The h Bar

12:33 AM

@0celouvsky somewhat

@G.Bergeron It is a peak time of inactivity, clearly ;)

@ACuriousMind I think I figured it out. The problem wants me to integrate over incoming angles as well

It didn't state that

1:48 AM

Hello

can anyone help me with a physics problem?

@BalarkaSen Let $\tilde\mu$ be the multiplication on $\tilde G$, and $\tilde \mu'$ be the multiplication after switching the pair. $\tilde G$ is Abelian iff $\tilde \mu=\tilde \mu'$ as maps. If $\mu$ is the multiplication on $G$, consider the map $f:\tilde G\times \tilde G$ given by $f=\mu(p\times p)$. Now $f_*=\mu_*(p_*\times p_*)$, so $$f_*\pi_1(\tilde G\times \tilde G,(\tilde e,\tilde e))=\mu_*(p_*\pi_1(\tilde G,\tilde e),p_*\pi_1(\tilde G,\tilde e)).$$
It is known that group multiplication on paths agrees with path composition in $\pi_1$ (Munkres page 335), so in fact

@Retweetfire Just ask your question, if anyone wants to answer it, they will

If a bullfrog starts on a horizontal log and leaps with a velocity of 4.40 m/s at an angle of 37.0 degrees to the horizontal, what distance can it cover? I am really having trouble since I cant figure out the initial y.

you don't know the height of the log?

Guys String theory is deep

2:01 AM

yes

It is sooo connected it is rediculous

@Retweetfire are you talking to me or the farm animal?

To you 0ce

what do you mean by yes

do you know it or not

2:03 AM

the farm animal is not offended

I swear this strings business is insane!!!

M5

I dont know it

Thats is why Im stuck

@0celouvsky still studying anal, but been looking at stuff during breaks :D

just make sure you use lots of lube

hehehe

@Retweetfire you're either missing something or just assume it's level with the ground

@ACuriousMind this is probably a really stupid question

remember when we were talking about de Rham with complex coefficients?

2:10 AM

*ridiculous

just corrected a spelling error. Honest mistake!

I think I am going to go dark for a few months and study

@Retweetfire If you don't know it (and there is nothing else in the problem that you haven't told us), just leave the initial height as a variable, I'd say

@0celouvsky yes

@ACuriousMind I'm confused by real/complex dimension. Is $H^k$ with real coefficients really $H^{k/2}$ with complex ones?

Rather, do complex cohomologies only give even degree real ones?

No

And what power exterior bundle of a complex manifold is trivial

Is it the real dimension?

So twice the complex dimension?

You'll need to be a bit more explicit here because right now I'm confused about what you're confused about

I mean, I think I get a general idea of what the problem is, but can you try to formulate it a bit more verbosely?

2:16 AM

@ACuriousMind What the fuck is $H^{2k}(X,\Bbb C)$, where $X$ is a complex manifold

It's...the $2k$-th cohomology with complex coefficients.

what cohomology

and is that 2k real dimensions or 2k complex dimensions

Of the constant $\mathbb{C}$-sheaf

So is it really 4k?

dmckee

@Retweetfire The problem probably presupposes that the frog is jumping from the log to the log and will land at the same height it started from. That's why they told you the log was horizontal.

2:18 AM

@ACuriousMind It should just be de Rham, I don't want to muddle my thoughts with sheaves right now

@0celouvsky It's not "dimensions". The $k$-th cohomology is the $k$-th cohomology of a complex, it does not denote a "dimension"

@ACuriousMind I mean which two things is it a quotient of

explicitly

I know it's not a dimension

@0celouvsky Of closed $2k+1$-forms with complex coefficients modulo exact $2k$-forms with complex coefficients. "$k$-forms" in the ordinary real sense.

So I'm thinking it's the sheaf cohomology of the resolution $$0\to\Bbb C\to \mathcal E_X^\bullet,$$where $\mathcal E_X^\bullet $ makes no reference to holomorphic forms or that $(p,q)$ splitting or whatever. Is that correct?

Yes

2:21 AM

One determines $\mathcal E_X^\bullet$ without any reference to the complex structure of $X$.

@ACuriousMind Ok, so am I right that for a complex manifold of complex dimension $k$, it's the forms of degree $2k+1$ that have to vanish?

If you're doing holomorphy or whatever you're using the complex structure of $X$ and probably Dolbeault cohomology, but (deRham) cohomology with complex coeffficients exists regardless of whether $X$ has a complex structure or not

@0celouvsky yes

Ok, good

2:34 AM

@ACuriousMind So is $H^\bullet (X,\Bbb R)\otimes \Bbb C$ the complex version?

Or is there something more sinister

I think it might have to be a tensor product over the sheaf of complex valued functions, though.

@0celouvsky It works out to that, yes. That's why there's really not difference between (co)homology with real or complex coefficients.

The interesting kind of coefficients other than real are integral or various cyclic $\mathbb{Z}_p$

@ACuriousMind The tensor product is just $\otimes_\Bbb R$?

@0celouvsky Sure, $H^\bullet$ is just a bunch of real vector spaces

@ACuriousMind Can that be seen directly from the definition?

What, exactly?

And which definition? :D

2:40 AM

Can $H^\bullet(X,\Bbb R)\otimes \Bbb C\cong H^\bullet(X,\Bbb C)$ be seen from the sheaf-theoretic definition?

@0celouvsky Not directly, at least not without some algebra: You basically do the abstract version of the universal coefficient theorem and observe that it's just an isomorphism between the two for $\mathbb{C}$.

Nooooo

This is why I like analysis

None of this dumb stuff

@ACuriousMind Why should I care about the universal bundle of a Grassmann space?

@0celouvsky I don't even know what that is.

Which part?

I meant Grassmann manifold