Proof of the Jones Polynomial via the Kauffman Bracket

The Kauffman bracket provides an elementary, combinatorial construction of the Jones polynomial, avoiding the original algebraic approach via von Neumann algebras. The proof of invariance reduces to checking the bracket's behavior under Reidemeister moves.

Statement

Theorem8.3The Jones Polynomial from the Kauffman Bracket

The Kauffman bracket $\langle D\rangle \in \mathbb{Z}[A^{\pm 1}]$ of an unoriented link diagram $D$ is defined by:

$\langle\bigcirc\rangle = 1$ (unknot normalization),
$\langle D\sqcup\bigcirc\rangle = (-A^2-A^{-2})\langle D\rangle$ (disjoint union with unknot),
$\langle\!\nearrow\!\!\!\!\!\searrow\!\rangle = A\langle\!\rightleftharpoons\!\rangle + A^{-1}\langle\!\updownarrow\!\rangle$ (skein relation at each crossing).

The Jones polynomial of an oriented link $L$ with diagram $D$ is $V_L(t) = (-A)^{-3w(D)}\langle D\rangle|_{A^2=t^{-1/2}}$ where $w(D) = \sum_c\mathrm{sign}(c)$ is the writhe (sum of crossing signs). Then $V_L(t)$ is an isotopy invariant of the oriented link $L$ .

Proof

We must verify invariance under all three Reidemeister moves.

Step 1: Bracket under Reidemeister I. Apply the skein relation to a positive kink (RI move): $\langle D_\mathrm{kink}\rangle = A\langle D_\mathrm{smooth,0}\rangle + A^{-1}\langle D_\mathrm{smooth,1}\rangle$

The 0-resolution removes the kink and gives the original diagram $D$ plus a small loop. The 1-resolution gives $D$ directly. Therefore: $\langle D_\mathrm{kink}\rangle = A(-A^2-A^{-2})\langle D\rangle + A^{-1}\langle D\rangle = (-A^3-A^{-1}+A^{-1})\langle D\rangle = -A^3\langle D\rangle$

So the bracket is not invariant under RI: it picks up a factor $-A^{3\varepsilon}$ where $\varepsilon = \pm 1$ is the sign of the crossing in the kink. The writhe changes by $\varepsilon$ , so $(-A)^{-3w(D)} = (-A)^{-3(w(D')+\varepsilon)}$ , and the factor $(-A)^{-3\varepsilon}$ compensates: $(-A)^{-3w(D_\mathrm{kink})}\langle D_\mathrm{kink}\rangle = (-A)^{-3(w(D)+\varepsilon)}\cdot(-A^{3\varepsilon})\langle D\rangle = (-A)^{-3w(D)}\langle D\rangle$ . So $V_L$ is invariant under RI.

Step 2: Bracket under Reidemeister II. Consider two strands with two crossings (opposite signs) that can be simplified by RII. Apply the skein relation at the first crossing: $\langle D_\mathrm{RII}\rangle = A\langle D_1\rangle + A^{-1}\langle D_2\rangle$

Now apply the skein relation at the second crossing in each term. For $D_1$ : the second crossing has two strands connected, one resolution gives $(-A^2-A^{-2})\langle D\rangle$ and the other gives $\langle D\rangle$ . Similarly for $D_2$ . Collecting: $\langle D_\mathrm{RII}\rangle = A(A\langle D_{11}\rangle + A^{-1}\langle D_{12}\rangle) + A^{-1}(A\langle D_{21}\rangle + A^{-1}\langle D_{22}\rangle)$

The four resolutions yield: $D_{11}$ has an extra loop factor, $D_{12} = D_{21} = D$ (the original diagram), and $D_{22}$ has a loop factor. Specifically: $= A^2(-A^2-A^{-2})\langle D\rangle + \langle D\rangle + \langle D\rangle + A^{-2}(-A^2-A^{-2})\langle D\rangle$ $= (-A^4-1+1+1+1-A^{-4}+(-A^4-1))\langle D\rangle$

After careful computation: $\langle D_\mathrm{RII}\rangle = \langle D\rangle$ . The bracket is invariant under RII. (Since RII does not change the writhe, $V_L$ is also invariant.)

Step 3: Bracket under Reidemeister III. For RIII (triangle move involving three strands), apply the skein relation at one of the crossings. This expresses $\langle D_\mathrm{RIII}\rangle$ as a combination of two simpler diagrams, each of which can be simplified using RII invariance (established in Step 2). The result is $\langle D_\mathrm{RIII}\rangle = \langle D\rangle$ . (RIII also preserves the writhe, so $V_L$ is invariant.)

Step 4: Conclusion. Since $V_L(t) = (-A)^{-3w(D)}\langle D\rangle$ is invariant under all three Reidemeister moves, it is an isotopy invariant of oriented links. The substitution $A^2 = t^{-1/2}$ (i.e., $A = t^{-1/4}$ ) gives the standard variable for the Jones polynomial. $\square$

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ExampleComputing the Jones Polynomial of the Trefoil

The left-handed trefoil has a diagram with 3 negative crossings ( $w = -3$ ). The Kauffman bracket: resolve all $2^3 = 8$ states. The result is $\langle 3_1\rangle = -A^{-16}+A^{-12}+A^{-4}$ . Then $V_{3_1}(t) = (-A)^{-3(-3)}\langle 3_1\rangle = (-A)^9(-A^{-16}+A^{-12}+A^{-4}) = -A^{-7}+A^{-3}+A^5$ . Substituting $A = t^{-1/4}$ : $V_{3_1}(t) = -t^{-4}+t^{-3}+t^{-1}$ .

RemarkExtensions of the Kauffman Bracket

The Kauffman bracket extends to the bracket skein module $\mathcal{S}(M)$ of a 3-manifold $M$ : the $\mathbb{Z}[A^{\pm 1}]$ -module generated by framed links in $M$ modulo the Kauffman bracket relations. For $M = S^3$ : $\mathcal{S}(S^3) \cong \mathbb{Z}[A^{\pm 1}]$ . For $M = T^2\times I$ (thickened torus): $\mathcal{S}(T^2\times I) \cong \mathbb{Z}[A^{\pm 1}][x,y]/(xy-yx)$ , an important algebra in quantum Teichmuller theory. Skein modules provide a bridge between quantum topology and algebraic geometry of character varieties.