Lecture 9 (26.10.2015)

The link invariants we have discussed so far were integers (crossing number, linking number, the determinant), of “yes-no type” (3-colourability, $p$-colourability), or polynomials (the Alexander polynomial). Today we introduce another invariant, which is more abstract – corresponding to each diagram, we will define a group.

Since no extensive background on group theory is required on your side, we start with an example:

The group ${\mathbb Z}^m$ (where $m$ is an integer.)

The set ${\mathbb Z}^m$ consists of all vectors with $m$ integer components.

  • We can add such vectors (component wise), i.e. have a map ${\mathbb Z}^m\times{\mathbb Z}^m\to{\mathbb Z}^m$, defined as $$\boldsymbol{x},\boldsymbol{y}\mapsto\boldsymbol{x}+\boldsymbol{y}.$$
  • There is the zero vector (the “neutral element”) $\boldsymbol{0}\in{\mathbb Z}^m$, adding which to some $\boldsymbol{x}\in{\mathbb Z}^m$ does not change it: $$\boldsymbol{x}+\boldsymbol{0}=\boldsymbol{x}\,,\qquad \boldsymbol{x}\in{\mathbb Z}^m.$$
  • To each $\boldsymbol{x}\in{\mathbb Z}^m$, there exists an “inverse element” $-\boldsymbol{x}$, such that $$\boldsymbol{x}+(-\boldsymbol{x})=\boldsymbol{0}.$$

A set together with an (associative) composition map satisfying these properties is called a group.
${\mathbb Z}^m$ is an abelian group, which means that the order of summation is irrelevant: $\boldsymbol{x}+\boldsymbol{y}=\boldsymbol{y}+\boldsymbol{x}$ for all $\boldsymbol{x},\boldsymbol{y}\in{\mathbb Z}^m$.

 

In the context of knots, we will be concerned with very special groups, the building blocks of which can be described as follows: Take $n\in{\mathbb Z}$, and define for $x\in{\mathbb Z}$ $$[x]_n:=x+n{\mathbb Z}=\{x,x\pm n,x\pm 2n,x\pm 3n,…\,\}.$$ The set of all these classes (they are actually equivalence classes) is denoted ${\mathbb Z}/n{\mathbb Z}$. Classes can be added similar to numbers, by defining $$[x]_n+[y]_n:=[x+y]_n\,,\qquad x,y\in {\mathbb Z}.$$ One then quickly checks that ${\mathbb Z}/n{\mathbb Z}$ is also a group.

The remarkable property of ${\mathbb Z}/n{\mathbb Z}$ is the following: If you take the class $[1]_n$, and you add it $n$ times to itself, then you get $$\begin{align*}
[1]_n+[1]_n+…+[1]_n
&=
[n]_n\qquad \text{(}n\text{ times)}
\\
&=
\{n,n\pm n,n\pm 2n,…\,\}
\\
&=
\{0,\pm n,\pm 2n,…\,\}
\\
&=
[0]_n\,.
\end{align*}$$

That is, in ${\mathbb Z}/n{\mathbb Z}$ you can ” add 1 $n$-times to itself and then you get 0″, when interpreted as adding the classes $[1]_n$. ${\mathbb Z}/n{\mathbb Z}$ is called the cyclic group of order $n$. Sometimes it is also written as ${\mathbb Z}_n$, which should not be confused with ${\mathbb Z}^n$.

We can do something very similar in $m$ dimensions. If $C$ is an $(m\times m)$ matrix with integer entries, then we can consider the classes $$[\boldsymbol{x}]_C:=\boldsymbol{x}+C\,{\mathbb Z}^m,$$ where $\boldsymbol{x}\in{\mathbb Z}^m$. The collection of all these classes is denoted ${\mathbb Z}^m/C{\mathbb Z}^m$. As before, you can check that ${\mathbb Z}^m/C{\mathbb Z}^m$ is an abelian group with the composition law $[\boldsymbol{x}]_C+[\boldsymbol{y}]_C:=[\boldsymbol{x}+\boldsymbol{y}]_C$.

Example: Let $C$ be diagonal, i.e.

$$
\begin{align*}
C=\left(
\begin{array}{ccc}
c_1\\ &\ddots\\&&c_m
\end{array}
\right)
\end{align*}$$
Then $$
\begin{align*}
[\boldsymbol{x}]_C
&=
\boldsymbol{x}+C\,{\mathbb Z}^m
\\
&=
\left(\begin{array}{c} x_1\\\vdots\\ x_m\end{array}\right)
+
\left(\begin{array}{c} 0,\pm c_1,\pm 2 c_1,…\,\\\vdots\\
0,\pm c_n,\pm 2 c_m,…\end{array}\right)
\\
&=
\left(\begin{array}{c} [x_1]_{c_1}\\\vdots\\ \,[x_m]_{c_m}\end{array}\right)
\end{align*}$$
Hence, in this diagonal case, $${\mathbb Z}^m/C\,{\mathbb Z}^m\cong{\mathbb Z}/c_1{\mathbb Z}+…+{\mathbb Z}/c_m{\mathbb Z},$$ where $\cong$ means a group isomorphism: A bijective map $\varphi$ with $\varphi(g+h)=\varphi(g)+\varphi(h)$ for all group elements $g,h$.

To clarify the structure of ${\mathbb Z}^m/C\,{\mathbb Z}^m$ when $C$ is not diagonal, we can diagonalise it:

Lemma 2.13: Let $C$ be an $(m\times m)$-matrix with integer coefficients. Then there exist three $(m\times m)$-matrices $L,E,R$ with integer coefficients, such that

  • $C=L\cdot E\cdot R$ (matrix product)
  • $E$ is diagonal
  • $\det L=1$, $\det R=1$

In particular, in this case we have $\det C=\det E=e_1\cdot e_2\cdots e_m$, where the $e_j$ denote the diagonal entries of $E$, and $${\mathbb Z}^m/C{\mathbb Z}^m\cong{\mathbb Z}^m/E{\mathbb Z}^m\cong{\mathbb Z}/e_1{\mathbb Z}+…+{\mathbb Z}/e_m{\mathbb Z},$$ where the $e_j$ are the diagonal entries of $E$.

The proof of this lemma is at the same time the calculational scheme for computing the diagonal entries $e_1,…,e_m$, which will be important in the context of knots. It works as follows.

Starting from the matrix $C$, you can perform four types of operations on $C$:

  • switch two columns, but change the sign of one of them.
  • add an integer multiple of one column to another (different one)
  • switch two rows, but change the sign of one of them.
  • add an integer multiple of one row to another (different one)

One can check that each of these operations on $C$ corresponds to multiplying $C$ from the left or right by a matrix with integer coefficients, with determinant 1. By carrying out finitely many such operations, $C$ can always be put into diagonal form.

  • First, order rows and columns in such a way that a nonzero entry of smallest absolute value goes into the top left corner.
  • Then subtract the first row/column from the remaining rows/columns to reduce the other entries in the first row/column.
  • After doing these two steps finitely many times, you arrive at a situation where in the first row/column, the only non-zero entry is the one in the top left corner.
  • Now proceed on the $(m-1)\times(m-1)$ matrix in the lower right corner in the same manner.
  • Proceeding like this, you obtain a diagonal matrix $E$ after finitely many steps.

Example: We start with the matrix

$$\begin{align*}
C=\left(\begin{array}{rrr}2&-1&0\\-1&2&-1\\0&-1&2\end{array}\right)
\end{align*}
$$
which, by the way, is a colouring matrix of the link
6knots

Now we switch the first two columns and multiply the (former) second one by -1, to put a non-zero entry of minimal modulus in the upper left. This gives
$$
\begin{align*}
\left(\begin{array}{rrr}1&2&0\\-2&-1&-1\\1&0&2\end{array}\right).
\end{align*}
$$

Now we already have a 1 in the upper left, which is most convenient. We subtract twice the first column from the second one, which gives
$$
\begin{align*}
\left(\begin{array}{rrr}1&0&0\\-2&3&-1\\1&-2&2\end{array}\right).
\end{align*}
$$

Now we add twice the first row to the second row, and subtract the first row from the third row. This gives
$$
\begin{align*}
\left(\begin{array}{rrr}1&0&0\\0&1&3\\0&-2&-2\end{array}\right).
\end{align*}
$$
The only non-zero entry in the first row and column is now the upper left corner. Therefore we can proceed to work with the remaining $2\times 2$ block in the lower right corner. We add twice the second row to the third one,
$$
\begin{align*}
\left(\begin{array}{rrr}1&0&0\\0&1&3\\0&0&4\end{array}\right),
\end{align*}
$$
and, finally, substract three times the second column from the third column. This yields the diagonal matrix
$$
\begin{align*}
E=\left(\begin{array}{rrr}1&0&0\\0&1&0\\0&0&4\end{array}\right),
\end{align*}
$$

with the diagonal entries $e_1=1$, $e_2=1$, $e_3=4$. In particular, the determinant of $C$ is $4$.

 

 

Now we want to apply this technique to studying a new invariant, the colouring group of a link. 

Definition 2.14: Let $D$ be a diagram with no closed loops as arcs. We define the colouring group of $D$, written as Col$(D)$, as the abelian group which has all but one of the arcs of $D$ as generators, with the crossing relations “2a=b+c” at each crossing. Formally, $$\text{Col}(D)=\left\langle x_1,…,x_{|D|-1}\,:\,A\boldsymbol{x}=0\,,\;x_{|D|}=0\right\rangle.$$

Here we decided to delete the arc number $|D|$ by setting $x_{|D|}$ to zero. The pointed bracket expression $\langle x_1,x_2, … \,:\, …\rangle$ means: The entries to the left of the “:” are the generators of the group, i.e. we look at all combinations of them under the group composition. The entries to the right of “:” mean that we impose the relations that are written there; in our case, the usual crossing relations.

This definition is fairly abstract. Nonetheless, one can show that it defines a link invariant which is easily computable.

Theorem 2.15: Let $D\cong D’$ be two diagrams without closed loops as arcs. Then Col$(D)\cong$Col$(D’)$, i.e. the two colouring groups are isomorphic.

Similar to what we saw with the Alexander polynomial, the colouring groups of two equivalent diagrams are not quite “the same”, but “essentially” they are identical, in the sense of being isomorphic.

Of course the question emerges how to check if two colouring groups are isomorphic. Relating to the first part of this lecture, first one finds that one always has $$\text{Col}(D)\cong{\mathbb Z}^{|D|-1}/A^T{\mathbb Z}^{|D|-1},$$ where $A^T$ is the transpose of a colouring matrix of $D$.

Diagonalizing $A^T$ and using Lemma 2.13, one finds further $$\text{Col}(D)\cong{\mathbb Z}^{|D|-1}/A^T{\mathbb Z}^{|D|-1}\cong{\mathbb Z}/e_1{\mathbb Z}+…+{\mathbb Z}/e_{|D|-1}{\mathbb Z},$$ where the $e_j$ are the diagonal entries of the diagonal matrix $E$ obtained from $A^T$ by diagonalization.

The (absolute value of) the product of all these integers is the determinant of the diagram $D$, $$\det D=|e_1\cdots e_{|D|-1}|.$$ But whereas we knew the determinant before, now we have a finer splitting of $\det D$ into a product, which carries more information. To evaluate this additional information, we restrict ourselves to the cases of integer entries different from 0 (with ${\mathbb Z}/0{\mathbb Z}\cong {\mathbb Z}$ or $1$ (with ${\mathbb Z}/1{\mathbb Z}\cong\{0\}$).

If $L\cong L’$ are two equivalent links, and $E, E’$ the diagonal matrices obtained from some of their colouring matrices (from some labeling) by diagonalisation, then the following is true: If $E$ has an entry $k$ on its diagonal, which is not 0 and not 1, and which appears exactly $N$ times on the diagonal of $E$, then also $E’$ must have $k$ exactly $N$ times on its diagonal.

For example, suppose you have two links $L,L’$ with determinants $\det L=\det L’=45$. Then we cannot be sure if $L\cong L’$ or not by looking at the determinant. But if we have computed, say

$$E=\left(\begin{array}{rrr}3&\\&3\\&&5\end{array}\right)$$

for the first link, $L$, and

$$E’=\left(\begin{array}{rrr}1&\\&5\\&&9\end{array}\right)$$

for the second link, $L’$, then we see that $L, L’$ are not equivalent because $E$ has two 3’s on its diagonal, whereas $E’$ does not.

Example: The diagonalisation example above shows that the 2-link $L$ with diagram

6knots

has colouring group Col$(L)\cong{\mathbb Z}/{\mathbb Z}+{\mathbb Z}/{\mathbb Z}+{\mathbb Z}/4{\mathbb Z}$. (Since ${\mathbb Z}/{\mathbb Z}=\{0\}$, this group is the same as ${\mathbb Z}/4{\mathbb Z}$.)