# Is the normalised Kauffman bracket more powerful than the Kauffman bracket?

The Kauffman bracket polynomial for a knot diagram $D$ is a Laurent polynomial $\langle D \rangle \in \mathbb{Z}[A, A^{-1}]$. Although it is invariant under Reidemeister moves of type II and III, performing a type I move on $D$ will change $\langle D \rangle$ by a factor of $-A^{\pm 3}$. So $\langle D \rangle$ is not a knot invariant.

One solution to this is to define a normalised bracket polynomial: $$X(D) = (-A^3)^{-w(D)} \langle D \rangle \quad \textrm{where w(D) is the writhe of D}.$$ This is invariant under type I moves too and so is a knot invariant (it also turns out to be a reparametrisation of the Jones polynomial).

However an alternate solution is to simply say that the bracket polynomial is only defined up to multiplication by a power of $-A^3$. Although this appears to be disregarding some information, it is still possible to use it to show that many pairs of knots are distinct. For example, as $$A^{16} - A^{12} - A^4 \neq (-A^3)^k \times 1 \quad \textrm{for any k \in \mathbb{Z}}$$ we can still conclude that the trefoil is not equal to the unknot. However, is this actually a weaker knot invariant? That is

Are there two distinct knots which the normalised bracket polynomial can tell apart but which the Kauffman bracket polynomial cannot?

I've checked all the knots listed on KnotInfo (which covers prime knots up to 12 crossings) and found no such pair.

It is also worth noting however that if we allow ourselves to work with oriented links then: if we reverse the orientation of one component $k$ of an oriented link diagram $D$ to obtain a new diagram $D'$ then $$\langle D' \rangle = A^{12 \lambda} \langle D \rangle \quad \textrm{where \lambda = \textrm{linknum}(k, D - k).}$$ Hence, for example, the normalised bracket polynomial can distinguish between the compatibly and incompatibly oriented Hopf links (L2a1{0} and L2a1{1}) but the bracket polynomial cannot.

• Rather than say that the Kauffman bracket is not a knot invariant, I would say that it is an invariant of framed knots. (Which, in my biased opinion, is nicer/better than being an invariant of unframed knots.) – Kevin Walker Aug 31 '13 at 14:01