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The generalized Lax conjecture was a central topic of study in the field of hyperbolic polynomials. Hyperbolic polynomials in general serve as a bridge between the seemingly disparate fields of algebraic geometry and convex optimization, as each hyperbolic polynomial gives rise in a natural way to a convex cone. The generalized Lax conjecture in particular gave a characterization of the possible convex cones that could arise in this way. The conjecture was disproven in October of 2026 by an internal model from OpenAI, which provided an explicit counterexample to the stated characterization. In a sense, the conjecture being false is the more interesting possibility, as it suggests that there is more richness to the theory of hyperbolic polynomials than previously thought. This post is an attempt to understand this counterexample and its generalizations.

A homogeneous polynomial $p \in \R[x_1, \dots, x_n]$ is said to be hyperbolic with respect to $v \in \R^n$ if $p(v) \neq 0$ and for each $x \in \R^n$, the univariate polynomial $p(x-tv)$ has only real roots. The hyperbolicity cone associated to such a polynomial is then

The prototypical example of a hyperbolic polynomial is the determinant of an $n\times n$ symmetric matrix, which is hyperbolic with respect to the identity matrix due to the spectral theorem, and whose hyperbolicity cone is the set of positive semidefinite matrices. The generalized Lax conjecture essentially states that every hyperbolicity cone can be expressed as the intersection between a positive semidefinite cone and a linear subspace [1]. Such a cone is said to be spectrahedral [3].

There are two steps to disproving the generalized Lax conjecture: firstly, one must find a polynomial $p$ and show that it is hyperbolic, and secondly one must show that $\Lambda_p$ is not spectrahedral. Both of these steps are nontrivial. While showing that a polynomial is hyperbolic is not always easy, there have been many interesting constructions of hyperbolic polynomials in the literature [2]. None of these constructions had been shown to have nonspectrahedral hyperbolicity cones in the past (though in light of the disproof of the generalized Lax conjecture, one may wish to revisit some of these earlier constructions to see if they also are counterexamples to the conjecture). So one suspects that showing that the hyperbolicity cone is not spectrahedral will be the hard part. The proof given by OpenAI actually reduces the issue of showing that the cone to the well known fact that there exist positive operators which are not completely positive, a distinction that has been studied in the context of quantum mechanics for many years [4].

A positive operator is a linear map $\Phi :\S^n \rightarrow \S^m$ that sends positive semidefinite matrices to positive semidefinite matrices, and a completely positive such operator is one that takes the form of a `Kraus operator’, which is to say it is of the form

where $A : \R^m \rightarrow \R^n \times \R^k$ for some $k$.

It is noteworthy that there is an equivalence between such linear operators and biquadratic forms, meaning polynomials that are separately homogeneous of degree 2 in two separate blocks of variables, so that a positive operator corresponds precisely a nonnegative biquadratic form, and a completely positive one corresponds to a sum-of-squares biquadratic form. In this way, the distinction between positive operators and completely positive operators is related to the distinction between sums-of-squares polynomials and nonnegative polynomials.

The most general construction of a hyperbolicity cone that can be derived from the OpenAI paper shows that for any nonnegative triquadratic form, which can be associated to a quadratic family of positive maps, there is an associated hyperbolicity cone. The OpenAI paper further shows that if this triquadratic form is constructed from a extremal nonnegative biquadratic form that is not sums-of-squares, then the resulting hyperbolicity cone is not spectrahedral.

Another paper in the same OpenAI release shows that there is a hyperbolic polynomial whose hyperbolicity cone is not even a spectrahedral shadow. We will not discuss this example here, except to note that the relevant hyperbolicity cone that serves as a counterexample to that construction also arises from the same construction applied to a certain nonnegative triquadratic form. Also, the OpenAI paper that produced the counter example to the generalized Lax conjecture applied some additional ideas not mentioned here that reduce the number of variables and the size of the counter example.

This document is constructed so that the proofs of each major result immediately follows the statement, but it is hidden underneath a dropdown menu. This is meant to give the reader the ability to go over the argument at a high level, and then pass through the details afterwards.

The Construction: Generalized Schur Complements and Positive Operators

The hyperbolicity cones that are ultimately shown to be nonspectrahedral arise from a natural and fairly general construction. To motivate the construction, we will recall the Schur complement characterization of the positive semidefinite cone:

We will note a few features of the map $(X,Y) \mapsto Y^{\T}X^{-1}Y$:

1. It is linear in $X^{-1}$.
2. It is quadratic in $Y$.
3. For any $Y$ and any $X \succeq 0$, $Y^{\T}X^{-1}Y \succeq 0$.

We will say that a linear map $\Phi : \S^n \rightarrow \S^m$ is positive if for every $X \succeq 0$, $\Phi(X) \succeq 0$. We will then define a quadratic family of positive maps (QFPM) to be a polynomial function $\Phi : \S^n \times \R^k \rightarrow \S^m$ so that $\Phi(X,Y)$ is linear in $X$, quadratic in $Y$, and for every $Y$, the linear map $\Phi(\cdot, Y)$ is positive.

We then come to the construction of the relevant hyperbolicity cone.

Theorem 1

Let $\Phi(X,Y)$ be a QFPM. Then the following is a hyperbolicity cone,

$$ H_{\Phi} = \overline{\{(X,Y,Z) \in \S^m \times \S^n \times \R^k : X \succ 0,\ Z \succ \Phi(X^{-1},Y)\}}, $$

for the polynomial

$$ p_{\Phi}(X,Y,Z) = \det\big(\det(X)Z - \Phi(\adj(X), Y)\big), $$

which is hyperbolic with respect to $(I,0,I)$. Here $\adj(X) = \det(X) X^{-1}$ for invertible $X$ is the adjugate of $X$.

Remark

This definition is remarkably close to that of a Siegel cone for the PSD cone, which amounts to the special case of this construction when $n = 1$. Siegel cones are of interest because every homogeneous cone can be constructed by applying the Siegel cone operation repeatedly a one dimensional cone [5].

Spectrahedral Cones and Completely Positive Maps

There are in fact many hyperbolicity cones that result from this construction. We now want to find one such cone which is not spectrahedral. Or, to put it another way, we want to find necessary conditions that $\Phi$ needs to satisfy in order for this cone to be spectrahedral, and then show that some $\Phi$ yields a hyperbolicity cone that does not satisfy those conditions.

One sufficient condition for the cone to be spectrahedral is for $\Phi$ to be completely positive in the following sense: there exist some positive operator $E : \S^n \rightarrow \S^{n'}$ and linear map $A : \R^k \rightarrow \R^{n' \times m}$ so that

in which case

The first thing that the OpenAI paper [8] did to show that there is a nonspectrahedral hyperbolicity cone is to show a partial converse to this result.

Recall that a positive operator is unital if $\Phi(I) = I$.

Theorem 2

Let $\Phi(X,Y)$ be a QFPM, and suppose that $H_{\Phi}$ is spectrahedral. Then there exist unital positive operators $E : \S^n \rightarrow \S^{n'}$ and $G: \S^m \rightarrow \S^{m'}$, together with a linear map $F : \R^k \rightarrow \R^{n' \times m'}$ so that

$$ H_{\Phi} = \left\{(X,Y,Z) \in \S^n \times \R^k \times \S^m : \begin{pmatrix} E(X) & F(Y) \\ F(Y)^{\T} & G(Z) \end{pmatrix} \succeq 0\right\}. $$

Thus, if $H_{\Phi}$ is spectrahedral, then it must look like the construction applied to a completely positive $\Phi$. In particular, this suggests that if a $\Phi$ which is not completely positive is passed through this construction, then $H_{\Phi}$ will not be spectrahedral. While we will not show that this is the case, we will show that if $\Phi$ is a positive map that is not completely positive and also `extremal’ in a certain sense, then $H_{\Phi}$ is not spectrahedral.

It may not seem obvious that there exists such a positive QFPM that is not completely positive. However, there is a fairly general class of constructions that will supply us with these QFPMs. For this, it is helpful to pass to an equivalent characterization of QFPMs: nonnegative triquadratic forms.

QFPMs and Nonnegative Triquadratic Forms

Preliminaries on Positive Operators and Nonnegative Biquadratic Forms

Associated to a linear map $\Phi : \S^n \rightarrow \S^m$ is the following biquadratic form $B : \R^m \times \R^n \rightarrow \R$:

By biquadratic, we mean that $B(\lambda u, \mu v) = \lambda^2 \mu^2 B(u,v)$ for any $\lambda, \mu \in \R$ and $u,v \in \R^n \times \R^m$. In fact, by dimension counting, we can see that for any biquadratic form $B$, there is some $\Phi$ so that $B(u,v) = u^{\T}\Phi(vv^{\T})u$. Thus, a biquadratic form is equivalent to a linear map $\Phi : \S^n \rightarrow \S^m$.

One reason to study $B(u,v)$ is that it is pointwise nonnegative if and only if $\Phi$ is a positive operator. This follows directly from the fact that the extreme rays of $\S^m$ are spanned by rank $1$ matrices.

One sufficient condition for $B(u,v)$ to be nonnegative is that it is a sum-of-squares of lower degree polynomials (which will necessarily be bilinear forms). A linear map associated to a sums-of-squares biquadratic form is said to be completely positive. Note that $\Phi$ being completely positive is equivalent to the statement that the map $\Phi$ is of the form $A^{\T}(I\otimes X)A$ for some $A$.

An equivalent characterization is that a linear map $\Phi$ is completely positive if and only if for every $k$, the linear map $I_k \otimes \Phi$ is positive. Such maps have been studied intently in quantum mechanics because they correspond to quantum communication channels that are respect entanglement relations, while more general positive maps do not. In particular, it is known that there are positive maps that are not completely positive, or in otherwise nonnegative biquadratic forms that are not sums-of-squares whenever $n,m > 2$.

An important example is the Choi–Lam form [6], defined as

This is an example of a nonnegative biquadratic form that is not a sum of squares. It is worth noting that this particular nonnegative biquadratic form is extreme, meaning that it cannot be written in a nontrivial way as a sum of other nonnegative biquadratic forms [7]. This is the key fact that the ultimate nonspectrahedrality argument relies on. However, it is also worth noting that any extremal nonnegative biquadratic form will suffice for this argument. Indeed, it is possible to deduce the existence of an extremal nonnegative biquadratic form that is not a sums-of-squares just from the fact that there exists any biquadratic form which is nonnegative but not sums-of-squares and general convex geometry.

Triquadratic forms and QFPMs

In analogy with the above relationship between linear maps between symmetric matrices and biquadratic forms, we can associate to each QFPM a triquadratic polynomial

Again, there is a bijective relationship between triquadratic forms and quadratic families of linear maps between spaces of symmetric matrices. This polynomial is pointwise nonnegative exactly when $\Phi$ is a QFPM, and $\Phi$ is completely positive if and only if $T$ is a sum of squares.

There is a way to turn a general biquadratic form into a triquadratic one that preserves nonnegativity and the property of being a sum of squares. Given a biquadratic form $B : \R^{\ell} \times \R^k \rightarrow \R$, let $f : \R^n \times \R^m \rightarrow \R^{\ell}$ be a bilinear map. Then $B(f(u,v), y)$ is a nonnegative triquadratic form.

For the purposes of the construction below, we will consider the case in which $n = m$ and the bilinear $f(u,v) = u_0 v' - v_0 u'$, where $u = (u_0, u')$ and $v = (v_0, v')$. In particular, there is a QFPM $\Phi$ that is associated to the triquadratic form obtained by composing the Choi-Lam form with this bilinear map, and this is precisely the $\Phi$ that the OpenAI paper shows leads to a nonspectrahedral hyperbolicity cone.

Note that more general bilinear maps $f$ and indeed triquadratic maps $T$ that do not arise in this way could be used to obtain more general classes of nonspectrahedral hyperbolicity cones, but we will focus on these cases for simplicity.

A Nonspectrahedral Hyperbolicity Cone

Theorem 3

Let $B : \R^{m-1} \times \R^{k}\rightarrow \R$ be an extremal nonnegative biquadratic form that is not a sum of squares. Let $u = \begin{pmatrix}u_0\\u'\end{pmatrix}$ and $v = \begin{pmatrix}v_0\\v'\end{pmatrix}$, and let $\Phi$ be the QFPM associated to the nonnegative triquadratic form

$$ T(u,v,Y) = B(u_0 v' - v_0 u', Y). $$

Then the convex cone $H_{\Phi}$ is not spectrahedral.

Here are two proofs that I only recently thought of as being related to each other (despite the fact that they are both named after Issai Schur). They are both about proving that a certain matrix is positive semidefinite (PSD), meaning that they have nonnegative eigenvalues, and we will prove them by finding that matrix inside a much larger PSD matrix.

It is actually not too hard to show that if you take a principal submatrix of a PSD matrix, that will also be PSD. But we can make use of a slightly stronger theorem to strengthen our results.

Theorem (Cauchy Interlacing Theorem)

Let $X$ be an $n \times n$ symmetric matrix, and let $Y$ be a $k\times k$ principal submatrix of $X$ (that is, we obtain $Y$ by deleting $n-k$ rows of $X$ and their corresponding columns). Let $\lambda_i(\cdot)$ denote the $i^{th}$ smallest eigenvalue function. Then, for each $i = 1,\dots,k$, \begin{equation*} \lambda_{i}(X) \le \lambda_i(Y) \le \lambda_{i+n-k}(X). \end{equation*}

Disclaimer: There are almost certainly better proofs of both of these results, but I think highlighting their similarities is interesting enough for us. It will also allow us to prove what I think is a previously unknown fact about Schur complements.

The Schur Product Theorem

We all know how to multiply square matrices. If $X$ and $Y$ are two matrices, then we just multiply them entriwise, so that each entry of $X \cdot Y$ is the product of the corresponding entries of $X$ and $Y$.

This style of multiplication, called the Hadamard product, is sort of weird, but what is weirder is that it actually interacts well with linear algebra, which usually use a different kind of multiplication.

Theorem (Schur Product Theorem)

Let $X$ and $Y$ be PSD matrices, then the Hadamard product, $X \cdot Y$ is also PSD.

How can we go about proving this? The difficulty is that the Hadamard product seems to depend a lot on the basis of the underlying vector space, whereas the eigenvalues of a matrix are naturally basis independent. What we might try to do is to relate the basis dependent $X \cdot Y$ with a naturally basis independent thing.

To do this, we will need the notion of a tensor product. There are actually two related notions here: the tensor product of vector spaces, and of matrices. The tensor product of vector spaces $V$ and $W$ is the vector space spanned by simple tensors of the form $v \otimes w$, where $v \in V$ and $w \in W$, together with some natural relations that make the tensor product behave like the usual product.

The tensor product of matrices $X$ and $Y$, $X \otimes Y$, is given as follows: it is an $n^2 \times n^2$ matrix where, for a tensor product vector $v \otimes w$, $(X \otimes Y)(v\otimes w) = (Xv)\otimes(Yw)$.

If $v$ is a vector satisfying $Xv = \lambda v$, and $w$ is a vector satisfying $Yw = \mu w$, then $v \otimes w$ satisfies $(X\otimes Y)(v \otimes w) = \lambda\mu v\otimes w$. So, we can see that the eigenvalues of $X \otimes Y$ are all of the form $\lambda \mu$, where $\lambda$ and $\mu$ are eigenvalues of $X$ and $Y$ respectively.

In particular, it is easy to see that if $X$ and $Y$ are PSD, then $X \otimes Y$ is also PSD. This proof was relatively straightforward, since the object that we are looking at $X \otimes Y$ is basis invariant.

Now we relate $X \otimes Y$ to $X \cdot Y$ by looking at the entries of $X \otimes Y$ in the natural basis consisting of vectors of the form $e_i \otimes e_j$ for basis vectors $e_i$ and $e_j$. We can find the entries of this matrix using the natural inner product of $V \otimes W$, given by $\langle v_1 \otimes w_1, v_2 \otimes w_2\rangle = \langle v_1, w_1 \rangle \langle v_2, w_2\rangle$, in which case, \begin{equation*} (e_i \otimes e_j)^{\intercal} (X \otimes Y) (e_{\ell}\otimes e_k) = X_{i,\ell}Y_{j,k}, \end{equation*} That is, the entries of $X \otimes Y$ in this basis are really just pairwise products of entries of entries of $X$ and $Y$, which sounds suspiciously like the definition of the Hadamard product. Indeed, let take the principal submatrix of $X \otimes Y$ that you get from just looking at basis elements of the form $e_i \otimes e_i$, in which case, you get that \begin{equation*} (e_i \otimes e_i)^{\intercal} (X \otimes Y) (e_j\otimes e_j) = X_{i,j}Y_{i,j} = (X \cdot Y)_{i,j}. \end{equation*} That is, if you delete all of the rows and columns of $X\otimes Y$ that are indexed by basis elements of the form $e_i \otimes e_j$ for $i \neq j$, then you get $X \cdot Y$. So, $X \cdot Y$ is actually a principal submatrix of $X \otimes Y$!

In particular, since $X \otimes Y$ is ‘obviously PSD’, that means that $X \cdot Y$ is PSD! We can even get bounds on the eigenvalues of $X \cdot Y$ in terms of the eigenvalues of $X$ and $Y$. Let $v(X,Y)$ denote the $n^2$ long vector that you get from taking all of the pairwise products of eigenvalues of $X$ and $Y$, and sorting them from smallest to largest. Then, by Cauchy interlacing,

\begin{equation*} v(X,Y)_i \le \lambda_i(X \cdot Y) \le v(X,Y)_{i + n^2 - n}. \end{equation*}

Schur Complements

I have heard it said that the difference between a pure mathematician and an applied mathematician is that an applied mathematician knows what a Schur complement is (and likes it). When I was an undergrad studying mathematics, I had vaguely heard of a Schur complement, and dismissed it as an ugly trick. Now, I suppose I have turned over a new leaf.

Let $X$ be an $n\times n $matrix, and let $T \subseteq [n]$ index a subset of the columns of $X$. Also, assume that $X_{T,T}$ is invertible. Then \begin{equation*} X \setminus T = X_{T^c,T^c} - X_{T^c, T} X_{T,T}^{-1} X_{T,T^c}. \end{equation*} Here, $X_{S,T}$ is the submatrix of $X$ whose columns are indexed by $S$ and whose rows are indexed by $T$.

One nice thing about Schur complements is that they interact nicely with other Schur complements, and also the operations of taking submatrices.

Precisely, let $S$ and $U$ be disjoint from $T$, then \begin{equation*} (X \setminus S)\setminus T = X\setminus (S \cup T), \end{equation*} and \begin{equation*} (X_{S\cup T,U\cup T})\setminus T = (X\setminus T)_{S,U}. \end{equation*} (Yes, the notation is quite terrible, but again, this is applied math, so that is perhaps to be expected.)

We also have the determinant formula for Schur complements, which we will use without proof.

Theorem (Determinant Formula)

$\det(X) = \det(X_{T,T})\det(X \setminus T)$.

Using the above properties, it can be seen that the entries of the Schur complement satisfy

\begin{equation*} (X \setminus T)_{i,j} = \frac{1}{\det(X_{T,T})}\det(X_{T+i, T+j}). \end{equation*}

We should expect that the following theorem hold then:

Theorem (Schur Complement Theorem)

Let $X$ be a PSD matrix, then $X \setminus T$ is PSD for any $T \subseteq [n]$.

This is often stated as a consequence of a more powerful fact, known as Haynesworth inertia additivity, named after Emilie Virginia Haynsworth, who I think should be a better known mathematician. Let’s give a proof more in line with our previous proof.

To do this, we will make use an object called the exterior power of a vector space, which is related to the tensor product, but slightly more complicated.

Given a vector space $V$, $\bigwedge^{(k)}V$ is a vector space spanned by elements of the form \begin{equation*} v_1 \wedge v_2 \dots \wedge v_k, \end{equation*} where $v_i \in V$ for each $i$. This exterior power also satisfies two relations:

• Multilinearity \begin{equation*} (v_1+\lambda w) \wedge v_2 \dots \wedge v_k = v_1 \wedge v_2 \dots \wedge v_k + \lambda (w \wedge v_2 \dots \wedge v_k) \end{equation*}

• Alternativity For any $i \in [k]$, \begin{equation*} v_1 \wedge v_2 \dots \wedge v_i \wedge v_{i+1} \dots \wedge v_k = - v_1 \wedge v_2 \dots \wedge v_{i+1} \wedge v_{i} \dots \wedge v_k \end{equation*}

These two properties make the exterior power related to the determinant of a matrix. To see this, we can define the wedge power of a matrix in a similar way to the tensor power: \begin{equation*} (\bigwedge^{(k)}X)(v_1 \wedge v_2 \dots \wedge v_k) = Xv_1 \wedge Xv_2 \dots \wedge Xv_k. \end{equation*}

In this case, it can be seent that $\bigwedge^{(n)} V$ is a 1-dimensional vector space spanned by $e_1\wedge e_2\dots \wedge e_n$, and that \begin{equation*} (\bigwedge^{(n)}X)(e_1 \wedge \dots \wedge e_n) = \det(X) e_1 \wedge \dots \wedge e_n. \end{equation*}

Generally, let $S \subseteq [n]$ be of size $k$, and let \begin{equation*} e_S = e_{s_1} \wedge e_{s_2} \wedge \dots e_{s_k}. \end{equation*} Then $\bigwedge^{(k)} V$ is an $\binom{n}{k}$ dimensional vector space with $\{e_S\}$ as a basis.

There is also a natural inner product on wedge powers given by \begin{equation*} \langle v_1 \wedge v_2 \dots \wedge v_k, w_1 \wedge w_2 \dots \wedge w_k \rangle = \det \begin{bmatrix} \langle v_1, w_1 \rangle & \langle v_1, w_2 \rangle & \dots & \langle v_1, w_k\rangle\\ \langle v_2, w_1 \rangle & \langle v_2, w_2 \rangle & \dots & \langle v_2, w_k\rangle\\ &&\dots&\\ \langle v_k, w_1 \rangle & \langle v_k, w_2 \rangle & \dots & \langle v_k, w_k\rangle \end{bmatrix}. \end{equation*} The fact that this is an inner product is implied by the Cauchy-Binet Theorem.

Now, we can perform the same trick as before: let’s look at the eigenvalues of this matrix, and also look at its entries.

If $v_1, \dots, v_k$ are eigenvectors of $X$, with eigenvalues $\lambda_1, \dots, \lambda_k$, then \begin{equation*} (\bigwedge^{(k)}X)(v_1 \wedge v_2 \dots \wedge v_k) = Xv_1 \wedge Xv_2 \dots \wedge Xv_k = \prod_{i=1}^k \lambda_i (v_1 \wedge \dots \wedge v_k). \end{equation*} So, it is clear that the eigenvalues of $\bigwedge^{(k)}X$ are precisely $\prod_{i=1}^k \lambda_i$, where $\lambda_i$ are distinct eigenvalues of $X$. In particular, if $X$ is PSD, then so is $\bigwedge^{(k)}X$.

On the other hand, we can look at the entries of $\bigwedge^{(k)}X$ as follows: let $S = \{i_1, \dots, i_k\}$, and let $U = \{j_1, \dots, j_k\}$, then \begin{equation*} (e_{i_1} \wedge e_{i_2} \dots \wedge e_{i_k})^{\intercal} (X \otimes Y) (e_{j_1} \wedge e_{j_2} \dots \wedge e_{j_k}) = \det(X_{S,U}). \end{equation*}

Now, we can connect this back to Schur complements: let $k = |T| + 1$, consider the submatrix of $\wedge^{(k)}X$ indexed by basis vectors of the form $e_{t_1} \wedge e_{t_2} \wedge e_{t_{k-1}} \wedge e_{j}$, for some $j \not \in T$, then then from our earlier properties,

\begin{equation*} (X \setminus T)_{j,k} = \frac{1}{\det(X_{T,T})}\det(X_{T+j,T+k}) = \frac{1}{\det(X_{T,T})} (\bigwedge^{(k)} X)_{T+i, T+k}. \end{equation*} Thus, we have discovered $(X \setminus T)$ as a principal submatrix of $\frac{1}{\det(X_{T,T})}\bigwedge^{(k)} X$, and so is PSD.

In fact, we have a slightly stronger fact, from the Cauchy interlacing theorem. To see this, note that for $1\le i \le n-|T|$, the $i^{th}$ smallest eigenvalue of $\bigwedge^{(|T|+1)}X$ is \begin{equation*} \lambda_{i}(\bigwedge^{(|T|+1)}X) = (\prod_{j=1}^{|T|}\lambda_j(X)) \lambda_{i+|T|+1}(X), \end{equation*} \begin{equation*} \frac{\prod_{j=1}^{|T|}\lambda_j(X)}{\det(X_{T,T})}\lambda_{i+|T|+1}(X) \le \lambda_i(X\setminus T). \end{equation*} Note that Cauchy interlacing in this case also implies that $\frac{\prod_{j=1}^{|T|}\lambda_j(X)}{\det(X_{T,T})} \le 1$, so this lower bound is somewhat smaller than $\lambda_{i+|T|+1}(X)$.

The upper bound from the Cauchy interlacing inequality is not very useful here, since it is obvious from the definition that \begin{equation*} \lambda_i(X\setminus T) \le \lambda_{i+|T|+1}(X), \end{equation*} which is already smaller than the upper bound given by the Cauchy interlacing inequality.

and the polynomial moments:

It is a general fact that these generate the ring of symmetric functions as an algebra, which is a fancy way of saying that if $p(x_1, \dots, x_n)$ is a polynomial which does not change if I permute the variables, then there is some other polynomials $q(y_1, \dots, y_d)$ and $r(y_1, \dots, y_d)$, so that

Here, I have suppressed the dependence of the $e_i$ and $\mu_i$ on $x_1, \dots, x_n$.

In particular, since the elementary symmetric polynomials and the polynomial moments are both symmetric, we see that there should be some polynomial relations between these polynomials. The Newton identities give such a relation recursively. Specifically,

and so on.

Even for fixed $n$, this can be extended to an infinite sequence of identities by simply letting $x_{n+1} = x_{n+2} = \dots = 0$, and noticing that this implies that for $k > n$, $e_k(x_1, \dots) = 0$.

I wanted to give a proof of this identity using generating functions, because I think it is slightly cooler than the version given on Wikipedia.

First, we define the moment generating function of $x_1, \dots, x_n$ to be

This is a power series in $t$, and we can simplify it just using the definition of $\mu_k$:

Let’s simplify this last term by giving it a common denominator:

Now, notice that

and we also have that

If we let $p(t) = \sum_{i=1}^n (-1)^k e_k(x_1, \dots, x_n)t^k$, we obtain the equation

If we expand out this product and equate terms of the same degree, we obtain the Newton identities!

There are a lot of places in algebraic geometry where convexity appears in surprising places. One well known result coming from quadratic programming is a theorem of Brickman.

Theorem (Brickman)

Let $n \ge 2$. Let $Q_1, Q_2$ be homogeneous quadratic maps on $\mathbb{R}^{n+1}$, and let $Q : \mathbb{R}^{n+1} \rightarrow \mathbb{R}^2$ be given by $Q(x) = (Q_1(x),Q_2(x))$. Let $S^{n} = \\{x \in \mathbb{R}^{n+1} : \|x\|_2 = 1\\}$. Then $Q(S^{n}) \subseteq \mathbb{R}^2$ is convex.

There are a few well known proofs of this theorem (see for example, Barvinok’s A Course on Convexity), which tend to involve passing to some associated convex set of positive semidefinite matrices, and then showing that the image of that convex set under a linear projection agrees with the image of the sphere under these quadratic maps.

Our goal is to give a short, geometric proof of this fact, which uses only some elementary observations.

Proof

We base our proof on two facts:

• The image of the unit circle $S^1 = \\{(x,y) : x^2+y^2 = 1\\}$ under a homogeneous quadratic map $Q : \mathbb{R}^2 \rightarrow \mathbb{R}^n$ is also the image of the unit circle under an affine map. We will call the image of $S^1$ under an affine map an ellipse. We prove this below the fold.

  Click to expand

      This surprising geometric fact is the main reason the result holds.
      To see why it is the case, notice that each point on the circle can be written as $(\cos(t), \sin(t))$, and that any homogeneous quadratic function $a\cos(t)^2 + b \cos(t) \sin(t) + c\sin(t)^2$ can be expressed as a linear combination of the functions $\cos(2t)$, $\sin(2t)$ and 1.
  
      This implies that if we apply a quadratic function to the unit circle, the image can be parameterized as $v_1 \cos(2t) + v_2 \sin(2t) + v_3$ for some $v_1, v_2, v_3 \in \mathbb{R}^n$, and it is therefore the image of $S^1$ under the affine map $(x,y) \mapsto v_1 x + v_2 y + v_3$.
  

  $\Box$

• If $X$ is a simply connected topological space (for example $S^{n}$ for $n \ge 2$), $\gamma$ is a closed curve in $X$, and $f : X \rightarrow Y$ is a continuous map, then the image of $\gamma$ is contractible in the image of $f$.

With these facts, we can show the result.

Let $a = Q(x)$, and $b = Q(y)$, where $x, y\in S^n$. We want to show that the line segment from $a$ to $b$ is in the image of $S^n$ under $Q$.

Consider the great circle from $x$ to $y$ in $S^n$, which is the image of $S^1$ under the linear map $(u,v) \mapsto u \hat{x} + v \hat{y}$, where $\\{\hat{x}, \hat{y}\\}$ is an orthonormal basis for the linear span of $x$ and $y$. Now, if we compose $Q$ with this linear map from $S^1$, we obtain a quadratic map from $S^1$ to $\mathbb{R}^2$, whose image we have seen is an ellipse, which contains both $a$ and $b$.

The ellipse divides $\mathbb{R}^2$ into two pieces: the compact inside part, and the unbounded outside part. The compact inside part is convex, and therefore, it contains the line segment from $a$ to $b$. Because the $S^{n}$ is simply connected, it must be the case that every point in the inside part of the ellipse must be contained in the image. (To see this, say that some subset of the inside part of the ellipse is missing from the image, then the ellipse would not be contractible in the image set.)

Therefore, the line segment from $a$ to $b$ is contained in the image, and we are done.

We’re going to talk about a needlessly complicated proof of this fact that uses some pretty cool math.

Categorification and Inequalities

When we see an inequality like

we might want to write down an equation for the binomial coefficients like $\binom{n}{k} = \frac{n!}{(n-k)!k!}$, and then do some algebra. This might be fast, but it’s hardly satisfying.

Maybe more satisfying would be to first think about what $\binom{n}{k}$ means; if $S_k$ is the number of sets of size $k$ in a set of $n$ elements, then $\binom{n}{k}$ is the size of $S_k$. So, another way of saying the inequality is that $S_k$ has fewer elements than $S_{k+1}$.

One way to show that one set has more elements than another is to find an injective map $f$ from one set to the other, and such maps are usually a lot more satisfying than a computational proof.

It turns out that this is kind of tricky in this case. For instance, we might naturally want to choose, for each $k$ element set $S$, an element $t(S) \in [n] \setminus S$, and then set $f(S) = S \cup \{t(S)\}$. We might hope that if $t(S)$ is well chosen, then $S$ can be recovered from $f(S)$ uniquely. There are in fact ways to do this, (see this), but it’s actually a little tricky to find such an injection. We’ll see in the sequal that this is because there is too much symmetry here; to find such a $t(S)$, we need some clever way of choosing an element of $[n] \setminus S$, and all elements of $[n] \setminus S$ basically look the same to me!

As a first towards getting this bijective proof we want, we can apply an an idea that shows up a lot in combinatorics (that we’ll also do in a fancier way in the sequel): instead of assigning to each $S \subseteq [n]$ with $|S| = k$, we can form a relation, which we might think of as a bipartite graph. This will let us get around the symmetry issue.

The relation here is just the subset relation. For each element $S$ of $S_k$, there are $(n-k)$ elements of $[n] \setminus S$ that we can add to $S_k$ to get an element of $S_{k+1}$. For each element of $S_{k+!}$, there are $k+1$ elements in $S_{k+1}$ that we can remove to get a set of size $k$. So, by a double counting argument, we know that $\frac{|S_{k}|}{|S_{k+1}|} = \frac{k+1}{n-k} < 1$. This doesn’t use the equation for binomial coefficients, but requires an unacceptably large amount of computation! Let’s see if we can do better.

Symmetry and vector spaces

Usually, we think of symmetry as being our friend, but in the above paragraph, we saw that symmetry was actually kind of an obstruction to getting a nice map. A slight modification of the above ideas though will let us get symmetry on our side again.

Instead of thinking of $\binom{n}{k}$ as the number of elements in some set, let’s think of it as the dimension of the vector space spanned by elements of $S_k$, say $V_k$. This is the set of formal linear combinations of elements of $S_k$. Once again, if we have an injective map from $V_k$ to $V_{k+1}$, then linear algebra tells us that $\binom{n}{k} \le \binom{n}{k+1}$.

The value of doing this is that we can now turn the relation that we had above into an honest map of vector spaces; for each $S \in S_k$, let

That $f_k(S)$ is the sum of all sets of sets of size $k+1$ that contain $S$, and is sort of the ‘vectorization’ of the above relation. If we can show that this map is injective, then we are done, and we will have satisfied our desire for a synthetic proof of this fact.

In fact, this is true; the map is injective and it can be shown for example by looking at the matrix associated with this map, computing its Gram matrix and showing that is positive definite (see for example this). Once again though, this is an unacceptable amount of computation that you can’t just handwave away. We must go deeper!

Playing around with commutators

There are a few steps we need to give a bijective proof `without computations’.

Firstly, the map we defined above has a transpose, specifically, this is a map $g_{k+1} : V_{k+1} \rightarrow V_k$ given by

Secondly, we need to assemble all of these $V_k$ into a single (graded) vector space, namely

We can define $f$ and $g$ as follows: if $v \in V_i$, then $f(v) = f_i(v)$, and similarly, $g(v) = g_i(v)$.

Now, we can do something a little strange; we can try computing the comutator between this map $g$ and $f$ for some $S \in V_k$:

The way you get this is by noticing that if $S' \in S_k$, and $S' \neq S$, then the coefficient in front of $S'$ in $g_{k+1}(f_k(S))$ is the number of elements in $S_{k+1}$ that contain both $S$ and $S'$, but there’s only one of these, namely $S \cup S'$. A similar idea for $f_{k-1}(g_k(S))$ gives us that the coefficient in front of $S'$ will also be 1, so these cancel, and the only surviving term is proportional to $X$, and again, it’s not that hard to figure out what this coefficient should be.

This is really nice! It says that the commutator of $f$ and $g$ acts diagonally on the $V$. Let’s call this commutator $h = gf - fg$.

We also have other commutation relations, which can also be verified `by hand waving’ as above: $[h, f] = -2f$ and $[h,g] = 2g$.

These relations are important because they define a representation, which we’ll look at next:

Lie algebras and $sl_2$

There are in fact $2 \times 2$ matrices which satisfy these same commutation relations, namely

These form what is called a Lie algebra, which is a vector space with an antisymmetric, bilinear map called the Lie bracket, denoted $[-,-]$. There are some additional properties that you need, but let’s not say what they are here. The name of this Lie algebra is $sl_2$.

Between two Lie algebras, both of which have an associated Lie bracket, a morphism is just a linear map $\phi : L_1 \rightarrow L_2$ so that $[\phi(x), \phi(y)] = \phi([x,y])$. A representation of a Lie algebra $L$ is a map from $L$% to $gl_n$, the Lie algebra of matrices with the normal matrix commutator.

The interesting thing about representations about $sl_2$ is that if you look at $\phi(h')$, we can say some important things about the eigenvectors of $h$, namely that if $v$ is an eigenvector of $\phi(h')$ with eigenvalue $\lambda$, then $\phi(f')v$ and $\phi(g')v$ are also eigenvectors of $h$ with eigenvalues $\lambda+2$ and $\lambda-2$ respectively. In particular, because $\phi(h')$ has only finitely many eigenvalues, $\phi(h')$ and $\phi(g')$ are nilpotent, and so the eigenvalues of $h$ must all be integers.

If $gl_n$ acts on an $n$ dimensional vector space $V$, and $\phi$ is a representation of $sl_2$ on $gl_n$, then we have a grading of $V$ according to the eigenvalue of $\phi(h)$, i.e.

where $V_i$ is an eigenspace of $h$. Notice that this is exactly the position we found ourselves in the previous section.

The important thing is that people have studied representations of $sl_2$ for a really long time. Specifically, they know that the irreducible representations have the following property: the dimension of $V_i$ when $i > 0$ is always at most the dimension of $V_{i+1}$! Because this is true for all irreducible representations of $sl_2$, and all representations are direct sums of irreducible ones, any representation of $sl_2$ will have this property! In addition, this theory will give us the fact that $f$ and $g$ above are injective, just because they satisfy these commutation relations, so we get our combinatorial proof, and all it took us was a few years worth of representation theory.

Applications to Betti Numbers and Hodge Theory

The representation theory of $sl_2$ actually appears in a lot of different fields. In particular, it shows up in the cohomology theory of complex projective varieties. I don’t understand this very well, but here’s the gist of it from my point of view.

Every complex projective variety has something called a Kahler structure, which, if the variety is smooth, we can think of as a differential 2-form that lives on the manifold. This differential 2-form $\omega$ (called the symplectic form) is closed, and therefore defines an element of the De Rham cohomology for the manifold.

The cohomology rings of a manifold have an attached ring structure, whose multiplciation is given by the cup product, and multipliciation by $\omega$ in this ring structure gives us a linear map from the cohomology ring to itself. Let’s call this map $f$, suggestively. There is also an inner product on the cohomology ring given by the Riemannian metric structure of the manifold, and so $f$ has an adjoint, which we might call $g$, again suggestively. As you might expect, $h = [f, g]$ turns out to act diagonally on the cohomology ring, its eigenspaces are the graded pieces of the cohomology ring, and the triple $f, g, h$ defines a representation of $sl_2$ on the cohomology ring of the variety.

The dimension of the degree $k$ graded piece of the cohomology ring is called the $k^{th}$ betti number $\beta_k$. So, what we get in total is that for $k > \frac{n}{2}$, $\beta_k \le \beta_{k+1}$.

This theory was used by Richard Stanley in the case of toric varieties (whose paper on unimodal sequences forms the substance of this post) to show that the number of facets of simplicial polytopes satisfy the least upper bound conjecture, which I don’t really get, but am assured is very cool.

This is a project that I worked on for the course math 7018, Fall 2020. Source image is from here.

Source Codes and Monoids

Basics of Coding Theory

The basic goal in Source information theory is to represent a collection of states using symbols in as efficient a way as possible. It is clear that the total number of symbols required to represent a collection of $k$ state is at least $k$. As an example, a stop sign has 3 possible states, either red green or yellow, and this is enough to convey whether a person should stop, slow or go. To communicate more information, like whether turning left is allowed, more symbols are needed.

To represent a possibly infinite number of states, one typically writes down a sequence of characters coming from some fixed alphabet, and in doing this, one typically wants to represent those states using as few characters as possible. As an example, each English word is represented as a sequence of characters from the fixed 26 character alphabet, and each English sentence is represented as a sequence of words from a fixed 200 thousand word dictionary.

In one of the basic problems of coding theory, one is interested in representing a sequence of characters from one alphabet in terms of a second alphabet. A typical example of this is ASCII encoding: most English text is composed of a sequence of about characters from a 200 character alphabet; in order to record a text in binary, one needs to represent each of these characters as a binary string. To represent about 200 characters, one needs at least 8 binary characters per English character, and indeed, this is the length of the representation of every English character in ASCII. In a sense, we are trying to simulate a full English alphabet of 26 characters using only 2 characters at a time.

In mathematical terms, we have some collection of states $X$, which we want to represent, and a base alphabet $A$. To each state $x \in X$, we want to assign an encoding $\phi(x) \in A^\*$, where $A^\*$ is the set of all strings made using the alphabet $A$. If our base alphabet $A$ consists of $n$ characters, we can represent $n^k$ possible states with $k$ character strings from $A$, so if all states in $X$ have representations of the same length in $A$, each character in $X$ needs to have an encoding of length at least $\log_n(X)$. If $X$ is itself a set of strings over some alphabet $B$, so that $X = B^\*$, then we would also like the encoding to preserve the concatenation operation, i.e. if we have two characters in $X$, $a, b \in X$, and we form the string $ab$, we want $\phi(ab) = \phi(a) + \phi(b)$.

Variable Length Codes

The ASCII encoding is nice because each character’s encoding has the same length; this makes it relatively easy to read an ASCII encoded string, you simply break the string into 8 character chucks, turn each chunk into a Latin character. However, this encoding is somewhat wasteful. Notice that the character e has the same length as the character q, even though e appears significantly more often in English than q does. For example, consider the opening lines of A Tale of Two Cities: “It was the best of times, it was the worst of times.” The letter e appears 5 times, whereas q does not appear at all. Imagine a code where e was represented by 7 characters, and q was represented by 9; for almost all text, this would result in a savings in terms of the number of bits needed to express them.

Nowadays, when bits are plentiful, we do not care so much about these minor savings, but back when people used telegrams, they cared a lot more. So for example, in Morse code, e takes up only 1 dot, whereas q takes up 4. This means that it is a lot faster to transmit e than to transmit q, which in practice is good, because you want to transmit an e a lot more often.

So, say we have finite alphabets $B$ and $A$, and want an encoding $\phi : B^* \rightarrow A^*$. In order for this code to be usable, it had better be the case that different elements of $B^*$ map to different elements of $A^*$. For example, if we had an encoding that looked like

\begin{align} a \rightarrow 011\ b \rightarrow 101\ c \rightarrow 01 \end{align}

then the strings cb and ac would be encoded by the same string 01101. If we try to decode this string, we would be faced with ambiguity.

A code where each encoded string has exactly one decoding is called uniquely decodable. This is the same as saying the map $\phi : B^\* \rightarrow A^\*$ is injective.

In practice, the goal is to find a code that minimizes the average length of a codeword, when the characters are drawn from a fixed distribution, like that of the English language. The uniquely decodable assumption puts constraints that make it difficult to make the code short. For example, we cannot give each character in the alphabet the code 0, even though that would make all of the codewords very short, because that would make it impossible to retrieve the information from the code. Intuitively, if we try to make the codewords too short, then for a large alphabet $B$, the codewords will have to start colliding when you concatenate a lot of characters together.

The formal bound expressing the intuition that short codes are difficult to create is known as Kraft’s inequality.

Lemma 1

Suppose that we have a uniquely decodable code $\phi : B^\* \rightarrow A^\*$, and that $|A| = n$. Use $|s|$ to denote the length of a string in $s \in B^\*$.

$$ > \sum_{b \in B} n^{-|\phi(b)|} \le 1 > $$

The proof of Kraft’s inequality for uniquely decodable codes is one of my favorite proofs, and we’ll do it in slightly greater generality in a moment.

As a slight abuse of language, if we have some numbers $\ell_i$, then we say that $\ell_i$ satisfies Kraft’s inequality if

It is a theorem that if $\ell_i$ satisfy Kraft’s inequality, then there is in fact a uniquely decodable code so that $|\phi(i)| = \ell_i$.

Relations in monoids

We now want to generalize some of our previous observations about codes. We say that a set $M$ is a monoid if there is a binary operation $\cdot : M \times M \rightarrow M$ so that $a \cdot (b \cdot c) = (a\cdot b)\cdot c$, and for some $e \in M$, $e \cdot m = m$ for all $m \in M$. If we have two monoids, $M_1$ and $M_2$, a map $\phi : M_1 \rightarrow M_2$ is said to be a homomorphism if $\phi(a\cdot b) = \phi(a) \cdot \phi(b)$.

The easiest example of a monoid is probably the free monoid over a set $B$, $B^\*$, where the set consists of all strings of characters in $B$, and the binary operation is string concatenation. We can reformulate our previous language as follows: a code is a monoid homomorphism between two free monoids, and a code is uniquely decodable if and only if that monoid homomorphism is injective.

All monoids can be thought of in terms of sequences of strings, but where some of the strings are considered to be equivalent. In a free monoid, if we write down one sequence of strings $x_1x_2\dots x_n$, and someone else writes down another sequence $y_1y_2\dots y_m$, then these two strings are equal if and only if $m = n$ and $x_i = y_i$ for each $i$ from 1 to $n$. This is pretty obviously something you can do this when dealing with binary strings, but we might be interested in slightly more exotic examples of monoids.

A classic example of this difficulty is Russian cursive: it is famously difficult to read Russian cursive, because many of the characters look the same.

In particular, it is very difficult, for foreigners like myself, when there are strings of similar appearing characters. We call such a thing a relation between characters.

A monoid (not necessarily free) is like the set of strings in some alphabet, but written down using very bad handwriting, so that some strings are indistinguishable from others, even though they might have been made by different characters. Nevertheless, Russians continue to write in cursive, and they are able to have a functioning society, indicating that it should still be possible to represent information using such an alphabet.

To do this, we want to understand the notion of source coding in a more general setting. Technically, in a general monoid, we do not have a notion of how “long” an element of the monoid is, because we cannot tell which strings are supposed to be length 1. Because of that, we will actually want our monoid $A$ to be finitely presented, meaning that we can write it in the following form:

where $a_1, \dots, a_n \in A$, and $s_1,t_1 \dots, s_k, t_k \in A^\*$.

This denotes the fact that every element in $A$ can be written as a product of characters $a_1, \dots, a_n$, but that the strings $s_i$ and $t_i$ are identified for each $i$. Given such a presentation, we can simply declare every generator to have length 1, and then declare the length of any element $x \in A$ to be minimum length of a string of generators resulting in $x$. In this case, we denote the length of $x$ by $|x|$.

Now, if $B$ is a finite alphabet, and $M$ is a finitely presented monoid, then a code for $B$ over $M$ is a monoid homomorphism $\phi$ from $B^*$ to $M$, and that code is said to be uniquely decodable if $\phi$ is injective. The length spectrum of a code $\phi$ is the list $(|\phi(b)|)_{b \in B}$.

We are interested in codes whose length spectrum is “small” in some sense, but once again, we might suspect that being uniquely decodable poses some restrictions on the length spectra on codes. In particular, if the length spectrum of $\phi$ consists of only small numbers, then there must be `a lot’ of short elements of $M$.

Formally, let $ M_k = \{m \in M : |m| = k\}$ be the set of all monoid elements of length $k$. There is a generalized Kraft inequality, given as follows:

Theorem 1

Let $M$ be a monoid, and let $\alpha \in \mathbb{R}$ be any number so that $|M_k| = O(\alpha^k)$ for all $k \ge 0$. Let $\phi : B^\* \rightarrow M$ be a uniquely decodable code, then

$$ > \sum_{b \in B} \alpha^{-|\phi(b)|} \le 1. > $$

Proof

We think of $g$ in the following expression to be a type of generating function for the lengths of encoded characters.

Then take $g^{\ell}$ for some $\ell \in \mathbb{N}$. We can expand this out to be a sum over all strings in $B$:

Notice that $|\phi(b_1)| + \dots + |\phi(b_{\ell})|$ is an upper bound for $|\phi(b_1+\dots+b_{\ell})|$. Thus,

Let $L_i = \{s \in B^\* : \phi(s) \in M_i\}$. First, note that $\phi$ is injective, so $|L_i| \le |M_i| = O(\alpha^i)$.

Next, we reindex this summation to first sum over $i$:

Where we have used the upper bound that any string of length $\ell$ in $B^\*$ maps to a string of length at most $\ell \max_{b \in B}|\phi(b)|$. Hence,

Note, if $g > 1$, then $g^{\ell}$ grows exponentially for large $\ell$, yet the right hand side grows only linearly in $\ell$. So for $\ell$ large enough, this inequality could not hold. Thus, $g \le 1$, and this concludes the proof.

One somewhat remarkable aspect of Kraft’s inequality is that it only uses asymptotic information about growth rate of elements in a monoid, yet, the result is fully quantitative.

Next, we will seek to apply this inequality in the case of a very special class of monoids: namely the monoid of braids.

Braids

One type of monoid, whose structure has been studied in depth in the field of topology, is the braid monoid. We think of a braid as a collection of strings, with fixed endpoints, which have been tangled with each other in some complicated way.

Formally, a braid on $n$ strands is a collection of $n$ paths $p_i : [0,1] \rightarrow \mathbb{R}^3$, where there exists a permutation $\pi \in S_n$, so that $p_i(0) = (i, 0, 0)$, $p_i(1) = (\pi(i), 0, 1)$, and for each $t$ and $i$, $p_i(t)_3 = t$. We consider two braids to be equivalent of one can be continuously deformed into the other. The permutation involved in the definition is crucial, and we will denote by $p(b)$ the underlying permutation of a braid $b$.

We will say that a braid is positive if each time $p_i(t)_1 = p_j(t)_1$, and $i > j$ we have that $p_i(t)_2 > p_j(t)_2$. That is to say, that each crossing of strands in the braid has the left strand crossing over the right one. Thus, a positive braid can be represented as a 2 dimensional picture, where we forget about the second coordinate.

This collection of braids turns into a monoid via concatenation, where we concatenate each of these paths with each other, say by letting

We will use the braid monoid to refer to the monoid of positive braids, as this will be the main object of our study.

There is a natural monoid homomorphism that sends the braid monoid to the symmetric group which sends each braid to the permutation of the endpoints that it induces.

Despite the fact that this definition of the braid monoid is continuous in nature, the braid monoid is in fact finitely presented. The Artin presentation for the positive braid monoid is given by

Geometrically, the generator $\sigma_i$ represents the operation of twisiting strands $i$ and $i+1$. The relations reflect basic operations that one can do on the level of braids which preserve homotopy equivalence. (See for example \cite{gonzalez2011basic}).

Codes over the braid monoid have an appealing physical interpretation: if we had a code $\phi : X^\* \rightarrow B_n$ one can imagine taking $n$ strands of string and braiding them together to encode messages in the alphabet $X$. As an example, it is not hard to see that if we take $X = \{0, 1\}$, and let $\phi(0) = \sigma_1$ and $\phi(1) = \sigma_2$, we would get a uniquely decodable binary code, and we could encode binary strings as braids. This shows that if $n > 3$, braids allow for representations that are at least as efficient as simple binary codes. Can we do better?

Kraft’s inequality indicates that we should begin by considering the growth rate of the braid monoid. This is the subject of much study, and has been completely solved in the most important of cases. We will recount the details of this study here, which show, somewhat surprisingly, that in fact the growth rate of all braid codes is bounded from above by a constant. This implies that while it is possible to obtain codes which are more efficient than binary codes using braids, it is not possible to be more efficient than a 4 character alphabet.

Also, the study the structure of which elements in a braid monoid generate a free monoid is well studied, but difficult, indicating that there is still much to learn in this field.

Growth Rates of Braid Codes

Let $M = B_n$ be the braid monoid on $n$ strands, together with the Artin presentation given above. Let $M_k$ be the set of elements of $M$ with length $k$. The growth function of $M$ is defined to be

This is a generating function for the number of elements of size $k$ in $M$.

A result of Deligne that this growth function is in fact rational (indeed, it is the reciprocal of a polynomial).

We will first give a proof of this fact (found in \cite{epstein1992word}) in terms of the so called `automatic structure’ of the braid monoid, as it will set up some interesting further questions related to alternative presentations of the braid monoid, and it will also give a taste of some of the proof techniques used in this field.

Then, we will give a more detailed examination of the growth function of the braid codes using the results in \cite{flores2018growth}. This will initially rely intimately on the so-called Garside structure for the braid monoid.

Automatic Monoids

The results described in this section are found in \cite{epstein1992word}

For our purposes, a finite automaton $F$ is a triple $(X, G, \ell, v_0)$, defined as follows. $X$ is some finite alphabet, and $G$ is a directed graph (possibly with parallel edges and loops). $\ell : E(G) \rightarrow X$ is a labelling of the edges of $G$, so that for each vertex $v \in G$, and $x \in X$, there is at most one directed edge incident to $v$ with the label $x$. $v_0$ is some distinguished vertex of $G$.

We will think of this kind of finite automaton as being a simplified model of computation. A string $w = w_1w_2\dots w_n$ is accepted by $F$ if there is a walk $v_0, v_1, \dots, v_n$ in $G$ so that $\ell(v_i, v_{i+1}) = w_{i+1}$ for each $i \in \{0, \dots, n\}$. Such a walk, if it exists, is unique. (As a note, we think of strings in this context as being terminated by a special character that signals that the string has ended. This does not impact the counting applications that we describe next.)

The regular language defined by $F$ is defined to be the set of strings which are accepted by $F$. A set of strings is said to be a regular language if it is accepted by some finite automaton.

For our counting related purposes, we will consider a finitely presented monoid $M$, generated by elements $g_1, \dots, g_k$, to be automatic if there is a regular language $L$ with alphabet $\{g_1, \dots, g_k\}$, with the property that the map $f : L \rightarrow M$ given by sending $w_1 w_2 \dots w_n \in L$ to the product $w_1\cdot w_2 \dots w_n$ is bijective. In this case, for any $b \in M$, we say that $f^{-1}(b)$ is the canonical word representing $b$. In the literature, there are some additional requirements for being automatic, which are defined in \cite{epstein1992word}.

The key fact which is relevant for counting purposes is that the growth rate of a regular language is highly constrained:

Lemma 2

If $L$ is a regular language, and we let $L_k$ be the set of elements of $L$ of length $k$, then

$$ > g(t) = \sum_{i=0}^{\infty} |L_k|t^k > $$

is the reciprocal of a polynomial.

Proof

Let $F$ a finite automaton recognizing $L$. The number of strings of length $k$ in $L$ are in bijection with walks in $G$

We can count such walks in terms of $A$, the adjacency matrix of $G$ . Indeed, if $A$ is this adjacency matrix, then the number of walks from $v_0$ to $v'$ of length $k$ is $A^k_{v_0, v'}$. Thus,

where crucially, $f$ is linear. Hence, we see that

The entries of $\frac{1}{1-At}$ are all rational functions of $t$, and so, we are done.

This is especially useful for computing growth rates.

Lemma 3

If $g(t) = \sum_{k \ge 0}a_k t^k$ is the reciprocal of a polynomial, then there is a constant $c$ so that $M_k = \Theta(poly(k) c^k)$, where $c$ is smallest norm of all of the roots of $\frac{1}{g(t)}$.

Thus, for an automatic monoid, the growth function is rational. The motivation for considering automatic monoids should be clear:

Lemma 4

The positive braid monoid is automatic.

The formal proof is given in \cite{epstein1992word}. The proof of this theorem requires some ideas that are relevant to the Garside theory of the braid monoid (which is also useful for doing computations with the braid monoid).

Firstly, we need the notion of a simple braid. A positive braid $b$ is simple if for pair of strands crosses at most once. We can imagine that if the $b$ induces a permutation $\pi$, that the strand starting from the $i^{th}$ starting point goes directly to the the $\pi(i)^{th}$ ending point without becoming entangled with any other strands. As it turns out the map $p$ sending a braid $b$ to its induced permutation restricts to a bijection on simple braids. Thus, for fixed $n$, there are $n!$ simple braids.

Positive braids are ordered under the prefix ordering, under which $b_1 \prec b_2$ if there is some positive braid $a$ so that $ab_1 = b_2$. The perhaps surprising thing about this ordering is that it forms a lattice; this implies that for any $b_1, b_2$, there is some element $b_1 \lor b_2$ so that if $b_1 \prec a$ and $b_2 \prec a$, then $b_1 \lor b_2 \prec a$. In this ordering, the meet of all simple braids is a special element known as the Garside element of $M$. In this case, the Garside element has a physical meaning: we can imagine taking all of the strands, and then performing a full twist, reversing the ordering of the strands. This lattice structure, together with this Garside element, produces a so-called Garside structure on the monoid of positive braids.

We can now sketch a proof that the braid monoid is automatic.

Proof

For each simple braid $s$, assign an arbitrary word $\overline{s}$ representing that braid to be its canonical form.

Once we have canonical words for each simple braid, we can produce a canonical form of each braid, known as the left-greedy form of the braid. In this form, we write

where each $d_1$ is simple and satisfies the following condition: $d_k$ is the maximal simple braid so that $d_k \prec b$, and if $b = ad_k$, then $d_{k-1}$ is the maximal simple braid so that $d_{k-1} \prec a$, and so on. Note that at each step, this maximal braid exists, due to the lattice structure.

It turns out that in order for a word to be in left-greedy form, it suffices for $d_i$ and $d_{i+1}$ be compatible (in a sense we do not specify) for each $i$, and for each $d_i$ to be written in its canonical form.

Now, we need to recognize the string in right greedy form. We want to read a string from left to right and find where the `breaks’ are between consecutive $d_i$, and also verify that each subsequent $d_i$ are compatible with the previous one.

Hence, let $G$ be the graph whose vertices are pairs of the form $(s_{curr}, s_{next})$, where $s_{curr}$ and $s_{next}$ are simple braids. Our starting vertex is $v_0 = (e, e)$, where $e$ is the identity braid.

We think of $s_{curr}$ as representing $s_i$ for some $i$, and $s_{next}$ as representing the starting part of $s_{i+1}$ in the above decomposition.

On an input character $c$, and at state $(s_{curr},s_{next})$, we make the following transitions: if $s_{next}c$ is a simple braid, then we transition to state $(s_{curr},s_{next}c)$. On the other hand, if $s_{next}c$ is not a simple braid, then we consider a couple cases:

In this case, we would need to show that $s_{next}$ and $s_{curr}$ are compatible, $s_{next}$ is written in its canonical form. If these conditions are not met, then we will fail to transition, and the string will be rejected. If they are met, then we will continue.

This gives a finite automaton recognizing left-greedy form, and we are done.

Computing the Growth Function of Braids

\cite{flores2018growth} offers a more efficient approach to obtaining the growth function for the positive braid monoid in terms of lexicographic representatives.

We will consider the generators of $B_n$ to be ordered, so that $\sigma_1 < \sigma_2 < \dots < \sigma_n$. Given a positive braid $b$, we let the word $w = w_1w_2\dots w_k$ (where each $w_1$ is one of the Artin generators) be the lexicographically first word representing the braid $b$. That is, out of all words $w$ which represent the braid $b$, we choose one where $w_1$ is as small as possible, and out of all such words, we choose one where $w_2$ is as small as possible, and so on.

As it happens, the set of lexicographically first words representing braids is also a regular language, though we will not show this here.

We divide $M$ into pieces depending on what generators appear in the lexicographically first representative of $b$.

Let

We see that if $b \in U^{(i,j)}\_k$, then the lexicographically first word representing $b$ starts with either $\sigma_j$ or $\sigma_{j+1}$ (since $\sigma_{j}$ divides $b$, and $\sigma_{j+2}$ through $\sigma_{n}$ do not). Moreover, $\sigma_i,\dots, \sigma_j \prec b$ is equivalent to saying that $\bigvee_{\ell= i }^j \sigma_{\ell} \prec b$ from the definition of the meet. Hence,

We also define

A few quick notes: there are two reasons why $U^{(i,j)}$ is of interest. Firstly, $\bigvee_{\ell= i }^j \sigma_{\ell}$ is of a simple form: the monoid generated by $\{\sigma_{i}, \sigma_{i+1}, \dots, \sigma_{\ell}\}$ is isomorphic to a smaller braid monoid (indeed, these generators for the submonoid are precisely the Artin generators for the submonoid). This implies that $\bigvee_{\ell= i }^j \sigma_{\ell}$ has the same length as the Garside element for $B_{j - i}$, which has length $\binom{j-i+2}{2}$.

Secondly, we have that for $i \ge j+2$, $\sigma_j$ commutes with $\bigvee_{\ell= i }^j \sigma_{\ell}$, which will also be useful for us.

Now, we will relate the $U^{(i,j)}_k$ to $M^{(i)}_k$ in two ways, which will allow us to derive a recurrence relation for $|M^{(i)}_k|$, which will turn allow us to compute $|M_k| = |M^{(n)}_k|$.

Lemma 5

$|U_k^{(i,j)}| = |M_{k - \binom{j-i+2}{2}}^{j+1}|$

Proof

Recall the definition

$\bigvee_{\ell= i }^j \sigma_{\ell} \prec b$ if and only if $b = \left( \bigvee_{\ell= i }^j \sigma_{\ell}\right)c$ for some $c$. Moreover, $\left( \bigvee_{\ell= i }^j \sigma_{\ell} \right) c \in M_k$ if and only if $|c| = k - |\bigvee_{\ell= i }^j \sigma_{\ell}|$, where $|\bigvee_{\ell= i }^j \sigma_{\ell}| = \binom{j-i+2}{2}$ We will let $m = k - |\bigvee_{\ell= i }^j \sigma_{\ell}|$.

Also, because $\sigma_{j+2}$ commutes with $\sigma_{\ell}$ for all $\ell \in [i, j]$, and therefore, $\sigma_{j+2}$ divides $\left( \bigvee_{\ell= i }^j \sigma_{\ell} \right) c$ if and only if $\sigma_{j+2} \prec c$. Thus, we have that

as desired

Lemma 6

$|M_{k}^{i+1}| - |M_{k}^{i}| = \sum_{\ell=i}^n (-1)^{\ell - i}|U^{i, \ell}_k|$

Proof

Note that $M_{k}^{i} \subseteq M_{k}^{i+1}$, so that $|M_{k}^{i+1}| - |M_{k}^{i}| = |L_k^i|$, where

Next, note that

We can divide $U^{(i,i)}$ into those elements where $\sigma_{i+1} \prec b$ and those where $\sigma_{i+1}\not \prec b$, in which case

In particular, $L_k^i \subseteq U^{(i,i)}_k$, and we can reason about the set difference.

More generally, applying the same kind of breaking into cases:

and

If we let

Then $U^{(i,j)} = L^{i,j}_k \sqcup L^{i,j+1}_k$.

Thus, the sum

which is what we wanted.

The previous two results indicate that there is a linear recurrence amongst the $M_i$:

Lemma 7

$|M_{k}^{i+1}| - |M_{k}^{i}| = \sum_{\ell=i}^n (-1)^{\ell - i}|M_{k - \binom{\ell-i+2}{2}}^{\ell+1}|$

Note that the amount of memory in this linear recurrence is $\binom{n}{2}$.

By writing out the matrix representing this linear recurrence explicitly, and then row reducing, the authors of \cite{flores2018growth} are able to produce an explicit $n+1 \times n+1$ matrix with entries depending on $t$ whose determinant is equal to $\frac{1}{g_M(t)}$.

Asymptotic Growth Rate of the Braid Monoid

The growth rate of a monoid $M$ is given by

The authors of \cite{flores2018growth} then use this linear recurrence to obtain asymptotics about the the number of braids with $k$ twists.

Let $M_k(n)$ be the elements of the $n$ strand braid monoid $B_n$ with length exactly $k$. As above, we will want to consider sets of the form

The interesting thing about these sets is that they stabilize as $n$ goes to infinity, i.e. there is a set $M^i_k(\infty)$ so that for all $n$ large enough, $M^{(i)}_k(n) = M^{i}_k(\infty)$. This can be seen because if we require that $\sigma_i$ appear in a representation for $b$, and if $\sigma_j$ appeared in $b$ for $j$ large enough, then we would be able to find a lexicographically larger representation for $b$.

It turns out that the sets $M^i_k(\infty)$ can be thought of as subsets of an infinite braid monoid. Specially, let $B_{\infty}$ be the braid monoid on infinitely many strands, or else, the inverse limit of all braid monoids under the obvious inclusions. Then

The authors use some analysis results to show that for any $i$,

That is, rather than considering the growth rates for all of the braid monoids separately, it is sufficient to consider this growth rate associateed with this infinite braid monoid.

Moreover, we have a recurrence relation on the sizes of $M^i_{k+1}(\infty)$.

where we note that this sum is in fact finite, since for $\ell$ large enough, the summands become 0.

Finally, the authors consider the generating function

They show that this is in fact a root of a bivariate generating function known as the partial theta function:

and

From this, and some ideas in analysis, this turns out to imply that

Giving a complete solution to this problem.

Simple Braid Generators

We say above that if we measure the length of an element in $B_n$ with respect to the Artin generators, we cannot obtain especially efficient codes. This makes some sense: suppose that $\sigma_1 = \phi(x)$ for some $x \in B$. If $y \in B$, and $\phi(y)$ does not involve either $\sigma_1$ or $\sigma_2$, then $\phi(y)$ will commute with $\phi(x)$, meaning that our code will not be injective. In general, if the codewords are all very short, then they must all lie in $B_n$ where $n$ is small.

In some senses, the Artin generators measure how much time it takes to generate a braid if we are only allowed to twist one pair of adjacent strands at a time. On the other hand, we might be more interested in the `space’ complexity of a braid, namely, how long do the strands need to be in order to produce enough codewords.

This is to some extent captured by the simple braid generators, which are those elements in the positive braid monoid where no two strands cross more than once. An algorithm for computing the growth function of the positive braid monoid relative to the simple braids is described in \cite{GEBHARDT2013232}.

In particular, the results of this paper imply that the number of braids in $B_n$ which can be written as the product of $k$ simple braids grows like $f(n)^{k}$. We would be interested in this case in obtaining a lower bound of the form $f(n) =\Omega(n)$.

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