# The Unapologetic Mathematician

## Commutativity in Series I

We’ve seen that associativity may or may not hold for infinite sums, but it can be improved with extra assumptions. As it happens, commutativity breaks down as well, though the story is a bit clearer here.

First we should be clear about what we’re doing. When we add up a finite list of real numbers, we can reorder the list in many ways. In fact, reorderings of $n$ numbers form the symmetric group $S_n$. If we look back at our group theory, we see that we can write any element in this group as a product of transpositions which swap neighboring entries in the list. Thus since the sum of two numbers is invariant under such a swap — $a+b=b+a$ — we can then rearrange any finite list of numbers and get the same sum every time.

Now we’re not concerned about finite sums, but about infinite sums. As such, we consider all possible rearrangements — bijections $p:\mathbb{N}\rightarrow\mathbb{N}$ — which make up the “infinity symmetric group $S_\infty$. Now we might not be able to effect every rearrangement by a finite number of transpositions, and commutativity might break down.

If we have a series with terms $a_k$ and a bijection $p$, then we say that the series with terms $b_k=a_{p(k)}$ is a rearrangement of the first series. If, on the other hand, $p$ is merely injective, then we say that the new series is a subseries of the first one.

Now, if $\sum_{k=0}^\infty a_k$ is only conditionally convergent, I say that we can rearrange the series to give any value we want! In fact, given $x\leq y$ (where these could also be $\pm\infty$) there will be a rearrangement $b_k=a_{p(k)}$ so that

$\displaystyle\liminf\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nb_k=x$
$\displaystyle\limsup\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nb_k=y$

First we throw away any zero terms in the series, since those won’t affect questions of convergence, or the value of the series if it does converge. Then let $p_n$ be the $n$th positive term in the sequence $a_k$, and let $-q_n$ be the $n$th negative term.

The two series with positive terms $\sum_{k=0}^\infty p_k$ and $\sum_{k=0}^\infty q_k$ both diverge. Indeed, if one converged but the other did not, then the original series $\sum_{k=0}^\infty a_k$ would diverge. On the other hand, if they both converged then the original series would converge absolutely. Conditional convergence happens when the subseries of positive terms and the subseries of negative terms just manage to balance each other out.

Now we take two sequences $x_n$ and $y_n$ converging to $x$ and $y$ respectively. Since the series of positive terms diverges, they’ll eventually exceed any positive number. We can take just enough of them (say $k_1$ so that

$\displaystyle\sum_{k=0}^{k_1}p_k>y_1$

Similarly, we can then take just enough negative terms so that

$\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l

Now take just enough of the remaining positive terms so that

$\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l+\sum\limits_{k=k_1+1}^{k_2}p_k>y_2$

and enough negatives so that

$\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l+\sum\limits_{k=k_1+1}^{k_2}p_k-\sum\limits_{l=l_1+1}^{l_2}q_l

and so on and so forth. This gives us a rearrangement of the terms of the series.

Each time we add positive terms we come within $p_{k_j}$ of $y_j$, and each time we add negative terms we come within $q_{l_j}$ of $x_j$. But since the original sequence $a_n$ must be converging to zero (otherwise the series couldn’t converge), so must the $p_{k_j}$ and $q_{l_j}$ be converging to zero. And the sequences $x_j$ and $y_j$ are converging to $x$ and $y$.

It’s straightforward from here to show that the limits superior and inferior of the partial sums of the rearranged series are as we claim. In particular, we can set them both equal to the same number and get that number as the sum of the rearranged series. So for conditionally convergent series, the commutativity property falls apart most drastically.

May 8, 2008 Posted by | Analysis, Calculus | 4 Comments

## Quantum Knot Mosaics

Today, Sam Lomonaco and Louis Kauffman posted to the arXiv a paper on “Quantum Knots and Mosaics”. I had the pleasure of a sneak preview back in March. Here’s what I said then (I haven’t had a chance to read the paper as posted, so some of this may be addressed):

About half the paper consists of setting up definitions of a mosaic and the Reidemeister moves. This concludes with the conjecture that before you allow superpositions the mosaic framework captures all of knot theory.

The grading by the size of the mosaic leads to an obvious conjecture: there exist mosaic knots which are mosaic equivalent, but which require arbitrarily many expansions. This is analogous to the same fact about crossing numbers.

Obviously, I’d write these combinatorial frameworks as categories with the mosaics as objects and the morphisms generated by the mosaic moves. Superpositions just seem to be the usual passage from a set to the vector space on that basis. See my new paper for how I say this for regular knots and Reidemeister moves.

Then (like I say in the paper) we want to talk about mosaic “covariants”. I think this ends up giving your notion of invariant after we decategorify (identify isomorphic outputs).

The only thing I’m wondering about (stopping shy of saying you two are “wrong”) is the quantum moves. The natural thing would be to go from the “group” (really its a groupoid like I said before) of moves to its linearization. That is, we should allow the “sum” of two moves as a move. This splits a basis mosaic input into a superposition.

In particular, the “surprising” result you state that one quantum mosaic is not quantum equivalent to the other must be altered. There is clearly a move in my view taking the left to the right. “Equivalence” is then the statement that two quantum mosaics are connected by an *invertible* move. I’m not sure that the move from left to right is invertible yet, but I think it is.