*Lecture 7 (19.10.2015)*

The notion of $p$-colourability of a link lead us last time to consider certain linear equations modulo a prime $p$. Before proceeding, we recall some properties of working “mod $p$”:

Let $a,b,c\in\{0,1,…,p-1\}$. Then $a=b\mod p$ means that $a-b$ is an integer multiple of $p$, that is, $a-b=n\cdot p$ for some $n\in\mathbb Z$.

We can define addition and multiplication in such a way that we do not leave the set $\{0,1,…,p-1\}$. Namely, $a+ b$ is defined to be $(a+b)\mod p$, that is, when “dividing with remainder” by $p$, $a+b$ is the remainder of that division. It is like reading a clock, which is mod 12 (or mod 24).

Example: Let $p=5$. Then $2+4=6=5+1=1\mod 5$.

Add the numbers as you are used to, then keep subtracting/adding $p$ until you are back into $\{0,1,…,p-1\}$.

The very same is true for multiplication, where we define $a\cdot b$ to be $a\cdot b\mod p$.

Example: Let $p=5$. Then $2\cdot 4=8=5+3=3\mod 5$.

For

divisionit is important that $p$ is prime. Namely, when $$a\cdot b=a\cdot c\mod p\qquad \text{and}\qquad a\neq 0 \mod p,$$then $$b=c\mod p.$$

Because by the first equation above, $a\cdot(b-c)$ is a multiple of $p$, and by the second equation above, $a$ is not a multiple of $p$. As $p$ is prime, this implies that $b-c$ is a multiple of $p$, which means precisely $b=c\mod p$.

Let us now come back to the example of the labeled diagram of the figure of eight knot:

We had translated the linear equations of Rule 2 of Def. 2.4 into the matrix equation

$\begin{align*}\left(\begin{array}{rrrr}-1&2&-1&0\\2&-1&0&-1\\-1&0&2&-1\\0&-1&-1&2\end{array}\right)\left(\begin{array}{r}x_1\\x_2\\x_3\\x_4\end{array}\right)=0\,\mod p\,,\end{align*}$

where $0$ denotes the 0 vector with four components, and mod $p$ refers to each component.

We have to keep in mind that we also have to obey Rule 1, which states that we do not want a “monochromatic” or “constant” solution of the form $x_1=x_2=…=x_{|D|}$. To remove this constraint, observe the following:

- Suppose the above matrix equation has a non-constant solution $x$. It also (always) has the constant solution $y$, with $y_1=y_2=…=y_{|D|}=1$. Since the system is linear, also $x+n\cdot y$ is a solution, for any integer $n$. We can always choose $n$ in such a manner that $x_1+n=0\mod p$, or $x_i+n=0\mod p$ for some $i\in\{1,…,|D|\}$. The solution $x’:=x+n\cdot y$ will have two properties: a) $x_i’=0$ and b) $x’$ is not constant.

Putting the $i$-th comonent to zero in the above matrix equation means that we delete column $i$ from our matrix, and entry $i$ from our vector. - Conversely, suppose we take the $|D|\times|D|$ matrix above and delete some (arbitrarily chosen) column, say number $i$, and suppose we find a non-zero solution of the system $Bx=0\mod p$, where $B$ is the $|D|\times(|D|-1)$ matrix obtained by removing column $i$. Then, by inserting a $0$ in the $i$-th component of that solution, we end up with a solution of the original system $\tilde{A}x=0\mod p$ which is not constant. (Here $\tilde{A}$ denotes the original $|D|\times|D|$-matrix.)

This means: The “reduced” system $Bx=0\mod p$ has *non-zero* solutions if and only if the original system $\tilde{A}x=0\mod p$ has *non-constant* solutions.

In the example of the figure eight knot, let us delete column 3. Then

$\begin{align*}\tilde{A}=\left(\begin{array}{rrrr}-1&2&-1&0\\2&-1&0&-1\\-1&0&2&-1\\0&-1&-1&2\end{array}\right)\,,\qquad B=\left(\begin{array}{rrrr}-1&2&0\\2&-1&-1\\-1&0&-1\\0&-1&2\end{array}\right).\end{align*}$

Now we observe that in the matrix $B$, the rows are linearly dependent: They sum to zero. Hence (linear algebra) the system $Bx=0\mod p$ has a non-zero solution if and only if the system $Ax=0\mod p$ has a non-zero solution, where $A$ is produced from $B$ by deleting some arbitrarily chosen *row*. In the example, let’s say we delete row 2 from $B$. Then

$\begin{align*}A=\left(\begin{array}{rrr}-1&2&0\\-1&0&-1\\0&-1&2\end{array}\right).\end{align*}$

This $A$ is again a square matrix, one size smaller than $\tilde{A}$. From linear algebra, we know that $Ax=0$ has a non-zero solution if and only if $A$ is not invertible if and only if $\det A=0$. Since we are working mod $p$ here, we find that $Ax=0\mod p$ has a non-zero solution if and only if $\det A=0\mod p$.

Computing the determinant, we get

$\begin{align*}\det A=\det\left(\begin{array}{rrr}-1&2&0\\-1&0&-1\\0&-1&2\end{array}\right)=\det\left(\begin{array}{rrr}-1&2&0\\0&-2&-1\\0&-1&2\end{array}\right)=-\det\left(\begin{array}{rr}-2&-1\\-1&2\end{array}\right)=5.\end{align*}$

Note that in all these computations the prime $p$ we are working mod $p$ with has not appeared explicitly. This comes now: We saw that the figure of eight knot is $p$-colourable if and only if $\det A=0\mod p$. By our above calculation, this is equivalent to $5=0\mod p$, i.e. $5$ has to be divisible by $p$. This is possible if and only if $p=5$. Hence the figure of eight knot is $5$-colourable, and not $p$-colourable for any other prime $p$.

Let us now look at the **general procedure**, i.e. not at the figure of eight knot, but rather a general link diagram. To begin with, let us assume that we have a diagram $D$ which has the same number of crossings and arcs (this is for example not true for the unknot, we will discuss these exceptional cases later.)

We proceed in several steps.

- Take a diagram $D$ with the same number $|D|$ of crossings and arcs. Label the crossings $1,2,…,|D|$ and the arcs $x_1,x_2,…,x_{|D|}$ in some arbitrary order.
- Write down the $|D|\times|D|$ matrix $\tilde{A}$ which represents all the linear equations of Rule 2 of Def. 2.4. Here each row corresponds to a crossing, and each column corresponds to an arc.
- Remove one arbitrarily chosen row and one arbitrarily chosen column from $\tilde{A}$. The resulting $(|D|-1)\times(|D|-1)$ matrix $A$ is called
*a colouring matrix of*$D$. - Compute the determinant of $A$.
- $D$ is $p$-colourable if and only if $\det A=0 \mod p$.

In the example, we saw that the step from $\tilde{A}$ to $A$ precisely translated $p$-colourability of $D$ into the condition that $Ax=0\mod p$ has non-zero solutions. This step had two aspects: First, we removed a column, exploiting the linearity of the system, and translated non-constant solutions of $\tilde{A}x=0\mod p$ to non-zero solutions of $Bx=0\mod p$, where $B$ was the matrix obtained from $\tilde{A}$ by removing one column. This works in just the same manner for any diagram.

Second, we observed that the the rows of $B$ summed to zero. This is not true in general, but it is always true that the rows of $B$ are linearly dependent (see, for example, the book of Livingston). Hence also in general, we may remove a column of $B$ (and call the resulting matrix $A$) without loss of information, and find that $D$ is $p$-colourable if and only if $Ax=0\mod p$ has a non-zero solution.

We therefore define

Definition 2.6.:The determinant of a diagram $D$ is defined as $$\det(D):=|\det(A)|\,,$$ where $A$ is some colouring matrix of $D$.

Note that although we made several choices for computing the determinant (how to label the crossings and arcs, which row and column to delete), these choices affect the determinant only by a sign $\pm1$. Since we take the absolute value in Def. 2.6., it is clear that the determinant of a diagram is well-defined, i.e. independent of the choices made.

Our discussion has shown:

Theorem 2.7:A diagram $D$ is $p$-colourable if and only if $\det(D)=0\mod p$.

Actually, a little more is true. To prepare this, we have to worry about the exceptional diagrams which have arcs that are closed loops, like the unknot, the 2-unlink, etc. We agree to define the determinant of a link as follows.

Definition 2.8.:The determinant of a link $L$ is defined as follows: First represent $L$ by a diagram $D$ which does not have arcs that are closed loops. Then define $\det(L):=\det(D).$ Here, the determinant of a $0\times 0$ matrix is defined to be 1.

Slightly stronger than Theorem 2.7, there holds:

Theorem 2.9:The determinant is a link invariant.

We will not give a proof of this theorem (see, for example, the book of Livingston).

Let us rather look at three more **examples**:

*The unknot.*We first have to choose a diagram of the unknot which does not have closed loops, like the one on the right in this image:

There is just one arc “1” and crossing $x_1$, and the condition at that crossing is $2x_1-x_1-x_1=0\mod p$. Hence we have $$\tilde{A}=\left(0\right).$$ Deleting one row and column gives us the zero times zero matrix, whose determinant is 1 by definition. Hence the determinant of the unknot is 1. Besides the unknot, there exist other knots, not equivalent to the unknot, with determinant 1. Hence the determinant is not a complete invariant. We will come back to this point next time.*The trefoil*. We label its standard diagram as

and read off the matrix $\tilde{A}$ to be

$$\tilde{A}=\left(\begin{array}{rrr}2&-1&-1\\-1&-1&2\\-1&2&-1\end{array}\right).$$ To arrive at a colouring matrix, we have to delete one row and column, let’s say row three and column three. Then $$A=\left(\begin{array}{rr}2&-1\\-1&-1\end{array}\right).$$ This matrix has determinant $-3$, so the determinant of the trefoil is $3$.This means that the trefoil is $3$-colourable (because $3=0\mod 3$), a fact that we knew before. Now we learned in addition that the trefoil is not $p$-colourable for any prime $p\neq 3$.*Splittable links.*As far as determinants of links are concerned, one “extreme situation” is a link with determinant 1. As also the unknot has determinant 1, we can not distinguish such links from the unknot by looking at determinants only. The other “extreme situation” are links with determinant 0. If a link $L$ has determinant zero, then $\det L=0\mod p$ for any prime $p$. Hence such links are $p$-colourable for any prime $p$.

This happens in particular when the link is*splittable*, a term that was already discussed in the context of linking number. Recall that $L$ being splittable means that $L$ has a diagram consisting of two disjoint parts, one each on the right and left side of the plane, with no connecting arcs. In a rough picture, this looks as

For the sake of the argument, we have not drawn proper diagrams here, and also not indicated over- and undercrossings properly. Why is a link of such a form $p$-colourable for any $p$? Well, independent of $p$, you always have the variables $0$ and $1$ in your set $\{0,1,…,p-1\}$ (As a prime, $p$ is larger than 1). Assign $0$ to any arc in the left part, and assign 1 to any arc in the right part of the diagram. Then the labeling is not monochromatic (because two different values appear), and also Rule 2 of Def. 2.4 is satisfied: At each crossing, $2a-b-c=0\mod p$ with the usual meaning of $a,b,c$, because only arcs with the same label meet. Hence $L$ is $p$-colourable for any $p$, i.e. $\det L$ is divisible by all primes $p$. This implies $\det L=0$. This observation, as well as the fact that there exist non-trivial knots with determinant 1, are shortcomings of the determinant as a link invariant. We will need to find “stronger” invariants. As an aside, we mention that there also exist links which are not splittable but still have zero determinant.

Problem 6b is an exercise on computing determinants of knots.