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

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Michael Albanese

9:01 PM

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

homeomorphism type or homotopy type?

Michael Albanese

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

Sorry.

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.

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

Chris's sis the artist

9:03 PM

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

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.

@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 :|

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

Chris's sis the artist

@r9m :-)))))))

Michael Albanese

9:06 PM

Yes, edited.

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

@Semiclassic: Kasman is an old friend/colleague.

Michael Albanese

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

oh? neat

Yup @MichaelA

Chris's sis the artist

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

9:09 PM

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

Goodnight, @MikeM

Morning.

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

Michael Albanese

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

That's good, @MikeM.

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

9:11 PM

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

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.'

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

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

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

9:12 PM

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

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

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?

yeah.

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

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

Michael Albanese

9:14 PM

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

OK, cool.

We were talking about sphere bundles a long time ago.

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

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

quixotic, perhaps

Michael Albanese

Yes, I remember.

9:16 PM

LOL, that makes me happy, @MikeM?

Michael Albanese

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

@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.

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

glares @MikeM

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

9:17 PM

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

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 ...

nods

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

heh, okay

@MichaelA: Getting closer to your orals?

9:19 PM

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.

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

calculus real-analysis sequences-and-series

@r9m @Hippalectryon ^^^

Michael Albanese

Yeah. That's one way of putting it.

Is he your advisor?

9:20 PM

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

@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).

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

@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.)

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

9:23 PM

@MichaelA: I see.

Yes.

Compact, oriented?

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

Chris's sis the artist

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

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

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 $S^4$.

9:27 PM

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

Ok, see ya.

Chris's sis the artist

@Semiclassical No. I miss anything in the picture?

bah

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

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

9:28 PM

any integer you want, @Semiclassic

like $-3$? :P

Same group as for $n=+3$.

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

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

eh, fair enough

@Chris'ssistheartist mea culpa, i misread it

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

Chris's sis the artist

9:30 PM

@Semiclassical hehe, OK :-)

@Chris'ssistheartist@TedShifrin Hi

Chris's sis the artist

@Gato Hi

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.

hi @Gato

Ok, really see ya.

9:33 PM

@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. :-)

@Chris'ssistheartist No

Chris's sis the artist

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

@hippa are you on irc

He's definitely not on here, @Ramanewb

9:39 PM

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

You still in Ireland?

@ted yep

cool

@TedShifrin

Yes?

9:42 PM

can you spare 5 minutes

for what?

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"

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 $\sum x_i = k$.

hold on i need to activate latex

nothing is working today :(

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.

2 hours later…

11:59 PM

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

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