come back a bit later.
14:54
@DanielFischer Right, but since none of those functions are bounded by $1$ along the positive real axis they are fine. OK.
Sounds like a nice conclusion to me: under the relatively weak assumption on the growth rate (being $|f(z)| \leq C\exp(|z|^{c|z|^\alpha})$ for some $\alpha < 1/2$), any nonconstant entire function has to grow unboundedly along any ray from the origin.
15:14
Is this the symmetric version of the Dirichlet eta function that satisfies the functional equation symmetric_eta(z)=symmetric_eta(1-z)? I think it is incorrect.
By the same token/condition would not the symmetrized functional equation for the zeta function be simply: $\zeta(1-s)\zeta(s)$
?
So it can't be can it?
15:44
@BalarkaSen Did you once say that isomorphic vector bundles are diffeomorphic?
@0celo7 Maybe. Why?
Unless I'm missing something, it doesn't seem at all obvious
Unless you know a proof right now, I'll think about it some more
Do you think it's true?
$p_i : E_i \to X$ be bundles over $X$, $i = 1, 2$. $f : E_1 \to E_2$ is a bundle morphism means it's a continuous map such that the triangular diagram $p_1, p_2, f$ commutes and $f$ sends fibers to fibers, yes? Then a bundle isomorphism is a morphism $f$ such that an inverse bundle-morphism $g$ exists.
Then $f : E_1 \to E_2$ is a continuous map with $g : E_2 \to E_1$ a continuous inverse. That means the total spaces are homeomorphic.
Am I missing anything?
@BalarkaSen I'll get back to you after lunch!
 
2 hours later…
17:22
@BalarkaSen I want my maps to be smooth, too.
I like my maps like I like my pick up lines-smooth
that makes little sense, I guess
welp
17:43
Can someone explain to me how to fold and glue the sides of an octagon to form the genus 2 variant of the torus shape?
17:55
@0celo7 But that's not an issue if you're working with smooth vector bundles?
'Cause then bundle morphisms are automatically smooth, by definition.
@SoumyoB (1) Can you build a torus by gluing edges of a square? (2) Can you build a punctured torus (torus minus a small disk on it) by gluing edges of a pentagon? (3) Can you get a 2-torus out of gluing two punctured torii along the boundary of the puncture?
If you can do all of those, you should also be able to see how to build a 2-torus out of an octagon.
This must be what I am searching for.
Or is it?
18:16
@BalarkaSen I've been wondering about integrals over product manifolds...
Consider $M\times M$ with $\pi$ the projection onto the first factor and $\rho$ the projection onto the second factor
Can I write $\int_{M\times M}\pi^*\omega\wedge \rho^*\tau$ as a product of $\int_M\omega$ and $\int_M\tau$?
I've never seen that before, so I doubt it
$M$ is compact
@0celo7 Fubini's theorem, basically. Yes, you can, provided $\omega$ and $\tau$ are integrable.
Oh jeez... I never realized how nasty of a thing I was just trying to do. How does one measure the length of a geodesic in $SO(n)$?
And you have the right orientation on the product. Add a factor of $-1$ if you have the wrong orientation.
It's a one-parameter subgroup, so it's some $p(\cdot)$, but each $p(t)$ is a matrix.
@0celo7 As DanielF said, it is Fubini's theorem.
(Sorry, I was away when you asked that)
18:35
Upon deeper introspection, I realize just how dumb of a question that was.
I need sleep.
If you want to think about it abstractly, then it's also a consequence of distribution law of cap product with cup product.
@DanielFischer Hmm, I thought that.
@BalarkaSen uhhh
I'm afraid I don't see how it's Fubini's theorem
$\omega,\tau$ are forms.
Compute on a chart.
I probably don't know what Fubini's theorem is.
18:48
Ok, I do know what it is
I don't see what iterated integrals do here
00:00 - 19:0020:00 - 00:00

« first day (2204 days earlier)      last day (3113 days later) »