00:00 - 13:0013:00 - 16:00

1:00 PM
Yeah

Blowup is a nice local construction. I spent days trying to understand it, and it felt great when I actually did.

So any comment on my Euler sequence map?

The spirit is definitely correct, but I don't know if I wrote it down correctly.

What's the version of Euler sequence you're working on? Mine is $0 \to O(-1) \to \Bbb C^{n+1} \to T\Bbb P^n \otimes O(-1) \to 0$.

1:04 PM
I twist by $O(1)$

Can you please write down the sequence you have in mind explicitly?

Just twist yours by $O(1)$, so tensor everything by $O(1)$

Ah, got it.

$0\to \mathcal O\to\bigoplus_{j=0}^n \mathcal O(1) \to \mathcal T_{\Bbb P^n}\to 0$

You're proving $T\Bbb P^n$ is stably isomorphic to direct sum of a bunch of $O(1)$. Got it.
Let's see what your map is now then.

1:07 PM
If it splits, I guess?
But the Euler sequence doesn't split

I meant topologically :)

What do you mean by topologically? I'm not so well-versed in the terminology...

It splits as real bundles. Don't worry about it.

Oh, right okay.

I am just translating your thing in the way I find it easier to parse, because I am more comfortable there.
What's $\hat{z}$ in what you wrote?

1:13 PM
$z/|z|$

OK, I see.

What I don't like is that somehow it feels like if I can use a normalized version of $\hat z$ I shouldn't need to keep track of the norm of $z$ at all, making the twist unnecessary---which would be wrong.
But I don't see a real problem with the actual map I wrote down...

What does the norm of $z$ have to do with the twist? I don't see your objection.

What if I'd just say I map $([z],v)\mapsto d_{\hat z}\pi(v)$, eliminating the dependence on the norm of the representative by always picking the norm-1 one? Is it somehow not possible to keep track of the norm of the representative of $[z]$?

I doubt that's a smooth map.

1:22 PM
@BalarkaSen Hmm... But you think the one I wrote down is okay?

Yes, I do.
It seems fine.

coolbeans

Hello,

Basic statistics question.

I am trying to figure out how I can begin to solve a probability that is setup as

$P(X = k | X + Y = n)$ where $X$ and $Y$ are independent.

I am trying to figure out if I can infer something just from $P(X | X + Y) = \frac{P(X \cap X + Y)}{P(X + Y)}$
Is there some property that I can leverage. I thought that maybe we know that the probability of $X$ given $X + Y$ would always be at least 1

1:50 PM
@BalarkaSen So how is your analysis going? What stuff are you doing?

Did some cool analysis today.
Conformal maps, etc.

Complex analysis? Nice
I wish I knew more about it

Yeah, complex analysis. It's a nice branch of mathematics.

Are you following a book?

Yes. Stein-Shakarchi mostly.

1:54 PM
A professor here is doing a seminar on "special topics in complex analysis" this semester... Could be cool if I had time to spare!

Nice.

A four-book series!

I'm doing the complex analysis book.

@BalarkaSen it's an interesting choice of ordering of the volumes

The running theme in their books is Fourier analysis, yes.
Also I agree with their choice of ordering of complex analysis and real analysis.

2:09 PM
Heh, really?

Real analysis is drier than complex, yes (from whatever I have studied of both and what people - actual analysts - have told me)

Yeah man
Whenever I tried to read a bit, it got SO boring :P

Does somebody happen to know the fourier transform of $\frac1{p_1^2-p_2^2+m^2}$ in off the top of their head?
Mathematica can't do it.. :(

2:34 PM
Morning @MikeMiller

morning

Saw your message. Interesting that there are no counterexamples over spheres.

@s.harp The Fourier transform of $1/|p|^2$ is $1/|x|$, if that helps

At least in a specific dimension.

Woops, of course---in $\Bbb R^3$!
I assumed $\Bbb R^3$ because that thing looked like a propagator in quantum field theory, where the momenta $p$ are vectors in $\Bbb R^3$

2:40 PM
I am not sure anymore if that comment was directed to me or Danu.

lol

Wait, it wasn't at me?
Surprisingly accurate, then!
No, it has to be at me :P

What is meant by $\frac{\partial(x, y)}{\partial(u,v)}$ if we have a function $\mathbf{r}(u,v)$?

I don't see what you mean by "in a specific dimension", because we're fixing the dimension of the base sphere (assuming of course that it was to me and not Danu).

Does that mean to sum the partial derivatives of u and v with regards to x and y?

2:47 PM
@Lozansky jacobian matrix

@BalarkaSen It has to be at me :P

okay thanks arctic tern

@Danu Shrug.

3:08 PM
@BalarkaSen I demanded n-disc bundle, not k-disc bundle.

Ah. I am already assuming rank of the bundle is dimension of the base for simplicity, hence the confusion.

Hello all
@MikeMiller What would you say is the significance of the following result in Holomorphic Functional Calculus:
If $a \in \mathcal{A}:= \text{Banach Algebra}$ and $f \in \mathcal{Hol}(a)$. Then $$\sigma(f(a)) = f(\sigma(a)).$$

Why me?
In any case, let's start easier. What's the significance of the continuous function calculus for normal operators?

How is the derivative applied in the inverse function Theorem's Left hand side?

My lecture note says $(rs)^{t}/r^{t}=(rs)^{t}(1/r)^{-t}$ for real number $t$. Shouldn't it be $(rs)^{t}(1/r)^{t}$?

3:20 PM
@MikeMiller I discussed operator theory with you some time ago so I assumed you might know. We are being presented with Banach Algebra definitions and theorems and then Holomorphic Functional Calculus and then Continuous Function Calculus...We have not covered Continuous Function Calculus yet.

@ItachíUchiha Eh? Can you elaborate what you mean by that?

OK, that might be easier to understand. But, for instance, the holomorphic functional calculus allows you to say that you can take power series of operators and their spectrum behaves like you would expect it to.
I vaguely remember it being useful when working with resolvents of compact operators.
But I think I told you before I'm hardly expert at this.

If f(x) and g(x) are inverse function, f'(g(x))=1/g'(x)

What does the left hand side means? umm,,Derivative of f(g(x))?

@BalarkaSen

Continuous functional calculus vs. holomorphic one.

It means derivative of f at the point g(x).

3:25 PM
@MikeMiller I just want an idea of the importance. You are referring to result which states: If $f \in \text{Hol}(\sigma(a))$ then $f(z) = \sum c_k(z- \lambda)^k$ and $f(a) = \sum c_k(a- \lambda)^k$.
What do you mean their spectrum behaves like you want it to.?

Eh, I'm just not going to be the right person to answer this.
Maybe @Frank?

I want to ask a question about math PhDs in Europe. Is it obligatory to teach in the PhD life? Or is there any requirement to correct students' answers?
No, I'm now more deviating from analysis.

@MikeMiller Okay no prob :)

I only remember that I took a course in functional analysis the last year and it contained some kind of continuous functional analysis to prove spectral theorem.

3:29 PM
@BalarkaSen Thank you very much. :)

@Alex Try posting on main, then. There are some very good operator theorists who post there.

@MikeMiller Yeah will do.

The main thing I remember the continuous functional calculus being used for is to show that positive operators have a unique square root.
unique positive sqrt at least
@FrankScience Try asking iwriteonbananas, who is in a Masters in Europe and might have some insight.
You're in China, right?

Now I am in France. But I have no idea where to go next.

iwrite hasn't been here for a while.

3:32 PM
I see. Are you finishing a masters?

This year, I hope.

Good luck with that and the application process.

Thanks. It heavily depends on what I can do this year, I mean, the research work.

Yeah, I understand. What are you thinking about?

What do you mean by thinking about?