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For a reduced connected alternating knot diagram, its knot signature is $s_A(D)-c_{+}(D)-1$, where $s_A$ counts circles in the all-$A$ bracket smoothing state and $c_{+}$ counts positive crossings. The convention is that the $A$-smoothing at a positive braid crossing is the oriented smoothing, and a positive trefoil knot has knot signature $-2$. Reversing the global knot signature convention negates the formula.

Resolve the two crossings in a skein relation using the Kauffman bracket, then correct by the writhe of a link diagram. The other smoothing cancels, leaving the displayed coefficient of the oriented smoothing. This relation and the value one on the unknot determine the Jones polynomial: crossing changes give descending diagrams, which are unlinks, while the smoothing terms have fewer crossings.

Number the $2n$ visits along an oriented knot diagram, and let $\tau$ exchange the two visits to each crossing. At an oriented smoothing, an incoming strand follows the outgoing strand of the other visit. Thus the cycles of the displayed permutation are exactly the Seifert circles. Their count $s$ gives the Seifert algorithm genus $(1-s+n)/2$ without requiring a separate drawing of the smoothed diagram.

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.

There are two common normalizations. I will define both explicitly, because the numerical conclusion in the question uses the unreduced one.

Set $\delta =-A^2-A^{-2}$. The reduced Kauffman bracket of a nonempty link diagram iswhere $s$ runs over all bracket smoothings, $a(s),b(s)$ count $A$- and $B$-smoothings, and $|s|$ is the number of resulting circles. Locally,Fix the usual Kauffman bracket convention in which the $A$-smoothing at a positive braid crossing is its oriented smoothing; reflecting a crossing exchanges $A$ and $A^{-1}$. The unknot has reduced bracket $1$, and adjoining another circle multiplies the bracket by $\delta$. The unreduced Kauffman bracket uses ${\delta }^{|s|}$ instead and assigns the empty diagram $1$. Thus for a nonempty diagram

For completeness, the bracket calculations for all three Reidemeister moves can be carried out in the Temperley-Lieb diagram algebra. Let $I$ be the identity two-strand tangle, and let $e$ be the cap-cup tangle. Composition gives $e^2=\delta e$. The two crossings correspond toConsequentlywhich proves invariance under the second Reidemeister move. On three strands, the cap-cup diagrams satisfy $e_1{e}_2{e}_1=e_1$, $e_2{e}_1{e}_2=e_2$, and $e_i^2=\delta e_i$. Expand the two sides of the third move. Their difference isThus the third Reidemeister move also preserves either bracket normalization. Reflected forms of these moves follow by replacing $A$ with $A^{-1}$.

A positive curl contributes $A\delta +A^{-1}=-A^3$, and a negative curl contributes $A^{-1}\delta +A=-A^{-3}$. The writhe of a link diagram $w(D)$ changes by $+1$ or $-1$ in exactly these cases. Therefore the corrected bracketis invariant under the first Reidemeister move as well; the second and third moves leave the writhe of a link diagram unchanged. Substituting $t=A^{-4}$ defines a Jones polynomial. For multiple components half-integer powers of $t$ may occur. The reduced version $V_L^{\mathrm{red}}$ has $V_U^{\mathrm{red}}=1$, and the Unreduced Jones polynomial is

At $A=1$, switching a crossing does not change the bracket: both smoothing coefficients are $1$. It changes the writhe of a link diagram by $2$, so the correction factor $(-1{)}^{-w}$ is unchanged. Hence either Jones polynomial at $t=1$ is unchanged by a crossing change. By switching crossings, any $l$-component link can be made an unlink. Its crossing-free diagram has $l$ circles and zero writhe of a link diagram, givingThus the printed conclusion $|V_L(1)|=2^l$ holds for the unreduced normalization. For the reduced Jones polynomial, the correct conclusion is $\boxed{|V_L^{\mathrm{red}}(1)|=2^{l-1}}$; the unknot alone already rules out $2^l$ in that convention.

Apply the given skein relation to a positive and a negative kink in a diagram of the unknot. Both crossed link diagrams represent the unknot, whereas the oriented smoothing represents the two-component unlink. ThusFactor the left-hand side and cancel the nonzero factor in the Laurent polynomial ring:The positive sign follows from the skein relation as printed in this paper. It differs from the frequently used Jones polynomial convention whose two-component unlink has value $-(q^{1/2}+q^{-1/2})$. With the printed convention, an $r$-component link has invariant $(-1{)}^{r-1}$ times that convention's Jones polynomial.

Insert a kink on any component of an oriented link diagram for $L$. Switching the kink's crossing gives another link diagram of $L$; its oriented smoothing separates off a small unknot in a ball disjoint from the rest of $L$. The skein relation therefore givesCancel the nonzero factor $q^{1/2}-q^{-1/2}$ in the Laurent polynomial ring to obtain

Reflection through the projection plane reverses the orientation of $S^3$ and exchanges the overpassing and underpassing strands at every crossing. Thus switching all crossings gives the mirror of a link, with the component orientations transported by the reflection.

Define $W(L)(q)=V(\overline{L})(q^{-1})$. Reflection exchanges $D_{-}$ with $D_{+}$ and preserves the oriented smoothing. Apply the printed skein relation to the reflected triple at parameter $q^{-1}$:Multiplication by $-1$ gives the original skein relation for $W$. Also $W(U_1)=1$. The stated uniqueness of the Jones polynomial now gives $W=V$, hence

Let $H_{+}$ be the positively oriented Hopf link obtained by closing a two-strand braid with two positive crossings. Switching one crossing gives the two-component unlink; the oriented smoothing gives the unknot. The printed skein relation givesRepresent the right-handed trefoil knot by the closure of the two-strand braid with three positive crossings. Switching one crossing gives the unknot, while its oriented smoothing gives $H_{+}$. Consequentlyso the two Jones polynomials areThe second expression follows from the mirror of a link identity. They are unequal Laurent polynomials, so invariance under link isotopy proves that the right-handed and left-handed trefoils are not isotopic.

Apply the oriented smoothing to every crossing of an oriented knot diagram, fill the resulting Seifert circles by disjoint disks at suitable heights, and reconnect them by twisted bands at the original crossings. The boundary is the original knot. If the connected surface has $s$ disks and $c$ bands, its Euler characteristic is $s-c$ and its genus is $(1-s+c)/2$.

A Seifert circle is a component of the crossing-free diagram obtained by oriented smoothing at every crossing. These circles bound the disks used in the Seifert algorithm.