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5:01 PM
What does Cons stand for here?
 
consistency of
(formalizing Con(ZFC) in the language of ZFC is long and nontrivial)
 
and a blunder by carlsen! the eval is now +1.3 for anand
 
Naively, why shouldn't Cons(ZFC) = ZFC if ZFC is consistent?
 
but anand failed to find the winning move and it is now 0 again!
 
Or is that not what Cons() is?
 
5:03 PM
It boils down to coding statements and proofs as integers in a robust way, and then you say something like "for no integer $n$, $n$ codes a proof of $0=1$".
 
@anakhro Cons(ZFC) is a sentence (in the language of set theory) saying that ZFC is consistent
 
Cons(ZFC) is the assertion "ZFC is consistent"
 
OH
Okay, makes more sense.
 
yay I've converted Alessandro to the Cons camp :P
and they agreed to a draw!
 
@AlessandroCodenotti There can be all sort of subtle weirdness going on in this coding though. If the model has nonstandard integers, it will have things that the model believes to be codes for formulas, but they don't actually correspond to "real" formulas externally
 
5:07 PM
@Leaky I changed it cuz everyone else seems to have their real name
and I am a sheep
 
lol
 
@BalarkaSen for Hatcher's zigzag comb, the contraction to a point is just contracting the bristles to the zigzag, and then the zigzag is R so it is contractible, or is there something that breaks there?
 
@Lukas if I want to take the second exam do I just register for the first and then not go?
 
Hatcher has some distinction between strong deformation retractions and deformation retractions with subtle differences between contractible and deformating retracting to a point iirc, be careful
 
@AlessandroCodenotti yeah I am not talking about deformation retractions with this, though.
Just contractibility.
 
5:13 PM
 
@anakhro That's a valid contraction, yes
 
can someone help me parse the sentences of this example? it's from "An Introduction to Linear Algebra" by Andrew D. Hwang, page 82.
 
@anakhro I see nothing wrong with it then
 
@sevdaicmis where do you first run into issues with parsing?
 
no problem until second "let".
 
5:17 PM
@AlessandroCodenotti do you have chess books recommendations?
 
then the definition of $E_i^j$ comes, and its complicated.
 
@LeakyNun I've never read one, even though I should do so eventually
 
@AlessandroCodenotti so how did you learn chess?
 
there are $mn$ pairs, right? let me take $(1,1)$ for example.
 
Playing and doing puzzles mostly. I watched some video lessons on youtube too, from the St. Louis chess club iirc, they have classes about various openings
 
5:19 PM
Yes there are m*n pairs.
So you have m*n linear transformations $E_i^j$.
 
that is, we are defining $E_1^1$
 
Sure.
 
now, there comes $k$.
 
@AlessandroCodenotti just noting that virtually every chess book can be accessed in the site one wouldn't explicitly mention here
 
but, you see, it didn't mention $k$ before.
 
5:25 PM
Maybe write it as $$E_i^j(v_k) = \begin{cases}w_i,& k=j,\\ 0^W,& k\neq j\end{cases}$$
That's what they are using k for.
It's like a dummy variable.
 
oh, i see
 
Because it suffices to define linear transformations on a basis (or spanning set), they are just defining how it acts on a basis element $v_k$. The first case they do is when $k=j$, and the last case it's $k\neq j$.
 
@LeakyNun This is not surprising
 
But instead of using $v_k, k=j$, they just write $v_j$.
 
so, this answers another question in my mind, i.e. "how does it define a linear transformation with two equations?"
it really gives $n$ equations.
@anakhro yeah, got it.
thanks
 
5:29 PM
Yeah, and you are comfortable with the notion of defining linear maps on a basis, right?
 
yeah, it's so intuitive.
 
Great, it's an important concept.
 
if you give the rule for transforming the basis, then the rule for transforming all others are implicitly given.
rectifying semantically problematic sentences are fun, btw :)
 
:)
If you find that fun, then you will enjoy the rest of mathematics. :P
 
you misunderstood me, i meant rectifying my own sentence.
parsing the sentence in the example was also fun, though.
 
5:38 PM
Oh, I missed the original sentence before rectification, then. :P
 
This reminds me of question I wondered about recently, but haven't been able to answer: If $V$ is an $F$-vector space, can we show $Z(GL(V))=F\cdot\mathrm{id}_V$ without choice? If yes, how?
 
Alright, so I don't have much of a math background, but I got a question of interest
Why is it that "or" over GF(2) or intersection over sets have commutativity, but minus for natural numbers doesn't?
Obviously they aren't strictly equivalent, but what would a strict equivalent to or and to union be in natural numbers?
 
5:57 PM
is Sawtooth wave actually function because you have on the same x argument multiple values .. or does the straight has a slope and it has different values that are too small to see?
The sawtooth wave (or saw wave) is a kind of non-sinusoidal waveform. It is so named based on its resemblance to the teeth of a plain-toothed saw with a zero rake angle. The convention is that a sawtooth wave ramps upward and then sharply drops. However, in a reverse (or inverse) sawtooth wave, the wave ramps downward and then sharply rises. It can also be considered the extreme case of an asymmetric triangle wave.The piecewise linear function x ( t ) = t − ⌊...
As far as I know that a function can not assign multiple values to the same argument
 
"minus" is better thought of as a unary operation rather than a binary operation
That picture is not the sawtooth wave, but a cut-off of its Fourier series, see the animation further down
 
@EdwardEvans depends on the lecturer some want you to actively fail the first exam
 
@Thorgott 1) is the sawtooth function not connected ? IE disconnected and making jumps?! 2) is the approximation using fourier as shown fulfil the def of function (return to first question)
 
Given a dynamical system $x'(t)=f(x(t),y(t))$ and $y'(t)=f(x(t),y(t))$ what are you supposed to solve for? $x(t)$ and $y(t)$?
but since they are coupled you have to solve them simultaneously?
 
@Thorgott the only part to show is when $\dim V = \infty$?
 
6:07 PM
@Thorgott How could it be thought of as unary if it takes a second argument?
 
@JohnnyApplesauce: $a-b = a+(-b)$.
 
He means $x\mapsto -x$
 
Connectedness is a property of topological spaces, not of functions. If you mean discontinuous, yes, it is. The approximation by Fourier series is not the definition of the sawtooth wave, it is a (non-trivial) property. A possible definition of the function is given in the article.
@anakhro Yeah, the only way in which I can prove the result is by constructing certain linear maps by defining them on a basis and that doesn't work without choice.
 
At any rate, @JohnnyApplesauce, not all operations are symmetric. Subtraction is an example of an anti-symmetric operation: $b-a = -(a-b)$.
 
6:11 PM
If $X,Y$ are topological spaces and $f:X→Y$ is a homeomorphism and $d$ is a metric on $X,$ then you can transport the metric via $f$ to get a metric $d'$ on $Y;$ this metric will induce the topology on $Y$ and $f$ will then be a homeomorphism (in fact, an isometry) by construction. What are the details of: "transport the metric via $f$?"
 
Hey @TedS, do you perhaps know the answer to this?
@Ultradark Write down the definition of what it would mean for $f$ to be an isometry and you pretty much have the definition for $d^{\prime}$
 
@Lukas lol wtf
 
@EdwardEvans wanna play?
 
I guess if I speak to Kasten and Vogel and just explain the situation they might be fine with it
:P
 
at this rate I will have read (yay future perfect) more chess books than maths books
 
6:17 PM
@Leaky alas, I am about to make myself some fodo
food
Don't tempt me fodo
 
chess is my food
that sounds wrong
 
funny situation, I need IELTS for my masters and I will likely get a good IELTS score, but the certificate will arrive ~2 days after the deadline for the masters application ends
so yeah I need to call someone at the British Council or from the masters
 
@LukasHeger just show them that you can play the English opening
(1. c4)
 
lol
 
6:25 PM
I would just speak to the admissions board lol
 
I'm not even applying for an anglophone country
 
18

A diffeomorphism is typically presented as a smooth, differentiable, invertible map between manifolds (or rather, between points on one manifold to points on another manifold). For example, take two sheets of paper and curl one of them up. There exists a diffeomorphism that relates points on the two sheets. So if $X,Y$ are manifolds and if $(u,v) \in X$ and $(\exp(u),\exp(v))\in Y$ then is the correct way to write the corresponding diffeomorphism as such?: $\psi:X \to Y$ s.t. $(u,v)\mapsto (\exp(u),\exp(v))$
 
What is $\exp$ here
 
exponential map, maybe?
 
$e^x$
 
6:28 PM
Is this in a Lie theory chapter?
 
yes
I think that it's a special case of the exponential map, for when the group is simply the multiplicative group of the real numbers. $X$ is all points in quadrant $III$ and quadrant $I$
 
It's just the exponential map in Lie theory, then.
In the theory of Lie groups, the exponential map is a map from the Lie algebra g {\displaystyle {\mathfrak {g}}} of a Lie group G {\displaystyle G} to the group, which allows one to recapture the local group structure from the Lie algebra. The existence of the exponential map is one of the primary reasons that Lie algebras are a useful tool for studying Lie groups. The ordinary exponential function of mathematical analysis is a special case of the exponential map when...
 
@LeakyNun shall have read :D
 
@TedShifrin why?
 
heya @TedShifrin
 
6:32 PM
hi @Lukas
 
I readn't any chess books
 
I hope I don't get rejected from my masters due to a pure formality
 
$\det(\text{Shifrin})$
 
I shall do something is the simple future; I will is a more definitive future. Oddly, it reverses for second and third person.
What pure formality, @Lukas?
 
@LukasHeger write to the responsible person/commitee as soon as possible, better if 10 minutes ago
 
6:34 PM
@Lukas that sounds familiar
 
I say just call them.
 
at least I have round seals @Edward
 
You get more done on the phone.
 
hahaha
 
I don't think there's anyone at the office on a weekend
 
6:35 PM
@LukasHeger lol that was so dumb
 
I'll call some responsible professor on monday
 
@Thorgott I say you just consider the subset corresponding to $M-W$ in your parametrization. This notion really doesn't depend on parametrization, anyhow. Measure 0 makes sense in a smooth manifold independent of parametrizations (because smooth maps take sets of measure 0 to sets of measure 0).
 
so if the metric on $X$ is $ds^2=dx^2+dy^2$ then can I get a hint on how to transport the metric to $Y$?
 
@TedShifrin I will receive my IELTS certificate ~2 days after the deadline for the application
 
@Leaky: See this. Colored uses.
 
6:38 PM
"You shall not pass! (speaker's command)" nice to see Lord of the Rings making an appearance on a grammar article
 
@Lukas: If you alert them ahead of time and plead for leniency, you should be fine. Of course, German bureaucracy being what it is :D
 
it's strange that I need an IELTS anyway, given that I'm not even applying for an anglophone country
@Ted it's not necessarily German bureaucracy I have to deal with
 
I'm surprised by the extent to which English is the universal language in math graduate programs in Europe.
 
Heidelberg has German maths masters
 
if $p$ is a point in $X$ then I can say $p\mapsto \exp(p)$
 
6:40 PM
I think the head of the committee is from the Netherlands or something
but it's most likely an international committee
Technically, they don't explicitly ask for an IELTS certificate, only for an IELTS score and I can get that on time
 
and $\exp(p)$ should have a similar metric to the Euclidean metric but I am not sure how to prove this
 
No, you need a Riemannian structure (or at least a connection) on a manifold to define the exponential. It is given by following geodesics. This is related to the exponential on Lie groups because if you put the right metric on a Lie group you get the same result.
 
Do you need compactness for such a right metric to exist?
 
Ok, thanks
 
Say I'm working in $\Bbb R.$ but only in the first and third quadrant?
 
6:45 PM
No, although it suffices, @Lukas.
 
the metric is the standard euclidean metric. but I'm not really sure what happens to the metric after the exponential map is applied to points in the manifold,
 
right here means bi-invariant
 
but it doesn't work for any Lie group, right? I think you need a biinvariant metric and that doesn't always exist
partially sniped
 
Yes, agreed.
I was just explaining terminology overlap.
 
@TedShifrin is there any other good uses for invoking a metric other than stuff with the normal bundle, characteristic classes, etc.?
By good, I mean cool.
 
6:48 PM
@LukasHeger Italian one is worse, believe it or not
 
@Alessandro oh wow
good to know given that I want to study in Padua for one year
 
"Other good uses"? You mean for non-geometric purposes? I guess existence of convex neighborhoods is another standard one.
 
@LukasHeger I know, that's why I warned you
 
But one of the beautiful stories in Riemannian geometry is the implications of various sorts of curvature on the topology of the manifold. (Remember, however, that I'm not a Riemannian person.)
 
We are the masters of inefficiency and unnecessary complications
 
6:52 PM
Oh yeah, I was also wondering if you knew of any good references for characteristic classes on topological manifolds.
 
Other than Milnor-Stasheff? I'm the wrong person to ask that.
 
Oh okay.
 
Not to your question, but I'm fond of Hirzebruch's Topological Methods in Algebraic Geometry.
 
What is that book like?
I was reading how the constant on the Pontryjagin classes are there to force equality of the classes on topological manifolds.
Or something like that.
 
How 'bout the constant is to make the class integral?
 
6:54 PM
Read that one, too.
Tu says that one, Morita says the other.
 
It all goes back, classically, to Grassmannians, Schubert cycles, and Poincaré duality.
Tu is being sloppy.
Or Morita. I can't tell who said what.
 
Morita said the topological manifolds one.
Tu said the integral class one.
I was surprised, Tu's table of contents was basically identical to Morita's.
 
OK, then Morita is probably being sloppy.
Stiefel-Whitney classes were defined for topological manifolds, but I believe that Chern and Pontryagin did stuff smoothly before those had a purely topological definition. My history may be rusty, but I don't think so.
 
Were the constants always there, even in Pontryjagin's original paper?
I suppose I could look up the original paper and answer that myself.
 
Oh yeah, Chern worked carefully on the constants.
Chern classes, too. Euler classes, too.
But if you understand trying to get the Chern class right for the tautological line bundle on $\Bbb CP^n$, then that forces powers of $(i/2\pi)$ immediately.
 
7:00 PM
By getting it right, you mean integral?
 
Well, again, these things are supposed to represent intersection numbers, so one knows what the integrals should be.
 
I haven't seen that interpretation of them yet.
I also need to learn to do them in principal G-bundles.
 
LOL, you have ignored my notes.
 
I found it difficult to follow them.
 
Fair enough.
 
7:03 PM
I think perhaps they were more written for you to lecture from than for someone to learn the theory from?
I am not sure, maybe it was just stuff at the start that you left out that scared me off, but the details are all there later.
 
Certainly there were additional pearls of wisdom and insight in my lectures ... I can only presume. :D
I would have to work reasonably hard to make those into distributed lecture notes, yes. And I don't intend to do it.
Even the undergrad notes ultimately increased in length by 30% from their original version to what's there now.
 
Nonetheless, Morita eventually does it on principal G-bundles, but I have just been combing through some of the details I was unsure about.
And then I think Morita has an entire book dedicated to characteristic classes.
 
You certainly don't need to do the principal bundle setting to do Chern-Weil, etc.
I have taught it both ways, I guess.
 
Yeah, he does Chern-Weil with vector bundles.
 
Am I missing something to say that would get this guy over his hurdle? I finally just threw up my hands in frustration.
He doesn't seem to settle for the fact that he knows a vector if he knows its components (with respect to the standard basis).
 
7:12 PM
Quite the stream of comments.
 
I thought I was clear and patient, but after the fourth repetition, I gave up.
But if you can think of something to help him, by all means, say it. I think he wants a derivation working entirely with vector values, but the only way to get Stokes's Theorem for vector-valued forms is to work with components (or dot with an arbitrary vector).
 
Reminds me of a situation in a tutorial yesterday in which I had to convince a student that if he knows the probability distribution of a random point in the plane, he also knows the probability distributions of its coordinates.
 
Yeah, I am not sure what he is being tripped by. Perhaps you could ask him to clarify his position and where exactly he is running into an issue?
 
Yeah. I mean sometimes it's nice to be coordinate-independent, I grant. But in the case I'm talking about, the definition of the gradient in a vector calculus setting is usually done with respect to coordinates.
@anakhro: He stated it. He wants to recover the vector equation with the $\vec i$ in there, etc. shrug
 
I like coordinate independent as long as you show the most intuitive way first.
Because coordinate-independent stuff can often seem like magic and ends up killing intuition.
 
7:19 PM
Yeah, I'm actually quite partial to the intuitive argument I gave him using the usual continuity estimate on a small thing.
(Pun not unintended.)
 
Heh.
This is Morita's book on characteristic classes specifically.
 
Yes, I looked at that book years ago. Bott's vanishing theorem is nice.
 
I have not heard of like any of this.
Only mapping class group, flat bundles, and foliations.
Oh beloved foliations. <3
Do you think it wouldn't be a bad read for the sake of just learning the material? Or do you think it is awfully specific and more of a thing to look at for reference?
 
I don't know the book well enough to advise. Is this stuff you need to use in your research program?
 
Reading doesn't hurt, but sometimes choosing to read one thing over another thing does.
No, I don't know what I should be doing for research. I should maybe ask.
 
7:26 PM
Probably time to focus a bit.
 
I think maybe I should do more symplectic/contact.
 
Well, certainly characteristic classes show up there, too, but you probably should start working on a problem and then see what you bump into.
 
Should I find a research supervisor before working on finding a problem? They don't give you a supervisor until the summer here.
I suppose the goal is to have the supervisor be able to help you with your research in some manner.
 
Have you been attending classes or seminars and interacting with anyone who interests you? You can ask for advice on papers to read.
You should ask ...
 
Yeah, I have a few people I am in contact with.
I think the only person who has interest in contact geometry in I am not in contact with.
 
7:30 PM
how ironic
 
That seems ill-fated.
 
Should I just go see them and introduce myself for the sake of putting myself out there?
 
Absolutely.
Take a look at a little of said person's work beforehand.
Perhaps say you'd like to do a directed reading going through one of his papers.
 
Pretty good idea.
How did you usually organize the grading for your grad level courses, Ted?
Like exam/midterm/assignments/presentation/paper kind of breakdown
 
Does this make sense? $\Bbb R^{1,1} \mapsto \Bbb \exp(\Bbb R^{1,1})$
The exercise is to transport the metric and show that they are equivalent after the transformation
$\Bbb M^{1,1}=\Bbb R^{1,1}$
 
7:49 PM
I've lost track of what you are doing
 
$\Bbb R^{1,1}$ is the minkowski diagram with rectangular hyperbolas
I want to transport the metric corresponding to $\Bbb M^{1,1}$ over to $\exp(\Bbb M^{1,1})$
 
$\mathbb R^{1,1}$ is a Lie algebra?
 
the metric for $\Bbb M^{1,1}$ should be the lorentz metric: $ds^2=dx^2-dt^2$
 
Or this is a Riemann exponential map?
That is to ask, what is this map $\exp$ to you?
 
$\Bbb R^{1,1}$ is a pseudo-euclidean space with a pseudo-euclidean metric
 
7:57 PM
So what is exp?
 
so I think that $\exp$ is just a diffeomorphism
 
What diffeomorphism?
 
that maps points on $\Bbb R^{1,1}$ to points on $\exp(\Bbb R^{1,1})$
 
What is $\exp(\mathbb R^{1,1})$
 
Its the shorthand notation for saying that all points in $\Bbb R^{1,1}$ are exponentiated. For example, $(x,y)\mapsto (e^x,e^y)$
exponentiating all points in the original manifold
 
8:00 PM
Exponential in the usual sense where e is Euler's constant?
 
yeah
 
You can just do the pullback metric with the inverse, can't you?
 
yeah that's my question. I keep getting stuck trying to do that
 
So first, what do you recall the definition for the pullback metric is?
 
If you have a metric $dY$ on $Y$ you can pull it back to $X via f:X→Y$ by setting $dX(a,b)$ to be $dY(f(a),f(b)).$
but I have a metric on the original manifold but not the target manifold
so is that why you are saying use the pullback metric with the inverse?
 
8:14 PM
Are you not doing metrics as manifold metrics?
 
Do the same thing but backward. $dY(a,b)=dX(f^{-1}(a),f^{-1}(b))$
 
so the natural log is the inverse of the exponential
 
Yes.
But the metric, is this on the tangent space, or as a metric space?
If it is the former, then your definition needs tweaking.
 
on the tangent space
 
So then what are you missing? Because $f^{-1}$ and $f$ are not defined on tangent vectors.
 
8:22 PM
okay I guess I don't know whether the metric is on the tangent space or as a metric space
I just guessed the tangent space
 
It's really important that you know what your mathematical objects are before trying to work with them.
Both for doing it yourself, and for getting help.
Easy way to answer your question is if the metric is a distance function on points of the actual manifold/space, or if it is like an inner product on the tangent space at a point.
 
oh it's actually the heading just above that
so I guess it's just a bit different than the euclidean inner product
 
Yes, so it's on the tangent space.
So you are missing something from your definition of the pullback metric.
 
okay, I think I need to find the definition of the pullback for a pseudo-euclidean metric instead of a euclidean metric
 
It will be the same.
Or, so I bet.
 
8:35 PM
so then I don't get it. the pullback will just be undefined?
 
You are missing something from your pullback.
 
do you already know what I'm missing?
 
Yes.
The setup is basically as follows: you have a smooth map $f\colon Y\to X$ and a metric $g$ on $X$. You want to pullback your metric on $X$ to a metric on $Y$ via $f$.
To do this, you suggested defining $g'(x,y) := g(f(x),f(y))$.
 
but wait, don't I have the metric on $Y$
 
No, we are doing it with the inverse.
 
8:39 PM
oh
 
The problem was that your metric is not on points of $X$ or $Y$. It's on the tangent spaces.
 
it's on tangent spaces of the geodesics?
 
So it should be $g'_p(v_p,w_p)$ that you want to define for $v_p,w_p\in T_pY$
 
okay
 
But then $f(v_p)$ and $f(w_p)$ doesn't make sense. So what do you do to remedy this?
 
8:42 PM
define new coordinates
 
No.
Why doesn't $f(v_p)$ make sense?
 
it doesn't make sense, because ln(x) is not defined for negative x
 
Try again.
What is $v_p$?
 
the tangent space at $p$
 
No, that's not what $v_p $ is
That's where $v_p$ lives.
 
8:45 PM
oh it's the tangent point
it's a specific point in the tangent space
it's a unit tangent vector at point $p$
in the tangent space of $Y$
 
It doesn't need to be unit.
It's just a tangent vector.
 
okay, $v_p$ is a tangent vector in $Y$
 
So note that $f$ is a function $Y\to X$. Why does $f(v_p)$ make no sense?
 
thinking
writing things down
because
it's backwards. You should take evaluate the function at the tangent vector in $X$ instead, because
 
9:02 PM
Evaluate what function?
 
$f$
 
Does $f$ take tangent vectors as input?
 
no it just takes scalars
 
You mean points $p\in Y$.
 
okay yeah $f$ takes points $p\in Y $ as input
 
9:05 PM
So how could you ever evaluate $f(v_p)$?
 
@anakhro Morita's book is very nice but I doubt it will be relevant to your work. It's a pleasant read if you have time for pleasant reads.
 
@MikeMiller thanks for the information! Perhaps I will look at it at a later date, then.
 
The kind of tangent bundle a topological manifold has is a "topological microbundle". Stiefel-Whitney classes can be defined for these (the definition in terms of the Thom isomorphism and Steenrod squares works without modification).
Rational pontryagin classes, too, may be defined as invariants of topological microbundles. But this is a highly non-trivial theorem (which I do not know the proof of).
Who knows about Chern classes (I can make a guess about a definition of complex topological microbundle, but I don't know a def of Chern class which extends straight away to complex topological microbundles), but since there's not really such a thing as a "complex topological manifold", I'm not too bothered.
 
@anakhro I don't know :(
 
@Ultradark well you can't. That's the problem. That's why it doesn't make sense.
@MikeMiller research-wise, is knot theory considered low-dimensional topology?
 
9:13 PM
yeah
 
Is low-dimensional topology still an okay research field, or do people find it hard to make progress in it?
That might be a vague/broad question.
Well, I guess, it is.
 
@anakhro so I have to use some other technique?
 
@Ultradark you have to use some other function in place of $f$ here.
 
Is the following derivation correct: ibb.co/w63zfzT?
 
Note that you want a function $?$ which you can apply to a tangent vector $v_p\in T_pY$ and get a tangent vector $?(v_p)\in T_{f(p)}X$.
Do you know of such a function $?\colon T_pY\to T_{f(p)}X$?
 
9:24 PM
@anakhro I see about 3 papers that should be called "low-dim topology" per day in arXiv.GT. People still seem to get jobs. But if you want to know this sort of thing before picking a project you need to look at people's advisors and students and where/if they get hired.
 
Yeah. It's just that I really enjoyed the Legendrian knot theory part of my M.Sc. thesis and I wonder if it is a good idea to read more into that.
I am not sure if that would still be low-dimensional topology, or it would be grouped under symplectic/contact.
 
it's a mistake to think that areas of research are cleanly separated like that
 
Yeah, but the flavour of "Legendrian" does change the interests of those who are giving out grants for that research field, I would think.
Or whatever flavour.
 
symplectic and contact stuff is sometimes more geometrical and sometimes more topological in flavor, and a lot of it is in dim 3/4. it would be silly to not say that this is also "low dimensional topology"
people who study legendrangian tori in contact 5-manifolds are not doing low dimensional topology in the same way
but they're still really not very far
the fields are defined more by the tools they use than the problems they solve
 
@anakhro I don't know know of such a function, do you ?
 
9:31 PM
@Ultradark indeed. Maybe you need to go refresh yourself on basic theory of smooth manifolds.
@MikeMiller that's a good distinction, between tools/problems. Thanks.
In that vein, would you say I should seek to use the tools I enjoyed working with, rather than to solve the problems like those I enjoyed working on?
 
I don't see how that's different --- if you enjoyed working on it that presumably means you enjoyed the things you did while working on it. Accepted problems in an area are often precisely those which we think are accessible to our tools
 
@anakhro Yeah I do. What is the function?
 
@Ultradark hint: it's related to $f$.
@MikeMiller I guess that is true.
 
If I can't find it, can I come back and ask you in a few hours?
 
Sure, but I won't just tell you the answer. You really should take this as an opportunity to learn from a book about manifolds.
I recommend Tu's "Introduction to Manifolds".
The answer is in Chapter 3 of said book, but if you are this uncomfortable with manifolds, I highly recommend reading both 2 and 3.
 
 
1 hour later…
10:59 PM
I've been looking for an integral identity that plays with a function inverse on the integrand limits, but I'm not sure I know how to search for it, I know it looks something like $\int_{f(b)}^{f(a)}x dx=\int_{b}^{a}xf'(x) dx$ but I'm not quite sure..
anyone know which one is the correct one or maybe a source, I know you can get this identity by integrating a function on the y axis and equating that with the inverse of a function on the x axis, they give the same area, thus they're the same..
 
$\displaystyle \int_a^b f(x)\, \mathrm{d}x = b f(b) - a f(a) - \int_{f(a)}^{f(b)} f^{-1} (x) \,\mathrm{d}x$
@Rainb Is it that?
 
I've not seen that before I think, I'll save it, cause it seems these identities are not googleable. lol
do you have a source of a proof or something?
 
In mathematics, integrals of inverse functions can be computed by means of a formula that expresses the antiderivatives of the inverse f − 1 {\displaystyle f^{-1}} of a continuous and invertible function f {\displaystyle f} , in terms of f − 1 {\displaystyle f^{-1}} and an antiderivative of f {\d...
@Rainb I didn't recall it from memory either, I searched it.
I remember trying to show this equality: for $a,b > 0$
$\displaystyle\int_0^1 \sqrt[a]{1-x^b}\,\mathrm{d}x=\int_0^1 \sqrt[b]{1-x^a}\,\mathrm{d}x$
I guess the best way to remember it is that picture in the Wiki page.
 
@EdwardEvans Lol
That's not actually true, though, is it?
…Also I guess sometime between posting that and now you changed your username
 
I actually found it, (I had to recreate it myself) I know I saw it somewhere else on some textbook but meh, here it is $ \int_{f^{-1}(b)}^{f^{-1}(a)}xf'(x)dx = \int_{b}^{a}f^{-1}(x)dx $
 
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