Mathematics - 2015-07-06 (page 4 of 4)

Michael Albanese

21:01

Thanks. Basically, you fix $0 \leq i_1< i_2 < \dots < i_m \leq n$ and want to determine the homeomorphism type of $\{[x_0, \dots, x_n] \in \mathbb{CP}^n \mid x_{i_1}, \dots, x_{i_m} \neq 0\}$. Here is the question: math.stackexchange.com/q/1351073/39599

Ted Shifrin

homeomorphism type or homotopy type?

Michael Albanese

The OP wanted both and I answered both (or at least attempted to).

Sorry.

Ted Shifrin

So the complement of a finite number of hyperplanes. That's an old classic question. Not that I remember how to do it.

Michael Albanese

Wanted homotopy type and whether it was diffeomorphic to something more concrete.

Semiclassical

@ted on the off-chance you're curious, this is the paper: Tau-functions, Grassmannians and rank one conditions.

Chris's sis the artist

21:03

@r9m Well, yeah, some things might be hard to accept like the whole elementary evaluation of series in Flajolet and Salvy. :-)

Ted Shifrin

Ah, intersection is easier than union, isn't it, @MichaelA?

Yes, at first glance, your answer seems correct. I was thinking union originally.

Of course it's independent of the particular multiindex.

Michael Albanese

Ah yes, union seems much harder.

r9m

@Chris'ssistheartist that's not what I made the face for ... it sounds to me like 'hey! I met Florence yesterday' when I have no idea who Florence is :|

Ted Shifrin

@MichaelAlbanese No, you mean union is harder :P It's a classic question, too.

Chris's sis the artist

@r9m :-)))))))

Michael Albanese

21:06

Yes, edited.

Ted Shifrin

LOL. But, yeah, I think your answer is right.

@Semiclassic: Kasman is an old friend/colleague.

Michael Albanese

The union would be $\{[x_0, \dots, x_n] \in \mathbb{CP}^n \mid x_i \neq 0\ \text{for some}\ 0 \leq i \leq m - 1\}$.

Semiclassical

oh? neat

Ted Shifrin

Yup @MichaelA

Chris's sis the artist

Did I show this one?$$\int_0^{\pi/2} \sin (x) \text{Li}_2(\csc (x)) \, dx$$

Mike Miller

21:09

@MichaelA: Did you see the message I sent you about $\text{Homeo}(S^n)$?

Ted Shifrin

Goodnight, @MikeM

Mike Miller

Morning.

Don't worry, I'm not already drinking.

Michael Albanese

Hi @MikeMiller. I did but I forgot to reply.

Ted Shifrin

That's good, @MikeM.

Mike Miller

Two more talks. One of them is Ciprian about his big theorem.

Ted Shifrin

21:11

Even though you've heard the talk dozens of times, you'll be a dutiful student and go and applaud :P

Mike Miller

OK, @MichaelA. Just checking. I hope I'm not the only one excited by these things.

Michael Albanese

I was a bit confused by 'I've finally got a result I'm satisfied with.'

Ted Shifrin

Get used to it, @MichaelA. Mike gets confusing a lot of the time. :)

Mike Miller

I haven't heard it once... I haven't even read the paper yet.

Ted Shifrin

Oh, in that case, you really better pay detention, @MikeM.

Semiclassical

21:12

@ted i'm hoping to adapt the main result there (matrices satisfying a rank $r=1$ condition) $\to$ KP tau-functions) to matrices with $r\gtrsim 1$

Mike Miller

@MichaelA: As in, I've found a result that I'm happy with. It gives me information in a form I like.

Ted Shifrin

Generally, @Semiclassic, in linear algebra and functional analysis, rank $>1$ is much harder to get a hold of than rank $1$.

Michael Albanese

Ah. This was just in relation to a general discussion we previously had or a particular problem?

Semiclassical

yeah.

my main examples are with $r=2,3$ but with arbitrarily large matrix size

Mike Miller

I just wanted to know what Homeo(S^n) looked like. It turned out to be delightfully complicated.

Michael Albanese

21:14

Don't get me wrong, I enjoy hearing about such things, I was just unsure of the context.

OK, cool.

Mike Miller

We were talking about sphere bundles a long time ago.

Semiclassical

one hope is that i can relate my matrix--which, as far as i know, isn't almost-intertwined with any operators--to one which does

Mike Miller

@TedS: You'll be happy to know he proves it with spectra.

Semiclassical

quixotic, perhaps

Michael Albanese

Yes, I remember.

Ted Shifrin

21:16

LOL, that makes me happy, @MikeM?

Michael Albanese

By the way Ted, thanks for checking out my answer.

Ted Shifrin

@Semiclassic: I mentioned to Malcolm Adams that you were interested in KP stuff. He said you were welcome to get in touch if you wanted.

Mike Miller

Or not. Seeing you at either end of the spectrum brings me glee, you see.

Ted Shifrin

glares @MikeM

Mike Miller

Vancouver is not so nice after a forest fire. Air is unpleasant.

Semiclassical

21:17

indeed? then i'll try to think of something useful

Ted Shifrin

they had a forest fire? CA just had a bad one near Vacaville.

@Semiclassic: He also said he might not be much help. But nevertheless ...

Semiclassical

nods

Ted Shifrin

He is traveling a bunch this month, so don't expect anything fast :P

Semiclassical

heh, okay

Ted Shifrin

@MichaelA: Getting closer to your orals?

Mike Miller

21:19

Yes. The roof of the world cup stadium was closed because of it.

Michael Albanese

Yeah. Still a while to go yet (about two months). Met with LeBrun this morning.

Ted Shifrin

Exciting, @MichaelA.

Chris's sis the artist

0

Q: Computing $\sum _{k=1}^{\infty } \frac{\Gamma \left(\frac{k}{2}+1\right)}{k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$

What tools other than beta function you might like to use here? $$\sum _{k=1}^{\infty } \frac{\Gamma \left(\frac{k}{2}+1\right)}{k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$$

@r9m @Hippalectryon ^^^

Michael Albanese

Yeah. That's one way of putting it.

Mike Miller

Is he your advisor?

Ted Shifrin

21:20

I'd enjoy being a fly on the wall for it. I miss this stuff, @MichaelA.

Semiclassical

@Chris'ssistheartist isn't that just a fancy way of saying $\frac{2}{k^2(k+3)}$?

Michael Albanese

The way it works here is that you formally get an advisor after you pass orals, but I am doing my major topic with him because I want to work with him (and he said he would take me on).

Ted Shifrin

Yeah, it doesn't really make sense to do it differently if the orals involve advanced topics leading to research.

Mike Miller

@Ted: Here's a problem you might hate. Can you write down a $(4k+3)$-manifold with $H_i = 0$, $i \leq 2k$, and $H_{2k+1}= \mathbb Z/n$? You can do this for $k = 0, 1$. You can't do the analogous statement for $4k+1$-folds: the torsion linking form provides an obstruction.

(I don't dare suggest you might like a problem.)

Ted Shifrin

The $0$ is meant to be an $=$?

Mike Miller

21:23

@MichaelA: I see.

Yes.

Ted Shifrin

Compact, oriented?

Semiclassical

$\mathbb{Z}/n=\mathbb{Z}/n\mathbb{Z}$?

Chris's sis the artist

@Semiclassical Is it? Let me see, I'm pretty tired now and I might do some mistakes.

Semiclassical

it's a ratio of gamma functions, with their usual functional equation $\Gamma(z+1)=z\Gamma(z)$

Mike Miller

I want both closed and oriented. The given condition automatically implies oriented for $k>0$ ($H_1 = 0 \implies$ oriented). For $k=0$ there aren't even any closed non-orientable manifolds with finite fundamental group.

So it's automatically oriented in every dimension we're talking about.

In dimension 3 you can do this realization with lens spaces. In dimension 7 you can do the realization with $S^3$ bundles over $S^4$.

Semiclassical

21:27

@MikeMiller what's $\mathbb{Z}/n$ supposed to be?

Mike Miller

Ok, see ya.

Chris's sis the artist

@Semiclassical No. I miss anything in the picture?

Semiclassical

bah

Ted Shifrin

What you said, @Semiclassic. It's a standard abbreviation.

Semiclassical

kk. i'm also not sure what $n$ was supposed to be

Ted Shifrin

21:28

any integer you want, @Semiclassic

Semiclassical

like $-3$? :P

Ted Shifrin

Same group as for $n=+3$.

Semiclassical

hmph, yes. (though 0 and 1 really don't make sense)

Ted Shifrin

They make sense. Just a silly way to write $\Bbb Z$ and $0$, respectively.

Semiclassical

eh, fair enough

@Chris'ssistheartist mea culpa, i misread it

thought the first one was $k/2+1/2$

Chris's sis the artist

21:30

@Semiclassical hehe, OK :-)

Gato

@Chris'ssistheartist@TedShifrin Hi

Chris's sis the artist

@Gato Hi

Mike Miller

Side note: don't try to do this with $S^{2k+1}$-bundles over $S^{2k+2}$. $\pi_{2k+1} SO(2k+2)$ is usually not rich enough for this to work. That it does for those small $k$ is a fluke.

Ted Shifrin

hi @Gato

Mike Miller

Ok, really see ya.

Gato

21:33

@Chris'ssistheartist Do you have a series that can be computed with complex analysis ?

Chris's sis the artist

@Gato Is it a trap question? Even the hardest series by Ramanujan are done with complex analysis. :-)

Gato

@Chris'ssistheartist No

Chris's sis the artist

@Gato I think I can handle with all my stuff using real analysis.

Ramanewbie

@hippa are you on irc

Ted Shifrin

He's definitely not on here, @Ramanewb

Ramanewbie

21:39

@ted Indeed I don' t see him on the list of the connected people

Ted Shifrin

You still in Ireland?

Ramanewbie

@ted yep

Ted Shifrin

cool

Danny

@TedShifrin

Ted Shifrin

Yes?

Danny

21:42

can you spare 5 minutes

Ted Shifrin

for what?

Danny

0

Q: Some notation is confusing me (sufficiency principle)

This example(from my book) illustrates the sufficiency principle and I will write it down so that you get some context : Let $\boldsymbol{X} = (X_1,X_2..,X_n)$ be a sample of independent bernoulli variables then the probability function is: $p(\boldsymbol{x};\theta) = \prod\limits_{i=1}^{i=n}\t...

self-study inference sufficient-statistics

i don't know if that is your "area"

but you know everything so maybe you could "say something"

Ted Shifrin

No, it's not my area. But I think it's just shorthand for $\mathbf X = \mathbf x$, i.e., $X_i = x_i$ ... all with $\sum x_i = k$.

Danny

hold on i need to activate latex

nothing is working today :(

Lucas

Can somebody remind me what a transformation of the form $f(x, y) = (x, y + h(x))$ is called?

Nevermind, I was looking for the term "shear mapping", which refers to a linear version of this.

leo

23:59

@Chris'ssistheartist I want a copy :-). Ping me when it's out so I can order one

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