Mathematics - 2021-03-04 (page 3 of 3)

Mike Miller

20:02

Surely using so many explicit indices makes everything much worse.

I'd write x in R^k and y in R^{n-k} so that points in R^n are written (x,y).

Ted Shifrin

Yes, @MikeM, I do that also, but I was trying to fit in with the notation of the posted question. Being a good Chern student, I also use index ranges: $\alpha$ for the submanifold and $\mu$ for the normal space.

I follow your suggestion, in fact, and write the implicit function theorem with $\partial \mathbf f/\partial\mathbf y$ (both in book and in lectures).

Mike Miller

Oh, I agree that you should match the notation of the question. I just think the author of that book has chosen obscuring notation.

Google says it's O'Neill, Semi-Riemannian Geometry. So of course he uses indices.

Ted Shifrin

Well, not that I don't agree, but note that Guillemin & Pollack don't do what you and I recommend. I'm not remembering about Hirsch.

I don't think Spivak does, either, in his magnum opus.

Mike Miller

Yes. When I teach out of GP I suggest people do what I say here, too. :)

A variable for each thing going on, not necessarily each number

Ted Shifrin

If I'm following a textbook, I generally try to adopt its notation just to make it easier on students, but maybe I should have followed your advice on this one. As I say, when I wrote my book, I did it right. (And I always hated Buck's Advanced Calculus because he wrote everything out explicitly in components, never used any vector notation. Ugh.)

Yes, and while we're ranting, all books and all teachers should try students to choose $x\in X$ and $y\in Y$ and let letter choices guide what's living where. I'm stunned by how many do it randomly.

Mike Miller

20:11

I was shocked when 1/4 of students said "Let U be a closed set" on a midterm!

Ted Shifrin

LOL. Yes, I do not allow that.

Mike Miller

They parsed this as just "variable for a set"

Ted Shifrin

Well, who are you and I to say $U$ has to denote an open?

Mike Miller

I tend to match notation when I think there will be a serious issue for the students otherwise. I match notation every time I teach calculus, say, and 99% of the time I match notation linear algebra. (The textbook I'm using has absolutely atrocious notation for change-of-coordinates, so I modified it.) In more advanced classes I give a dictionary between notation but do not make as much effort to match.

Ted Shifrin

We're back to leslie's symbol-pushing. If you're pushing symbols consistently, who cares what they are?

Mike Miller

20:12

But we are social creatures, and I might have trouble following the symbols.

Ted Shifrin

That's your problem.

Astyx

Well, U rotated resembles C, which stands for closed right ?

Ted Shifrin

As long as I'm being presumptuous, see my comment here.

@Astyx: But damn you, you rotated negatively.

Mike Miller

I know nothing about MC but I understand the OP's question.

Ted Shifrin

Oh, I guess that's OK, because you're starting with C and rotating positively.

Astyx

20:14

The multivariate Cauchy formula requires integrating on a polytorus right?

Ted Shifrin

Yes, I understand OP's question, too, but if you actually want to do geometry with these homogeneous spaces, MC is essential. As opposed to pushing symbols.

Mike Miller

I would be tempted to tell him that he should delete his question and work very hard to understand those identifications himself.

Astyx

It's obvious but I never thought about it

Ted Shifrin

Yes, @Astyx.

That's why it's about holomorphic $n$-forms :P

Astyx

I had done very little multivariate complex analysis, and always assumed it went circle -> sphere -> hypersphere

But it turns out the product of circles is not a sphere

Ted Shifrin

20:15

@MikeM: It also comes back to my pet peeve that books don't talk about the tangent bundle as an explicited associated bundle (associated to the adjoint rep). Indeed, that OP seems not to know what the tangent bundle is.

@Astyx: There are different analogues of a residue in higher dimension. But if you're working with a complete intersection, then the polytorus you're observing is what shows up.

Mike Miller

Agree, though I worked that out myself as a boy and have understood it in that language ever since.

I think I remember having conversations with. you about this a half decade ago.

Astyx

It becomes much harder to visualize

Ted Shifrin

Yeah. I was never taught it. I only figured it out when I taught grad diff geo at UGA (the third quarter of it!).

I worked with Stiefel manifolds and frame bundles very comfortably, but never thought much about associated vector bundles.

Well, @Astyx, you're not going to visualize higher dimensional varieties, anyhow :P

user2103480

@Astyx criminal

Astyx

It's very clear what it means for a singularity to be in a circle in $\Bbb C^1$. I have no clue how to think about it being in a polytorus in $\Bbb C^2$

Ted Shifrin

20:22

Well, think it as arising as the intersection of two curves (intersecting transversely) in $\Bbb C^2$. For example, $f(x,y)=xy=0$. Take tubular neighborhoods of the two component curves and intersect.

I'm using "curve" in the algebraic sense (i.e., one-dimensional complex submanifold).

Astyx

right

OneCold Ruben

Can I ask math questions in here instead of posting a question in an effort to have a conversation about it? If so: If we have a function from R to R, x= x^2 what would we say about the image of the interval of [-1,1) ? would we say the it doesn't exist because the inverse function function of -1 doesn't exist in the reals.

Or would we only list the real part of it?

Ted Shifrin

You mean $f(x)=x^2$. And you mean inverse image, which has almost nothing to do with an inverse function.

Call it preimage if you're going to think of inverse functions.

Jakobian

I guess my problem was that I didn't realize f^k(x_1, ..., x_m) can be defined on much larger set than just W, even though f^k is defined by inverting (x_1, ..., x_m) on W

let me see if that makes sense

Ted Shifrin

20:38

You are restricting the chart to the submanifold but the chart is more global.

OneCold Ruben

not the inverse image, the image is given as [-1,1) thus the output of the function is [-1,0)

Semiclassical

@TedShifrin the fact that $f^{-1}(\{x\})$ exists for a given $x$ even if $f^{-1}(x)$ doesn't is sorta weird to me. i get why the notation works like that, to be clear

OneCold Ruben

sorry [-1,1)

Semiclassical

but notationally it sorta makes me frown

erk

Ted Shifrin

Just do what I said and use the chart to put you in Euclidean space to start with .

Mike Miller

20:39

Any function g on R^k extends to a function on R^n by f(x_1, ..., x_n) = g(x_1, ..., x_k).

You're just not using the extra variables since they aren't important.

I maintain this author's notation makes things worse.

Ted Shifrin

What you're saying doesn't make sense, @OneColdRuben.

OneCold Ruben

This is a function from the reals to the reals. I cant imagine how that makes any

This is why i thought is just DNE

Ted Shifrin

Yes, from reals to reals. The image means we evaluate on the points in $[-1,1)$ and end up with $[0,1]$.

OneCold Ruben

So it is saying what interval gives the image [-1,1)

Ted Shifrin

The preimage means we ask what $x$ values give us $f(x)\in [-1,1)$.

You are talking about preimage or inverse image, as I said, then.

Nobody says you have to hit every point in there. You just want everything that lands in there.

OneCold Ruben

20:43

Well i am telling you the image so yes we are saying the same thing backward from eachother.

Thorgott

Ahh, I haven't seen this before, but it makes perfect sense. This captures the intuition I was trying to get at before, but I wasn't able to make it precise. Since bundles over the circle are either trivial or trivial + Möbius, this illustrates why the Möbius band is the quintessential example of non-orientability once again. Of course, if you don't have these loops, fix an orientation at one point and just transfer it to any other point via arbitrary path (the pullback bundle on the interval is trivial, so an orientation of one fiber unambiguously determines an orientation of the others) a…

Ted Shifrin

I'm using the language correctly.

OneCold Ruben

And i am a fool.

Ted Shifrin

It is not saying that the image has to be all of that interval. It's important to understand.

OneCold Ruben

Thus the questions, to be less foolish.

Ted Shifrin

20:46

@Thorgott Mike is nicely making “no orientation-reversing loop” more tangible with the bundle. I like it.

OneCold Ruben

So it is only the interval that covers most of the image given.

Ted Shifrin

It might not be an interval. It might be unions of intervals.

Mike Miller

@Thorgott Your dragging buys you a map $[0,1] \times D^n \to M$, and taking df_{t,0} gives a bundle iso from [0,1] x R^n to pullback bundle of TM along [0,1] x {0}. Since when you glue this up you get the non-trivial bundle, df_{0,0} and df_{1,0} must be oriented differently.

This can probably be phrased more elegantly.

Ted Shifrin

You want all $x$ so that $f(x)$ is in the desired set. Try $(0,1)$.

OneCold Ruben

Why would you need to specify that a function from R to R could ne a union of intervals?

Ted Shifrin

20:49

Damn. Where did I say that?

OneCold Ruben

:57239892

Ted Shifrin

Using language carefully is the first thing you have to do.

OneCold Ruben

Oops.

Ted Shifrin

I never changed your function.

What $x$ map to the interval $(0,1)$?

OneCold Ruben

It would be [0,1) -> [-1,1) so the answer is [-1,0)

Mike Miller

20:52

also, Thorgott, consider the homomorphism w: pi_1(M) -> Z/2 given by w(gamma) = 1 for gamma an orientation reversing loop and 0 otherwise.

1) Notice that ker(w) is the subgroup corresponding to the oriented double cover. 2) Homomorphisms pi_1 X -> A correspond bijectively to cohomology classes H^1(X; A). w now corresponds to a class called w_1(M), the second Stiefel Whitney class. 3) Elements in H^1(M; Z/2) correspond bijectively to real line bundles. w corresponds to the real line bundle "det(TM)", whose transition functions are the determinants of those of TM.

In terms of the oriented double cover M' it's the bundle M' x_{Z/2} R, with Z/2 acting by negation on R

Ted Shifrin

I have no idea what you just said. For what values of $x$ is $x^2$ in the interval $(0,1j$?

OneCold Ruben

But I find that weird. because nothing goes to -1 using the function F(x) = x^2

Mike Miller

I like how all of this matches up so cleanly. If you have Poincare duality these also correspond to a closed codimension 1 submanifold S --- the zero set of a transverse section of det(TM), for which w(gamma) = #(gamma cap S)

And this is the divisor picture of line bundles you learn in complex geometry

Thorgott

@MikeMiller Ah nice. You're saying the differentials constitute the clutching function $S^0\rightarrow\mathbb{R}^n$ for a non-trivial $S^1$-bundle, so must be reversely oriented.

OneCold Ruben

F( [-1,1) ) = [0,1) but that is only a subset of [-1,1), using the function F(x) = x^2

Thorgott

20:57

uh, GL(n)*

Mike Miller

Yup

OneCold Ruben

sorry corrected
F( (-1,1) ) = [0,1) but that is only a subset of [-1,1), using the function F(x) = x^2 so we are saying that only a sub set is still the correct answer for the question about the preimage of [-1,1) of F(x)=x^2 in an function defined from R to R.

Thorgott

21:14

1) is clear. Let me think about 2). The first Stiefel-Whitney class of a line bundle should be the same as the mod 2 Euler class. So that is the cohomology class that takes a homology class, pushes it forward via zero section to the tangent bundle, then takes a representative transversal to the zero section and counts intersections mod 2, right?
So what you're saying is that an oriented loop generically intersects an even number of times, but an unoriented one an odd number of times. Looking at the standard pictures of the trivial and the Möbius bundle over a circle and the obvious sections…

Mike Miller

Sure, that's the perspective I wrote down below "If you have Poincare duality..."

I probably would not have defined w_1 the way you did but it's probably standard. It's also probably not how I would prove the relevant claim. I would prove the relevant claim in RP^n for large n, since every line bundle on an n-dim CW complex is pulled back from the tautological bundle on RP^n.

Thorgott

truth is I just don't know a good definition of Stiefel-Whitney classes, so I defaulted to using the Euler class

I'm still not really versed in the intersection viewpoint, but this all sounds very reasonable

how does the correspondence between H^1 and line bundles work? I don't think I've seen that before

Mike Miller

Do you know the relation to EM spaces?

Actually, since this is H^1, I can skip down to knowing how K(G,1)'s work

You know that <X, K(G,1)> = Hom(pi_1 X, G), right?

So H^1(X; A) = <X, K(A,1)> = Hom(pi_1 X, A). But also, for A = Z/2, K(Z/2, 1) = RP^inf, and <X, RP^inf> = line bundles over X

Thorgott

21:38

ah, interesting

that's a nice unified perspective on why orientable <=> w_1=0 <=> det(TM) trivial

Semiclassical

@MikeMiller what does EM mean here, out of curiousity?

Thorgott

Eilenberg-MacLane

Alessandro Codenotti

eilenberg maclane

Thorgott

sniped

Semiclassical

kk

google assumes em spaces means typography

which, tbf, would usually be the right answer

Thorgott

21:41

but man, there's way too many equivalent ways of characterizing orientability

I can probably give at least 10 equivalent definitions of orientability and explain why they're equivalent, but I still don't feel like I get the essence. Truly <insert von Neumann quote here>

Edward Evans

if u twist is differnet

2

robjohn

22:06

@Ted ha!

I haven't really gotten past the third word.

10 Replies

an integer not divisible by 3, must be (3n)-1 or (3n)-2, right?

robjohn

@10Replies yes

10 Replies

do I need to prove this?

robjohn

I guess that depends on the course and the instructor

10 Replies

I'm writing a proof for proving that if n can't be evenly divided by 3, then n squared-1 must be divisible by 3

robjohn

22:10

I'd accept that, the $3n-1$ or $3n-2$ part, without proof if I were teaching that class.

10 Replies

ok, thanks :)

robjohn

However, if there is doubt, including a proof wouldn't hurt.

10 Replies

what would that proof look like?

I don't think I'll include it, but I'm curious

user2103480

@Thorgott >flag as inappropriate

To rub it in like this.......

robjohn

@10Replies that depends on what you are assumed to know

10 Replies

22:15

How do I find that out

Thorgott

Time to finish this job. Let's do a more general scenario for simplicity: if $M,N$ are manifolds with boundary and $f\colon\partial M\rightarrow\partial N$ is a diffeomorphism, $M\cup_fN$ can be made a smooth manifold without boundary so that the canonical inclusions are embeddings. If $M,N$ are oriented and $f$ is orientation-reversing, $M\cup_fN$ can be oriented compatibly with $M,N$. I claim the diffeomorphism type of $M\cup_fN$ only depends on the isotopy type of $f$, so let $f,g$ be isotopic. Then $gf^{-1}$ is isotopic to the identity on $\partial N$ and can be extended to a diffeotopy…

I'm explicitly relying on the fact that the smooth structure on $M\cup_fN$ is unique up to diffeomorphism, which is not obvious from the construction, but whatever

robjohn

@10Replies you could start by looking at division by $3$ and noting that the remainder has to be at least $0$ and less than $3$

Thorgott

I guess I should figure out why CP^2#CP^2 and CP^2#\overline{CP^2} are different for completeness sake

robjohn

That leaves $0,1,2$

dc3rd

One is as old or as young as one chooses to be @robjohn........I've chosen my fixed age as 28. (there may be lot's of variation in that choice, but I'm sticking to it :) )

robjohn

22:30

@dc3rd It is interesting that I did not get pinged by the @robjohn there. Perhaps it has to do with the dots after it.

Mike Miller

@Thorgott Hm? I think it's clear enough, you construct an atlas (which depends on f, but no additional choices)

Well done, IMO going through this is very very instructive

Thorgott

the construction I have in mind works by choosing collars

I guess you should be able to construct an atlas completely explicitly, but it seems very annoying

dc3rd

22:48

I'm guessing that is probably what it is, which would be odd.....

user2103480

23:25

@Semiclassical if I have the positions and velocities of particles in a fluid under some hamiltonian dynamics at hand, what is the stress tensor?

Ted Shifrin

23:46

@robjohn Oh, now I get why you ha!ed me.

robjohn

35? Give me a break!

Of course, I wonder what the average is here.

Ted Shifrin

In chat or on MSE?

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$Mathematics$ Mathematics