Mathematics - 2016-08-16 (page 2 of 2)

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Semiclassical

20:03

@BalarkaSen gotcha, that's all pretty straightforward

Balarka Sen

@Semiclassical I also learnt friction and elasticity, by the way.

Semiclassical

ah, yes.

friction's pretty easy, though making sure one is using it correctly in problems is where the rubber meets the road

Balarka Sen

Heh, yes. It's pretty fun though.

Semiclassical

bonus question: is that idiom an example of static or kinetic friction? :>

Balarka Sen

Semiclassical

20:06

elasticity is pretty simple as well, though it depends on the student of course

my pet peeve is when people write Hooke's law as $F=-k x$

I'd much prefer $\Delta F = - k \Delta x$.

Balarka Sen

It is indeed, but our textbook has a couple of weird problems where one gives extra information to confuse, and a couple where the problem has less information than required.

Apparently they confused themselves.

Semiclassical

yeah.

my main reason for insisting on the Delta form of that is that 1) it's closer to $k=-\frac{dF}{dx}$ which is probably the best version of Hooke's law, 2) it doesn't implicitly suggest that $F(0)=0$.

Balarka Sen

I haven't started on much of rotational motion yet (not rotation of rigid bodies: just the rotational kinematics). I should start doing that soon

Semiclassical

right.

Balarka Sen

@Semiclassical Right.

0celo7

20:10

@Semiclassical What's the trace of a matrix in bra-ket notation

taking QM in two days

Semiclassical

depends on how you're writing your basis

0celo7

$|i\rangle$ for simplicity.

Semiclassical

mmkay.

then it'd just be stuff like $\text{tr}\hat{M}=\sum_n \langle n | \hat{M} |n \rangle$

0celo7

right

Semiclassical

though you don't see trace a ton in intro QM, if memory serves

the main places I remember it are 1) in the context of density matrices

0celo7

20:13

it's a 2 semester graduate course

Semiclassical

and related to that, 2) in the context of defining the QM partition function as $Z=\text{tr }{e^{-\beta H}}$.

0celo7

Yes, I know both of those.

Semiclassical

mmkay.

the usual workhorse for quantum is the representations of the identity, i.e. $\mathbf{1}=\sum_n | n \rangle \langle n |$

which is just familiar linear algebra: the sum of all the projection matrices is the identity matrix

0celo7

@Semiclassical I actually needed that result for cohomology, not QM :P

Semiclassical

lol

0celo7

20:15

I do linear algebra in bra-ket notation

Balarka Sen

...

0celo7

so much better

Semiclassical

I tend to do it either in index notation or coordinate-free

I guess it's a habit thing, that I tend to only use bra-ket for when I'm doing quantum

0celo7

once you have duals, bra-ket is amazing

Semiclassical

right.

0celo7

20:16

I'm trying to muddle through some Poincare duality stuff

Semiclassical

If I can get away with thinking of duals as row vectors, then the usual matrix notation is fine. but that's not a great idea in quantum

0celo7

I found a neat integral representation for $\mathrm{tr}\,H^q(M)$, @BalarkaSen

I'm trying to exploit it

Balarka Sen

No time now, I gotta go

0celo7

cheerio

Alessandro

good evening

0celo7

20:19

AHA

I have computed the Poincare dual of a graph (de Rham theory)

Semiclassical

Poincare duality is something that actually has some fairly concrete applications in condensed matter

Namely, if two graphs are Poincare duals, then the partition functions defined on them will also be dual. (or something like that, it's been a while since I read this stuff)

which is especially neat when a graph happens to be self-dual, since in that case the two partition functions are usually of the same form but at different temperatures.

0celo7

$$\eta_\Gamma=\sum_{ij}\left((-1)^{(\mathrm{deg}\tau_i+\mathrm{deg}\omega_j) \mathrm{deg} \omega_i }\int_M\tau_i\wedge f^*\omega_j\right)\pi^*\omega_i\wedge\rho^*\tau_j$$

This might be useful.

Semiclassical

right, Kramers-Wannier duality

0celo7

Now let's integrate that formula.

oh sweek jesus

you get a Dirac delta

AND A TRACE

Semiclassical

lol

0celo7

20:26

@Semiclassical I'm trying to prove that the homological Euler characteristic and the intersection-theoretic one are the same.

Semiclassical

mmkay

0celo7

I'm also failing at finding graphene

But our postdoc basically solved my oxide issue

0celo7

20:45

I proved the Lefschetz fixed point theorem oh my god

0celo7

20:57

@BalarkaSen You're gonna be impressed when I'm done

Balarka Sen

I'll be asleep when you're done.

0celo7

you will dream about me

Huy

you should already be asleep

0celo7

and be impressed in the dream

Semiclassical

that's juuuust a bit creepy

Balarka Sen

20:59

Do not identify me with one of your imaginary girlfriends.

I am sure even the real one does not do that.

0celo7

She dreams of me sometimes

But it's always weird, not sure what to make of that.

Huy

it's cuz ur weird

0celo7

oriented intersection theory is the worst.

Balarka Sen

@Huy In my defense I was studying chemistry.

0celo7

@BalarkaSen What's the Poincare dual of a disjoint union? Is it just the sum of Poincare duals of the components?

Balarka Sen

21:02

Poincare dual of a disjoint union of two submanifolds, you mean? Yes.

0celo7

Yes.

Good.

Balarka Sen

Disjoint union of submanifolds represent sum of homology classes each of the components they represent.

And dual of sum is sum of dual because Poincare duality morphism is a homomorphism.

0celo7

Hmm, no, not good. I got that the Lefschetz number is the sum of fixed points.

Balarka Sen

Which is false, boo.

0celo7

Yeah.

Balarka Sen

21:04

You need to take care of multiplicity around each fixed point.

But you seem close.

0celo7

@BalarkaSen I was able to prove that the integral $\int_\Delta\eta_\Gamma$ is indeed the Lefschetz number. Using this I was able to prove the Lefschetz fixed point theorem.

Where $\eta_\Gamma$ is the Poincare dual of the graph of $f$ in $M\times M$.

and $\Delta$ is the diagonal

Balarka Sen

I see: good to hear.

0celo7

(co)homological lefschetz number

now I want to connect that integral to GP's Lefschetz number

after that the Euler characteristic is easy to get

@BalarkaSen Does orientation change the sign of the Poincare dual at all?

Balarka Sen

It does.

0celo7

Aha, so the Poincare dual of the points has to take into account the signs

Balarka Sen

21:13

Yes.

0celo7

Yo I'm pretty sure this gives the intersection number directly.

I'll have to think on it.

@BalarkaSen Oh.

I just realized there's a way easier proof of this :P

It uses Morse theory, but it's easier. You do Milnor's proof of Poincare-Hopf and compare with GP's proof. Milnor's method computes the homological one and GP's computes the intersection one.

And they both equal the index so they must be equal.

One line proof ftw.

2 hours later…

The Quantum Physicist

23:10

Hey, guys, I have a quick question about matrices exponentials and eigenvalues. Why is it valid to do $\exp(U (-i\Lambda) U^H)=U \exp(-i\Lambda) U^H$ in the following answer? Is this correct for any matrix $A$ or just for hermitian and anti-hermitian $A$?

http://scicomp.stackexchange.com/questions/1234/matrix-exponential-of-a-skew-hermitian-matrix-with-fortran-95-and-lapack

Semiclassical

I'm about to go so I can't look at that question, but I imagine the key point is that if $U$ is unitary (i.e. $U U^H=I$) then $$[U(-i \Lambda)U^H]^2=U(-i \Lambda)U^H U(-i \Lambda)U^H=U(-i \lambda)^2 U^H$$

and more generally $[U(-i\Lambda)U^H]^n = U(-i\Lambda)^n U^H$. That's enough to imply that result for the matrix exponential since that's an infinite sum of such terms

The Quantum Physicist

@Semiclassical hmmmmmmm

Semiclassical

okay, gotta go

The Quantum Physicist

@Semiclassical Thanks

That helps!