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Crossing matrices of positive braids

Mauricio Gutierrez magutierrez002@gmail.com  and  Zbigniew Nitecki Department of Mathematics
Tufts University
503 Boston Avenue
Medford, MA 02155
znitecki@tufts.edu
(Date: August 16, 2025)
Abstract.

The crossing matrix of a braid on NN strands is the N×NN\times N integer matrix with zero diagonal whose i,ji,j entry is the algebraic number (positive minus negative) of crossings by strand ii over strand jj . When restricted to the subgroup of pure braids, this defines a homomorphism onto the additive subgroup of N×NN\times N symmetric integer matrices with zero diagonal–in fact, it represents the abelianization of this subgroup. As a function on the whole NN-braid group, it is a derivation defined by the action of the symmetric group on square matrices. The set of all crossing matrices can be described using the natural decomposition of any braid as the product of a pure braid with a “permutation braid” in the sense of Thurston, but the subset of crossing matrices for positive braids is harder to describe. We formulate a finite algorithm which exhibits all positive braids with a given crossing matrix, if any exist, or declares that there are none.

Key words and phrases:
geometric braid, permutation braid, positive braid, crossing matrix, (virtually) totally blocked matrix
2010 Mathematics Subject Classification:
Primary 57M25, 20F36; Secondary 05B20, 57M04

1. Introduction

The notion of a crossing matrix was formulated in [3]: for a geometric braid bb with NN strands, the crossing matrix 𝒞(b)\mathcal{C}(b) is an N×NN\times N matrix whose i,ji,j entry is the (algebraic) number of crossings of strand ii over strand jj. This matrix is invariant under the braid relations on geometric braids and hence is well-defined as a function on the braid group 𝔅N\mathfrak{B}_{N}. In section 2 we review the basic properties of this function and the relatively easy characterization of its range. The current paper focuses on the deeper problem of characterizing those matrices that arise as crossing matrices of positive braids–that is, geometric braids whose crossings all go in the same direction (left over right). The reason for asking for such a characterization is that in this case there is no cancellation of a left-over-right crossing with a right-over-left crossing by the same strands–all crossings “show up” in the matrix. Examples show that a crossing matrix with all entries non-negative may nonetheless not represent any positive braid. In this paper, we explain several obstructions to being the crossing matrix of a positive braid, and give an algorithm which displays all the positive braids with a given matrix as their crossing matrix (including deciding when there are none).111This algorithm has been successfully implemented in a Mathematica program, which can handle examples up to about size 7×77\times 7 on a MacBook Pro with 8GB of memory. We have not been able to come up with a conceptual characterization of which matrices arise as crossing matrices of positive braids in general, although we conjecture a characterization for pure positive braids, resting on a very specific (conjectured) lemma.

2. Crossing Matrices of Braids

2.1. Definition of crossing matrices

We think of a braid with NN strands as the homotopy class, modulo endpoints, of a geometric braid: an ensemble of NN differentiable paths pi(t)p_{i}\!\left(t\right), i=1,,Ni=1,\dots,N in the plane (the strands of the braid) whose tangent is never horizontal; the set of (distinct) initial positions {pi(0):i=1,,N}\{p_{i}\!\left(0\right)\ :\ i=1,\dots,N\} and the set of final positions {pi(1):i=1,,N}\{p_{i}\!\left(1\right)\ :\ i=1,\dots,N\} differ only in their common vertical coordinate, which we take to be 11 for the initial and 0 for the final points; it is not required that the initial and final positions of any particular strand be horizontally aligned. We assume for convenience that these paths are pairwise transverse, so that there are only finitely many crossings between strands, and each such crossing is assigned crossing data: it is regarded as positive (resp. negative) depending on whether it is left-over-right or right-over-left as we move down the path.

Given a geometric braid bb, we define its crossing matrix as the N×NN\times N matrix 𝓒(𝒃)\boldsymbol{\mathcal{C}(b)} whose i,ji,j entry is the algebraic crossing number (number of positive crossings minus number of negative crossings) for strand ii crossing over strand jj. By definition, the diagonal entries of 𝒞(b)\mathcal{C}(b) are zero.

It is easy to check that the braid relations (which do modify some crossings) don’t affect the crossing matrix. So the crossing matrix 𝒞(b)\mathcal{C}(b) can be thought of as defined on the braid represented by bb. The description of a geometric braid in terms of crossings of strands is a point of view used by Thurston [4], and contrasts with the point of view of Artin (in terms of generators and relations) in his original expositions of the braid group 𝔅N\mathfrak{B}_{N} [1, 2].

2.2. The permutation associated to a braid

Attached to each geometric braid is the permutation on the set of horizontal coordinates of the starting points of strands which takes (the horizontal coordinate of) the starting point of each strand to its ending point.

We use Greek letters to denote permutations, regarded as rearrangements–that is, permutations act on positions rather than elements: a permutation πΣN\pi\in\Sigma_{N} will be specified by the word222 Permutations will act on words on the right, denoted by superscript. (12N)π=π1π2πN(12\cdots N)^{\pi}=\pi_{1}\pi_{2}\cdots\pi_{N} in the numbers 1,2,,N1,2,\dots,N resulting from the action of π\pi on the word 12N12\cdots N. This contrasts with regarding a permutation as a bijective mapping π:{1,2,,N}{1,2,,N}\pi\!\colon\!\left\{1,2,\dots,N\right\}\negthinspace\rightarrow\negthinspace\left\{1,2,\dots,N\right\} on a set of NN elements and denoting it by the NN-tuple (π(1),π(2),,π(N))(\pi\!\left(1\right),\pi\!\left(2\right),\dots,\pi\!\left(N\right)) of images of the individual elements 1,2,,N1,2,\dots,N under this mapping. In fact the word π(1)π(2)π(N)\pi\!\left(1\right)\pi\!\left(2\right)\cdots\pi\!\left(N\right) in our “rearrangement” notation denotes the inverse permutation333 The inverse of a permutation or braid will be denoted by an overbar: π¯\bar{\pi}. . For example, the rearrangement πΣ4\pi\in\Sigma_{4} which acts on the positions 1,2,3,41,2,3,4 via

π(1)=3,π(2)=1,π(3)=4,π(4)=2\pi\!\left(1\right)=3,\quad\pi\!\left(2\right)=1,\quad\pi\!\left(3\right)=4,\quad\pi\!\left(4\right)=2

takes the word abcdabcd to

(abcd)π=bdac(e.g., (1234)π=π1π2π3π4=2413)(abcd)^{\pi}=bdac\quad\textit{(e.g., }(1234)^{\pi}=\pi_{1}\pi_{2}\pi_{3}\pi_{4}=2413\text{)}

while

(π(1),π(2),π(3),π(4))=(3,1,4,2).(\pi\!\left(1\right),\pi\!\left(2\right),\pi\!\left(3\right),\pi\!\left(4\right))=(3,1,4,2).

We extend the “rearrangement” action of permutations to matrices: If A=((aij))A=((a_{ij})) is an N×NN\times N matrix and πΣN\pi\in\Sigma_{N}, then AπA^{\pi} is the matrix obtained by rearranging the rows as well as the columns of AA according to π\pi.

When a matrix is the crossing matrix of a braid, say A=𝒞(a)A=\mathcal{C}(a), then the permutation πa\pi_{a} associated to aa can be read off of AA: every (positive) crossing of strand ii over another strand moves its position one place to the right, while every (positive) crossing of another strand over the ithi^{th} moves it to the left one space. From this it follows that πa\pi_{a} is defined (as a mapping πa:{1,2,,N}{1,2,,N}\pi_{a}\!\colon\!\left\{1,2,\dots,N\right\}\negthinspace\rightarrow\negthinspace\left\{1,2,\dots,N\right\}) by adding to ii the sum of the ithi^{th} row and subtracting the sum of the ithi^{th} column of AA:444For an arbitrary matrix, this formula does not necessarily define a permutation, only a mapping–which might even map {1,2,,N}\left\{1,2,\dots,N\right\} to a different set of integers. However, we shall show that in the context we consider it is always a permutation (Remark 7).

(1) πA(i)=i+j=1NAijk=1NAki.\pi_{A}\!\left(i\right)=i+\sum_{j=1}^{N}A_{ij}-\sum_{k=1}^{N}A_{ki}.

2.3. Crossing matrix of a product of braids

With this notation we can explain the relation between the crossing matrices 𝒞(a)\mathcal{C}(a), 𝒞(b)\mathcal{C}(b) of two braids and the crossing matrix 𝒞(ab)\mathcal{C}(ab) of their product in the braid group, which we think of as represented by the geometric braid aa followed by the geometric braid bb. The main observation is that the numbering of the strands of bb is changed when we premultiply by aa: the ithi^{th}strand of bb becomes an extension of the strand of aa which landed at the ithi^{th} position–that is, in abab it continues the π¯(i)th\bar{\pi}\!\left(i\right)^{th} strand. From this it follows that a crossing of the ithi^{th} strand of bb over its jthj^{th} strand appears in abab as a crossing of the π¯(i)th\bar{\pi}\!\left(i\right)^{th} strand over the π¯(j)th\bar{\pi}\!\left(j\right)^{th} strand. With this renumbering of strands in bb, the crossings add, so we have

Proposition 1.

For any two braids a,b𝔅Na,b\in\mathfrak{B}_{N},

𝒞(ab)=𝒞(a)+𝒞(b)πa.\mathcal{C}(ab)=\mathcal{C}(a)+\mathcal{C}(b)^{\pi_{a}}.

We will denote this crossing product operation on (crossing) matrices by a circled asterisk:

AB:=A+BπA.A\circledast B:=A+B^{\pi_{A}}.

2.4. Order reversal sets

In [4, §9.1], Thurston defined the order reversal set of a permutation πΣN\pi\in\Sigma_{N}:

OR(π):={(i,j){1,2,,N}×{1,2,,N}|i<j but π(i)>π(j)}.OR\!\left(\pi\right):=\left\{(i,j)\in\left\{1,2,\dots,N\right\}\times\left\{1,2,\dots,N\right\}\,|\,i<j\text{ but }\pi\!\left(i\right)>\pi\!\left(j\right)\right\}.

He characterized the sets S{1,2,,N}×{1,2,,N}S\subset\left\{1,2,\dots,N\right\}\times\left\{1,2,\dots,N\right\} which are order reversal sets for some permutation πΣN\pi\in\Sigma_{N} via two properties which are easily seen to be necessary, and with a little more work are sufficient:

Proposition 2 (Thurston).

A subset S{1,2,,N}×{1,2,,N}S\subset\left\{1,2,\dots,N\right\}\times\left\{1,2,\dots,N\right\} equals OR(π)OR\!\left(\pi\right) for some πΣN\pi\in\Sigma_{N} if and only if the following properties both hold:

  1. (1)

    If (i,j)S(i,j)\in S, then for every kk between ii and jj, either (i,k)S(i,k)\in S or (k,j)S(k,j)\in S (or both).

  2. (2)

    Given i<k<ji<k<j, if (i,k)S(i,k)\in S and (k,j)S(k,j)\in S, then (i,j)S(i,j)\in S.

When these properties hold, the permutation π\pi is uniquely determined by SS.

Proof.

The necessity of these two conditions for an order reversal set is more easily seen when they are replaced with their contrapositives:

contra(1):

If, for some kk between ii and jj, neither (i,k)(i,k) nor (j,k)(j,k) belongs to SS, then neither does (i,j)(i,j),

contra(2):

If (i,j)S(i,j)\not\in S given i<k<ji<k<j, then at most one of (i,k)(i,k) and (k,j)(k,j) belongs to SS.

For the first, given i<k<ji<k<j, if π(i)<π(k)\pi\!\left(i\right)<\pi\!\left(k\right) and π(k)<π(j)\pi\!\left(k\right)<\pi\!\left(j\right), then of course π(i)<π(j)\pi\!\left(i\right)<\pi\!\left(j\right); for the second, π(i)<π(j)\pi\!\left(i\right)<\pi\!\left(j\right) means we can’t have both π(k)<π(i)\pi\!\left(k\right)<\pi\!\left(i\right) and π(k)>π(j)\pi\!\left(k\right)>\pi\!\left(j\right).

To establish sufficiency, we consider the mapping σ:{1,2,,N}{1,2,,N}\sigma\!\colon\!\left\{1,2,\dots,N\right\}\negthinspace\rightarrow\negthinspace\left\{1,2,\dots,N\right\} defined by the analogue of Equation 1,

(2) σ(i)=i+#{k|(i,k)S}#{k|(k.i)S}.\sigma\left(i\right)=i+\#\left\{k\,|\,(i,k)\in S\right\}-\#\left\{k\,|\,(k.i)\in S\right\}.
1. (i,j)Sσ(i)>σ(j)\boldsymbol{(i,j)\in S\Rightarrow\sigma\left(i\right)>\sigma\left(j\right)}:

Given (i,j)S(i,j)\in S, the first hypothesis insures that

#{k<j|(i,k)S}+#{k>i|(k,j)S}>ji\#\left\{k<j\,|\,(i,k)\in S\right\}+\#\left\{k>i\,|\,(k,j)\in S\right\}>j-i

since every kk between ii and jj appears in at least one of these two sets, and jj (resp. ii) appears in the first (resp. second) set. This inequality can be rewritten as

i+#{k<j|(i,k)S}\displaystyle i+\#\left\{k<j\,|\,(i,k)\in S\right\} >j#{k>i|(k,j)S}.\displaystyle>j-\#\left\{k>i\,|\,(k,j)\in S\right\}.
Now, the second hypothesis tells us that for k>jk>j, (j,k)S(j,k)\in S implies (since (i,j)S(i,j)\in S) that also (i,k)S(i,k)\in S, hence
#{k>j|(i,k)S}\displaystyle\#\left\{k>j\,|\,(i,k)\in S\right\} #{k>j|(j.k)S}\displaystyle\geq\#\left\{k>j\,|\,(j.k)\in S\right\}
and similarly, for k<ik<i if (k,i)S(k,i)\in S then also (k,j)S(k,j)\in S, hence
#{k<i|(k,i)S}\displaystyle-\#\left\{k<i\,|\,(k,i)\in S\right\} #{k<i|(k,j)S}.\displaystyle\geq-\#\left\{k<i\,|\,(k,j)\in S\right\}.

Adding these inequalities, we obtain

i+#{k<j|(i,k)S}+#{k>j|(i,k)S}#{k<i|(k,i)S}>j#{k>i|(k,j)S}+#{k>j|(j.k)S}#{k<i|(k,j)S}i+\#\left\{k<j\,|\,(i,k)\in S\right\}+\#\left\{k>j\,|\,(i,k)\in S\right\}-\#\left\{k<i\,|\,(k,i)\in S\right\}\\ >j-\#\left\{k>i\,|\,(k,j)\in S\right\}+\#\left\{k>j\,|\,(j.k)\in S\right\}-\#\left\{k<i\,|\,(k,j)\in S\right\}

or

σ(i):=i+#{k|(i,k)S}#{k|(k,i)S}>j#{k|(k,j)S}#{k|(k,j)S}:=σ(j).\sigma\left(i\right):=i+\#\left\{k\,|\,(i,k)\in S\right\}-\#\left\{k\,|\,(k,i)\in S\right\}\\ >j-\#\left\{k\,|\,(k,j)\in S\right\}-\#\left\{k\,|\,(k,j)\in S\right\}:=\sigma\left(j\right).
2. i<j&(i,j)Sσ(i)<σ(j)\boldsymbol{i<j\ \&\ (i,j)\not\in S\Rightarrow\sigma\left(i\right)<\sigma\left(j\right)}:

Replacing the first (resp. second) hypothesis with its contrapositive, which is to say arguments parallel to the above yield the opposite inequalities and hence the desired conclusion, that for i<ji<j, if (i,j)S(i,j)\not\in S then σ(i)<σ(j)\sigma\left(i\right)<\sigma\left(j\right).

The definition of σ(i)\sigma\left(i\right) immediately guarantees that 1σ(i)N1\leq\sigma\left(i\right)\leq N, and the two inequalities above show that σ\sigma is injective, hence a permutation. Finally, it is easy to see that different permutations have different order reversal sets, giving the uniqueness statement in Proposition 2. ∎

Thurston then formulated the notion of a permutation braid:

Definition 3.

A permutation braid is a positive braid in which no pair of strands crosses more than once.

Given a permutation πΣN\pi\in\Sigma_{N}–which is to say, given the ending position of each strand, we can construct a geometric braid π+\pi^{+} by joining (i,1)2(i,1)\in\mathbb{R}^{2} to (π(i),0)(\pi\!\left(i\right),0) by a straight line segment and making all crossings positive (in case this yields more than two such line segments crossing at the same point, we can perturb a little to get only pairwise crossings). It is clear that no pair of strands crosses more than once. This shows that every πΣN\pi\in\Sigma_{N} is the permutation associated to some permutation braid; moreover any permutation braid can be “straightened out” so as to be a π+\pi^{+}. Thus

Remark 4.

The mapping p:ΣN𝔅Np\!\colon\!\Sigma_{N}\negthinspace\rightarrow\negthinspace\mathfrak{B}_{N} taking a permutation πΣN\pi\in\Sigma_{N} to the braid π+\pi^{+} defined above is a bijection onto the set of permutation braids.

We caution the reader that this map is not a homomorphism: for example a braid with a single crossing is a permutation braid, but its square is not.

2.5. Characterization of crossing matrices

The crossing matrix 𝐑π\mathbf{R_{\pi}} of the permutation braid π+\pi^{+} is clearly the same as the matrix RR with i,ji,j entry 11 if (i,j)OR(π)(i,j)\in OR\!\left(\pi\right) and 0 otherwise. Since all pairs (i,j)OR(π)(i,j)\in OR\!\left(\pi\right) have i<ji<j, RR is strictly upper triangular (Rij=0R_{ij}=0 for iji\geq j). The conditions in Proposition 2 characterizing the orientation-reversing set of a permutation can be reinterpreted as conditions on the matrix RR, which we formulate in

Definition 5.

We say a square matrix AA is 𝐓𝟎\boldsymbol{T0} (resp. 𝐓𝟏\boldsymbol{T1}) if

T0:

For any triple of indices i<k<ji<k<j, if Aik=0A_{ik}=0 and Akj=0A_{kj}=0, then Aij=0A_{ij}=0.

T1:

For any triple of indices i<k<ji<k<j, if Aik0A_{ik}\neq 0 and Akj0A_{kj}\neq 0, then Aij0A_{ij}\neq 0.

Note that both conditions refer only to entries above the diagonal of AA, and can be viewed as limitations on the distribution of zero entries (above the diagonal) in AA.

Definition 6.

An 𝐑\boldsymbol{R}-matrix is a strictly upper triangular matrix, all of whose nonzero entries are 11, and which is both T0T0 and T1T1.

For any strictly upper triangular N×NN\times N matrix RR, the positions of its nonzero entries form a set SS of pairs (i,j)(i,j) with 1i<jN1\leq i<j\leq N, and Proposition 2 tells us when SS is an OR set. The sufficiency argument in the proof of Proposition 2 amounts to saying that, for an RR-matrix, Equation 2 defines a permutation σΣN\sigma\in\Sigma_{N}. It is easy to see that, for an RR-matrix, the term in Equation 2 which is added to ii equals the ithi^{th} row sum

#{k|(i,k)S}=#{k>i|(i,k)S}=k=1NRik\#\left\{k\,|\,(i,k)\in S\right\}=\#\left\{k>i\,|\,(i,k)\in S\right\}=\sum_{k=1}^{N}R_{ik}

and the subtracted term equals the ithi^{th} column sum

#{k|(k.i)S}=#{k<i|(k.i)S}=k=1NRki\#\left\{k\,|\,(k.i)\in S\right\}=\#\left\{k<i\,|\,(k.i)\in S\right\}=\sum_{k=1}^{N}R_{ki}

so Equation 2 really is the same as Equation 1 applied to the RR-matrix RR. therefore does not change the permutation defined by Equation 1. Thus, a corollary of Proposition 2 is

Remark 7.

For any N×NN\times N matrix A=S+RA=S+R, where SS is symmetric and RR is an RR-matrix, the formula

πA(i)=i+j=1NAijk=1NAki\pi_{A}\!\left(i\right)=i+\sum_{j=1}^{N}A_{ij}-\sum_{k=1}^{N}A_{ki}

defines a permutation πAΣN\pi_{A}\in\Sigma_{N}.

To characterize the matrices which arise as crossing matrices (of some, not necessarily positive, braid) we note two necessary conditions:

Proposition 8.

For any crossing matrix A=((aij))=𝒞(b)A=((a_{ij}))=\mathcal{C}(b), b𝔅Nb\in\mathfrak{B}_{N},

  1. (1)

    the diagonal entries are all zero:

    aii=0 for i=1,,N;a_{ii}=0\text{ for }i=1,\dots,N;
  2. (2)

    each entry above the diagonal is either equal to its symmetric twin, or exceeds it by one:

    aijaji{0,1} for 1i<jN.a_{ij}-a_{ji}\in\left\{0,1\right\}\text{ for }1\leq i<j\leq N.
Proof.
  1. (1)

    The first equation is the observation that no strand crosses itself.

  2. (2)

    Suppose 1i<jN1\leq i<j\leq N. Since strand ii starts to the left of strand jj, if any crossings of these two strands occur, the first one mst move ii to the right of jj either via a positive crossing of ii over jj, or via a negative crossing of jj over ii; the effect of this is to contribute an increase by one to the difference aijajia_{ij}-a_{ji}. A second crossing must move ii back to the left of jj, either via a negative crossing of ii over jj or via a positive crossing of jj over ii; this decreases the difference aijajia_{ij}-a_{ji} by one.

    As long as crossings of ii with jj continue, these two situations will alternate strictly, so the difference aijajia_{ij}-a_{ji} will oscillate between 0 and 11.

A braid is bb called pure if its permutation πb\pi_{b} (as defined in §2.3) is the identity permutation. It is easy to see that this condition can be expressed by saying that the ithi^{th} row sum equals the ithi^{th} column sum for i=1,,Ni=1,\dots,N. If for each pair of indices i,ji,j with 1i<jN1\leq i<j\leq N we set xij=aijajix_{ij}=a_{ij}-a_{ji}, this condition becomes the system of N1N-1 equations in (N1)(N2)2\frac{(N-1)(N-2)}{2} unknowns of the form j=i+1Nxij=0\sum_{j=i+1}^{N}x_{ij}=0, i=1,,N1i=1,\dots,N-1. This system has many integer solutions, for example the matrix

[020102110]\left[\begin{array}[]{ccc}0&2&0\\ 1&0&2\\ 1&1&0\end{array}\right]

has each row sum equal to the corresponding column sum, but if we also throw in the earlier requirement that aijaji:=xij{0,1}a_{ij}-a_{ji}:=x_{ij}\in\left\{0,1\right\} we see that the only solution is xij=0x_{ij}=0 for all i<ji<j. Thus

Lemma 9.

The crossing matrix of every pure braid is symmetric.

We note that the pure braids form a subgroup 𝔓N\mathfrak{P}_{N} of the braid group 𝔅N\mathfrak{B}_{N}, and that the restriction of the crossing matrix mapping to pure braids is a homomorphism to the additive group 𝔖N0[𝐙]\mathfrak{S}_{N}^{0}[\mathbf{Z}] of symmetric N×NN\times N integer matrices with zero diagonal. This map is also surjective; to see this, note that each of the symmetric matrices SijS_{ij} whose only nonzero entries are a “11” in the i,ji,j and j,ij,i positions is the crossing matrix of the braid sijs_{ij} in which strand ii crosses over all intermediate strands, “hooks” strand jj, and then crosses back over all the intermediate strands555 (Note that the latter set of crossings is negative.) to return to its initial position.

Since the N×NN\times N matrices 𝒞(sij)=Sij\mathcal{C}(s_{ij})=S_{ij} for 1i<jN1\leq i<j\leq N generate the additive group 𝔖N0[𝐙]\mathfrak{S}_{N}^{0}[\mathbf{Z}], this shows

Proposition 10.

The crossing matrix map takes the subgroup 𝔓N\mathfrak{P}_{N} of pure NN-strand braids onto the additive group 𝔖N0[𝐙]\mathfrak{S}_{N}^{0}[\mathbf{Z}] of N×NN\times N symmetric integer matrices with zero diagonal:

𝒞(𝔓N)=𝔖N0[𝐙].\mathcal{C}(\mathfrak{P}_{N})=\mathfrak{S}_{N}^{0}[\mathbf{Z}].

To extend our characterization of crossing matrices to all braids, we note that if b𝔅Nb\in\mathfrak{B}_{N} is a braid with permutation πb\pi_{b}, then the braid s(b):=b(πb+)1s(b):=b(\pi_{b}^{+})^{-1} is a pure braid, and so we have a unique factoring b=s(b)πb+b=s(b)\pi_{b}^{+} as a pure braid followed by a permutation braid. It follows that we can write the crossing matrix of bb as

𝒞(b)=𝒞(s(b)πb+)=SR=S+R,\mathcal{C}(b)=\mathcal{C}(s(b)\pi_{b}^{+})=S\circledast R=S+R,

where S=𝒞(s(b))𝔖N0[𝐙]S=\mathcal{C}(s(b))\in\mathfrak{S}_{N}^{0}[\mathbf{Z}] (so πS=id\pi_{S}=id) and R=RπbR=R_{\pi_{b}} is an RR-matrix. Since RR-matrices are upper triangular, the expression A=S+RA=S+R is uniquely determined (if it exists) for any matrix; we will refer to it as the 𝐒𝐑\boldsymbol{SR} decomposition of AA.. We can therefore complete our characterization of crossing matrices for (general) braids:

Theorem 11.

An N×NN\times N matrix AA is the crossing matrix of some braid if and only if it has an SRSR decomposition.

We denote the set of all N×NN\times N integer matrices with zero diagonal which have an SRSR decomposition by 𝓢𝓡𝑵\boldsymbol{\mathcal{SR}_{N}}.

It can sometimes be difficult to immediately visualize the SRSR decomposition of an integer matrix, so we have adopted a tableau notation to make this clearer: given an N×NN\times N matrix with an SRSR decomposition A=S+RA=S+R, we exhibit each entry of AA i,ji,j strictly above the diagonal (1i<jN1\leq i<j\leq N) as the sum of the corresponding entries sijs_{ij} of SS and rijr_{ij} of RR, in the form “sijS+rijRs_{ij}S+r_{ij}R”; note that

sij\displaystyle s_{ij} =aji\displaystyle=a_{ji}
rij\displaystyle r_{ij} =aijaji.\displaystyle=a_{ij}-a_{ji}.

For example, if

A=[0131000120000110]=[0020000120010110]+[0111000000010000]A=\left[\begin{array}[]{cccc}0&1&3&1\\ 0&0&0&1\\ 2&0&0&0\\ 0&1&-1&0\end{array}\right]=\left[\begin{array}[]{cccc}0&0&2&0\\ 0&0&0&1\\ 2&0&0&-1\\ 0&1&-1&0\end{array}\right]+\left[\begin{array}[]{cccc}0&1&1&1\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0\end{array}\right]

then the 𝐒𝐑\boldsymbol{SR} tableau encoding this data is

SRT(A)=1R2S+RR20S3S+R4SRT\!\left(A\right)=\begin{array}[]{cccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}1}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 2S+R\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}2}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-4}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}3}\hfil\lx@intercol&\vrule\lx@intercol\hfil-S+R\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-4}\cr&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}4}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{4-4}\cr\end{array}

To complete this picture, we note that the crossing matrix map is not injective. First, we observe that the restriction of the crossing matrix map to the subgroup 𝔓N\mathfrak{P}_{N} is the abelianization of that group; it follows that its kernel is the commutator subgroup of 𝔓N\mathfrak{P}_{N}. In general, two braids aa and bb with the same crossing matrix will have the same permutation (according to the formula in Equation 1) so c=a¯bc=\bar{a}b is a pure braid with zero crossing matrix; it follows that b=acb=ac, where cc belongs to the commutator subgroup of the pure braid group–that is, cc can be written as a product of finitely many braids of the form [p,q]:=piqipi¯qi¯[p,q]:=p_{i}q_{i}\bar{p_{i}}\bar{q_{i}}. We formalize this observation:

Proposition 12.

Two braids a,b𝔅Na,b\in\mathfrak{B}_{N} satisfy 𝒞(a)=𝒞(b)\mathcal{C}(a)=\mathcal{C}(b) if and only if a=bca=bc, where cc belongs to the commutator subgroup of 𝔓N\mathfrak{P}_{N}; that is, c=[p1,q1][pk,qk]c=[p_{1},q_{1}]\cdots[p_{k},q_{k}] where pi,qi𝔓Np_{i},q_{i}\in\mathfrak{P}_{N} for i=1,,ki=1,\dots,k.

3. Crossing Matrices of Positive Braids

We turn now to the focus of this paper, the crossing matrices of positive braids; we will adapt all of our notation to this case by using a superscript “plus” to denote positivity: 𝔅N+\mathfrak{B}_{N}^{+} (resp. 𝔓N+\mathfrak{P}^{+}_{N}) will denote the positive (resp. pure positive) braids. Again, our interest in this special case is prompted by the fact that there is no cancellation of crossings in the crossing matrix: for b𝔅N+b\in\mathfrak{B}_{N}^{+}, every crossing is accounted for in 𝒞(b)\mathcal{C}(b).

3.1. Two examples

It would be natural to expect, in view of Theorem 11, that 𝒞(𝔅N+)\mathcal{C}(\mathfrak{B}_{N}^{+}) consists of all matrices in 𝒮N\mathcal{SR}_{N} with all entries non-negative. However, this is false; the following are examples of elements of 𝒮N\mathcal{SR}_{N} which, while they have non-negative entries and are crossing matrices of some braids, are not crossing matrices of any positive braids.

It will prove easier to understand these and other examples using their SRSR tableaux:

SRT(G)=1RS020S3R4SRT\!\left(G\right)=\begin{array}[]{cccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}1}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}2}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-4}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}3}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-4}\cr&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}4}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{4-4}\cr\end{array}

and

SRT(K)=10S002RR03RS405SRT\!\left(K\right)=\begin{array}[]{ccccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}1}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}2}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-5}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}3}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-5}\cr&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}4}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{4-5}\cr&&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}5}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{5-5}\cr\end{array}

We will explain why these two matrices are not crossing matrices of any positive braids after developing some properties of such crossing matrices.

3.2. The positive realization problem

The positive braids form a semigroup which includes all permutation braids, so any product of permutation braids, while it may no longer be a permutation braid, will be a positive braid. Conversely, since the standard generators of the braid group are the permutation braids for transpositions of adjacent strands and a positive braid is given by a positive word in these generators, any realization of a matrix AA as the crossing matrix of some positive braid can always be expressed as a product of permutation braids, and this corresponds to a factoring of AA into a product (with respect to the crossing product operation \circledast) of RR-matrices.

Remark 13.

An N×NN\times N matrix AA is the crossing matrix of some positive braid if and only if it can be factored as a product

A=Rπ1Rπ2RπkA=R_{\pi_{1}}\circledast R_{\pi_{2}}\circledast\cdots\circledast R_{\pi_{k}}

of RR-matrices.

Our problem, then, is how to tell, given a matrix AA, whether such a factoring is possible. Since any permutation braid is a product of single-crossing braids (i.e., any permutation is a product of transpositions) we can focus on factorization into crossing matrices of transpositions.666 By abuse of notation, we will refer to “factoring into transpositions”.

We note that in addition to the fact that all entries in the crossing matrix of a positive braid are non-negative integers, there is an immediate further restriction on such crossing matrices:

Remark 14.

If a braid is positive, b𝔅N+b\in\mathfrak{B}_{N}^{+}, then its crossing matrix 𝒞(b)\mathcal{C}(b) must

  1. (1)

    have non-negative integer entries and

  2. (2)

    satisfy the T0T0 property (Definition 5)

To see the second requirement, note that the argument for T0T0 in the context of RR-matrices (the necessity of the first condition in Proposition 2) is based only on the assumption that permutation braids are positive. By contrast, the argument for T1T1 in Proposition 2 is based on the assumption that in a permutation braid no pair of strands crosses more than once; this is not necessarily the case for general positive braids.

In view of Remark 14, we define 𝓢𝓡𝑵+\boldsymbol{\mathcal{SR}^{+}_{N}} to be the set of (non-negative integer, zero diagonal) T0T0 matrices which can be written as the sum of a (non-negative integer) symmetric matrix and an RR-matrix:

𝒮N+:={A+|A is T0 and A=S+R, where S𝔖N0[𝐙] and R is an Rmatrix}\mathcal{SR}^{+}_{N}:=\left\{A\in\mathbb{Z}^{+}\,|\,A\text{ is }T0\text{ and }A=S+R,\text{ where }S\in\mathfrak{S}_{N}^{0}[\mathbf{Z}]\text{ and }R\text{ is an }R-\text{matrix}\right\}

We caution that the T0T0 property for AA does not necessarily require SS, the symmetric part of the decomposition, to be T0T0 on its own (unless, of course, R=0R=0, so the braid is pure).

Note that the two examples in Section 3.1 fit this description: G𝒮4+G\in\mathcal{SR}^{+}_{4} and K𝒮5+K\in\mathcal{SR}^{+}_{5}.

3.3. Left division by permutations

To study the factorization problem posed by Remark 14, we formulate a partial inverse to the crossing product operation \circledast. If we solve the equation

A=RπB:=Rπ+BπA=R_{\pi}\circledast B:=R_{\pi}+B^{\pi}

for BB in terms of AA, we obtain a “division on the left”, defined by

B\displaystyle B =Rπ\A:=(ARπ)π¯\displaystyle=R_{\pi}\backslash A:=(A-R_{\pi})^{\bar{\pi}}
which we will usually refer to as left division by 𝝅\boldsymbol{\pi}:
B\displaystyle B =π\A.\displaystyle=\pi\backslash A.

If AA and BB are to be crossing matrices of positive braids, we must check conditions which insure, for A𝒮N+A\in\mathcal{SR}^{+}_{N}, that B𝒮N+B\in\mathcal{SR}^{+}_{N}.

The first requirement is the obvious one, that all entries of the matrix ARπA-R_{\pi} are non-negative; this is the same as requiring that RπAR_{\pi}\leq A entrywise. When this is the case, we will say that the permutation π\pi is subordinate to AA.

The second requirement is that BB be T0T0. To understand how this can fail, we take a detour and develop a way to “localize” some of the effects of multiplication (RπBR_{\pi}\circledast B) and division (π\A\pi\backslash A) on a matrix B𝒮N+B\in\mathcal{SR}^{+}_{N} (resp. A𝒮N+A\in\mathcal{SR}^{+}_{N}).

3.3.1. Configurations

Given a braid bb on NN strands, we can pick out a proper subset \mathcal{I} of m<nm<n strands and study the relations just between these selected strands by considering the “subbraid” b𝔅mb_{\mathcal{I}}\in\mathfrak{B}_{m} which results from erasing all the other strands of bb. The (crossing) matrix analogue of this is, given A𝒮N+A\in\mathcal{SR}^{+}_{N}, to look at the m×mm\times m submatrix AA_{\mathcal{I}} consisting of elements whose indices are both in \mathcal{I}. Note that if A𝒮N+A\in\mathcal{SR}^{+}_{N} then A𝒮m+A_{\mathcal{I}}\in\mathcal{SR}^{+}_{m}.

We can abstract this:

Definition 15 (Configuration).

Given an N×NN\times N matrix AA and an index set

={k1,k2,,km}{1,2,,N}(mN),\mathcal{I}=\left\{k_{1},k_{2},\dots,k_{m}\right\}\subset\left\{1,2,\dots,N\right\}\quad(m\leq N),

the configuration of AA on \mathcal{I} is the m×mm\times m matrix 𝐀𝓘\boldsymbol{A_{\mathcal{I}}} consisting of the rows and columns of AA whose indices belong to \mathcal{I}.

When AA is acted on by a permutation π\pi (for example as part of a product or left division), the entries of AA_{\mathcal{I}} change position, in two ways: the set of new positions is the image π()\pi\!\left(\mathcal{I}\right) of \mathcal{I} (as a mapping of {1,2,,N}\left\{1,2,\dots,N\right\} to itself), and their relative order may be scrambled; the scrambling action is given by the permutation π\pi_{\mathcal{I}} whose order-reversal set is the intersection of OR(π)OR\!\left(\pi\right) with ×\mathcal{I}\times\mathcal{I}:

OR(π)=OR(π)(×).OR\!\left(\pi_{\mathcal{I}}\right)=OR\!\left(\pi\right)\cap(\mathcal{I}\times\mathcal{I}).

We refer to the permutation π\pi_{\mathcal{I}} (by abuse of terminology) as the restriction of π\boldsymbol{\pi} to \mathcal{I}.

It is easy to confirm that the total effect of π\pi on configurations is given by

Remark 16.

For any N×NN\times N matrix AA, any subset {1,,N}\mathcal{I}\subset\left\{1,\dots,N\right\} and any permutation π\pi of {1,,N}\left\{1,\dots,N\right\},

(Aπ)π()=(A)π.(A^{\pi})_{\pi\!\left(\mathcal{I}\right)}=(A_{\mathcal{I}})^{\pi_{\mathcal{I}}}.

The point here is that, although the set of indices \mathcal{I} is not in general invariant under π\pi, it is still true that if ki<kjk_{i}<k_{j} are two elements of \mathcal{I}, then π(ki)>π(kj)\pi\!\left(k_{i}\right)>\pi\!\left(k_{j}\right) only if π(i)>π(j)\pi_{\mathcal{I}}\!\left(i\right)>\pi_{\mathcal{I}}\!\left(j\right).

Remark 16 lets us trace a configuration of crossings corresponding to a sub-braid through operations like “division by a permutation” without regard to how crossings outside that configuration are affected.

Remark 17.

Given a permutation π\pi and an index set =(k1,,km)\mathcal{I}=(k_{1},\dots,k_{m}),

Rπ=(Rπ).R_{\pi_{\mathcal{I}}}=(R_{\pi})_{\mathcal{I}}.

In particular, given a matrix AA, a permutation π\pi and the index set \mathcal{I}, the effect on the configuration AA_{\mathcal{I}} corresponding to \mathcal{I} of multiplying AA by R=RπR=R_{\pi} is given by

(3) [(RA)π]πi,πj=Ai,j+Ri,j for i,j[(R\circledast A)_{\mathcal{I}^{\pi}}]_{\pi_{i},\pi_{j}}=A_{i,j}+R_{i,j}\quad\text{ for }i,j\in\mathcal{I}

while the effect of left-dividing AA by π\pi (provided πA\pi\leq A) is expressed by

(4) [(π\A)π]π¯i,π¯j=Ai,jRi,j for i,j.[(\pi\backslash A)_{\mathcal{I}^{\pi}}]_{{\bar{\pi}}_{i},{\bar{\pi}}_{j}}=A_{i,j}-R_{i,j}\quad\text{ for }i,j\in\mathcal{I}.

A consequence of Remark 17 is that for any configuration AA_{\mathcal{I}} of AA, if the restriction π\pi_{\mathcal{I}} to \mathcal{I} of a permutation π\pi is the identity (that is, its order reversal set is disjoint from \mathcal{I}), then the configurations (RπA)π(R_{\pi}\circledast A)_{\mathcal{I}^{\pi}} and (π\A)π(\pi\backslash A)_{\mathcal{I}^{\pi}} both equal the configuration AA_{\mathcal{I}}–in other words, the relative positions of these entries do not change, even though this matrix may be embedded in the corresponding larger matrix RπAR_{\pi}\circledast A (resp. π\A\pi\backslash A) in different ways.

3.3.2. Multiplication and division by transpositions

Any permutation can be expressed as a product of transpositions of adjacent positions. We denote by 𝝉𝒊ΣN\boldsymbol{\tau_{i}}\in\Sigma_{N} (1i<n1\leq i<n) the permutation which interchanges positions ii and i+1i+1 and leaves every other position alone. Its crossing matrix 𝒕𝒊=Rτi\boldsymbol{t_{i}}=R_{\tau_{i}} is the matrix with 11 in position i,i+1i,i+1 and zero everywhere else. Note that τi\tau_{i} is its own inverse.777Caution: this is not true of the corresponding permutation braid.

Multiplication by τi\tau_{i}: Given A𝒮N+A\in\mathcal{SR}^{+}_{N}, tiAt_{i}\circledast A has rows (resp. columns) ii and i+1i+1 interchanged, in particular the subdiagonal entry of tiAt_{i}\circledast A in position i+1,ii+1,i is the same as the superdiagonal entry at i,i+1i,i+1 in A, while the i,i+1i,i+1 entry of tiAt_{i}\circledast A is one more than the i+1,ii+1,i entry of AA. In terms of the SRSR tableau, again aside from the i,i+1i,i+1 entry, multiplication by tit_{i} interchanges rows ii and i+1i+1 with each other, and columns ii and i+1i+1 with each other. In position i,i+1i,i+1, a zero (resp. R) in AA becomes an R (resp. S) in τiA\tau_{i}\circledast A.

Division by τi\tau_{i}: The transposition τi\tau_{i} is subordinate to the matrix A𝒮N+A\in\mathcal{SR}^{+}_{N} if and only if the entry ai,i+1a_{i,i+1} on the first superdiagonal of AA is nonzero, and in this case left division by τi\tau_{i} also interchanges rows (resp. columns) ii and i+1i+1 except that the (subdiagonal) i+1,ii+1,i entry of τi\A\tau_{i}\backslash A is one less than the (superdiagonal) i,i+1i,i+1 entry of AA. In terms of the SRSR tableau, division by τi\tau_{i} also interchanges the ithi^{th} an (i+1)st(i+1)^{st} rows (resp. columns) of A and, in the i,i+1i,i+1 position of the tableau, changes an S to an R or an R to a zero.

3.3.3. Mirror symmetry

It will simplify some arguments to note that the symmetry on m×mm\times m matrices defined by

(A)i,j:=A(m+1i),(m+1j)(A^{\star})_{i,j}:=A_{(m+1-i),(m+1-j)}

preserves realizability. This consists of reversing the order of the rows (and columns): it is the analogue of transpose, but “flips” the matrix about its antidiagonal instead of its diagonal. We will refer to AA^{\star} as the “mirror” of AA. It is useful to have this operation not just for the full N×NN\times N matrices, but for their sub matrices as well (which is why it is defined above for m×mm\times m instead of just N×NN\times N matrices).

If M=C(b)M=C(b), then M=C(b)M^{\star}=C(b^{\star}), where bb^{\star} is the geometric braid obtained by looking at bb from “behind”; equivalently, bb^{\star} is obtained from bb by numbering the strands right-to-left instead of left-to-right. (Note that a positive crossing remains positive if viewed from “behind”.) Thus, a matrix A𝒮N+A\in\mathcal{SR}^{+}_{N} is realizable as the crossing matrix of a (positive) braid if and only if AA^{\star} is.

3.3.4. Creating T0T0 violations

We want to understand how left division of A𝒮N+A\in\mathcal{SR}^{+}_{N} by a transposition π=τi\pi=\tau_{i} subordinate to AA results in π\A𝒮N+\pi\backslash A\not\in\mathcal{SR}^{+}_{N}, a situation which we have seen can only happen if B=π\AB=\pi\backslash A fails to be T0T0. A violation of T𝟎\boldsymbol{T0} means a 3×33\times 3 configuration in BB of the form

B()π=[00ba00ba0]B_{(\mathcal{I})^{\pi_{\mathcal{I}}}}=\left[\begin{array}[]{ccc}0&0&b\\ a&0&0\\ b^{\prime}&a^{\prime}&0\end{array}\right]

with b0b\neq 0. If π()={p1,p2,p3}\pi\!\left(\mathcal{I}\right)=\left\{p_{1},p_{2},p_{3}\right\} with p1<p2<p3p_{1}<p_{2}<p_{3} and A𝒮N+A\in\mathcal{SR}^{+}_{N} (so that the corresponding configuration AA_{\mathcal{I}} is different from Bπ()B_{\pi\!\left(\mathcal{I}\right)}), then π\pi, if it is a transposition, must interchange either p1p_{1} and p2p_{2}, or p2p_{2} and p3p_{3}. This requires that the interchanged pair of indices be adjacent. In short, there are only two possible ways that left dividing AA by a subordinate transposition can change AA_{\mathcal{I}}: either π=τ1\pi_{\mathcal{I}}=\tau_{1} (and p2=p1+1p_{2}=p_{1}+1) or π=τ2\pi_{\mathcal{I}}=\tau_{2} (and p3=p2+1p_{3}=p_{2}+1). Notice that these two situations are mirrors of each other, so we can concentrate on the case that π=τ1\pi_{\mathcal{I}}=\tau_{1} (and p2=p1+1p_{2}=p_{1}+1). Then it follows from Remark 17 that

A=τ1Bπ()=[010000000][00ba00ba0]=[0a+1000bab0].A_{\mathcal{I}}=\tau_{1}\circledast B_{\pi\!\left(\mathcal{I}\right)}=\left[\begin{array}[]{ccc}0&1&0\\ 0&0&0\\ 0&0&0\end{array}\right]\circledast\left[\begin{array}[]{ccc}0&0&b\\ a&0&0\\ b^{\prime}&a^{\prime}&0\end{array}\right]=\left[\begin{array}[]{ccc}0&a+1&0\\ 0&0&b\\ a^{\prime}&b^{\prime}&0\end{array}\right].

Since AA (and hence AA_{\mathcal{I}}) has an SRSR decomposition, we must have a=a=0a=a^{\prime}=0 and b=bb=b^{\prime} or b+1b^{\prime}+1; thus

A=[01000b0b0]A_{\mathcal{I}}=\left[\begin{array}[]{ccc}0&1&0\\ 0&0&b\\ 0&b^{\prime}&0\end{array}\right]

with b=bb=b^{\prime} or b=b+1b=b^{\prime}+1. This means the SRSR tableau of AA_{\mathcal{I}} is one of two possibilities:

SRT(A)=p1R0p2bSp3orp1R0p2bS+Rp3SRT\!\left(A_{\mathcal{I}}\right)=\begin{array}[]{ccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}$p_{1}$}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}$p_{2}$}\hfil\lx@intercol&\vrule\lx@intercol\hfil bS\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-3}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}$p_{3}$}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-3}\cr\end{array}\quad\text{or}\quad\begin{array}[]{ccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}$p_{1}$}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}$p_{2}$}\hfil\lx@intercol&\vrule\lx@intercol\hfil bS+R\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-3}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}$p_{3}$}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-3}\cr\end{array}

But the second tableau violates the requirement that the R-matrix in the SRSR decomposition must be T1T1, so only the first can belong to 𝒮N+\mathcal{SR}^{+}_{N}.

We therefore adopt the following

Definition 18.

A blockage in the matrix A𝒮N+A\in\mathcal{SR}^{+}_{N} is a 3×33\times 3 configuration of the form

SRT(A)\displaystyle SRT\!\left(A_{\mathcal{I}}\right) =iR0jbSk\displaystyle=\begin{array}[]{ccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}i}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j}\hfil\lx@intercol&\vrule\lx@intercol\hfil bS\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-3}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}k}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-3}\cr\end{array}
or its mirror image
SRT(A)\displaystyle SRT\!\left(A_{\mathcal{I}}\right) =ibS0jRk.\displaystyle=\begin{array}[]{ccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}i}\hfil\lx@intercol&\vrule\lx@intercol\hfil bS\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-3}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}k}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-3}\cr\end{array}.

with b0b\neq 0. In either of these situations, we say that the “R” entry is blocked (by the “S” entry) .

The preceding discussion shows that division of a matrix in 𝒮N+\mathcal{SR}^{+}_{N} by a subordinate transposition results in a T0T0 violation precisely if that transposition corresponds to a “blocked R” in the matrix:

Proposition 19 (Division by a Blocked “R”).

If B=τ\A𝒮N+B=\tau\backslash A\not\in\mathcal{SR}^{+}_{N} where A𝒮N+A\in\mathcal{SR}^{+}_{N} and τΣN\tau\in\Sigma_{N} is a transposition subordinate to AA, then AA has a blockage AA_{\mathcal{I}} whose “R” entry lies on the first superdiagonal of AA, at the position corresponding to τ\tau: either

  1. (1)

    τ=τi\tau=\tau_{i}, j=i+1j=i+1 and

    SRT(A)=iR0i+1bSk,SRT\!\left(A_{\mathcal{I}}\right)=\begin{array}[]{ccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}i}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}i+1}\hfil\lx@intercol&\vrule\lx@intercol\hfil bS\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-3}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}k}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-3}\cr\end{array},

    or

  2. (2)

    τ=τj\tau=\tau_{j}, k=j+1k=j+1 and

    SRT(A)=ibS0jRj+1SRT\!\left(A_{\mathcal{I}}\right)=\begin{array}[]{ccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}i}\hfil\lx@intercol&\vrule\lx@intercol\hfil bS\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-3}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j+1}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-3}\cr\end{array}

    .

In either case, the resulting T0T0 violation is

SRT(B)=i0bSj0k.SRT\!\left(B_{\mathcal{I}}\right)=\begin{array}[]{ccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}i}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil bS\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-3}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}k}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-3}\cr\end{array}.

Proposition 19 can also be understood geometrically, in terms of possible realizations of the matrix. Suppose A𝒮N+A\in\mathcal{SR}^{+}_{N} has a blockage, not necessarily embedded with the “RR” on the first superdiagonal of AA. Then, in any possible realization of AA as the crossing matrix of a positive braid, the subbraid corresponding to this configuration consists of three strands, with the middle strand “hooking” one of the outer strands but crossing the other outer strand only once. In such a (sub)braid, all the “hooks” must precede the crossing, because once the crossing has occurred, the formerly outside strand separates the formerly middle strand from the other outer strand (see Figure 1).

In the situation where the “R” does lie on the first superdiagonal, this reasoning shows that attempting at that stage to introduce the transposition corresponding to that “R” cannot lead to a realization of the matrix. However, if via other divisions we can move the blockage so that the “S” entry is on the first superdiagonal, then we can divide by that transposition, thereby destroying the blockage, and continue.

\braidikj
Figure 1. Geometric Interpretation of a Blockage

The two examples in Subsection 3.1 both have the property that every nonzero entry in the first superdiagonal is a blocked “RR: the configurations G124G_{124}, G134G_{134}, K235K_{235} and K134K_{134} are all blockages whose “RR” is embedded in the first superdiagonal of the respective matrix. Therefore, it is impossible to factor either of these matrices into permutation matrices–or equivalently, neither matrix is the crossing matrix of any positive braid. We call a matrix A𝒮N+A\in\mathcal{SR}^{+}_{N} with this property a totally blocked matrix: it is impossible to left divide it by a transposition subordinate to the matrix (see below).

Remark 20.

If A𝒮N+A\in\mathcal{SR}^{+}_{N} is totally blocked–that is, every nonzero entry in the first superdiagonal of AA is a blocked “RR”– then AA is not the crossing matrix of any positive braid.

As we pointed out above, the presence of a blockage somewhere in A𝒮N+A\in\mathcal{SR}^{+}_{N} does not a priori mean that AA is not the crossing matrix of some positive braid–in fact, there are factorizations (equivalently, sequences of left divisions by transpositions) which create blockages, but then these blockages move around the matrix until they land with the “SS” term on the first superdiagonal.

When the “S” entry lies on the first superdiagonal, division by the transposition subordinate to it yields a configuration of the form

i0RjR+(b1)Sj+1\begin{array}[]{ccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}i}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j}\hfil\lx@intercol&\vrule\lx@intercol\hfil R+(b-1)S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-3}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j+1}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-3}\cr\end{array}

(or its mirror image) and a further division by the same transposition yields

iR0j(b1)Sj+1\begin{array}[]{ccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}i}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j}\hfil\lx@intercol&\vrule\lx@intercol\hfil(b-1)S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-3}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j+1}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-3}\cr\end{array}

(resp. its mirror image); repeating this process 2(b1)2(b-1) more times ends in the configuration

i0RjRj+1\begin{array}[]{ccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}i}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-3}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}j+1}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-3}\cr\end{array}

thus eliminating the blockage.

However, a matrix A𝒮N+A\in\mathcal{SR}^{+}_{N} need not be totally blocked to fail to be a crossing matrix of some positive braid. Consider for example the 6×66\times 6 matrix

V=[011100101010000011100011001000001100]V=\left[\begin{array}[]{cccccc}0&1&1&1&0&0\\ 1&0&1&0&1&0\\ 0&0&0&0&1&1\\ 1&0&0&0&1&1\\ 0&0&1&0&0&0\\ 0&0&1&1&0&0\end{array}\right]

with SRSR tableau

SRT(V)=1SRS002R0R030SS4RS506.SRT\!\left(V\right)=\begin{array}[]{cccccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}1}\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}2}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-6}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}3}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-6}\cr&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}4}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{4-6}\cr&&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}5}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{5-6}\cr&&&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}6}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{6-6}\cr\end{array}\quad.

The two “RR”’s on the first superdiagonal are blocked (V236V_{236}, V145V_{145}). Thus the only allowed left division is by τ1\tau_{1}, and this results in

SRT(V1=τ1\V)=1RR0R02RS0030SS4RS506.SRT\!\left(V1=\tau_{1}\backslash V\right)=\begin{array}[]{cccccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}1}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}2}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-6}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}3}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-6}\cr&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}4}\hfil\lx@intercol&\vrule\lx@intercol\hfil R\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{4-6}\cr&&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}5}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{5-6}\cr&&&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}6}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{6-6}\cr\end{array}.

which is totally blocked (the “S” in position 2,42,4 blocks the “RR”’s in positions 1,21,2 and 4,54,5, while the “SS” in position 3,53,5 blocks the “RR” in position 2,32,3). One might say that the original VV (as well as V1V1) is virtually totally blocked–every sequence of allowed left divisions by transpositions eventually leads to a totally blocked matrix. Such a matrix clearly cannot represent a positive braid.

This leads to a brute-force scheme for determining whether a given matrix A𝒮N+A\in\mathcal{SR}^{+}_{N} is the crossing matrix of some positive braid: quite simply, one tries every allowable sequence of left divisions by transpositions until one reaches either the zero matrix (in which case the original matrix has been factored into RR-matrices, which exhibits a realization of AA) or one reaches a totally blocked matrix. Since left division by a transposition subordinate to A𝒮N+A\in\mathcal{SR}^{+}_{N} reduces N(A)N(A), the sum of the entries of AA, by 11, any string of allowable left divisions will terminate in one of these two possibilities after at most N(A)N(A) steps. Of course the number of allowable sequences of left divisions is a priori on the order of N(A)!N(A)!, but we have implemented this scheme in Mathematica code which can handle matrices of size up to about 7×77\times 7 on a MacBook Pro with 8GB of memory.888We record our deep thanks to our colleague Bruce Boghosian, who spent many hours helping us develop this code, as well as our former student Dan Fortunato, who helped us at the beginning of this process.

Of course, this scheme provides an algorithmic characterization of crossing matrices for positive braids:

Theorem 21.

The crossing matrix of any positive braid on NN strands belongs to 𝒮N+\mathcal{SR}^{+}_{N}: it is a non-negative integer matrix with an SRSR decompostion.

Conversely, every such matrix A𝒮NA\in\mathcal{SR}_{N} is either the crossing matrix of some (perhaps many) positive braid on NN strands, or it is virtually totally blocked (i.e., every sequence of left divisions by transpositions corresponding to positions in the first superdiagonals of successive left quotients terminates in a totally blocked matrix).

However, this fails to be the kind of conceptual characterization of such crossing matrices which we would like to see: a criterion which can be applied directly to a matrix without an exhaustive search through allowable sequences of left divisions. We have not succeeded in formulating even a conjectural version of such a criterion.

4. The realizability problem for symmetric matrices

The fact that the crossing product operation \circledast is simply matrix addition when restricted to 𝔖N0[𝐙]\mathfrak{S}_{N}^{0}[\mathbf{Z}] makes it easier to think about realizability in this case–for example,

Remark 22.

A sum of realizable symmetric matrices is automatically realizable as a composition of the individual realizations.

Also, we note that for any RR-matrix RR (which by the discussion in Subsection 2.4 is the crossing matrix RπR_{\pi} of some permutation πΣN\pi\in\Sigma_{N}) the symmetrization SπS_{\pi} of RR is realized by π+(π¯)+\pi^{+}(\bar{\pi})^{+}. Combined with Remark 22 this shows

Remark 23.

If S𝔖N0[𝐙+]S\in\mathfrak{S}_{N}^{0}[\mathbf{Z}^{+}] is T0T0 and T1T1, it is (positively) realizable.

Using these observations together with a case-by-case argument, we can show that for N=4N=4, any T0T0 matrix in 𝔖N0[𝐙+]\mathfrak{S}_{N}^{0}[\mathbf{Z}^{+}] is (positively) realizable;

Theorem 24.

A symmetric 4×44\times 4 non-negative integer matrix with zero diagonal is positively realizable if and only if it is T0T0.

It is natural, based on this and our experimental evidence using the algorithm resulting from Theorem 21, that Theorem 24 extends to all NN.

An approach to trying to prove the analogue of Theorem 24 for all NN might be via an induction on the number of nonzero entries in the matrix, using Theorem 24 to establish an initial case, and then the following idea, which we find plausible but have not succeeded in proving or disproving. We call an entry aija_{ij} of the T0T0 matrix AA fully supported (even if that entry is zero) if, for every kk between ii and jj, at least one of the entries aika_{ik} and akja_{kj} is nonzero.

Conjecture 25.

Suppose AA is a positively realizable symmetric matrix, and aij=0a_{ij}=0 but the position is fully supported in AA. Then there is some realization of AA in which strands ii and jj become adjacent somewhere, so that changing aija_{ij} from zero to one results in a positively realizable matrix.

We caution that the word “some” is necessary here: the tableau

1S002S03S4=1S0020030410002S030410002003S4\begin{array}[]{cccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}1}\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}2}\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-4}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}3}\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-4}\cr&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}4}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{4-4}\cr\end{array}\quad=\begin{array}[]{cccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}1}\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}2}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-4}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}3}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-4}\cr&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}4}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{4-4}\cr\end{array}\quad\oplus\begin{array}[]{cccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}1}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}2}\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-4}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}3}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-4}\cr&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}4}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{4-4}\cr\end{array}\quad\oplus\begin{array}[]{cccc}\hline\cr\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}1}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \hline\cr&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}2}\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol&\vrule\lx@intercol\hfil 0\hfil\lx@intercol\vrule\lx@intercol\\ \cline{2-4}\cr&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}3}\hfil\lx@intercol&\vrule\lx@intercol\hfil S\hfil\lx@intercol\vrule\lx@intercol\\ \cline{3-4}\cr&&&\vrule\lx@intercol\hfil\hbox{\pagecolor[gray]{.80}4}\hfil\lx@intercol\vrule\lx@intercol\\ \cline{4-4}\cr\end{array}

can be realized as a composition of three “hooks” in any order. If the middle one in the sum above occurs before or after both of the others (Figure 2), strands 11 and 44 come into adjacent positions,

\braid12341234
Figure 2. Middle hook after the others: strands 1 and 4 adjacent

but if it occurs between them (Figure 3), then at any stage either strand 2 or strand 3 separates these two.

\braid12341234
Figure 3. Middle between the others: strands 1 and 4 separated by strands 2 and 3

Given this conjecture as a lemma, an inductive argument goes as follows: in any T0T0 matrix, there are nonzero entries which do not support any other element (since all supports of an entry live in lower-numbered superdiagonals). “Erasing” one such entry yields a T0T0 matrix with a lower entry sum; by induction on this sum, the latter is realizable, and hence by the conjectured lemma, so is our given matrix.

References

  • [1] E. Artin, Theorie der Zöpfe, Abhandlungen der Mathematisches Seminar Universität Hamburg 4 (1925), no. 1, 47–72.
  • [2] by same author, Theory of braids, Annals of Mathematics 48 (1947), no. 2, 101–126.
  • [3] J. Burillo, M. Gutierrez, S. Krstić, and Z. Nitecki, Crossing matrices and Thurston’s normal form for braids, Topology and its Applications 118 (2002), 293–308.
  • [4] W. Thurston, Braid groups, Word processing in groups (D. Epstein, ed.), Jones and Bartlett, Boston, MA, 1992, pp. 181–209.