The h Bar

3:00 PM

@JohnRennie Or maybe you're just not ticklish ._.

@ACuriousMind What does this passage mean? Why would the Chern class of a differentiable complex line bundle not be in $\tilde H^2(X,\Bbb Z)$, I thought the whole point was that they are

paracetamol

@BernardoMeurer Wait, if you never knew him before, how'd you get him to autograph two books for you? O_o

Bernardo Meurer

@paracetamol I do know him, that is why I am bragging

paracetamol

^ Brilliant XD

Bernardo Meurer

I was asking if anyone knew who he was so that I could then brag. If no one knows who he is then the bragging is pointless

paracetamol

3:02 PM

Yeah... I figured that from your last comment ;)

So @JohnR, what's for lunch?

ACuriousMind

@0celouvsky It is not saying that the Chern class is not in $\tilde{H}^2$ - it's saying that there is information lost in the passage from $H^2$ to $\tilde{H}^2$, namely the torsion, and that the bundles are classified by $H^2$, not $\tilde{H}^2$, i.e. two different bundles can have the same class in $\tilde{H}^2$, but not in $H^2$.

John Rennie

@paracetamol Fruit. I'm eating healthily today.

Bernardo Meurer

How is no one freaking out :P

I freaked out

0celouvsky

@ACuriousMind So he's saying that the chern class is too coarse for that classification?

But why does he say that the they are determined by the Chern class in $H^2(X,\Bbb Z)$ when it naturally lives in $\tilde H^2$?

John Rennie

@BernardoMeurer I'm impressed :-)

ACuriousMind

3:05 PM

@0celouvsky Ah...it doesn't live "naturally" in $\tilde{H}^2$, it lives in $H^2$!

Bernardo Meurer

@JohnRennie You're the best John :P

0celouvsky

Proposition 4.3 says that for $E\to X$ a complex line bundle, $c_1(E)\in \tilde H^2$.

Bernardo Meurer

You even support my bragging

Danu

@ACuriousMind What's $\tilde H$ anyways?

0celouvsky

The proof makes sense modulo some cell decomposition that's not something I care to look into

ACuriousMind

3:06 PM

@Danu The image of $H^2(X,\mathbb{Z})\to H^2(X,\mathbb{R})$

Danu

Oh, okay.

Shitty notation :p

John Rennie

@BernardoMeurer The best John? How many Johns do you know? Just the one? :-)

Danu

Looks like Cech cohomology

Bernardo Meurer

@JohnRennie Do only Johns count or do Jons factor in too?

ACuriousMind

@Danu There's a difference between $\tilde{H}$ and $\check{H}$, you need to Cech it out carefully ;)

Danu

3:08 PM

@0celouvsky the claim is pretty simple: $c_1$ lives in $H^2(X;\Bbb Z)$. The map into real cohomology kills torsion, so you lose info. Am I missing something?

0celouvsky

@Danu There's a claim that $c_1$ lives in $\tilde H^2$.

Danu

So what is the definition of $c_1$?

0celouvsky

Chern-Weil theory

ACuriousMind

@0celouvsky This discussion is a bit pointless since it depends on what exactly you defined as $c_1$ - but the passage seems to clearly distinguish between the Chern class in $H^2$ and the real Chern class in $\tilde{H}^2$.

Danu

Right, so that's not the true definition

It's more like the useful definition :P

But Chern classes are supposed to be "normal" (i.e. integral) cohomology classes

0celouvsky

3:10 PM

@ACuriousMind He proves that the Chern class is real!

Danu

What does that mean

ACuriousMind

I don't know what that means

Heh

Danu

Of course you can map that class into the real cohomology

Nothing to prove there

0celouvsky

none of this makes sense

Danu

You have to just carefully read what the proposition says, I'm sure it'll become clear.

John Doe

3:13 PM

@ACuriousMind These phase factors have to be restricted to being functions of either $n$ or $t$ right?

ACuriousMind

What phase factors?

John Doe

@ACuriousMind Earlier I said: When writing the time independent Hamiltonian $$H(t)\psi_n(t) = E_n(t)\psi_n(t).$$ As I understand, we can consider that $\psi_n(t)$ is arbitrary up to a phase factor. That implies that $\psi_n(t)e^{i \phi}$ is considered equivalent to $\psi_n(t)$ in the context of being a stationary state.

Down Christopher

Ok so I am asking you guys if this challenge is clear. I know you guys may not be programmers so that is why I I am asking if the challenge is clear

ACuriousMind

@JohnDoe The phase factors aren't functions of anything. Given a vector $\lvert \psi\rangle$, the vector $\mathrm{e}^{\mathrm{i}\phi}$, for any $\phi\in\mathbb{R}$, represents the same state.

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