I can find nowhere what a purely atomic measure is. T.T

@Josué: a bunch of delta functions at isolated points

Hi @Ted.

hi @TedShifrin @BalarkaSen

a guy from University of toronto is coming here to give a talk about generalized complex geometry

what is generalized complex geometry lol

is it differential geometry on complex manifolds ?

@TedShifrin do you have an idea what that is ?

@SamuelYusim Sorry for not replying back sooner. I seem to have been (still am) ill for longer than I thought I would be. The essence of Borsuk-Ulam is this: you want to prove there's no antipodal preserving map $f : S^n \to S^{n-1}$. By definition of antipodal preserving, $f$ gives a map $\tilde{f} : \Bbb{RP}^n \to \Bbb{RP}^{n-1}$ of the corresponding quotient spaces.

$\tilde{f}$ also induces an isomorphism on $\pi_1$. To see this, take a generating loop $\gamma$ in $\Bbb{RP}^n$; that lifts to a path $\tilde{\gamma}$ joining two antipodal points in $S^n$. $f\tilde{\gamma}$ is a path joining antipodal points ('cause $f$ preserves antipodes), once again. Pushing that downstairs gives me a generating loop in $\Bbb{RP}^{n-1}$. Covering space theory tells us all this is precisely what $\tilde{f}$ does, so it's an isomorphism in $\pi_1$.

We thus get a map $\Bbb{RP}^n \to \Bbb{RP}^{n-1}$ which induces isomorphism in $\pi_1$. For $n = 2$ you can already see why that's garbage: $\pi_1\Bbb{RP}^1 \cong \Bbb Z/2$ and $\pi_1 \Bbb{RP}^1 \cong \Bbb Z$. There's no nontrivial map $\Bbb Z/2 \to \Bbb Z$.

For higher $n$ the fundamental groups coincide so one needs a different technique. The key fact is the cohomology ring structure of $\Bbb{RP}^k$: if $\alpha$ is a generator of $H^1(\Bbb{RP}^k; \Bbb Z/2)$, all the generators in $H^\ell(\Bbb{RP}^k; \Bbb Z/2)$ are $\ell$-times cup powers of $\alpha$, aka $\alpha^\ell$. If $\tilde{f}$ was an isomorphism in $\pi_1$, it'd be an isomorphism in $H^1(-; \Bbb Z/2)$. The map $H^*(\tilde{f})$ on cohomology will thus be isomorphism in grade 1.

By naturality of cup product, $\tilde{f}$ will be nontrivial on the topmost cohomology, but that's nonsense as $H^n(\Bbb{RP}^n)$ is nonzero whereas $H^n(\Bbb{RP}^{n-1})$ is zero.

I hope that's written in a comprehensible way.

Hi @Balarka @Adeek Karim

Hi again.

No, complex manifolds is complex geometry. It could be supergeometry (which shows up in physics). I don't know.

I wasn't here when you hi-ed the last time, Balarka :P

Sorry you're feeling crummy ... still ... I go to the new doctor tomorrow.

I still far prefer the winding number proof using differential topology for Borsuk-Ulam, @Balarka.

yeah, but I thought I'd give a hi again anyway. Hopefully you'll be better tomorrow.

Actually, I don't feel so horrible. After a week, the antibiotics have worn off and the side-effects have worn off too. Now there's just the question of killing off the annoying problem.

I guess Karim went far, far away

@TedShifrin I don't remember G-P's proof. Is it any different? I thought one would just do the thing about using intersection theory instead.

No, it's totally different. This is one of my all-time favorite proofs and I told you to study it :P

It uses the generalization of the argument principle from complex analysis, that winding number counts (with signs) roots inside.

Oh, yes, now I remember.

It's inductive, studying upper and lower hemispheres and reducing to the upper hemisphere.

If you can turn it into the usual algebraic topology proof, I'll give you a small prize.

I think I can. I'll have to reread it though, I didn't pay much attention to it the first time (as is obvious now...)

I don't think you see anything about the hyperplane class generating the cohomology ring mod 2 in this proof.

But maybe I haven't tried hard enough.

I'll start working on that. Good to know the antibiotics aren't misbehaving, by the way.

Not my usual thing.

@Balarka: Because I quit them a week ago?

Ah, I misread that bit.

I put a comment, @Jessy. Start with easier cases.

@Ted, I'm looking at that now.

Agh, I continue to be driven nuts by the number of people who just have to post a solution when they can see that several of us have given hints to the OP. Agh. Agh. Agh.

Reputations > help, as always :P

@TedShifrin, I posted a reply. Still very confused.

@TedShifrin I asked Mike a question yesterday, have you seen that? I haven't yet found an example I was looking for.

@Balarka: Nope, haven't seen it.

@Jessy: Answered.

@Balarka: As you know, plenty of professors out there don't think my style of teaching is good teaching. So shrug

What's an example of nonisomorphic vector bundles E, F such that Total(E) (ie total space) is homeomorphic to Total(F). One obstruction is the Euler class of the zero section - if they are not the same, Total(E) and Total(F) will have different compactly supported cohomology. So easiest potential example is probably TS^5 and S^5 x R^5. Why or why not do they have different total spaces?

Ah, I remember such a question driving me nuts for ages, @Balarka. There are definitely non-isomorphic bundles with diffeomorphic total spaces, but can I reconstruct one ... ?

That's a great question, btw.

@Ted, so I did what you said to do...

It's also true that tangent bundle of oriented 3-manifolds have zero Euler class, so I looked at some. Started with TRP^3 (which I know stably what it is), but of course that's trivial since RP^3 is Lie. Google says any orientable 3-manifold is parallelizable so that also fails. I don't know if there are any easier lower dimensional examples to look for without comparing with the trivial bundle.

OK, @Jessy, and it gives you an $x$ that works, right?

@TedShifrin Heh.

hi @TedShifrin

btw I really like your book @TedShifrin

your multi variable mathematics book

No need to say nice things, Karim ... I help you anyway :P

haha :)

btw I want to start doing research in algebraic geometry starting from next summer

@TedShifrin It has been infinitely more helpful for me to work through the problems you and Mike have given me instead of just reading through books. But I can also understand that some people are impatient and this method usually takes time.

I was wondering what is the pre req required to study that algebraic geometry book Hartshorne's?

Just so you know, I think your teaching style is great.

@Adeek Oh that books requires uncountably many prerequisites.

yeah I agree with @BalarkaSen every prof should teach like you.

I gave up on studying that when I was on algebraic geometry.

yeah it seemed a lot of things

You need solid commutative algebra, Karim. Atiyah-Macdonald, for example. Unless you want to do algebraic geometry the way I did, with more differential geometry, topology, and complex analysis. :P

Uncountably many? :)

Eisenbud's commutative algebra book is much more accessible, but infinitely longer than A-M.

I think far more algebra than A-M is needed for Hartshorne. Probably all of Eisenbud.

But that's just my impression.

I like your way @TedShifrin

It would help to study Riemann surfaces using analysis before even starting Hartshorne. Also, not necessarily to go linearly through Hartshorne.

I am really starting to dig complex analysis

what is a good book into riemann surfaces?

You can also read more casual introductions to algebraic geometry, like Myles Reid and Shafaverich.

I am gonna take commutative algebra next semester

I like Shafarevich.

I like Forster a lot. There are classical books, like Springer, from many years ago.

cool

Miles Reid made one of the most awful puns in mathemarics.

@Balarka: Most people don't have their own private me and Mike, either ... :P

lol

I don't think puns have the LUB property, @MikeM.

@MikeMiller What's what?

I'm curious.

Oh wait ... they don't have the Archimedean property.

lool

p-adic puns?

lol miles reid speaks japanese

I wonder if Mike's gonna tell.

just got Eisenbud seems like a nice book

A complex variety has its canonical bundle, $K_X$. If you pick a resolution of this variety, $f: X' \to X$, then you can pull back the canonical bundle on the base, $f^*K_X$. But there might be a difference between $K_{X'}$ and $f^*K_X$. What do you call it when there's no discrepancy?

I give up.

disjuncted.

I'll give Balarka a chance.

I'm hopeless.

Not good at making terminologies. "canonical resolution"?

is canonical in the answer?

I am watching some lectures into differential geometry given by a physics prof

Well, it's not discrepant. So you call it a crepant resolution, of course.

it is pretty good actually

the guy is awesome

lol

Oh, I've heard that before, @MikeM. I think it's become standard.

It has indeed. When I first read it I thought it was french until I went to google translate.

crépuscule = sunset

I didn't have a guess. I was curious, so I searched.

I knew it was one or the other ... just have forgotten everything.

Quite an imaginative terminology.

Usually in math we would say "undis..." ... this is better :P

Now I'm going to spend the next month trying to recreate the vector bundle example that I actually knew 30 years ago.

I'll probably fail.

haha

Example of what?

What Balarka asked you yesterday, he said.

Non-isomorphic vector bundles with homeomorphic total space.

I remember this drove me nuts ...

Surely Milnor gives an example somewhere.

Oh, I didn't think about it.

I thought about it for many years.

Balarka wants to rule out cohomological obstructions.

Or at least something along the lines of a general technique to tell when they're not homeomorphic when the bundles have same Euler class (eg TS^5 and S^5 x R^5 like I said).

Yeah, @Balarka, that's a non-unintelligent approach.

But I don't think I did that.

non-unintelligent? :P

I think I found some example where I could see the diffeomorphism.

Don't you have school today, Balarka?

I'll bunk school for the next few days.

Didn't you have another example you suggested?

I forget now.

Holidays are ahead, so nobody will budge.

@MikeMiller I tried with TRP^3 but 'course that's trivial. No oriented 3-manifold example can't work, google says, as they are all parallelizable.

So nope.

@Balarka: Sadly, this MO question is related but doesn't answer your question.

Yeah, google came up with that but not anything else.

Ah, but there's a paper referenced in there that allegedly has an theorem and counterexamples in certain dimensions: dl.dropboxusercontent.com/u/5188175/desapiowalschap.pdf.

Is there a good description of total space of direct sum of two vector bundles, in general? I doubt.

I swear I had a reasonable example years ago.

Yeah, that paper was too complicated so I didn't read it.

LOL

The example refers to a theorem of Hirsch, I see.

It is also pretty cool that they use differential geometry to give examples.

I'm pretty sure the fundamental group of the unit sphere bundle is a homeomorphism invariant of the vector bundle - it's the fundamental group at infinity. So you just need to break that.

They're getting examples from normal bundles of high-dimensional knotted spheres.

Codimension at least 3.

Fun enough.

Wow. Maybe I'm deluding myself that I came up with an example. I thought I used complex geometry somehow, but it's been way too long.

Well, the above is only true when the base is compact. But in any case it should work and it feels like it shouldn't be that hard to apply.

Sure, compact base.

@MikeMiller Hmm.

It will only help you with rank 2 bundles.

Eh. So it won't help.

But rank 2 real bundles are complex line bundles.

No anomaly with the Euler class can happen.

Watch your tongue.

oriented rank 2 bundles?

Oops.

Better to pick on him until he catches on.

At least, more fun.

He's sickly, though, so I have compassion.

Besides, he already knows more topology than I do.

And in my dementia haze I'm losing math apace. :P

That paper is only about bundles over homotopy spheres. So I still think there's an example with a more complicated base space.

What's the total space of the tautological line bundle? $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$ minus a line from the $\Bbb{CP}^2$ factor?

Over $\Bbb{CP}^1$ I mean.

$\Bbb C^2$ with a point blown up.

So that's what you said in a harder way :P

lol, yes.

OK, I just wanted an example to check Mike's statement. That seems to work fine.

@Balarka: Jim Belk's answer to that MO question seems very down-to-earth.

Well, I'm outta here. Get healthy, Balarka, and I'll have fun going to the new doctor tomorrow :)

Ah, sounds like that's the proof of Mike's statement.

Right.

Bubye.

Although it seems to prove only one direction, not both.

is this true

hm

@BalarkaSen still awake ?

Yep.

I wonder if the homotopy type of the unit bundle is a homeomorphism invariant. It seems that, at least under favorable conditions, the homotopy groups are.

why if $f : X \rightarrow Y$ is differentiable iff for all $C^{\infty}$ functions g on Y, $g \circ f$ is $C^{\infty}$ on X ?

X and Y here are manifolds.

@Adeek Think about charts and bump functions.

Sorry, I have to go for now. If you haven't solved it by the time I get back, I'll answer to it.

alright

thanks @MikeMiller I will think about those I was thinking I could use bump functions somehow

@Balarka You should post the vector bundle question on main. I think I can start to write an answer, but other people might have input too.

Ted got tired of trying to explain that to me, and I don't blame him.

@MikeMiller for going the forward direction my idea is as follow

we know that $F : X \rightarrow Y$ is differentiable

Let us say we have the following charts $\{h_j,U_j\}$ and $\{p_i,V_i\}$ for manifolds X and Y respectively

we know if F is differentiable then the composition $p_i \circ F \circ h_j^{-1}$ whenever it is defined for all i and j.

Let L be a vector. What is the physical meaning of equations: (L . \nabla)L = 0. The paper I am reading states that this means the line following L is straight.

for the forward direction my idea is if $x \in X$ I consider $(g \circ f)(x) = g(f(x))$ and rewrite it somehow in terms of charts?

@MikeMiller ?

when you come back @BalarkaSen can you check my question math.stackexchange.com/questions/1941757/…

there's an extra x-3 that you've multiplied the equation with in the 2nd pic in the second step

@KaumudiHarikumar: $x=3$ is not allowed, because $x-3$ is in the denominators. In fact, you should multiply by $x-3$ and there won't be any $x-3$ at all!!

But why did you do that?

Why not multiply both sides by $(x-7)(x+9)$ as well?

hey @MAFIA36790

and @KaumudiHarikumar it's actually a bad idea to multiply an equation you're solving for any variable with anything that has that variable

Really? The textbook is very wrong.

that is, you were solving for x, and you multiplied with x-3 and x+1 etc

Well, we're trying to clear denominators, @SoumyoB.

you can just take the denominators outside and keep them outside without having to multiply them and cancel them out

I mean, as I said, we could multiply by $x-7$ or $x+23$, just as easily. We'll get extraneous roots, i.e., numbers that are not roots of the original equation.

Yes, of course I understand, @SoumyoB. But for many students trying to find a common denominator is harder than clearing denominators.

No, you allow yourself to multiply only by the factors needed to clear out the denominators, and nothing more. Do you know how to put fractions over a common denominator? That's another way to do it. Put both fractions over the same denominator, but make that denominator have as few factors as possible.

To see something simpler: how would you solve $\dfrac{x}{3} = \dfrac{7}{2}$ for $x$?

Right, @Kaumudi. That is precisely what you did in the first, and it's excellent.

Your textbook apparently did not follow my rule for the second method — multiply only by what is necessary, nothing extra. If they are doing this to get an extraneous root and have you check that $x=3$ will not work, there's an object lesson there.

@TedShifrin multiplying by factors that apparently aren't extraneous could still cause problems later on

@Kaumudi. Right.

Checking for extraneous roots is something that becomes important in algebra (and in higher math) later on.

I don't believe you, @SoumyoB. :)

If you do your original method, @Kaumudi, there can't be anything extraneous. You've only rewritten algebra, not multiplied by anything :)

for example one of the roots might turn out to be $\alpha$ where $x-\alpha$ was something you had multiplied the equation with previously

No, if there were no extra factors, @SoumyoB, that can't happen.

40+ years of teaching, @Kaumudi :)

You'll be fine!

Sure thing.

@SoumyoB: The famous situation in which it arises is something like solving $x=\sqrt{x+2}$ by squaring both sides.

it's the same as multiplying by $(x+\sqrt{x+2})$

you mean when you rewrite it as $x-\sqrt{x+2} = 0$?

yes

nevertheless, you end up with an extraneous root :)

yes

which emphasizes the point I make in teaching beginning proofs that solving an equation is an iff problem, so one must make sure every step is iff. This isn't.

i was just pointing out that this case was not so different from the other case where they multiplied by $(x-3)$ too much

I understand, but the thing you multiplied by has no root, so it's a different situation :)

which seems contradictory ...

Ah, it depends on the branch of square root, of course.

it has

$-1$

Of course. I was being dumb.

Thanks. I'm doing 6 things at once :)

$6$

why did I mathjax this lol

ha ha

I can barely do 2 things at once

@MikeMiller I'll do that, but not right now.

I'm trying to get a few things done.

Ah, I guess @Balarka is allowed to be awake.

Even if he should be at school.

I fixed my sleep cycle yesterday.

By sleeping 12 hours and waking up at 6am today :P

We'll see how long you stay "fixed."

That's arguable, yes.

Who's getting fixed?

Of course.

Hello. The following is from Reed and Simon's Functional Analysis. They have not defined "open sets" but they give the following theorem. I would take (e) as the definition of open sets, but they leave it as a fact. So what might they mean by "open set"?

@MAFIA36790 I assume that's a consequence of their definition since it's listed as part (a) of the theorem

@MAFIA36790 They probably mean something like elimination.

@cap Have they defined metric space?

@TobiasKildetoft yes

@cap hmm, and yet they have not defined what it means for a subset of a metric space to be open at this point? Weird.

If you write down $AX = Y$ in terms of linear equations in terms of the rows, you get stuff like $\sum_i a_{ik} x_i = y_i$, for all $k$ corresponding to the columns.

Row transformations on $A$ makes eliminations to those set of equations, like we do in high school.

Eliminations of variables, that is.

OK, I see what they mean. Suppose you make the row transformation on $A$ by adding row $i$ to $c$ times row $j$.

Then you're essentially adding the equation $\sum_k a_{ki} x_k = y_i$ to $c$ times the equation $\sum_k a_{kj} x_k = y_i$.

That's all what they mean.

That's what they mean by "linear combination"

@MAFIA36790 but that just leaves one needing a definition of an open set, which is no easier

There's a typo in here though. That expression should have been $a_{ki} x_k$. And "for all $i$ corresponding to rows".

heya @Tobias ... Last time I looked there was no conversation in here.

@TedShifrin Hi

They must have defined open somewhere before this.

@TedShifrin You left UGA before Paul Sobaje came there as a postdoc, right?

Correct. This is my 2nd year absent.

ok

Of course, I'm still writing oodles of letters of recommendation. But that should stop eventually :)

Anyone here who can help me understand random variables and distribution functions? Is this the right place to talk about this?

@TedShifrin It will be a while before I need to do that

What does it mean to find a formula for distribution function of a random variable?

@MAFIA36790 So clearly multiplying by a constant is, right?

@MAFIA36790 yes, a linear combination of that equation

Let us agree that switching two equations is certainly linear combination.

It just says "instead of looking at this equation, look at the other".

@MAFIA36790 sure, so is this

math.stackexchange.com/questions/1941837/… anyone who can help me with this problem

Sure, why not. Linear combination of a bunch of equations is precisely what you did in high school.

Nothing new.

@MAFIA36790 Oh, I mean, that thing in scare quotes is an informal explanation of switching rows, in the case of the corresponding equations.

Not of linear combinations.

I didn't notice what message you were referring to.

Hi, do you guys know about something like a "countable powerset" of a set $X$, ie the countable subsets of $X$ (and why not the countable complement subsets). Does this thing usually or always have different cardinality from $\mathcal P(X)$ if $X$ is uncountable?

@s.harp Yes, it will be a union of $|X|$ sets of cardinality $< |X|$ and thus of cardinality $|X|$.

(assuming choice)

wait, that was nonsense

O_o

that would have shocked me

to be more clear, I think it should be false that the cardinality of this thing is $2^{|X|}$ whenever $X$ is uncountable

I am not sure if this runs into GCH or something like that actually

I'm not an expert on such fundamentals, but I think this is far too tame^

Hi @Kari

o/

@Danu Hey.

More questionssss

This one should be easy

Consider a compact Riemann surface, and the line bundles associated to the divisors $[x_i]$, which we will denote by $\mathcal O(x_i)$

If there are two distinct points such that $\mathcal O(x_1)\cong \mathcal O(x_2)$, I should conclude $X\cong\Bbb P^1$.