meh. disagreeing tastes. see starboard.

I'm fine with it for standard engineering calculus ...

I told an engineering student to learn vector identities in terms of tensor notation because it's super fast and easy.

tern didn't like it :P

Well, I'll tell you to learn it all with differential forms. :)

You tell me how to do curl in 4-D. :)

Ain't nobody got time for forms

I acknowledge the massive usefulness of coordinates, but their automatic usage in lieu of finding geometric explanations when possible can irk me.

I don't really have any comment on that student's issues.

I wasted about 6 hours of my life early in my career with a chemist who was convinced that he could do vector analysis in $\Bbb R^4$ without differential forms. I lectured him for about 3 hours. He came back the next week denying I had said anything and started with the identical stupid arguments.

mentally subtracts 3 from 6 to get the length of the next week's lecture

Sometimes invariant notation obscures, tern. Sometimes it clarifies. I think judgment is required.

@arctictern Did you get the obvious result?

Oh oh, @MikeM is here to raise hell.

Taught my first 225 today. They were uninspired by the inverse function theorem and mildly inspired by Banach manifolds.

@Mike has been a little grumpy over the past days :P

You're doing it again this year? How borrrrring. :P

@Danu: But Ted is always grumpy.

Okay, I'm back @Ted

John Duffield keeps on flagging my chat messages :P

So, back to exact sequences

So $\sum f(e_i)\otimes g_i=0$

@TedShifrin I always thought your multivariable text was more or less a direct response to H&H. Is that the book you are refering to?

Is that book really a "classic"?

@Ted Easiest job ever.

I'm willing to bet that my Quantum Mechanics 1 job is easier

@arctictern It's often very useful to have an explicit construction of certain things in coordinates even when you can intrinsically prove things.

not disagreeing

So much backup for me here haha

I guess it's only because the implicit overall agreement lies on your side tern :)

Oh, no, @PVAL, H&H is very, very modern (too much so). It was Buck's Advanced Calculus.

@TedShifrin I was for one thoroughly confused by H&H.

I told you my story, that one month into teaching out of it I realized it was not good and started writing my book to use second semester.

They're now in their infinite edition, and I'm still in first, so you can tell what the world thinks of me. :P

D'awwww

I think people want to brag, like Hubbard about himself, about his esoteric treatments. He is a smart, smart man, however.

Well Stewart's sold more copies than either of you so what does the world know.

Different audience, @PVAL, totally.

@Danu: Have you done injectivity? Any ideas for an approach?

@TedShifrin I feel ridiculous, but no! :\

Okay, let's see

Well, if all else fails (tern won't like this), take a basis for $E$ and a basis for $G$.

How do you get a basis for $E\otimes G$?

That's what I wanted to do haha

@TedShifrin Just pair em up

Well, now finish.

So once it's in terms of a basis the different terms cannot cancel each other, so they must all cancel individually

@Ted What in your opinion is the best book for the audience Stewart is usually taught to?

@arctictern: Do you have a nice "elegant" proof that tensor product is flat (working with vector spaces)?

@MikeMiller Typical people.

I guess Hallet-Hughes is probably better, but it still seems like a scam to me.

@PVAL: I'm not that fond of Stewart. He ripped off Edwards & Penney a lot. I loved their first and second editions, but they went downhill with every edition. I do not like Hughes-Hallet. I rather like Rogowski (from UCLA).

@TedShifrin I said "when possible." haters gonna hate.

I'm not hating. I'm serious.

I was telling Danu to suppose $\sum e_i\otimes g_i$ mapped to $0$, and then I realized I was stuck.

oh, then no

But there should be an invariant argument.

I hate when that happens lol

@TedShifrin lol

So it wasn't just me?

@Balarka I mean, I'm also excited by Banach manifolds.

Well, there's no good way to combine tensor products, of course. All I could think of was working with a basis for $G$ (not for $E$) and looking at the resulting equations.

Do you think inverse function theorem is boring? I doubt.

@Ted A metric on a surface is conformal to a constant curvsture metric. If the surface is noncompact, how do I know what the sign is?

I love the inverse function theorem and the Banach space proof. I recite it every chance I get. :)

@Ted I think at UCSD Rogawski was used for like normal math/engineering/cse majors (20 series) and Hughes-Hallet was used for other lower level math classes (10 series). I never taught the latter so didn't see much of that book, but I thought it was strictly more difficult than Stewart or Hughes-Hallet

I don't know the background of the people in the course you're teaching though, so I may be misinterpreting their reactions.

Hughes-Hallet had a lofty goal, but in my perusal of it (we tried it at UGA in several classes) I found it disappointing. Rogawski does a few idioscyncratic things, but I think it's solid. I never taught out of it, either, but worked through chapters of it with high school teachers learning BC calculus stuff to teach.

@MikeM: Is it complete?

Sure.

Actually, maybe not.

Oh, then it can't have positive curvature. Oh, then it can.

Is the best way really to look at us harmonic functions?

sub

I'm not remotely thinking of that. I'm thinking about surfaces in $\Bbb R^3$ with constant positive curvature.

@TedShifrin The edition I used to teach out of seemed like a relatively standard calculus text. I think it went under substantial revisions and changes from its first few editions.

They have to be spheres if we're complete.

Don't care about those. If you demand that the conformal metric is complete, the sign of a constant curvsture conformal metric is well-defined.

@PVAL: They were trying to be more conceptual and less rote-computational. I have no idea what happened after numerous editions.

But I don't know how to find the sign given a metric.

Anyways, @TedShifrin, I guess if $\sum_{ij}a_{ij}f(e_i)\otimes g_j=0$ then by injectivity of $f$ we have $a_{ij}=0$ for every $i,j$.

Why are there $a_{ij}$ in there, @Danu?

@TedShifrin I expanded stuff in a basis

Choose your basis for $F$ to be the image of the basis in $E$, extended if necessary. Keep life simple. (See, sometimes I side with tern!)

@MikeM: I haven't thought about this in a while, but can't you look at the scalar curvature and see how it transforms for a conformal change of metric? Won't that be relatively simple? If not, what about Ric?

Yeah, I can. I was talking to Balarka about solving the uniformization theorem earlier. But to do that, you need to start by knowing the sign.

@TedShifrin So up to notation you agree with me though, right?

I guess the easiest trichotomy is probably from subharmonic functions. But I should check how Mazzeo does it.

Not yet, @Danu.

Mazzeo is smart, @MikeM. (He was my first official student. :) )

@Danu: Are you going back to my original, or are your $e_i$ and $g_j$ known to be basis vectors?

@TedShifrin Basis

There seems to be multiple conversations going on here. What's the most hip one?

@TedShifrin So if I choose my basis of $F$ to be an extension of the image of the basis of $E$, i.e. set $f_i:=f(e_i)$, where $\{e_i\}$ are the basis for $E$, then $f\otimes \operatorname{id}_G(\sum_{ij}a_{ij}e_i\otimes g_j)=\sum_{ij}a_{ij}f_i\otimes g_j=0$

Not mine, @Balarka

Oh, then, your $f(e_i)$ is wrong. The matrix for your map is in terms of given bases for $E$ and $F$.

You should have $\sum a_{ij}f_j$.

But like you said, I can pick the basis of $F$ so that that reduces to just one summand

I was correcting the argument you gave. You have to remember what the matrix for a linear map means. It presupposes bases for both vector spaces. Then the formula needs to be right.

But, yes, you can simplify the argument.

So do you still consider this to be incorrect:

@Mike Doesn't $\Bbb R^2$ admit a flat complete metric and one with constant negative curvature coming from the hyperboloid.

Am I not understanding what your asking?

@PVAL The one you describe is either not complete or not conformal to the standard metric.

@Danu: Pedantically speaking, you need to say $\{f_i\otimes g_j\}$ forms a linearly independent set because it's a subset of a basis. But yes.

@PVAL: The hyperboloid ain't got constant negative curvature. Remember Hilbert.

There is no complete surface of constant negative curvature in $\Bbb R^3$.

@MikeMiller I am going to start - perhaps not finish - reading your answer on my question; better than whining about time issues. What's a proper homotopy equivalence? A homotopy equivalence which is proper?

But, anyhow, you can easily see non-constant negative curvature.

@Balarka No, a proper map that has a proper map in the other direction such that fg and gf are properly homotopic to the identity.

Aha, got it. Yes, of course.

$*$ and $\Bbb R$ are not proper homotopy equivalent.

I agree.

I don't remember ever encountering that term. (Not to be confused with tern.)

Just thinking about homotopy equivalence is trivial here because contract the vector space fibers, so you're thinking about this. OK.

@Ted It's what compactly aupported cohomology is incariant under.

Yeah, I get it. I wonder if Bott/Tu say that, cuz I've certainly been through it.

And locally finite homology, of course.

Essentially equivalent to homotopy equivalence of the one-point compactifications, aren't they not?

Doubt it.

@Balarka Not so sure about that

I don't think Bott/Tu mumble about homotopy equivalence at all. At least, nothing in the index.

In the pointed homotopy category maybe.

Ah, yes, rel the pt at infinity.

Otherwise I doubt it too.

Bye for now, all.

Byes.

You should also assume your spaces are locally compact so the compactification isn't crap.

good evening

Yeah.

Actually you're probably never right.

Your maps and homotopy equivalences can send other points to the point at infinity.

Oh well.

@TedShifrin Oh, bye!

I think the one point compactification of the Whitehead manifold is homotopy equivalent to $S^3$.

Hmm, well, it's S^3 mod the Whitehead continua isn't it.

I wonder if the obvious map from S^3 to that is isomorphism on homotopy.

It's not obvious to me that S^3 mod Whitehead continuum is simply connected.

So I read that a SES of holomorphic vector bundles is equivalent to a dual SES. In the courses that I took, we always used that a SES split to argue that the dual is also a SES. What do I use in this context?

@Danu: If $T\colon E\to F$ is injective (surjective), then $T^*\colon F^*\to E^*$ is surjective (injective).

@TedShifrin Okay, I'll think about that.

Huybrechts uses it in proving the existence of the Euler sequence

Also, I think he glosses over the surjectivity of the map $\bigoplus_{j=0}^n\mathcal O(1)\to \mathcal T_{\mathbb P^n}$

What is the map, even?

I'm trying to find a map of this sort somewhere in the book...

@Ted Have you read Kazdan-Warner?

@MikeM: Many, many years ago.

Was the series readable? I want to learn it, but I don't know if there are modern sources.

I don't know the answer to the latter. Both of them write well.

@Danu: Doing pullbacks of forms, you can see how things transform and see you need an $\mathscr O(-1)$. Six of one, half dozen of the other. I love forms, but you don't. The tangent vector argument is a bit more geometric. Think about how a cone collapses ... a vector further out is affected how?

@TedShifrin Where'd you get that I don't like forms? Also, the rest of this message sounds like some mystical prophecy to me.

@TedShifrin I just realized there are plenty of examples to Balarka's question.

@TedShifrin Also, what do you mean by "to make it well-defined"?

The map $\pi_*$ is... well-defined, no?

@TedShifrin Nevermind, no I don't. But I do have examples where the unit tangent bundles are diffeomorphic but the total spaces are not.

Are probably not.

@Danu: How do we pick which $z$ maps to $[z]$?

Well, there are many... Is that a problem?

Do you remember what well-defined means? :)

:^) math.stackexchange.com/questions/1944282/…

@TedShifrin That every element is only mapped to one element.

Also see my comment about geometry of a cone. Think about projecting a real cone to the sphere.

I.e., map defined independent of choices ...

Please flock to my question, thx

@TedShifrin Like what choices?

I never flock. Never.

So I am probably not understanding what map $\pi_*$ is. Is it not the differential?

@Danu, choice of $z$ in the line, of course.

Sure, derivative map.

If I have any map between smooth manifolds then the derivative map is immediately well-defined, no? o_0

You need sleep. Read all the stuff I've typed.

No, this is driving me crazy now. You are saying that $d\pi$ is not well-defined?

I'm saying $z$ is not well-defined.

I don't understand what you mean. For any $z,z'\in \ell_z$, $[z]=[z']$. What do you mean?

Or are you talking about $0\in \Bbb C^{n+1}$?

Does the derivative of $\pi$ at $z$ and at $z'$ map $v$ to the same thing? Reread my comments about cones ....

@TedShifrin I sadly don't understand the comment about cones.

@Ted Yes, ok, so interestingly take the unit disc bundles of the 4-plane bundles Milnor uses to define his exotic spheres. These are never diffeomorphic, even though the unit sphere bundles frequently are.

Seriously, Danu, get sleep and draw pictures tomorrow.

@TedShifrin Ok, so do you mean that if I take $\hat z_0$ and $10 \hat z_0$ and consider the curves which go through those points with velocity $1$ in the $\hat z_1$-direction, the derivative of $\pi$ is not the same? I can live with that.

Ok. So the map is not well-defined as it stands.

But the fact that $d_{\hat z_0}\pi(v)$ does not equal $d_{10\hat{z_0}}\pi(v)$ is... not what I would call ill-defined?

I need a tangent vector to $\Bbb P^n$.