## Braids and the Yang-Baxter equation

I recently gave a talk on the Yang-Baxter equation. The focus of the talk was to state the connection between the Yang-Baxter equation and the braid relation. This connection comes from a system of interacting particles. In this post, I’ll go over part of my talk. You can access the full set of notes here.

### Interacting particles

Imagine two particles on a line, each with a state that can be any element of a set $\mathcal{X}$. Suppose that the only way particles can change their states is by interacting with each other. An interaction occurs when two particles pass by each other. We could define a function $F : \mathcal{X} \times \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ that describes how the states change after interaction. Specifically, if the first particle is in state $x$ and the second particle is in state $y$, then their states after interacting will be

$F(x,y) = (F_a(x,y), F_b(x,y)) = (\text{new state of particle 1}, \text{new state of particle 2}),$

where $F_a,F_b : \mathcal{X} \times \mathcal{X} \to \mathcal{X}$ are the components of $F$. Recall that the particles move past each other when they interact. Thus, to keep track of the whole system we need an element of $\mathcal{X} \times \mathcal{X}$ to keep track of the states and a permutation $\sigma \in S_2$ to keep track of the positions.

#### Three particles

Now suppose that we have $3$ particles labelled $1,2,3$. As before, each particle has a state in $\mathcal{X}$. We can thus keep track of the state of each particle with an element of $\mathcal{X}^3$. The particles also have a position which is described by a permutation $\sigma \in S_3$. The order the entries of $(x,y,z) \in \mathcal{X}^3$ corresponds to the labels of the particles not their positions. A possible configuration is shown below:

As before, the particles can interact with each other. However, we’ll now add the restriction that the particles can only interact two at a time and interacting particles must have adjacent positions. When two particles interact, they swap positions and their states change according to $F$. The state and position of the remaining particle is unchanged. For example, in the above picture we could interact particles $1$ and $3$. This will produce the below configuration:

To keep track of how the states of the particles change over time we will introduce three functions from $\mathcal{X}^3$ to $\mathcal{X}^3$. These functions are $F_{12},F_{13},F_{23}$. The function $F_{ij}$ is given by applying $F$ to the $i,j$ coordinates of $(x,y,z)$ and acting by the identity on the remaining coordinate. In symbols,

$F_{12}(x,y,z) = (F_a(x,y), F_b(x,y), z),$
$F_{13}(x,y,z) = (F_a(x,z), y, F_b(x,z)),$
$F_{23}(x,y,z) = (x, F_a(y,z), F_b(y,z)).$

The function $F_{ij}$ exactly describes how the states of the three particles change when particles $i$ and $j$ interact. Now suppose that three particles begin in position $123$ and states $(x,y,z)$. We cannot directly interact particles $1$ and $3$ since they are not adjacent. We have to pass first pass one of the particles through particle $2$. This means that there are two ways we can interact particles $1$ and $3$. These are illustrated below.

In the top chain of interactions, we first interact particles $2$ and $3$. In this chain of interactions, the states evolve as follows:

$(x,y,z) \to F_{23}(x,y,z) \to Â F_{13}(F_{23}(x,y,z)) \to F_{12}(F_{13}(F_{23}(x,y,z))).$

In the bottom chain of interactions, we first interact particles $1$ and $2$. On this chain, the states evolve in a different way:

$(x,y,z) \to F_{12}(x,y,z) \to Â F_{13}(F_{12}(x,y,z)) \to F_{23}(F_{13}(F_{12}(x,y,z))).$

Note that after both of these chains of interactions the particles are in position $321$. The function $F$ is said to solve the Yang–Baxter equation if two chains of interactions also result in the same states.

Definition: A function $F : \mathcal{X} \times \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is a solution to the set theoretic Yang–Baxter equation if,

$F_{12}\circ F_{13} \circ F_{23} = F_{23} \circ F_{13} \circ F_{12}.$

This equation can be visualized as the “braid relation” shown below. Here the strings represent the three particles and interacting two particles corresponds to crossing one string over the other.

Here are some examples of solutions to the set theoretic Yang-Baxter equation,

• The identity on $\mathcal{X} \times \mathcal{X}$.
• The swap map, $(x,y) \mapsto (y,x)$.
• If $f,g : \mathcal{X} \to \mathcal{X}$ commute, then the function $(x,y) \to (f(x), g(y))$ is a solution the Yang-Baxter equation.

In the full set of notes I talk about a number of extensions and variations of the Yang-Baxter equation. These include having more than three particles, allowing for the particle states to be entangle and the parametric Yang-Baxter equation.

## Double cosets and contingency tables

Like my previous post, this blog is also motivated by a comment by Professor Persi Diaconis in his recent Stanford probability seminar. The seminar was about a way of “collapsing” a random walk on a group to a random walk on the set of double cosets. In this post, I’ll first define double cosets and then go over the example Professor Diaconis used to make us probabilists and statisticians more comfortable with all the group theory he was discussing.

### Double cosets

Let $G$ be a group and let $H$ and $K$ be two subgroups of $G$. For each $g \in G$, the $(H,K)$-double coset containing $g$ is defined to be the set

$HgK = \{hgk : h \in H, k \in K\}$

To simplify notation, we will simply write double coset instead of $(H,K)$-double coset. The double coset of $g$ can also be defined as the equivalence class of $g$ under the relation

$g \sim g' \Longleftrightarrow g' = hgk$ for some $h \in H$ and $g \in G$

Like regular cosets, the above relation is indeed an equivalence relation. Thus, the group $G$ can be written as a disjoint union of double cosets. The set of all double cosets of $G$ is denoted by $H\backslash G / K$. That is,

$H\backslash G /K = \{HgK : g \in G\}$

Note that if we take $H = \{e\}$, the trivial subgroup, then the double cosets are simply the left cosets of $K$, $G / K$. Likewise if $K = \{e\}$, then the double cosets are the right cosets of $H$, $H \backslash G$. Thus, double cosets generalise both left and right cosets.

### Double cosets in $S_n$

Fix a natural number $n > 0$. A partition of $n$ is a finite sequence $a = (a_1,a_2,\ldots, a_I)$ such that $a_1,a_2,\ldots, a_I \in \mathbb{N}$, $a_1 \ge a_2 \ge \ldots a_I > 0$ and $\sum_{i=1}^I a_i = n$. For each partition of $n$, $a$, we can form a subgroup $S_a$ of the symmetric group $S_n$. The subgroup $S_a$ contains all permutations $\sigma \in S_n$ such that $\sigma$ fixes the sets $A_1 = \{1,\ldots, a_1\}, A_2 = \{a_1+1,\ldots, a_1 + a_2\},\ldots, A_I = \{n - a_I +1,\ldots, n\}$. Meaning that $\sigma(A_i) = A_i$ for all $1 \le i \le I$. Thus, a permutation $\sigma \in S_a$ must individually permute the elements of $A_1$, the elements of $A_2$ and so on. This means that, in a natural way,

$S_a \cong S_{a_1} \times S_{a_2} \times \ldots \times S_{a_I}$

If we have two partitions $a = (a_1,a_2,\ldots, a_I)$ and $b = (b_1,b_2,\ldots,b_J)$, then we can form two subgroups $H = S_a$ and $K = S_b$ and consider the double cosets $H \backslash G / K = S_a \backslash S_n / S_b$. The claim made in the seminar was that the double cosets are in one-to-one correspondence with $I \times J$ contingency tables with row sums equal to $a$ and column sums equal to $b$. Before we explain this correspondence and properly define contingency tables, let’s first consider the cases when either $H$ or $K$ is the trivial subgroup.

### Left cosets in $S_n$

Note that if $a = (1,1,\ldots,1)$, then $S_a$ is the trivial subgroup and, as noted above, $S_a \backslash S_n / S_b$ is simply equal to $S_n / S_b$. We will see that the cosets in $S_n/S_b$ can be described by forgetting something about the permutations in $S_n$.

We can think of the permutations in $S_n$ as all the ways of drawing without replacement $n$ balls labelled $1,2,\ldots, n$. We can think of the partition $b = (b_1,b_2,\ldots,b_J)$ as a colouring of the $n$ balls by $J$ colours. We colour balls $1,2,\ldots, b_1$ by the first colour $c_1$, then we colour $b_1+1,b_1+2,\ldots, b_1+b_2$ the second colour $c_2$ and so on until we colour $n-b_J+1, n-b_J+2,\ldots, n$ the final colour $c_J$. Below is an example when $n$ is equal to 6 and $b = (3,2,1)$.

Note that a permutation $\sigma \in S_n$ is in $S_b$ if and only if we draw the balls by colour groups, i.e. we first draw all the balls with colour $c_1$, then we draw all the balls with colour $c_2$ and so on. Thus, continuing with the previous example, the permutation $\sigma_1$ below is in $S_b$ but $\sigma_2$ is not in $S_b$.

It turns out that we can think of the cosets in $S_n \setminus S_b$ as what happens when we “forget” the labels $1,2,\ldots,n$ and only remember the colours of the balls. By “forgetting” the labels we mean only paying attention to the list of colours. That is for all $\sigma_1,\sigma_2 \in S_n$, $\sigma_1 S_b = \sigma_2 S_b$ if and only if the list of colours from the draw $\sigma_1$ is the same as the list of colours from the draw $\sigma_2$. Thus, the below two permutations define the same coset of $S_b$

To see why this is true, note that $\sigma_1 S_b = \sigma_2 S_b$ if and only if $\sigma_2 = \sigma_1 \circ \tau$ for some $\tau \in S_b$. Furthermore, $\tau \in S_b$ if and only if $\tau$ maps each colour group to itself. Recall that function composition is read right to left. Thus, the equation $\sigma_2 = \sigma_1 \circ \tau$ means that if we first relabel the balls according to $\tau$ and then draw the balls according to $\sigma_1$, then we get the result as just drawing by $\sigma_2$. That is, $\sigma_2 = \sigma_1 \circ \tau$ for some $\tau \in S_b$ if and only if drawing by $\sigma_2$ is the same as first relabelling the balls within each colour group and then drawing the balls according to $\sigma_1$. Thus, $\sigma_1 S_b = \sigma_2 S_b$, if and only if when we forget the labels of the balls and only look at the colours, $\sigma_1$ and $\sigma_2$ give the same list of colours. This is illustrated with our running example below.

### Right cosets of $S_n$

Typically, the subgroup $S_a$ is not a normal subgroup of $S_n$. This means that the right coset $S_a \sigma$ will not equal the left coset $\sigma S_a$. Thus, colouring the balls and forgetting the labelling won’t describe the right cosets $S_a \backslash S_n$. We’ll see that a different type of forgetting can be used to describe $S_a \backslash S_n = \{S_a\sigma : \sigma \in S_n\}$.

Fix a partition $a = (a_1,a_2,\ldots,a_I)$ and now, instead of considering $I$ colours, think of $I$ different people $p_1,p_2,\ldots,p_I$. As before, a permutation $\sigma \in S_n$ can be thought of drawing $n$ balls labelled $1,\ldots,n$ without replacement. We can imagine giving the first $a_1$ balls drawn to person $p_1$, then giving the next $a_2$ balls to the person $p_2$ and so on until we give the last $a_I$ balls to person $p_I$. An example with $n = 6$ and $a = (2,2,2)$ is drawn below.

Note that $\sigma \in S_a$ if and only if person $p_i$ receives the balls with labels $\sum_{k=0}^{i-1}a_k+1,\sum_{k=0}^{i-1}a_k+2,\ldots, \sum_{k=0}^i a_k$ in any order. Thus, in the below example $\sigma_1 \in S_a$ but $\sigma_2 \notin S_a$.

It turns out the cosets $S_a \backslash S_n$ are exactly determined by “forgetting” the order in which each person $p_i$ received their balls and only remembering which balls they received. Thus, the two permutation below belong to the same coset in $S_a \backslash S_n$.

To see why this is true in general, consider two permutations $\sigma_1,\sigma_2$. The permutations $\sigma_1,\sigma_2$ result in each person $p_i$ receiving the same balls if and only if after $\sigma_1$ we can apply a permutation $\tau$ that fixes each subset $A_i = \left\{\sum_{k=0}^{i-1}a_k+1,\ldots, \sum_{k=0}^i a_k\right\}$ and get $\sigma_2$. That is, $\sigma_1$ and $\sigma_2$ result in each person $p_i$ receiving the same balls if and only if $\sigma_2 = \tau \circ \sigma_1$ for some $\tau \in S_a$. Thus, $\sigma_1,\sigma_2$ are the same after forgetting the order in which each person received their balls if and only if $S_a \sigma_1 = S_a \sigma_2$. This is illustrated below,

We can thus see why $S_a \backslash S_n \neq S_n / S_a$. A left coset $\sigma S_a$ correspond to pre-composing $\sigma$ with elements of $S_a$ and a right cosets $S_a\sigma$ correspond to post-composing $\sigma$ with elements of $S_a$.

### Contingency tables

With the last two sections under our belts, describing the double cosets $S_a \backslash S_n / S_b$ is straight forward. We simply have to combine our two types of forgetting. That is, we first colour the $n$ balls with $J$ colours according to $b = (b_1,b_2,\ldots,b_J)$. We then draw the balls without replace and give the balls to $I$ different people according $a = (a_1,a_2,\ldots,a_I)$. We then forget both the original labels and the order in which each person received their balls. That is, we only remember the number of balls of each colour each person receives. Describing the double cosets by “double forgetting” is illustrated below with $a = (2,2,2)$ and $b = (3,2,1)$.

The proof that double forgetting does indeed describe the double cosets is simply a combination of the two arguments given above. After double forgetting, the number of balls given to each person can be recorded in an $I \times J$ table. The $N_{ij}$ entry of the table is simply the number of balls person $p_i$ receives of colour $c_j$. Two permutations are the same after double forgetting if and only if they produce the same table. For example, $\sigma_1$ and $\sigma_2$ above both produce the following table

By the definition of how the balls are coloured and distributed to each person we must have for all $1 \le i \le I$ and $1 \le j \le J$

$\sum_{j=1}^J N_{ij} = a_i$ and $\sum_{i=1}^I N_{ij} = b_j$

An $I \times J$ table with entries $N_{ij} \in \{0,1,2,\ldots\}$ satisfying the above conditions is called a contingency table. Given such a contingency table with entries $N_{ij}$ where the rows sum to $a$ and the columns sum to $b$, there always exists at least one permutation $\sigma$ such that $N_{ij}$ is the number of balls received by person $p_i$ of colour $c_i$. We have already seen that two permutations produce the same table if and only if they are in the same double coset. Thus, the double cosets $S_a \backslash S_n /S_b$ are in one-to-one correspondence with such contingency tables.

### The hypergeometric distribution

I would like to end this blog post with a little bit of probability and relate the contingency tables above to the hyper geometric distribution. If $a = (m, n-m)$ for some $m \in \{0,1,\ldots,n\}$, then the contingency tables described above have two rows and are determined by the values $N_{11}, N_{12},\ldots, N_{1J}$ in the first row. The numbers $N_{1j}$ are the number of balls of colour $c_j$ the first person receives. Since the balls are drawn without replacement, this means that if we put the uniform distribution on $S_n$, then the vector $Y=(N_{11}, N_{12},\ldots, N_{1J})$ follows the multivariate hypergeometric distribution. Thus, if we have a random walk on $S_n$ that quickly converges to the uniform distribution on $S_n$, then we could use the double cosets to get a random walk that converges to the multivariate hypergeometric distribution (although there are smarter ways to do such sampling).

## The field with one element in a probability seminar

Something very exciting this afternoon. Professor Persi Diaconis was presenting at the Stanford probability seminar and the field with one element made an appearance. The talk was about joint work with Mackenzie Simper and Arun Ram. They had developed a way of “collapsing” a random walk on a group to a random walk on the set of double cosets. As an illustrative example, Persi discussed a random walk on $GL_n(\mathbb{F}_q)$ given by multiplication by a random transvection (a map of the form $v \mapsto v + ab^Tv$, where $a^Tb = 0$).

The Bruhat decomposition can be used to match double cosets of $GL_n(\mathbb{F}_q)$ with elements of the symmetric group $S_n$. So by collapsing the random walk on $GL_n(\mathbb{F}_q)$ we get a random walk on $S_n$ for all prime powers $q$. As Professor Diaconis said, you can’t stop him from taking $q \to 1$ and asking what the resulting random walk on $S_n$ is. The answer? Multiplication by a random transposition. As pointed sets are vector spaces over the field with one element and the symmetric groups are the matrix groups, this all fits with what’s expected of the field with one element.

This was just one small part of a very enjoyable seminar. There was plenty of group theory, probability, some general theory and engaging examples.

Update: I have written another post about some of the group theory from the seminar! You can read it here: Double cosets and contingency tables.

## Commuting in Groups

In July 2020 I submitted my honours thesis which was titled “Commuting in Groups” and supervised by Professor Anthony Licata. You can read the abstract below and the full thesis here.

## Abstract

In this thesis, we study four different classes of groups coming from geometric group theory. Each of these classes are defined in terms of fellow traveller conditions. First we study automatic groups and show that such groups have a solvable word problem. We then study hyperbolic groups and show that a group is hyperbolic
if and only if it is strongly geodesically automatic. We also show that a group is hyperbolic if and only if it has a divergence function. We next study combable groups and examine some examples of groups that are combable but not automatic. Finally we introduce biautomatic groups. We show that biautomatic groups have a solvable conjugacy problem and that many nilpotent groups cannot be subgroups of biautomatic groups. Finally we introduce Artin groups of finite type and show that all such groups are biautomatic.

## Sums of Squares – Part 2: Motzkin

In the previous blog post we observed that if a polynomial $g$ could be written as a sum $\sum_{i=1}^n h_i^2$ where each $h_i$ is a polynomial, then $g$ must be non-negative. We then asked if the converse was true. That is, if $g$ is a non-negative polynomial, can $g$ be written a sum of squares of polynomials? As we saw, a positive answer to this question would allow us to optimize a polynomial function $f$ by looking at the values of $\lambda$ such that $g = f - \lambda$ can be written as a sum of squares.

Unfortunately, as we shall see, not all non-negative polynomials can be written as sums of squares.

## The Motzkin Polynomial

The Motzkin polynomial is a two variable polynomial defined by

$g(x,y) = x^4y^2 +x^2y^4-3x^2y^2+1$.

We wish to prove two things about this polynomial. The first claim is that $g(x,y) \ge 0$ for all $x,y \in \mathbb{R}$ and the second claim is that $g$ cannot be written as a sum of squares of polynomials. To prove the first claim we will use the arithmetic mean geometric mean inequality. This inequality states that for all $n \in \mathbb{N}$ and all $a_1,a_2,\ldots, a_n \ge 0$, we have that

$\left(a_1a_2\ldots a_n\right)^{1/n} \le \frac{a_1+a_2+\ldots+a_n}{n}.$

We will apply this inequality with $n = 3$, $a_1 = x^4 y^2$, $a_2 = x^2 y^4$ and $a_3 = 1$. This gives that

$\left(x^4 y^2 x^2 y^4 1\right)^\frac{1}{3} \le \frac{x^4 y^2 + y^2 x^4 +1 }{3}$.

Simplifying the left hand side and multiplying both sides by $3$ gives that

$3x^2 y^2 \le x^4 y^2 + y^2 x^4 + 1$.

Since our Motzkin polynomial $g$ is given by $g(x,y) = x^4 y^2 + y^2 x^2 - 3x^2 y^2 +1$, the above inequality is equivalent to $g(x,y) \ge 0$.

## Newton Polytopes

Thus we have shown that the Motzkin polynomial is non-negative. We now wish to show that it is not a sum of squares. To do this we will first have to define the Newton polytope of a polynomial. If our polynomial $f$ has $n$ variables, then the Newton polytope of $f$, denoted $N(f)$ is a convex subset of $\mathbb{R}^n$. To define $N(f)$ we associate a point in $\mathbb{R}^n$ to each term of the polynomial $f$ based on the degree of each variable. For instance the term $3x^2y^4$ is assigned the point $(2,4) \in \mathbb{R}^2$ and the term $4xyz^3$ is assigned the point $(1,1,3) \in \mathbb{R}^2$. Note that the coefficients of the term don’t affect this assignment.

We then define $N(f)$ to be the convex hull of all points assigned to terms of the polynomial $f$. For instance if $f(x,y) = x^4y^3+5x^3y - 7y^3+8$, then the Newton polytope of $f$ is the following subset of $\mathbb{R}^2$.

Note again that changing the non-zero coefficients of the polynomial $f$ does not change $N(f)$.

Now suppose that our polynomial is a sum of squares, that is $f = \sum_{i=1}^n h_i^2$. It turns out the the Newton polytope of $f$ can be defined in terms of the Newton polytopes of $h_i$. In particular we have that $N(f) =2X$ where $X$ is the convex hull of the union of $N(h_i)$ for $i = 1,\ldots, n$.

To see why this is true, note that if $a$ and $b$ are the points corresponding to two terms $p$ and $q$, then $a+b$ corresponds to $pq$. Thus we can see that every term of $f$ corresponds to a point that can be written as $a+b$ for some $a,b \in N(h_i)$. It follows that $N(f) \subseteq 2X$.

Conversely note that if we have terms $p, q, r$ corresponding to points $a,b, c$ and $pq = r^2$, then $a+b = 2c$. It follows that if $c$ is a vertex in $X$ corresponding to the term $r$ in some polynomial $h_i$ and $p,q$ are two terms in a polynomial $h_j$ such that $pq = r^2$, then $p=q=r$. This is because if $r$ was not equal to either $p$ or $q$, then the point $c$ would not be a vertex and would instead lie on the line between the points assigned to $p$ and $q$.

It follows that if $c$ is a vertex of $X$ with corresponding term $r$, then $r^2$ appears with a positive coefficient in $f = \sum_{i=1}^n h_i$. It follows that $2X \subseteq N(f)$ and so $2X = N(f)$.

## Not a sum of squares

Let’s now take another look at the Motzkin polynomial which is defined to be

$g(x,y) = x^4y^2 + x^2 y^4 - 3x^2y^2 + 1$.

Thus the Newton polytope of $g$ has corners $(4,2), (2,4), (0,0)$ and looks like this

Thus if the Motzkin polynomial was a sum of squares $g = \sum_{i=1}^n h_i^2$, then the Newton polytope of each $h_i$ must be contained in $\frac{1}{2}N(g)$. Now $\frac{1}{2}N(g)$ looks like this

The only integer points contained in $\frac{1}{2}N(g)$ are $(0,0), (1,1), (1,2)$ and $(2,1)$. Thus each $h_i$ must be equal to $h_i(x,y) = a_i x^2 y+b_i xy^2 + c_i xy + d_i$. If we square this polynomial we see that the coefficient of $x^2 y^2$ is $c_i^2$. Thus the coefficient of $x^2y^2$ in $\sum_{i=1}^n h_i^2$ must be $c_1^2+c_2^2+\ldots +c_n^2 \ge 0$. But in the Motzkin polynomial, $x^2y^2$ has coefficient $-3$.

Thus the Motzkin polynomial is a concrete example of a non-negative polynomial that is not a sum of squares.

## References

As stated in the first part of this series, these two blog posts are based off conversations I had with Abhishek Bhardwaj last year. I also used these slides to remind myself why the Motzkin polynomial is not a sum of squares.

## Sums of Squares – Part 1: Optimization

About a year ago I had the pleasure of having a number of conversations with Abhishek Bhardwaj about the area of mathematics his PhD is on. This pair of posts is based on the fond memories I have of these conversations. Part two can be found here.

## Optimization and Summing Squares

Optimization is a huge area of both pure and applied mathematics. Many questions and problems can be reduced to finding the minimum or maximum value of some function. That is we have a function $f : \mathbb{R}^n \to \mathbb{R}$ and we wish to find the number $\lambda$ such that

$\lambda = \min \{ f(x_1,\ldots, x_n) \mid (x_1,\ldots, x_n) \in \mathbb{R}^n \}$,

or

$\lambda = \max \{ f(x_1,\ldots, x_n) \mid (x_1,\ldots, x_n) \in \mathbb{R}^n \}$.

By considering the function $-f$ we can reduce finding the maximum of $f$ to finding the minimum of another function. Minimizing a function isn’t always easy. Even when the function $f$ is a relative simple function such as a polynomial, it can be a very difficult problem.

A different way of thinking about the problem of minimizing the function $f$, is to instead think about the function $f - \lambda$ were $\lambda$ is a real number. The minimum of $f$ is the largest value of $\lambda$ such that the function $f - \lambda$ is non-negative for all values $x \in \mathbb{R}^n$. Thus we have a reduced our optimization problem into the problem of working out for which values of $\lambda$ is the function $f-\lambda$ non-negative.

## An example

Suppose our function $f$ was the following function that takes in two variables

$f(x,y) = 2x^2 + 4y^2-4xy-4x+4y+7$.

To minimize this function we can look at the function $g = f - \lambda$ which is equal to

$g(x,y) = 2x^2 + 4y^2-4xy-4x+4y+7 - \lambda$.

We wish to work for which values of $\lambda$ is this function $g$ non-negative. By doing the following algebraic manipulations we can rewrite $g$ like so

$g(x,y) = (x^2-2x+1)+(x^2+4y^2+1-4xy-2x+4y)+5-\lambda$,

which is in turn equal to

$g(x,y) = (x-1)^2+(-x+2y+1)^2 +(5-\lambda)$

Since $(x-1)^2$ and $(-x+2y+1)^2$ are both squares of real numbers, they are both non-negative. Furthermore at the point $(x,y) = (1,0)$, we have that $(x-1)^2+(-x+2y+1)^2=0$. Thus $g$ is non-negative if and only if $(5-\lambda) \ge 0$, that is if and only if $\lambda \le 5$. Thus we can conclude that the minimum of $f$ is $5$.

## Sums of Squares

Note that the number $5-\lambda$ can be written as $(\sqrt{5-\lambda})^2$ if and only if $\lambda \ge 5$. Thus, in the above example we have that $g$ is non-negative if and only if $g$ can be written as a sum of squares. That is $g$ is non-negative if and only if $g$ can be written as $\sum_{i=1}^n h_i^2$ for some polynomials $h_1, h_2, \ldots, h_n$.

In general, if a polynomial can be written as a sum of squares of other polynomials, then the polynomial must be non-negative. A natural question to ask is if every non-negative polynomial can be written as a sum of squares. This would make our optimization problem a lot easier. To minimize $f$, we can check for which values of $\lambda$ can $f-\lambda$ be written as a sum of squares.

This question of whether every non-negative polynomial can be written as a sum of squares was answered in the negative by David Hilbert in 1888. However, Hilbert’s proof was non-constructive, it didn’t give an explicit example of such a polynomial. The proof only showed that such polynomials exist out there somewhere. It took nearly nearly 80 years for the first explicit example to be given by the mathematician Theodore Motzkin in 1967. We will take a look at Motzkin’s polynomial in the next blog post.

## The Outer Automorphism of S6

This blog post is about why the number 6 is my favourite number – because the symmetric group on six elements is the only finite permutation group that contains an outer automorphism. In the post we will first look at why there are no other symmetric groups with outer automorphisms and then we will define an outer automorphism of $S_6$.

## Inner and Outer Automorphisms

If $G$ is any group, then we can construct a new group $\text{Aut}(G)$ consisting of all bijective group homomorphism $\phi : G \to G$. This set forms a group under function composition. There is a natural map from $G$ to $\text{Aut}(G)$. A group element $g \in G$ is taken to the function $\gamma_g : G \to G$ given by $\gamma_g(h) = ghg^{-1}$. Thus $\gamma_g$ is conjugation by $g$.

The kernel of $g \mapsto \gamma_g$ is called the center of $G$ and is denoted by $Z(G)$. That is $Z(G)$ is the subgroup consisting of group elements that commute with every other group element. The image of this map is denoted by $\text{Inn}(G) \subseteq \text{Aut}(G)$. An automorphism which is in $\text{Inn}(G)$ is called an inner automorphism. One can check that the subgroup of inner automorphisms is a normal subgroup. Indeed if we have a group element $g \in G$ and an automorphism $\phi \in \text{Aut}(G)$, then $\phi \circ \gamma_g \circ \phi^{-1} = \gamma_{\phi(g)}$. Thus we can form the quotient $\text{Aut}(G)/\text{Inn}(G)$ which we will call the outer automorphisms of $G$ and denote by $\text{Out}(G)$.

Thus the inner automorphisms of $G$ are those which come from $G$ in the form of conjugation. The outer automorphisms of $G$ are equivalence classes of all the automorphisms of $G$. Two automorphisms $\phi, \phi'$ are equivalent in $\text{Out}(G)$ if there is an inner automorphism $\gamma_g$ such that $\gamma_g \circ \phi = \phi'$.

## The Symmetric Group

For a natural number $n \in \mathbb{N}$, $S_n$ will denote the group of permutations of the set $\{1,2,\ldots,n\}$. Any permutation $\sigma \in S_n$ can be written as product of disjoint cycles which we will call its cycle type. For instance, if $n=5$ and $\sigma(1)=3$, $\sigma(2)=5$, $\sigma(3)=4$, $\sigma(4)=1$ and $\sigma(5)=2$, then $\sigma$ can be written in cycle notation as $(1,3,4)(2,5)$. Cycle notation is very useful for describing conjugation in $S_n$. If we have two permutations $\sigma, \tau \in S_n$ and we know the cycle notation for $\sigma$, then the cycle notation for $\tau \sigma \tau^{-1}$ can be derived by applying $\tau$ to every entry in the cycle notation of $\sigma$. For instance, with $\sigma = (1,3,4)(2,5)$, we have

$\tau(1,3,4)(2,5)\tau^{-1} = (\tau(1),\tau(3),\tau(4))(\tau(2),\tau(5))$.

To see why this result about conjugation is true, note the following. If $\sigma(i)=j$ for some $i,j \in \{1,2,\ldots,n\}$, then

$(\tau\sigma\tau^{-1})\tau(i)=\tau(\sigma(\tau^{-1}(\tau(i))))=\tau(\sigma(i))=\tau(j)$.

Thus two permutations $\sigma, \sigma'$ are conjugate to one another if and only if $\sigma$ and $\sigma'$ have the same cycle type (ie they have the same number of cycles of any given size). This implies that for $n \ge 3$, the centre of $S_n$ must be trivial as for any non-identity $\sigma \in S_n$, there is at least one other element of the same cycle type. Hence the map $\sigma \mapsto \gamma_\sigma$ from $S_n$ to $\text{Aut}(S_n)$ is injective for $n \ge 3$.

## Counting Conjugacy Classes

Before constructing an outer automorphism for $S_6$, we will see why such a map is special. In particular we will learn that $\text{Aut}(S_n)$ consists entirely of inner automorphism for $n \neq 6$. The argument is based off studying the size of different conjugacy classes in $S_n$. This is important because if $\phi$ is an automorphism of $S_n$ and $\sigma,\sigma' \in S_n$ are two permutations, then $\sigma$ is conjugate to $\sigma'$ if and only if $\phi(\sigma)$ is conjugate to $\phi(\sigma')$.

In $S_n$ we know that two elements are conjugate if and only if they have the same cycle type. We can use this to count the size of different conjugacy classes of $S_n$. In particular the number of permutations with $n_k$ cycles of size $k$ for $k = 1,2,\ldots, n$ is

$\displaystyle \frac{\displaystyle n!}{\displaystyle \prod\limits_{k=1}^n n_k! k^{n_k}}$.

The transpositions $(j,k)$ for $1 \le j < k\le n$ generate $S_n$ and form a conjugacy class which we call $T$. For any automorphism $\phi \in \text{Aut}(S_n)$ , the set $\phi(T)$ must be a conjugacy class that is also of size $| T | = n(n-1)/2$. Since automorphism preserve order, the elements of $\phi(T)$ must all be of order two. Thus the cycle type of the conjugacy class $\phi(T)$ must contain only cycles of size two or size one. Also there must be an odd number of two cycles. If there was an even number of two cycles, then we would have that $\phi(T)$ is contained in the subgroup $A_n \subseteq S_n$. However this would imply that $\phi(S_n) \subseteq \phi(A_n)$ which contradicts the fact that $\phi : S_n \rightarrow S_n$ is a bijection. Thus the following lemma implies that $\phi(T)=T$ when $n \neq 6$.

Lemma: Fix $n \neq 6$ and an odd number $k$ such that $k > 1$ and $2k \le n$. Let $m_k$ denote the number of permutations $\sigma \in S_n$ which have $k$ cycles of size two and $n-2k$ cycles of size one, then $m_k \neq n(n-1)/2$.

Proof: This claim can be manually checked for $n < 6$. For $n > 6$, assume in order to get a contradiction that $m_k = \frac{n(n-1)}{2}$. Note that $m_k$ is equal to $\frac{n!}{k! 2^k(n-2k)! 1^{n-2k}}=\frac{n!}{k! (n-2k)! 2^k}$. Thus we have $\frac{n(n-1)}{2} = \frac{n!}{k!(n-2k)!2^k}$ and hence

$2^{k-1} = \frac{(n-2)(n-3)\ldots(n-2k+1)}{k!}$.

Now if $k \ge 5$, then $(n-4)(n-5)\ldots(n-2k+1)$ is a product of at least $k$ consecutive integers and hence divisible by $k!$ (see here). Thus

$2^{k-1} = \frac{(n-2)(n-3)\ldots(n-2k+1)}{k!} = (n-2)(n-3)\frac{(n-4)(n-5)\ldots(n-2k+1)}{k!}$

and hence $\frac{(n-2)(n-3)\ldots(n-2k+1)}{k!}$ is divisible by an odd number other than one (either $n-2$ or $n-3$). Thus $\frac{(n-2)(n-3)\ldots(n-2k+1)}{k!}$ cannot be a power of $2$, contradicting the above equation.

When $k = 3$, the above equation becomes:

$4= \frac{(n-2)(n-3)(n-4)(n-5)}{3!} = (n-2)\frac{(n-3)(n-4)(n-5)}{3!} = \frac{(n-2)(n-3)(n-4)}{3!}(n-5)$.

Thus if $n$ is odd, $n-2$ is an odd number other than $1$ that divides 4, a contradiction. If $n$ is even, then since $n > 6$, $n-5$ is an odd number other than $1$ that divides $4$, which is again a contradiction. Thus we cannot have any case when $m_k = \frac{n(n-1)}{2}$.

## $S_n$ does not have an outer automorphism if $n \neq 6$

With the above lemma we can now prove that all the automorphism of $S_n$ for $n \neq 6$ are inner. First note that the only automorphism of $S_2 = \{ \text{id}, (12)\}$ is the identity which is an inner automorphism. When $n \ge 3$, the map $\sigma \mapsto \gamma_\sigma$ from $S_n$ to $\text{Aut}(S_n)$ is injective. Thus there are exactly $n!$ inner automorphisms of $S_n$. We will show that for $n \neq 6$ there are exactly $n!$ automorphisms. Hence there can be no outer automorphisms.

Let $\phi$ be an automorphism of $S_n$. The symmetric group $S_n$ is also generated by the $n-1$ transpositions $\sigma_i = (i,i+1)$ (for $i =1, \ldots, n-1$). By the previous lemma we know that if $n \neq 6$, then each $\phi(\sigma_i)$ must be a transposition $(j_i,k_i)$ for some $j_i, k_i \in \{1,\ldots, n\}$ such that $j_i \neq k_i$. As the collection $\sigma_1,\ldots,\sigma_{n-1}$ generates $S_n$, the automorphism $\phi$ is determined by the transpositions $\phi(\sigma_i)= (j_i,k_i)$. We will now show that there are exactly $n!$ choices for these transpositions.

Since $\sigma_1\sigma_{2} = (1,2,3)$, we know that $\phi(\sigma_1)\phi(\sigma_2) = (j_1,k_1)(j_2,k_2)$ must be a three cycle. Thus one of $j_2,k_2$ must be equal to one of $j_1,k_1$ and the other must be distinct. By relabeling if necessary, we will require that $k_1=j_2$ and $j_1 \neq k_2$. Thus there are $n$ choices for $j_1$, $n-1$ choices for $k_1=j_2$ and $n-2$ choices for $k_2$. This gives us $n(n-1)(n-2)$ choices for the first two transpositions. Continuing in this manner, we note that $\sigma_1\sigma_3 = (1,2)(3,4)$ and $\sigma_2\sigma_3 = (2,3,4)$, we get that one of $j_3,k_3$ is equal to one of $k_2$ and the other one is distinct from $j_1,k_1=j_2,k_2$, thus giving us $n-3$ choices for the third transposition. In general we will have that $\sigma_{i-1}\sigma_i$ is a three cycle and $\sigma_l\sigma_i$ is a pair of disjoint two cycles if $l < i-1$. Thus one can show inductively that there are $n-i$ choices for the transposition $\phi(\sigma_i)$. Hence there are $n(n-1)(n-2)\ldots 1 = n!$ choices for all the transpositions $\phi(\sigma_i)$ and hence $n!$ choices for $\phi$. Thus, if $n \neq 6$, all automorphisms of $S_n$ are inner.

## The outer automorphism of $S_6$.

The key part of showing that $S_n$ didn’t have any outer automorphisms when $n \neq 6$ was showing that any automorphism must map transpositions to transpositions. Thus to find the outer automorphism of $S_6$ we will want to find an automorphism that does not preserve cycle type. In particular we will find one which maps transpositions to the cycle type $(a,b)(c,d)(e,f)$. There are $\frac{6!}{3!2^3} = 15$ such triple transpositions which is exactly $\frac{6\cdot 5}{2}$ – the number of single transpositions.

The outer automorphism, $\psi$, will be determined by the values it takes on $\sigma_1,\sigma_2,\sigma_3,\sigma_4$ and $\sigma_5$ (where again $\sigma_i = (i,i+1)$). Each of these transpositions will get to a different triple transposition $(a,b)(c,d)(e,f)$. To extend $\psi$ to the rest of $S_6$ it suffices to check the Coxeter relations $\psi(\sigma_k)^2 = (\psi(\sigma_i)\psi(\sigma_{i+1}))^3 = (\psi(\sigma_i)\psi(\sigma_j))^2=\text{id}$ for $k = 1,2,3,4,5$, $i =1,2,3,4$ and $j \ge i+2$.

The relation $\psi(\sigma_k)^2=\text{id}$ will hold for any triple transposition $\psi(\sigma_k)$. The other relations are a bit trickier. If we have two triple transpositions $\tau = (a,b)(c,d)(e,f)$ and $\tau' = (a',b')(c',d')(e',f')$, then $\tau\tau'$ will be a product of two three cycles if none of the transpositions of $\tau$ is a transposition of $\tau'$. If $\tau$ and $\tau'$ share exactly one transposition, then $\tau\tau'$ is a product of two two cycles. If $\tau$ and $\tau'$ share two or more transpositions they must actually be equal and hence $\tau\tau'$ is the identity.

With this in mind, one can verify that following assignment extends to a valid group homomorphism from $S_6$ to $S_6$:

$\psi(\sigma_1) = (1,2)(3,4)(5,6)$,
$\psi(\sigma_2) = (1,3)(2,5)(4,6)$,
$\psi(\sigma_3) = (1,5)(2,6)(3,4)$,
$\psi(\sigma_4) = (1,3)(2,4)(5,6)$,
$\psi(\sigma_5) = (1,6)(2,5)(3,4)$.

We now just need to justify that the group homomorphism $\psi$ is an automorphism. Since $S_6$ is a finite group, $\psi$ is a bijection if and only if $\psi$ is an injection. Recall that only normal subgroups of $S_6$ are the trivial subgroup, $A_6$ and $S_6$. Thus to prove injectivity of $\psi$, it suffices to show that $\psi(\tau) \neq \text{id}$ for some $\tau \in A_6$. Indeed $\psi(\sigma_1\sigma_3) = (1,6)(2,5) \neq \text{id}$ as required. Thus we have our outer automorphism!

This isn’t the most satisfying way of defining the outer automorphism as it requires using the generators of the symmetric group. There are many more intrinsic ways to define this outer automorphism which will hopefully be the topic of a later blog post!

## References

The proof I give that for $n \neq 6$, the group $S_n$ has no outer automorphisms is based off the one given in this paper On the Completeness of Symmetric Group.

## Combing groups

Two weeks ago I gave talk titled “Two Combings of $\mathbb{Z}^2$“. The talk was about some material I have been discussing lately with my honours supervisor. The talk went well and I thought it would be worth sharing a written version of what I said.

## Geometric Group Theory

Combings are a tool that gets used in a branch of mathematics called geometric group theory. Geometric group theory is a relatively new area of mathematics and is only around 30 years old. The main idea behind geometric group theory is to use tools and ideas from geometry and low dimensional topology to study and understand groups. It turns out that some of the simplest questions one can ask about groups have interesting geometric answers. For instance, the Dehn function of a group gives a natural geometric answer to the solvability of the word problem.

## Generators

Before we can define what a combing is we’ll need to set up some notation. If $A$ is a set then we will write $A^*$ for the set of words written using elements of $A$ and inverses of elements of $A$. For instance if $A = \{a,b\}$, then $A^* = \{\varepsilon, a,b,a^{-1},b^{-1},aa=a^2,abb^{-1}a^{-3},\ldots\}$ (here $\varepsilon$ refers to the empty word). If $w$ is a word in $A^*$, we will write $l(w)$ for the length of $w$. Thus $l(\varepsilon)=0, l(a)=1, l(abb^{-1}a^{-3})=6$ and so on.

If $G$ is a group and $A$ is a subset of $G$, then we have a natural map $\pi : A^* \rightarrow G$ given by:

$\pi(a_1^{\pm 1}a_2^{\pm 1}\ldots a_n^{\pm 1}) = a_1^{\pm 1}\cdot a_2^{\pm 1} \cdot \ldots \cdot a_n^{\pm 1}$.

We will say that $A$ generates $G$ if the above map is surjective. In this case we will write $\overline{w}$ for $\pi(w)$ when $w$ is a word in $A^*$.

## The Word Metric

The geometry in “geometric group theory” often arises when studying how a group acts on different geometric spaces. A group always acts on itself by left multiplication. The following definition adds a geometric aspect to this action. If $G$ is a group with generators $A$, then the word metric on $G$ with respect to $A$ is the function $d_G : G \times G \rightarrow \mathbb{R}$ given by

$d_G(g,h) = \min \{ l(w) \mid w \in A^*, \overline{w} = g^{-1}h \}.$

That, is the distance between two group elements $g,h \in G$ is the length of the shortest word in $A^*$ we can use to represent $g^{-1}h$. Equivalently the distance between $g$ and $h$ is the length of the shortest word we have to append to $g$ to produce $h$. This metric is invariant under left-multiplication by $G$ (ie $d_G(g\cdot h,g\cdot h') =d_G(h,h')$ for all $g,h,h' \in G$). Thus $G$ acts on $(G,d_G)$ by isometries.

## Words are Paths

Now that we are viewing the group $G$ as a geometric space, we can also change how we think of words $w \in A^*$. Such a word can be thought of as discrete path in $G$. That is we can think of $w$ as a function from $\mathbb{N}$ to $G$. This way of thinking of $w$ as a discrete path is best illuminated with an example. Suppose we have the word $w = ab^2a^{-1}b$, then

$w(0) = e,$
$w(1) = \overline{a}$
$w(2) = \overline{ab}$
$w(3) = \overline{ab^2}$
$w(4) = \overline{ab^2a^{-1}}$
$w(5) = \overline{ab^2a^{-1}b}$
$w(t) = \overline{ab^2a^{-1}b},$ $t \ge 5$.

Thus the path $w : \mathbb{N} \rightarrow G$ is given by taking the first $t$ letters of $w$ and mapping this word to the group element it represents. With this interpretation of word in $A^*$ in mind we can now define combings.

## Combings

Let $G$ be a group with a finite set of generators $A$. Then a combing of $G$ with respect to $A$ is a function $\sigma : G \rightarrow A^*$ such that

1. For all $g \in G$, $\overline{\sigma_g} = g$ (we will write $\sigma_g$ for $\sigma(g))$.
2. There exists $k >0$ such that for all $g, h \in G$ with $g \cdot \overline{a} = h$ for some $a \in A$, we have that $d_G(\sigma_g(t),\sigma_h(t)) \le k$ for all $t \in \mathbb{N}$.

The first condition says that we can think of $\sigma$ as a way of picking a normal form $\sigma_g \in A^*$ for each $g \in G$. The second condition is a bit more involved. It states that if the group elements $g, h \in G$ are distance 1 from each other in the word metric, then the paths $\sigma_g,\sigma_h : \mathbb{N} \rightarrow G$ are within distance $k$ of each other at any point in time.

## An Example

Not all groups can be given a combing. Indeed if we have a combing of $G$, then the word problem in $G$ is solvable and the Dehn function of $G$ is at most exponential. One group that does admit a combing is $\mathbb{Z}^2 = \{(m,n) \mid m,n \in \mathbb{Z}\}$. This group is generated by $A = \{(1,0),(0,1)\} = \{\beta,\gamma\}$ and one combing of $\mathbb{Z}^2$ with respect to this generating set is

$\sigma_{(m,n)} = \beta^m\gamma^n$.

The first condition of being a combing is clearly satisfied and the following picture shows that the second condition can be satisfied with $k = 2$.

## A Non-Example

The discrete Heisenberg group, $H_3$, can be given by the following presentation

$H_3 = \langle \alpha,\beta,\gamma \mid \alpha\gamma = \gamma\alpha, \beta\gamma = \gamma\beta, \beta\alpha = \alpha\beta\gamma \rangle$.

That is, the group $H_3$ has three generators $\alpha,\beta$ and $\gamma$. The generator $\gamma$ commutes with both $\alpha$ and $\beta$. The generators $\alpha$ and $\beta$ almost commute but don’t quite as seen in the relation $\beta\alpha = \alpha\beta\gamma$.

Any $h \in H_3$ can be represented uniquely as $\sigma_h = \alpha^p\beta^m\gamma^n$ for $p,m,n \in \mathbb{Z}$. To see why such a representation exists it’s best to consider an example. Suppose that $h = \overline{\gamma\beta\alpha\gamma\alpha\beta\gamma}$. Then we can use the fact that $\gamma$ commutes with $\alpha$ and $\beta$ to push all $\gamma$‘s to the right and we get that $h = \overline{\beta\alpha\alpha\beta\gamma^3}$. We can then apply the third relation to switch the order of $\alpha$ and $\beta$ on the right. This gives us that that $h = \overline{\alpha\beta\gamma\alpha\beta\gamma^3}=\overline{\alpha\beta\alpha\beta\gamma^4}$. If we apply this relation once more we get that $h = \overline{\alpha^2\beta^2\gamma^5}$ and thus $\sigma_h = \alpha^2\beta^2\gamma^5$. The procedure used to write $h$ in the form $\alpha^p\beta^m\gamma^n$ can be generalized to any word written using $\alpha^{\pm 1}, \beta^{\pm 1}, \gamma^{\pm 1}$.

The fact the such a representation is unique (that is if $\overline{\alpha^p\beta^m\gamma^n} = \overline{\alpha^{p'}\beta^{m'}\gamma^{n'}}$, then $(p,m,n) = (p',m',n')$) is harder to justify but can be proved by defining an action of $H_3$ on $\mathbb{Z}^3$. Thus we can define a function $\sigma : H_3 \rightarrow \{\alpha,\beta,\gamma\}^*$ by setting $\sigma_h$ to be the unique word of the form $\alpha^p \beta^m\gamma^n$ that represents $h$. This map satisfies the first condition of being a combing and has many nice properties. These include that it is easy to check whether or not a word in $\{\alpha,\beta,\gamma\}^*$ is equal to $\sigma_h$ for some $h \in H_3$ and there are fast algorithms for putting a word in $\{\alpha,\beta,\gamma\}^*$ into its normal form. Unfortunately this map fails to be a combing.

The reason why $\sigma : H_3 \rightarrow \{\alpha,\beta,\gamma\}^*$ fails to be a combing can be seen back when we turned $\overline{ \gamma\beta\alpha\gamma\alpha\beta\gamma }$ into $\overline{\alpha^2\beta^2\gamma^5}$. To move $\alpha$‘s on the right to the left we had to move past $\beta$‘s and produce $\gamma$‘s in the process. More concretely fix $m \in \mathbb{N}$ and let $h = \overline{\beta^m}$ and $g = \overline{\beta^m \alpha} = \overline{\alpha \beta^m\gamma^m}$. We have $\sigma_h = \beta^m$ and $\sigma_g = \alpha \beta^m \gamma^m$. The group elements $g$ and $h$ differ by a generator. Thus, if $\sigma$ was a combing we should be able to uniformly bound $d_{H_3}(\sigma_g(t),\sigma_h(t))$ for all $t \in \mathbb{N}$ and all $m \in \mathbb{N}$.

If we then let $t = m+1$, we can recall that

$d_{H_3}(\sigma_g(t),\sigma_h(t)) = \min\{l(w) \mid w \in \{\alpha,\beta,\gamma\}^*, \overline{w} = \sigma_g(t)^{-1}\sigma_h(t)\}.$

We have that $\sigma_h(t) = \overline{\beta^m}$ and $\sigma_g(t) = \overline{\alpha \beta^m}$ and thus

$\sigma_g(t)^{-1}\sigma_h(t) = (\overline{\alpha\beta^m})^{-1}\overline{\beta^m} = \overline{\beta^{-m}\alpha^{-1}\beta^m} = \overline{\alpha^{-1}\beta^{-m}\gamma^{m}\beta^m}=\overline{\alpha^{-1}\gamma^{m}}.$

The group element $\overline{\alpha^{-1}\gamma^{m}}$ cannot be represented as a shorter element of $\{\alpha,\beta,\gamma\}^*$ and thus $d_{H_3}(\sigma_g(t),\sigma_h(t)) = m+1$ and the map $\sigma$ is not a combing.

## Can we comb the Heisenberg group?

This leaves us with a question, can we comb the group $H_3$? It turns out that we can but the answer actually lies in finding a better combing of $\mathbb{Z}^2$. This is because $H_3$ contains the subgroup $\mathbb{Z}^2 \cong \langle \beta, \gamma \rangle \subseteq H_3$. Rather than using the normal form $\sigma_h = \alpha^p \beta^m \gamma^n$, we will use $\sigma'_h = \alpha^p \tau_{(m,n)}$ where $\tau : \mathbb{Z}^2 \rightarrow \{\beta,\gamma \}^*$ is a combing of $\mathbb{Z}^2$ that is more symmetric. The word $\tau_{(m,n)}$ is defined to be the sequence of $m$ $\beta$‘s and $n$ $\gamma$‘s that stays closest to the straight line in $\mathbb{R}^2$ that joins $(0,0)$ to $(n,p)$ (when we view $\beta$ and $\gamma$ as representing $(1,0)$ and $(0,1)$ respectively). Below is an illustration:

This new function isn’t quite a combing of $H_3$ but it is the next best thing! It is an asynchronous combing. An asynchronous combing is one where we again require that the paths $\sigma_h,\sigma_g$ stay close to each other whenever $h$ and $g$ are close to each other. However we allow the paths $\sigma_h$ and $\sigma_g$ to travel at different speeds. Many of the results that can be proved for combable groups extend to asynchronously combable groups.

## References

Hairdressing in Groups by Sarah Rees is a survey paper that includes lots examples of groups that do or do not admit combings. It also talks about the language complexity of a combing, something I didn’t have time to touch on in my talk.

Combings of Semidirect Products and 3-Manifold Groups by Martin Bridson contains a proof that $H_3$ is asynchornously combable. He actually proves the more general result that any group of the form $\mathbb{Z}^n \rtimes \mathbb{Z}^m$ is asynchronously combable.

Thank you to my supervisor, Tony Licata, for suggesting I give my talk on combing $H_3$ and for all the support he has given me so far.