Quantcast
Channel: math.NT – Annoying Precision
Viewing all articles
Browse latest Browse all 10

The man who knew elliptic integrals, prime number theorems, and black holes

$
0
0

I went to see The Man Who Knew Infinity yesterday. I have nothing much to say about the movie as a movie that wasn’t already said in Scott Aaronson‘s review, except that I learned a few fun facts during the Q&A session with writer/director Matthew Brown afterwards. Namely, it’s a little surprising the movie was able to get high-profile stars like Dev Patel and Jeremy Irons on board given that it was made on a relatively low budget. Apparently, Dev Patel signed on because he really wanted to popularize the story of Ramanujan, and Jeremy Irons signed on because he was hooked after being given a copy of Hardy’s A Mathematician’s Apology.

(Disclaimer: this blog does not endorse any of the opinions Hardy expresses in the Apology, e.g. the one about mathematics being a young man’s game, the one about pure math being better than applied math, or the one about exposition being an unfit activity for a real mathematician. The opinion of this blog is that the Apology should be read mostly for insight into Hardy’s psychology rather than for guidance about how to do mathematics.)

Anyway, since this is a movie about Ramanujan, let’s talk about some of the math that appears in the movie. It’s what he would have wanted, probably.

Elliptic integrals

There’s a moment in the movie where a Cambridge professor writes on the board (if memory serves) the complete elliptic integral of the first kind

\displaystyle K(k) = \int_0^{\frac{\pi}{2}} \frac{d \theta}{\sqrt{1 - k^2 \sin^2 \theta}}

and goads Ramanujan into stepping up to the board, presumably with the intent to embarrass him, whereupon Ramanujan immediately writes down the Taylor series expansion of the integral as a function of k. In the movie it’s a bit unclear whether this meant he worked out the answer off the top of his head or knew it already, but my inclination is to assume the latter based on the fact that this integral appears in Carr’s Synopsis, which Ramanujan famously studied in India.

In any case, how might we go about finding this Taylor series? A natural strategy is to first compute the Taylor series of the integrand, then integrate it term-by-term, especially if, in the spirit of Ramanujan, we’re willing to play fast and loose with issues like where the Taylor series converges and whether we can exchange infinite sums and integrals here. Using the general form of the binomial theorem, the integrand expands to

\displaystyle \left( 1 - k^2 \sin^2 \theta \right)^{- \frac{1}{2}} = \sum_{n=0}^{\infty} {- \frac{1}{2} \choose n} (-1)^n k^{2n} \sin^{2n} \theta

where

\displaystyle {-\frac{1}{2} \choose n} = \frac{\left( - \frac{1}{2} \right) \left( -\frac{1}{2}- 1 \right) \left(-\frac{1}{2} - 2 \right) \dots \left( -\frac{1}{2}- (n-1) \right)}{n!}

which simplifies to

\displaystyle (-1)^n \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2^n n!} = (-1)^n \frac{(2n)!}{ \left( 2^n n! \right)^2 }.

It follows that we have

\displaystyle K(k) = \sum_{n=0}^{\infty} \frac{(2n)!}{\left( 2^n n! \right)^2} \left( \int_0^{\frac{\pi}{2}} \sin^{2n} \theta \, d \theta \right) k^{2n}

so we’re left with computing the integral of \sin^{2n} \theta. This is straightforward to compute using Euler’s formula, which gives

\displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n} \theta \, d \theta = \int_0^{\frac{\pi}{2}} \left( \frac{e^{i \theta} - e^{-i \theta}}{2i} \right)^{2n} \, d \theta.

Using a second application of the binomial theorem, the k^{th} term is

\displaystyle \frac{(-1)^{n+k}}{4^n} {2n \choose k} \int_0^{\frac{\pi}{2}}  e^{i (2n - 2k) \theta} \, d \theta.

Since we know the answer is real, we can ignore the imaginary part of this integral and focus on its real part. The integral of the real part

\displaystyle \int_0^{\frac{\pi}{2}} \cos (2n-2k) \theta \, d \theta

vanishes unless k = n by symmetry considerations, so k =n is the only relevant term and we get

\displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n} \theta \, d \theta = \frac{1}{4^n} {2n \choose n} \int_0^{\frac{\pi}{2}} 1 \, d \theta

where the integral is just \frac{\pi}{2}. This gives the final answer

\displaystyle K(k) = \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{(2n)!^2}{(2^n n!)^4} k^{2n}

which Ramanujan wrote as

\displaystyle K(k) = \frac{\pi}{2} \left( 1 + \left( \frac{1}{2} \right)^2 k^2 + \left( \frac{1 \cdot 3}{2 \cdot 4} \right)^2 k^4 + \left( \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \right)^2 k^6 + \dots \right).

Ramanujan’s prime number theorem

Ramanujan claimed, in his letters to Hardy, that he had found a more or less exact formula for the prime counting function \pi(n) (the number of primes less than or equal to n). Upon closer inspection by Hardy and Littlewood, this formula was later shown to be incorrect. This is part of the ongoing tension in the movie between mathematical intuition, as exemplified by Ramanujan, and mathematical rigor, as exemplified by Hardy; Hardy emphasizes that situations like this are why intuition is not enough and Ramanujan needs rigor as well. But the movie never explains what, exactly, Ramanujan’s error was. So what was it?

Edit, 5/9/16: In fact Hardy writes about this error in The Indian Mathematician Ramanujan (beginning on page 150), as I learned from Alison Miller in the comments. It’s well worth reading everything Hardy has to say, but on the subject of \pi(n) in particular, he writes

Ramanujan’s theory of primes was vitiated by his ignorance of the theory of functions of a complex variable. It was (so to say) what the theory might be if the Zeta-function had no complex zeros. His method depended upon a wholesale use of divergent series… That his proofs should have been invalid was only to be expected. But the mistakes went deeper than that, and many of the actual results were false. He had obtained the dominant terms of the classical formulae, although by invalid methods; but none of them are such close approximations as he supposed.

So, it sounds like what happened is that Ramanujan found a version of the explicit formulas relating \pi(n) to the zeroes of the Riemann zeta function. However, he believed, incorrectly, that the zeta function had no complex zeroes, and so didn’t include the terms in the explicit formulas having to do with those zeroes; this simplifies the formulas but at the cost of introducing errors which Ramanujan did not do enough computations to notice. Hardy once said elsewhere of Ramanujan that he

had indeed but the vaguest idea of what a function of a complex variable was was

and says here, more specifically, that he

knew nothing at all about the theory of analytic functions

so this is perhaps unsurprising.

Black holes

At some point in the movie there is some text claiming that Ramanujan’s work is now being applied to understand black holes. It’s easy for such claims to be overblown: for example, when Grothendieck died, some articles claimed that his work had applications to subjects like cryptography, robotics, and genetics. These claims come from a combination of two claims:

  • Grothendieck’s work had a big impact on algebraic geometry.
  • Algebraic geometry is applied to cryptography, robotics, and genetics.

However, as far as I can tell, Grothendieck’s work in particular has no direct relevance to cryptography, robotics, or genetics, although I’d be happy to see evidence to the contrary.

But this black hole claim seems to check out, sort of. I believe it refers to Ramanujan’s work on mock modular forms, which he studied in the last year of his life, after leaving England and before he died (not shown in the movie). Ramanujan described various examples of such functions, but a general theory, including a general definition, was missing until surprisingly recently, when Zwegers showed in 2002 that they were related to harmonic Maass forms.

The connection to black holes comes from Dabholkar, Murthy, and Zagier, who showed that certain mock modular forms arise as generating functions of BPS states in certain supersymmetric string theories, which are relevant to the study of black holes from the perspective of quantum gravity. This ties Ramanujan’s mock modular forms to a rich interaction between physics and mathematics involving the AdS/CFT correspondence, also known as the holographic principle, and variants of Monstrous moonshine such as umbral moonshine.

A more famous example of this relationship comes from Monstrous moonshine itself, as follows. Perhaps the most famous non-mock modular form is the j-invariant, whose Fourier expansion begins

\displaystyle j(q) = \frac{1}{q} + 744 + 196884 q + 21493760 q^2 + \dots.

The story of Monstrous moonshine begins with McKay’s famous 1978 observation that the coefficients of j can be written as sums of the dimensions of the irreducible representations of the Monster group; for example, the dimension of its smallest nontrivial irreducible representation is 196883, and 196884 = 196883 + 1. Frenkel, Lepowsky, and Meurman later showed that this is because the j-invariant is the generating function for the dimensions of the graded pieces of a vertex operator algebra on which the Monster acts, and which describes a certain conformal field theory related to the Leech lattice.

Here the relationship to black holes comes from Witten (via John Baez), who suggested that the Monster conformal field theory might have something to do with 3d (really 2+1d; 2 space, 1 time) quantum gravity, and hence with 3d black holes, via the holographic principle. A tantalizing piece of numerical evidence for this conjecture comes from calculations of black hole entropy. The lightest black hole in one version of the theory has 196883 states, and so its entropy is

\displaystyle \ln 196883 \approx 12.19

whereas a semiclassical approximation to this entropy, using the Bekenstein-Hawking formula, gives

\displaystyle 4 \pi \approx 12.57.

These aren’t supposed to agree exactly because there are quantum corrections to the semiclassical approximation. There is a parameter k that can be varied in the theory, and as k \to \infty the agreement between the quantum and semiclassical answers becomes exact. This is proven using a known asymptotic for the coefficients of the j-invariant, namely that

\displaystyle \log [q^n] j(q) \approx 4 \pi \sqrt{n}.

It’s very curious to think that this 4 \pi \sqrt{n} might be related to black hole entropy.

Incidentally, the proof of this result (with good error terms) relies on the Hardy-Littlewood circle method, which was pioneered by Hardy and Ramanujan in the work on the asymptotics of the partition function. This is a major part of the movie which we’ll defer discussion of to a second post.


Viewing all articles
Browse latest Browse all 10

Trending Articles