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For a nonempty link with $\ell$ components, the Jones polynomial skein relation at $t=1$ makes a crossing change preserve the value. Crossing changes turn the link into an unlink with the same number of components. Repeated Jones polynomial of a split union gives the formula. In particular the Jones polynomial is never the zero polynomial.

This asks whether the Jones polynomial equal to one characterizes the unknot among knots. A zero breadth of a Laurent polynomial makes $V_K$ a monomial; the identities $V_K(1)=1$ and $V_K^{\prime }(1)=0$ then force that monomial to be one. Thus using Jones breadth as a strictly positive complexity for nontrivial knots requires this detection property. To prove the derivative identity, differentiate the Jones polynomial skein relation at one for a crossing change in a knot. Both crossed diagrams have value one, while the smoothed two-component link has value minus two by Jones polynomial evaluation at one, so $V_{K_{+}}^{\prime }(1)=V_{K_{-}}^{\prime }(1)$. Crossing changes reduce to the unknot, whose derivative is zero. The corresponding Alexander polynomial detection property is false, as nontrivial untwisted Whitehead doubles demonstrate. The Jones question is discussed in research on closed four-strand braids.

False. The zero polynomial is already a counterexample. The Jones polynomial evaluation at one of a nonempty link with $\ell$ components isTo prove this, specialize the Jones polynomial skein relation at $t=1$: switching a crossing preserves the value. Switch crossings until the diagram is an unlink without changing its component count. The Kauffman bracket loop rule, equivalently the Jones polynomial of a split union, gives the displayed evaluation. Hence arbitrary Laurent polynomials cannot occur.

True with the stated Conway normalization. Fix $t^{1/2}=i$ at $t=-1$. The Jones polynomial skein relation becomesThe Conway polynomial satisfies ${\nabla }_{L_{+}}(z)-{\nabla }_{L_{-}}(z)=z{\nabla }_{L_0}(z)$. Put $W_L=(-1{)}^{\ell (L)-1}{\nabla }_L(2i)$. An oriented smoothing changes the number of components by one, so these quantities satisfy precisely $W_{L_{+}}-W_{L_{-}}=-2i{W}_{L_0}$. Both $W$ and $V(-1)$ are one on the unknot and zero on every multi-component unlink: for $V$, the split-union factor vanishes at this substitution.

These skein relations and unlink values determine the invariants uniquely. Switching a first undesired crossing gives a diagram closer to a descending diagram, while the smoothing term has fewer crossings; induction first on crossing count and then on undesired crossings reduces everything to unlinks. Thus $W_L=V_L(-1)$. For a knot there is one component and the Conway-normalized Alexander polynomial givesThis is the signed Jones evaluation at minus one. Replacing ${\Delta }_K$ by an arbitrary Laurent unit multiple would destroy the signed conclusion.

Put $\delta =-A^2-A^{-2}$. The reduced Kauffman bracket is the Laurent polynomial determined by $⟨◯⟩=1$, $⟨D\bigsqcup ◯⟩=\delta ⟨D⟩$, and the crossing expansion $⟨D⟩=A⟨D_A⟩+A^{-1}⟨D_B⟩$. Fix the A smoothing so that a positive curl has multiplier $-A^3$; rotating the local crossing interchanges the smoothing descriptions. Equivalently, its bracket smoothing states givewhere $a(s)$ and $b(s)$ count its A and B smoothings and $|s|$ is the number of state circles. Resolving all crossings proves that the recursive rules define the same polynomial independently of the order of expansion.

Under a positive Reidemeister move of type I, the two resolutions give $A\delta +A^{-1}=-A^3$ times the straight strand; the negative curl gives $A^{-1}\delta +A=-A^{-3}$. Thus the bracket itself is not an unframed link invariant. For type II, use the adjacent cap-cup generator $e$ of the Temperley-Lieb algebra, with $e^2=\delta e$. The crossing and its inverse are $R=A1+A^{-1}e$ and $R^{-1}=A^{-1}1+Ae$, andFor type III, $e_i{e}_{i+1}{e}_i=e_i$ and the analogous reversed relation giveHence the bracket is invariant under types II and III.

For an oriented diagram, let $w(D)$ be its writhe of a link diagram, the sum of its signed crossings. Types II and III preserve writhe of a link diagram, while a positive or negative curl changes it by one or minus one. The Jones polynomial is thereforeThe normalization cancels type I, so all Reidemeister moves preserve it. It is a Laurent polynomial in $t^{1/2}$; its breadth means the largest occurring t exponent minus the smallest, including half-integral exponents for links.

The Jones polynomial skein relation follows directly from the same normalization. Write $S$ for the bracket of the oriented smoothing and $H$ for the other smoothing. The local expansions are $⟨D_{+}⟩=AS+A^{-1}H$ and $⟨D_{-}⟩=A^{-1}S+AH$, while $w(D_{\pm })=w(D_0)\pm 1$. ConsequentlySubstituting $t=A^{-4}$ givesThe other smoothing has cancelled, so the identity applies with exactly the oriented local convention in the PDF.

Use the Conway-normalized Alexander polynomial. Choosing $t^{1/2}=i$ in the Jones polynomial skein relation gives $V_{L_{+}}(-1)-V_{L_{-}}(-1)=-2i{V}_{L_0}(-1)$. The quantities $(-1{)}^{\ell -1}{\nabla }_L(2i)$ obey the same relation, because smoothing changes the component count by one. Their unlink values agree, so crossing-change induction proves equality for every link. For a knot, the sign is one and ${\nabla }_K(2i)={\Delta }_K(-1)$. This proves the signed identity, not only equality of knot determinants.