computing the Kauffman bracket with the given relation

Question

My Problem: Use the relation

to compute the bracket of the link diagram $D_n$ with $n$ components:

My attempt: It seems to me that raising the given equation to the $n^{th}$ power is the most sensible. (Note, that I interpret the power of a "knotted chunk," or the product of several of them, to be the respective connect-sum.)

So I would have

$$ \langle{D_n}\rangle = ((1-A^4)\langle x\rangle+A^{-2} \langle y\rangle )^n. $$

Is this correct?

Answer 1

You need to close it off at the end, since if I understand what you mean, $((1-A^4)\langle x\rangle + A^{-2}\langle y\rangle)^n$ will just be a tangle. So, you would expand this out, then close off $\langle x\rangle$ and $\langle y\rangle$, giving $\delta$ and $1$, respectively, where $\delta=-A^2-A^{-2}$.

The relations you need are

• $\langle x\rangle\langle x\rangle=\langle x\rangle$
• $\langle x\rangle\langle y\rangle=\langle y\rangle$
• $\langle y\rangle\langle x\rangle=\langle y\rangle$
• $\langle y\rangle\langle y\rangle=\delta\langle y\rangle$

Notice the horizontal composition of the brackets of these tangles is commutative, so you can expand your expression out like a polynomial: \begin{align} ((1-A^4)\langle x\rangle + A^{-2}\langle y\rangle)^n &= \sum_{k=0}^n \binom{n}{k}(1-A^4)^kA^{-2(n-k)}\langle x\rangle^k \langle y\rangle^{n-k} \\ &= (1-A^4)^n\langle x\rangle + \sum_{k=0}^{n-1} \binom{n}{k} (1-A^4)A^{2(k-n)} \delta^{n-k} \langle y\rangle \end{align}

Hence, \begin{align} \langle D_n\rangle &= (1-A^4)^n\delta + \sum_{k=0}^{n-1} \binom{n}{k} (1-A^4)A^{2(k-n)} \delta^{n-k}. \end{align}

That is, this is the bracket polynomial modulo any errors I might have made.

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Another approach I like is to see what the clasp does to $a\langle x\rangle+b\langle y\rangle$ when attached to one side. In the $\langle x\rangle,\langle y\rangle$ basis, this can be represented by the matrix $$\begin{pmatrix} 1-A^4 & 0 \\ A^{-2} & -A^4-A^{-4} \end{pmatrix}$$ This matrix is diagonalizable as $PDP^{-1}$ with \begin{align} P&=\begin{pmatrix} A^2+A^{-2} & 0 \\ 1&1\end{pmatrix} &D&=\begin{pmatrix} 1-A^4 & 0 \\ 0 & -A^4-A^{-4} \end{pmatrix} \end{align} Hence, we may take the $n$th power of the matrix by computing $PD^nP^{-1}$, which is \begin{align} PD^nP^{-1} &= \begin{pmatrix} \left(1-A^4\right)^n & 0 \\ \frac{\left(1-A^4\right)^n-\left(-A^4-A^{-4}\right)^n}{A^2+A^{-2}} & \left(-A^4-A^{-4}\right)^n \\ \end{pmatrix} \end{align} The first column of this is what $n$ clasps does to $\langle x\rangle$, which is what we want to then close off to get $\langle D_n\rangle$. Hence, \begin{align} \langle D_n\rangle &= \left(1-A^4\right)^n \delta + \frac{\left(1-A^4\right)^n-\left(-A^4-A^{-4}\right)^n}{A^2+A^{-2}}\\ &= \left(-A^4-A^{-4}\right)^n \end{align}