Hi Prof. Borcherds, I don't know if you will ever see this, but thanks for all the videos. It's really important for those who are learning on their own :) I truly believe this is somewhat the future of teaching!
Maybe one thing to add in conclusion would be the general formula for the dimension of the space of weight k modular forms, with the floor of k/12 and the distinction for k=2 mod 12 (I agree this is trivial from your last induction step, but I remember being quite impressed by this formula the first time I saw it!)
Can someone explain why the integral over the horizontal line @ 5:30 corresponds to an integral over a circle? If I plug in the values \tau=1/2 \pm i t I get -\exp{\mp \pi t}
@@itaypikaz8071 Why do you have \tau in the exponent? You should have t > 1 after plugging in. I think I solved it, just write \tau = - 1/2 + i t + \alpha, with \alpha in [0,1] parametrizing the horizontal line, plug it in and you get q=- e^{-2\pi t} e^{2\pi i \alpha}, which gives the correct circle (after correcting for the extra minus sign)
Hi Prof. Borcherds, I don't know if you will ever see this, but thanks for all the videos. It's really important for those who are learning on their own :)
I truly believe this is somewhat the future of teaching!
3:54 A note on why f(tau) does not have zero for Im(tau) >> 0:
View f(q) as a function of q. Then f does not have zero for 0
Very interesting topic and nice lecture
This is mind-blowing. It's like a rabbit being pulled from a hat.
Great lecture as always!
Maybe one thing to add in conclusion would be the general formula for the dimension of the space of weight k modular forms, with the floor of k/12 and the distinction for k=2 mod 12 (I agree this is trivial from your last induction step, but I remember being quite impressed by this formula the first time I saw it!)
Can someone explain why the integral over the horizontal line @ 5:30 corresponds to an integral over a circle? If I plug in the values \tau=1/2 \pm i t I get -\exp{\mp \pi t}
Plugging in both values gives -\exp{-\pi \tau}, because the imaginary part is constant, and the difference of the real parts is an integer.
@@itaypikaz8071 Why do you have \tau in the exponent? You should have t > 1 after plugging in. I think I solved it, just write \tau = - 1/2 + i t + \alpha, with \alpha in [0,1] parametrizing the horizontal line, plug it in and you get q=- e^{-2\pi t} e^{2\pi i \alpha}, which gives the correct circle (after correcting for the extra minus sign)
It should be \tau= \pm 1/2 + i t , there is a mistake in the video
yeeeeeeeeeeeeeee
Yeeeeeeeeeeeeee
The comment section should be more active.
Yee