Mathematics

12:00 AM

At each fixed point $z\in \Bbb C^{n+1}$, $d_z\pi$ is a linear map from $T_z\Bbb C^{n+1}$ to $T_{[z]}\Bbb P^n$, which well-defined. No?

Ted Shifrin

The exotic sphere story is mostly unknown to me, @MikeM.

So what, Danu?

Danu

This was what I was talking about in the above

This is what I meant when I was saying that stuff was not ill-defined. In what sense are you talking about ill-defined?

Ted Shifrin

Remember that we're starting at a point in $\Bbb P^n$. Everything's a bundle on it.

Danu

Waaait... You're talking about $\bigoplus_{j=0}^n \mathcal O_{\Bbb P^n}$??!!

Ted Shifrin

That's the trivial $\Bbb C^{n+1}$ bundle. Yes.

Danu

12:04 AM

When, in your original message, you wrote $T_z\Bbb C^{n+1}$ I just took that to mean the tangent space (at $z\in \Bbb C^{n+1}$!) to $\Bbb C^{n+1}$

Mike Miller

@TedShifrin What I said follows immediately from reading Wall's paper classifying (n-1)-connected 2n-manifold.s

So like hell if I know how to do this now. Surely the unit disc bundles of non-isomorphic vector bundles can be non-diffeomorphic.

Ted Shifrin

I'm no topologist, Mike :)

Danu

Are you saying that the map $\pi_*:T_z \Bbb C^{n+1}\to T_{[z]}\Bbb P^n$ induces one between the two bundles on $\Bbb P^n$, except that it's not well-defined?

Ted Shifrin

Yes

Now make it well-defined.

Danu

I thought you were saying that $\pi_*$ itself was ill-defined

Which is why I freaked out

Ted Shifrin

12:07 AM

I never said any such thing. Any time you work with equivalence classes, ...

Danu

46 mins ago, by Ted Shifrin

@Danu: I mentioned that to you earlier. You look at the derivative mapping $\pi_*: T_z(\Bbb C^{n+1}) \to T_{[z]}\Bbb P^n$ and see why you need the twist by $\mathscr O(1)$ to make it well-defined.

I thought this literally meant: Look at $\pi_*$ and see why you need [stuff] to make it well-defined.

Ted Shifrin

Yes, you're choosing $z$ given $[z]$. Duh.

Danu

So I'll draw a diagram to see how $\pi_*$ induces a map between those bundles.

Okay, so I first map $(z,v)\in T_z\Bbb C^{n+1}$ to $([z],v)\in \bigoplus \mathcal O_{\Bbb P^n}=\Bbb P^n\times \Bbb C^{n+1}$. Then I want to go to something like $([z],d_z\pi(v))$, but now we get the problem with dependence on representative of $[z]$, because $d_z\pi(v)\neq d_{z'}\pi(v)$.

Ted Shifrin

Right. We're on the same wavelength (or frequency). I leave you to it :)

Danu

Gosh, I'm happy I at least managed to parse that bit.

I'd still love to understand the cone thing...

So I want to say that I just need to keep track of the norm of $z$, so something like $z=\lambda\hat z$, and then keeping track of the $\lambda$ by adding the information contained in $\lambda \hat z^*$, where $\hat z^*$ is the linear map that projects along $\hat z$.

Anyways, thanks for the help @TedShifrin!

Now, I just need to keep track how how $d\pi$ depends on the norm of $z$.

Ah, it's linear :D

Ted Shifrin

12:34 AM

Not quite.

Danu

It's not linear, or the other stuff not quite?

Because I felt pretty sure it's linear

I mean, if I go out twice as far from the origin (horizontally) and draw vertical arrow that is also twice as long (double the tangent vector), I end up in the same ray

Ted Shifrin

No, that means it's antilinear as a function depending on $z$.

This is what I meant about the geometry of projecting a cone to the sphere.

If you write what you said, you have $d\pi_{tz} v = \frac 1t d\pi_z v$.

Danu

@TedShifrin Hmmm... I'm drawing a "cone" i.e. two lines in a plane.... Radially project to a circle?

@TedShifrin Yes, I agree with that

Ted Shifrin

12:49 AM

So that's not linear in $z$. :)

(in terms of scalar multiplication, that is)

Danu

Oh, lol Eh okay :P It's what I meant though :)

Ted Shifrin

OK, so what do you need to do to make there be a well-defined bundle map?

Danu

Can you perhaps try to elaborate on what you mean by projecting cones?

Ted Shifrin

I'm just talking about the radial projection map $x\rightsquigarrow x/\|x\|$ in the real case.

That's essentially what we're doing with projective space.

Danu

Sure

But how would this show me the thing about the projection being antilinear?

Ted Shifrin

12:51 AM

Think about resolving a vector into a piece along the ray and a piece orthogonal. So a sphere of radius $r$ maps how to the sphere of radius $1$?

Danu

@TedShifrin Yeah, I did the orthogonal thing. Right, so it should shrink it

Because the angle covered is smaller

Ted Shifrin

It has to ... by a factor of $1/r$.

Danu

Okay, so that's the same that I was doing, but I was thikning in terms of right-angled triangles

Ted Shifrin

OK. Done.

Danu

OK, so now all I need to do is keep track of the modulus

Ted Shifrin

12:52 AM

Well, over $\Bbb C$, the scalar multiple, yes.

Danu

in terms of a linear functional

and then use that to cancel the dependence

Ted Shifrin

well, you're tensoring all of $\Bbb C^{n+1}$ with $\mathscr O(1)$, but yes.

not linear functional, by the way, dual.

Danu

but dual is linear functional? Aren't we working with rays (i.e. vector spaces)?

Ted Shifrin

Huh? Oh, linear functional on the line, hence section of $\mathscr O(1)$, sure, sorry.

Danu

I spent the past several days grinding LINEAR FUNCTIONAL into my head :P

Enjoys Math

12:56 AM

Solved it :) math.stackexchange.com/a/1944351/26327

Danu

So, Ted, do you always denote the trivial bundle on $X$ of rank $n+1$ by simply $\Bbb C^{n+1}$?

I think that this was essentially what was throwing me off earlier

Ted Shifrin

No. Sometimes I write $\varepsilon^{n+1}$ if I'm doing bundles.

Danu

ok

Ted Shifrin

If I'm in alg geo land with sheaves, I write what you wrote earlier.

I never wrote the sequence of bundles earlier. If I had, I would have written $\Bbb P^n\times \Bbb C^{n+1}$ for the trivial bundle.

I need to eat dinner and go play bridge. Sleep well!

Danu

Okay---I'll just figure out the details of how to encode the linear thing in the right way

Danu

1:18 AM

Oh, this is pretty nice---I'm also getting a clear picture of why the image of $\mathcal O$ should be the kernel of the second map @TedShifrin: That inclusion comes from twisting the inclusion of the tautological line bundle into the trivial bundle by $\mathcal O(1)$, but that clearly only produces radial vectors in the $\Bbb C^{n+1}$-factor---precisely the kernel of $d\pi$!

Even better, it seems clear that the second map is surjective because every tangent vector to $\Bbb P^n$ comes from a tangent vector to $\Bbb C^{n+1}$!

Have a nice evening @Mike! Hope you're doing well

Ming

1:36 AM

morning all

user228700

3:56 AM

Hello everyone :-) I have a quick question. If, for a given quadratic equation in $x$, the determinant comes out to be 0, then apart from the fact that this equation has two complex conjugate roots, what else does this imply? For instance, does it imply that the equation is $>0$ for all $x$ belonging to the real set? My textbook seems to think so...is this correct?

arctic tern

4:10 AM

@KaumudiHarikumar if the discriminant (not determinant) is 0 then there is only one root (with multiplicity 2). if, in addition, the coefficients are real, then the solution is real, and then the outputs are either all nonnegative or all nonpositive.

user228700

4:29 AM

@arctictern Yes, discriminant-typo, sorry. Oh, crap. I just realized that I worded the question wrongly! I meant if the discriminant is less than zero. Crap. So sorry.

Alexander Bollbach

5:25 AM

how, in a normal range-mapping procedure, where the formula is output = output_start + ((output_end - output_start) / (input_end - input_start)) * (input - input_start) could I reverse the input range?

Sangchul Lee

5:39 AM

Hello everyone, I have a stupid question.

How would you pronounce the expression $\prod_{i=1}^{n} x_i$?

I would read it as "product i from 1 to n x i", but I'm not sure if this is the usual way...

Balarka Sen

6:24 AM

@Danu I'm sure you have discussed all this with Ted but the point is the map $d\pi_\tilde{z} : T_{\tilde{z}}(\Bbb C^{n+1} - 0) \to T_z\Bbb P^n$ requires a choice of a point $\tilde{z}$ in the preimage of $z$ by the quotient map $\Bbb C^{n+1} - 0 \to \Bbb P^n$. So to get a well-defined map to the tangent bundle $T\Bbb P^n$, you need to twist it by $O(-1)$

This thing also drove me mad a few days ago because I was coming up with a wrong exact sequence.

Tobias Kildetoft

7:17 AM

First time I have run into this: Latex not being able to compile because it could not read a full line due to buffer size

It is probably also the first time my .tex file is larger than the .pdf it creates

Mike Miller

@Balarka So the most obvious way to find an example to your question is to find examples where the unit disc bundle is diffeomorphic (which keeps us in the land of compact manifolds).

So as you did the obvious place to try is spheres. If I'm reading this paper of Wall right, unfortunately, two n-disc bundles over $S^n$ are diffeomorphic iff they're isomorphic.

Danu

7:48 AM

@BalarkaSen The main confusion that arose was that I didn't realize that Ted was talking about the induced map from the trivial bundle on $\Bbb P^n$---I thought he was literally talking about the map from the tangent bundle on $\Bbb C^{n+1}-0$, and I couldn't understand why he kept saying it wasn't well-defined.

In any case, I hope I did the right twist (the map I was talking about is $\bigoplus_{j=0}^n \mathcal O(1)\to\mathcal T_{\Bbb P^n}$). In the end I wrote down $\varphi([z],v,\hat z^*)\mapsto \hat z^*(z)d_z\pi(v)$, where $\hat z^*$ is the linear functional given by projection onto the $\hat z$-direction.

Then it you choose two representatives $z,z'=tz$ of $[z]$, you get $\hat z^*(z')d_{z'}\pi(v)=tt^{-1} d_z\pi(v)$, so a well-defined map, I hope.

I'm not happy with my own notation but I'm not sure how to improve it.

I think I should probably not be using $\hat z^*$---but I don't see any actual problem with what I wrote down. It just doesn't feel quite right.

Andrew Thompson

8:21 AM

How's everybody been doing lately?

Danu

Me, terrified by complex geometry

Tobias Kildetoft

@AndrewThompson More or less breaking my LaTeX compiler

Andrew Thompson

@Danu Geometry is *scary*. (Which is why I think of algebraic geometry as something purely categorical where one has a lot of polynomial rings involved for some reason.)

@TobiasKildetoft That's impressive.

Tobias Kildetoft

@AndrewThompson Not so much that it really broke, but I needed to increase the maximal length of lines it would read and then wait forever for it to even save the file

(increasing to 100k characters was not enough)

Andrew Thompson

What on earth are you writing?

Tobias Kildetoft

8:24 AM

Tables of twosided cells in type $B_6$

the .tex file is about 2MB

The .pdf is only about 500kb and 5 pages (thought these are 100x220inch pages)

Andrew Thompson

Are these personal notes or do you actually intend to put this beast up somewhere?

Tobias Kildetoft

@AndrewThompson I probably won't include these, but I am planning on putting together the tables for some smaller ranks and putting it on the arXiv as it might be useful to others (I could certainly have used that when I started trying to learn this stuff a few years ago)

This was just because it was the smallest case where a certain thing might happen and I needed to check if it did (I could just have asked the computer of course, but I was making the tables anyway for the other cases)

Tobias Kildetoft

9:11 AM

@AndrewThompson What about you? What are you up to?

Secret

Well, one easy example will be the constant function whoose domain is $[0,1]$. But then I still don't really get what we are doing in the definition of a topology. Suppose we have two topologies for the interval [0,1], $\tau_1=\{[0,1],\emptyset,[0,0.5),[0.5,1]\}$ and $\tau_2=\{[0,1],\emptyset,[0,0.9),[0.9,1]\}$, then these two $\tau_1$ are different topologies, but how are they different in the commonly said intuition that a topology connects things?
Basically, I still don't really understand concretely what I am doing when I define a topology for a space other than the definitions

Tht is what I am doing to the space when I define a topology, and why it has to be closed under union and intersections

Tobias Kildetoft

@Secret A topology does not connect things. It tells you which things are very close to each other

Secret

Tobias Kildetoft

@Secret What is your question about that?

Secret

(Other than because it violates the definition), why is this not a topology, what does it mean when $\{2\}$ is missing?

Tobias Kildetoft

9:17 AM

@Secret It violates the definition of a topology, hence it is not a topology.

Secret

you mention that topology tells how close two things are to each other, but on the last non example, it seems 1,2,3, 2 and 3 and 1 and 2 are close to each other, thus why missing 2 will still be a problem in defining the closiness of things in this space?

That is, I don't understand the motivation of the definition of a topology in why it has to be closed under union and intersections

mercio

it's not a topology because it violates the definition of a topology

user116211

@mercio: You were actually right! Law of composition is what magma is!!

mercio

?W?

Secret

Ok let me ask another way: We define a mathematical object called a topology to be a set of open subsets with the following properties: 1. The emptyset and the whole set is part of the topology, 2. The topology need to be closed under intersection and union.
What is the motivation of 2. in the definition, what issue might have happened if we don't have 2.?

mercio

9:25 AM

didn't you forget to ask what were the motivation of 1 ?

user116211

@mercio Bourbaki never mentioned of binary operation; what they mentioned is magma. Binary operation is a general term that doesn't need to have the same co-domain as the domain.

mercio

the issue is that then when you define continuous function, it probably won't work at all

@MAFIA36790 are you saying a magma is a kind of binary operation ?

user116211

@mercio yup where the operation is closed.

mercio

nononononono

a magma is a set together with an associative binary operation on that set (+ closed if you're not english)

user116211

Binary operation is a more general term.

mercio

9:28 AM

but a magma is not an operation

it's a set

with an operation on that set

user116211

@mercio A set with a law of composition.

mercio

yes

user116211

;D

Secret

@mercio That one is straightforward because the whole set and empty set is always the subset of the whole set itself, thus these ensure 1. has to be there since a topology is a set of subsets

mercio

but don't say that a magma is a law of composition

Secret

9:29 AM

@mercio Hmm ok, I will think about it

mercio

absolutely not @secret

i really don't see the logic in that

there are sets of subsets of $X$ which don't contain $X$ nor the empty set

user116211

@mercio yes; I know that; what i'm saying is that binary operation is more general term where there are two operands; that's it.

mercio

for exemple $\{\{2\} ; \{2;3\}\}$ is a set of subsets of $\{1;2;3;4\}$

basically you're trying to understand topologies the wrong way

but first being told an intuitive?? explanation of what it should be or what it should do and then reading the definition and then complaining

instead you have to read the definition, have faith in it, read the rest of the textbook, do a lot of exercise, do a lot of more exercises, and then you will begin to understand slowly intuituively what a topology should be or what it should do

and now i have to go

user116211

@mercio o/

Secret

ok, I will continue to read through and do the exercise until it hopefully clicks

Secret

10:01 AM

(The following screenie are not questions, please ignore them for now)

Tobias Kildetoft

@MAFIA36790 A binary operation on a set $A$ is by definition a map from $A\times A$ to $A$.

user116211

@TobiasKildetoft Bourbaki didn't define this in this way.

Tobias Kildetoft

@MAFIA36790 how do they define a binary operation?

user116211

@TobiasKildetoft Although I started with this definition.

Secret

(Ok so homoemorphism in topology is "stronger" than the notion of group homeomorphism as that one only require f to be surjective)

Tobias Kildetoft

10:13 AM

@Secret No, no definition of homeomorphism only requires surjectivity

Danu

@Secret There is no notion of group homeomorphism (unless your group happens to be considered as a topological group, but that's not what you're talking about).

You meant group homomorphism, maybe...

Secret

Sorry I got the wrong spelling as the words spelt similarly

Tobias Kildetoft

@Secret there is no requirement that group homomorphisms be surjective

Danu

None of the two is "stronger" in some obvious sense---these types of maps do not (in the most general case) make sense for the same types of objects at all.

Secret

I see

Danu

10:16 AM

Here it would be helpful to use some basic language from (some slightly simplified) category theory---every "type of objects" (e.g. topological spaces, or groups) defines a category with a natural type of maps between the objects, the so-called "morphisms".

user116211

Well, @tobias, let me check if I find it somewhere... well at $\mathsf{Pr}\infty\mathsf{fWiki}$...

Danu

A morphism is a map between two objects in the category that preserves the relevant structure.

In the case of the category of groups, the morphisms are maps that preserve the group structure, i.e. group homomorphisms.

In the case of topological spaces, the morphisms are maps that preserve the topological structure, i.e. continuous maps.

If there is a morphism from object $A$ to $B$, but also one from $B$ to $A$, such that the two are each other's inverses, then you say both maps are isomorphisms.

In the category of topological spaces, an isomorphism is a continuous map with continuous inverse, i.e. a homeomorphism.

So a homeomorphism is an isomorphism of topological spaces.

A group homomorphism is not necessarily an isomorphism of groups.

Do you have any more questions @Secret?

Secret

Yes, that is what I learnt in my linear algebra course

I see, so in general the prefixes homeo- and homo- are not related, despite both group homomorphism and homeomorphism for topological spaces both have that notion of preserving a structure?

Danu

@Secret The prefix homeo- is not a common prefix. As far as I know, it's only used for isomorphisms of topological spaces.

Secret

ok I will keep that in mind

5 mins ago, by Danu

Do you have any more questions @Secret?

Actually, as I went through the notes, I am generating questions at the rate of every 2 definition, example and theorems. Based on what the discussion with Tobias and Mercio, it seems a lot of these might be due to a false premise or lack of understanding, thus hopefully they will answer itself as I ran through the notes

Learning topology formally is a new thing to me that only just recently I have time to dig through, as back in my undergrad I never get the chance to take a topology course

Danu

10:30 AM

It's standard for students to get really confused for a few weeks when learning basic point-set topology. The best way out is to do many exercises.

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$Mathematics$ Mathematics