A Short Course on Modular Forms by Prof. M. Ram Murty, Lecture 1: q-Series

This is the first lecture in a series of lectures given by Prof. M. Ram Murty (Queen's University) at IISER Bhopal as part of the GIAN program titled "A short course on modular forms" held in 2016.

[@ethanjensen7967]

2 years ago I started reading your book on modular forms, and now I am a graduate student studying modular forms!! It has changed my life

[@SeniorMarsTries]

hey, someone posted this screenshot on discord and i thought it was so incredible that you posted your update on your life with math. it was really cool! good luck in grad school!

[@ethanjensen7967]

@@SeniorMarsTries Thank you! I wish the best for you too!!

[@NavdeepVarshney-ep4ck]

Could you express the beauty of subject or how it intitutively suggest more things

[@synthmath]

A beautiful introduction to a fascinating topic. In case anybody has was walked in the same shoes, I leave here some comments on possible typos and stuff unsaid: - 26:46 in the series the exponent is q^(n(n+1)/2), but he writes q^binomial(n, 2) justifying it's a matter of relabeling indices. However, as one can see from the book (Exercise 1.1.3) the factor q^binomial(n, 2) occurs when we are considering the infinite product for n = 0, 1, 2, ... of (1 + xq^n) (starting from 0). But in blackboard we are considering E_{1/q}(x), which is the product for n = 1, 2, ... of (1 + xq^n) (starting from 1). I think the exponent should really be q^binomial(n+1, 2). - 27:40 The theorem is correct, since the product starts from 0, and with a few computations this leads to the factor q^binomial(n, 2). - 44:20 The last identity before concluding Jacobi Triple Product formula is not exactly the Theorem with "star" label, nor the identity for E_q(x), because in both cases the factors 1 + xq^n (resp. 1 + x/q^n) have exponent 1, while now we need exponent -1. The result we need is Exercise 1.1.4 from the book, stating that for |q| < 1 the product for n = 0, 1, 2, ... of (1 + xq^n)^(-1) is equal to the sum for n = 0, 1, 2, ... of x^n / (q^n - 1)···(q - 1). - 54:12 In Corollary 2, the q-power in the series general term should be q^(k(k+1)/2), as also stated in Theorem 1.2.3 from the book. - 1:02:50 In the Jacobi Triple Product formula with x → a^4 q and q → q² the middle factor on the right hand side should be 1 + a^(-4) q^(4n-3) instead of 1 - a^(-4) q^(4n-3). [Corrected in 1:08:20] - 1:09:58 As far as I can tell, there's no typo in the book, the signs on the denominators of the series should be +, and the sign right before 4 should also be +. - 1:12:35 The product for n = 1, 2, 3,... of (1 - q^(2n-1))³(1 - q^(2n))³ cannot be rewritten with factors (1 - q^(2n-1))^4 (1 - q^(2n))^3. What's going on is to keep it with powers=3, then divide both LHS and RHS by the product he denoted with "product (...)(...)(...)". In the book (p. 9, proof of Theorem 1.3.1) it's stated that the latter can be rewritten in a way [1] such that the quotient on the LHS becomes the desired product for n = 1, 2, 3,... of (1 - q^(2n-1))^4 (1 - q^(2n))^2, and the RHS is now only 1 + the series. [1] I think it should be something like the following (here n = 1, 2, ...): since 1+q^(4n-1) and 1+q^(4n-3) include all the odd powers, the product could be replace the general term by (1 - q^(4n))(1 + q^(2n-1)), and the second factor is the quotient (1 - (q^(2n-1))²)/(1 - q^(2n-1)), so the general term is now (1 - q^(4n))(1 - q^(4n-2))/(1 - q^(2n-1)). Since the ones in the numerator include all the even powers, we may replace the general term by (1 - q^(2n))/(1 - q^(2n-1)).

[@ethanjensen7967]

This is awesome!!! I'm going through your analytic number theory book right now, and it's great!

[@Greg-xs5py]

Glad at 36 minutes the cameraman woke up…. and then zoomed right back out

[@xxxscodyxxx]

27:35 Why did the product index start from 0? Before this equation he found 1/q analogue of e(x) first as an infinite product and then as an infinite sum. When equating the infinite product to infinite sum he changed the starting point from 1 to 0 in the product index but did nothing about the summation.

[@kaidenpink1771]

45:40 There's a pretty big mistake here in the professors algebra. No disrespect just with something as algebra dense as this mistakes are bound to happen. The "star" identity he is referring to is not the sum that appears in his algebra as there is q^n in the sum not a q^n(n+1)/2, which is what appears in the identity he started the lecture with. Also if it did work like that, you would need the reciprocal of the product so that you can multiply the product to the other side to get the jacobi triple product. The star identity would not produce a reciprocal product. I have found a way to fix this, although as far as I can tell it can't be done with just the original identity. You need another similar identity that can be proven using very similar logic to what he did at the start. I would write down the proof and identity I found if youtube comments were nicer to math notation but I'm sure most people watching this lecture could fix his mistake on their own once made aware.

[@NilodeRoock]

16:16 Series starts at n=0 instead of 1.

[@xxxscodyxxx]

I'm trying to solve the excercises in his book and i'm very confused about the starting index value of the summation representation of q analogue of e(x). So i'm here to watch the lecture and happy to see your comment. You're right. In his book he writes eq(x)=1+[(n=1 to inf.)Sigma x^n/(q^n-1)...] and then he defines the empty product (n=0 case) as 1 and he rewrites the summation as eq(x)=(n=0 to inf.)Sigma x^n/[(q^n-1)...(q-1)].

[@diktakt1187]

07 58 Zéta

[@borisbryant.007]

is this intended to be followed or what ??? why does the camera man not zoom in on the text???

[@user-zr7gx4xb6n]

Wat

[@wdobni]

that was ghastly.....i don't attribute any malice to the lecturer but he rambled hither and yon and scribbled symbols and really made nothing clear and offered no explanations .... dry boring rambling disconnected

[@iconjack]

go away