03 March 2008

From products of infinite sums to subtraction and division

In some of the research I'm doing, I have occasion to write expressions like

A + ABA + ABABA + ABABABA + \dots = (A^{-1} - B)^{-1}

with no consideration of convergence.

Although in my setting I have (multiplicative) inverses and (additive) negatives, it nonetheless becomes interesting to ask what sorts of operations we can do with just plus and times, provided that we can do such things infinitely.

For example, with just positive integers and some questionable summations, we can get all negative fractions:

-1 = 1 + 2 + 4 + 8 + 16 + \dots
-m/n = m + (n+1)\times m + (n+1)^2\times m + \dots

and some irrationals:

\sqrt{2\pi} = 1 \times 2 \times 3 \times 4 \times 5 \times \dot

The first two sums are from the fact that 1+r+r^2+\dots=1/(1-r). Indeed, this comes from multiplying and shifting. If, e.g., we add 1 to the first equation, and then telescope the sum, we have

1+1 + 2 + 4 + 8 + 16 + \dots = 2+2 + 4 + 8 + 16 + \dots = 4+4 + 8 + 16 + \dots = 8+8 + 16 + \dots = etc.

The product is from \zeta-function regularization: we recognize the right hand side as the exponential of d\zeta(s)/ds at s=0.

Since we can multiply power series, we can get all fractions; for any given fraction, there are lots of ways of writing it as a series. For example,

1 = (-1)^2 = (1+2+4+8+\dots)^2 = 1 + (2+2) + (4+4+4) + (8+8+8+8) + \dots
= \sum (n+1)2^n.

Since \sum 2^n = -1, we clearly have \sum n2^{n-1} = 1 as well, which can be checked directly (it is d[\sum 2^n]/d2 = d[1/(1-2)]/d2 = 1/(1-2)^2.).

How about inverses of variables? Especially if you're not sure of the 2-adic sum, perhaps you'd rather use

-1/X = 1 + (1+X) + (1+X)^2 + (1+X)^3 + \dots

to get your negative and inverse Xs. This formula is hard to verify, since it still requires the same shifting tricks:

1 + ( X + X(1+X) + X(1+X)^2 + \dots ) = 1 + X + X(1+X) + X(1+X)^2 + \dots
= 1+X + X(1+X) + X(1+X)^2 + \dots
= (1+X)(1+X) + X(1+X)^2 + \dots
= (1+X)(1+X)^2 + \dots
= etc.

See, infinite sums are not associative --- the distance between terms matters --- but these manipulations are justified since infinite sums are finitely associative.

We get, either by squaring or differentiating,

1/X^2 = 1 + 2(1+X) + 3(1+X)^2 + 4(1+X)^3 + \dots

and hence all powers of X. Well, we get \pm 1/X^n; for, say, -X, the shortest is to use X^2(-1/X) = X^2 + X^2(1+X) + X^2(1+X)^2 + X^2(1+X)^3 + \dots, which telescopes in the same way.

These manipulations require that there be a number 1 which multiplies with X to X and which we can add to X. Which is true is X is, say, a matrix, or a CW complex, but not if X is a two-zero-tensor. If X is such a tensor, then X^{-1} is the tensor which convolves on both sides to the identity matrix. This is the setting of my original line (A is two-zero, and B is zero-two). In this situation, I don't have a good construction of X^{-1}.

12 February 2008

New Blog

If you do not subscribe to the RSS feed for this blog, then you have by now stopped checking it for updates. I intend to post occasionally, but school is absorbing my mathematical thoughts, rather than writing.

Since I have no extra time, I have started a new blog, sister to this one. Whereas Orange Juice Files is primarily for mathematical (and nonmathematical) essays, Local Seasoning will be a repository of recipes and meals from my kitchen. Enjoy.

08 January 2008

A composition law for discrete-time QM

One of the best ways to understand quantum mechanics — bear with me — is as a one-dimensional quantum field theory. No, it's not backwards, and we really should think of QM as inherently one-dimensional: there's one dimension of time. The configuration space of a particle is finite-dimensional; the size of the space of paths the particle could take — and Feynman says that a particle takes every possible path — is largely determined by number of "time" dimensions in the problem, since a path has a point in configuration space for every moment in time.

In any case, I'm going to think of it that way. And then I'm going to think about path integration: the transition amplitude between two configurations is some poorly-defined integral over the infinite-dimensional space of paths connecting those configurations. Rather than trying to compute in infinite dimensions, physicists since Feynman have formally expanded the integral asymptotically, and interpreted the coefficients combinatorially as diagrams. (And interpreted the diagrams as describing actual events, which is a matter for them physicists rather than for us mathematicians to discuss.)

These formal power series should — indeed, must, if the formalism is to make sense — satisfy a particular gluing axiom. But it seems that no one has gone to the trouble to verify that in fact they do; there is no proof available in the literature. In his fall-semester QFT class, Nicolai Reshetikhin suggested that someone try to fix this, and I volunteered. Silly me.

Long story short, after making progress on a related issue, and then avoiding the project for more than a month, I've worked the 0-dimensional analogue, where we approximate the time interval by a sequence of (equally-spaced) discrete points. If you would like to read about it, the paper is available here.

Update: The second, and likely final, installment is now available. It concludes with the gluing rules for arbitrary perturbative quantum field theories.