Modular forms: Theta functions

[@888Xenon]

"If the functions we're going to use didn't count as 'nice', there'd be something wrong with your definition of 'nice'" Brilliant lines in every video, cheers again!

[@MichaelFJ1969]

I agree. I crack up laughing in all his videos. I really like the humor.

[@ApePostle]

Prof borcherds, you have gone above and beyond with these lectures. Just want to say I am feeling extreme gratitude to have the opportunity to listen to you lecture

[@3.saar.a]

I remember seeing somewhere where the author put the period factor in the definition of inner products Namely on C_period(L) define := 1/L Integrate over a period f g* dx, and everything looks nice.

[@annaclarafenyo8185]

The reason for the physicist conventions to avoid 2\pi in the exponent is because Dirac explained the right place is in the measure dk. The volume of k-space is what is 2\pi sized. If you put the 2\pi in the exponent, you get 2\pi scale factor each time you differentiate a wavefunction with respect to x, which clearly will not do, it is a bad convention. When the 2\pi is in the k measure, all the formulas are as simple as possible, even when you do complicated things. Mathematicians don't use this convention because mathematicians are allergic to complicated things.

[@philipschloesser]

That's why mathematicians usually don't use k for the parameter you put into the Fourier transform, but \xi. But sure, by just using k = 2\pi \xi you get your convention, which I also quite like.

[@annaclarafenyo8185]

@@philipschloesser The issue I have with the claim that "That's just ... like... your opinion .... man!" with regard to 2\pi Fourier factors is that they appear in physical contexts. A one-loop graph in 4 dimensions goes like g^2/4(2\pi)^2, and those pi factors are actual honest to goodness pi's. The correction to the electron magnetic moment has a physical pi, it's alpha/2\pi, and alpha is defined without a reference to a circle. This means that the phase-space integrals over momentum actually do give honest 2\pi factors. The physicist convention for counting pi's is to stuff them in the measure, and in the delta-functions (anti-measures), so that the number of physical 2\pi's becomes obvious, because pi is really big, way too big to ignore. Two factors of pi is one order of magnitude.

[@varunachar87]

@@annaclarafenyo8185 you make a compelling argument, but I'm inclined to believe the factors of π don't really have to be considered part of "physical reality", but are instead really just artefacts of convention. Could it be, for example, that if the unit of action were consistently h bar instead of h, or if energies were measured in terms of angular frequencies instead of cycle frequencies, these π factors would all just get absorbed?

[@drakezhard]

What you have to replace Lebesgue measure dx on R^k is (2 π)^(-k/2) dx. Then you can drop the factor of 2pi from the exponent. But, this is unnatural as the volume of a sphere of radius 1 in R^k is π^{k/2}/Gamma(n/2 +1). Indeed the correct way to think about Lebesgue measure is in terms of dyadic cubes. So, I think the solution of putting the 2pi in the exponent is quite elegant because it does it in terms of the measure of the unit cube rather some fudge factor depending on something not quite equal to the volume of a sphere in R^k.

[@antoinebrgt]

Great lecture as always! For some more confusion, it should be noted that in the context of string theory the theta function described here is denoted theta_3 :D

[@rosieshen8431]

5:55 should be i/tau under the square root

[@drakezhard]

The Fourier transform F(f)(y) = ∫ e^{-2 π i yx} dx, it's missing a sign. That actually is the inverse Fourier transform when we have the convention of putting the 2pi inside the exponential.

[@abhijitkashyap3338]

Sir what type of technique do you use to show you face and at the same time what you are writting

[@ananyakaushik6631]

"For anyone who wants to copy the format of the lectures, I'm using a program called OBS studio with a webcam and document camera and lapel mic. I use sketchpad for the thumbnails. " - found this in the about section of his profile.

[@abhijitkashyap3338]

@@ananyakaushik6631 thank you.

[@Jaylooker]

They remind me of Gaussian integrals

[@rosieshen8431]

Okay, I agree that the integral involving theta does not converge for Re(s) large because of the n=0 term in theta; but in the equation 11:45, aren't you only summing over terms with n eq 0? In other words, in the formula 11:18, theta(ix)/2 should be (theta(ix)-1)/2 instead. So while the integral as written does not converge, the correct expression for zeta^*(x) should converge at least for Re(s) large.

[@migarsormrapophis2755]

ye

[@modern_genghis_khan0393]

π/2=log(i)/i , where i=√-1 prove this .This is my challenge 🤪

[@jacobschmid2898]

red meat makes my plegm develop when i dont eat red meat for a long time. it is important for the immune system.

[@jacobschmid2898]

liverwurst