Cohomology and fermions in supersymmetry

In summary, the conversation discusses the correspondence between the Hibert space of the symmetric Hamiltonian and the cohomology of a manifold in differential geometry. The connection is explained through the use of Hodge theorem and the Hamiltonian being equivalent to the Laplacian. The conversation also mentions the need for further clarification and resources on the topic.
  • #1
kakaho345
5
0
TL;DR Summary
https://www.youtube.com/watch?v=bdASx74y7oI&list=PL7aXC0jU4Qk7K778c5nmgQImd6VKKFMYu&index=9&ab_channel=KRaviteja
1:03:00
Hirosi claims that the hamiltonian hibert space corresponds to the cohomology on the manifold. I don't understand why
Hello,
I have been looking at some differential geometry and watching Hirosi's video lecture online:

At 1:03:00, I found that they claimed that there is a correspondence between the Hibert space of the symmetric Hamiltonian and the cohomology of the manifold.
I am super new to the subject and this is the best I can describe the problem. Would anyone explain to me why that correspondence is true?

If possible, can anyone point me to some lecture videos that explain in more details and clearer? I feel like Hirosi is teaching too fast for me.
(I know Nakahara is an excellent reference, but I am still finding for more resources.)
 
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  • #2
I probably misunderstood your question because I haven't watched the first one hour of the lecture and the previous seven lectures, so ignore my answer if it is off the mark. What he says is that by Hodge theorem (he doesn't say the theorem, but he probably covered it somewhere in the first lectures) a cohomology class is represented by a harmonic form i.e. a zero of the Laplacian. On the other hand the ground states are the zero energy states, which I guess amounts to the same thing.
 
  • #3
martinbn said:
I probably misunderstood your question because I haven't watched the first one hour of the lecture and the previous seven lectures, so ignore my answer if it is off the mark. What he says is that by Hodge theorem (he doesn't say the theorem, but he probably covered it somewhere in the first lectures) a cohomology class is represented by a harmonic form i.e. a zero of the Laplacian. On the other hand the ground states are the zero energy states, which I guess amounts to the same thing.
I see, thanks for replying. Your answer is super relevant and helpful. He didn't explicitly stated Hodge theorem, but he did talked about the Harmonic forms previously.

For future people interested, it is in his lecture 3 when he talks about representatives of cohomologies.
He mentioned the codifferential ##\delta## and the Laplacian ##\Delta = \delta d + d \delta## and that we can choose a solution requiring ##d\omega = 0## and it implies ##\Delta \omega = 0##, hence ##d\omega = 0## implying an element in the cohomology implies also a member of the harmonic form ##\Delta \omega = 0##. And the Hamiltonian is exactly the Laplacian. Hence proven the claim.

Then of course, I need to review why the Hamiltonian is the Laplacian.
The lecture did not show in details, but the general idea is that one can identify some supercharge ##Q# and ##Q^\bar## from the Lagrangian, and those supercharge are identified with ##\delta## and ##d##. And the anticommutator between the differential gets you the Hamiltonian.

P.S. anyone know why my latex code does not work properly?
 
Last edited:
  • #4
kakaho345 said:
I see, thanks for replying. Your answer is super relevant and helpful. He didn't explicitly stated Hodge theorem, but he did talked about the Harmonic forms previously.

For future people interested, it is in his lecture 3 when he talks about representatives of cohomologies.
He mentioned the codifferential ##\delta## and the Laplacian ##\Delta = \delta d + d \delta## and that we can choose a solution requiring ##d\omega = 0## and it implies ##\Delta \omega = 0##, hence ##d\omega = 0## implying an element in the cohomology implies also a member of the harmonic form ##\Delta \omega = 0##. And the Hamiltonian is exactly the Laplacian. Hence proven the claim.

Then of course, I need to review why the Hamiltonian is the Laplacian.
The lecture did not show in details, but the general idea is that one can identify some supercharge ##Q## and ##\bar{Q}## from the Lagrangian, and those supercharge are identified with ##\delta## and ##d##. And the anticommutator between the differential gets you the Hamiltonian.
 

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