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A diagram is A-adequate if changing a single smoothing of its all-A state joins two different state circles, and B-adequate analogously. It is adequate if both conditions hold. In an adequate diagram the top and bottom degree terms of the Kauffman bracket arise uniquely from these two states. Reduced alternating diagrams are adequate because their Tait graphs and duals have no loop edges.

A bracket smoothing state chooses one of the two unoriented smoothings at every crossing of a link diagram. Its weight is $A^{a-b}{\delta }^{s-1}$ in the reduced Kauffman bracket, where $a,b$ count the smoothing types and $s$ counts resulting circles. The all-$A$ and all-$B$ states choose the same smoothing type at every crossing.

Close a three-strand planar tangle and evaluate its reduced Kauffman bracket. The basis $1,e_1,e_2,e_1{e}_2,e_2{e}_1$ has the displayed trace values, with the last value also valid for $e_2{e}_1$. Writing braid generators as $A1+A^{-1}{e}_i$ turns knot-polynomial computations into a five-dimensional algebra calculation.

The Jones polynomial is obtained from the Kauffman bracket of an oriented diagram by the writhe normalization

For knots, matching bracket smoothing states across a connected sum of knots joins one state circle from each side. Their total circle count is therefore the sum of the two counts minus one. The reduced Kauffman bracket state weights multiply, and writhe of a link diagram adds, proving the formula.

Reflecting a link diagram interchanges its two bracket smoothings and negates its writhe of a link diagram. This replaces $A$ by $A^{-1}$ in the normalized Kauffman bracket, hence $t$ by $t^{-1}$. A knot whose Jones polynomial differs from its reciprocal cannot be an amphichiral knot.

Placing the two link diagrams in disjoint disks keeps all state circles separate. The reduced Kauffman bracket therefore acquires one extra loop factor $-A^2-A^{-2}$. Correcting writhe of a link diagram and setting $t=A^{-4}$ gives the displayed formula. It concerns a split link, not an arbitrary union of linked components.

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.

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.

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.

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, the sum of its signed crossings. Types II and III preserve writhe, 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.

Write $s_A,s_B$ for the all-A and all-B circle counts of a diagram with $n$ crossings. A state obtained from the all-A state by $k$ smoothing changes has at most $s_A+k$ circles, because each change either joins two circles or splits one. Its largest A exponent is consequently at most $n-2k+2(s_A+k-1)=n+2s_A-2$. The symmetric argument from the all-B state bounds every exponent below by $-n-2s_B+2$. Thus the Kauffman bracket breadth bound isFor a connected diagram, construct the Turaev surface by placing A-state circles above the projection sphere, B-state circles below it, inserting one saddle at each crossing, and capping the state circles. This is a connected embedded closed orientable surface; its disks and saddles give $\chi =s_A+s_B-n$. Since $\chi =2-2g_T\le 2$, $s_A+s_B\le n+2$. Writhe normalization changes only a monomial, and $t=A^{-4}$ divides breadth by four. Therefore

For a connected reduced alternating diagram, checkerboard color its complementary regions. Alternation makes all-A circles the boundaries of one color and all-B circles the boundaries of the other. Euler's formula for the connected four-valent projection graph gives $s_A+s_B=n+2$. Its Tait graph and planar dual have no loops: a loop in either, equivalently a loop or bridge in one, gives a nugatory crossing cut off by a simple closed curve. Reducedness excludes precisely such crossings. Changing one smoothing in the all-A or all-B state therefore joins distinct circles, so the diagram is an adequate link diagram.

This also proves that the extreme coefficients cannot cancel. If $k\ge 1$ smoothings are changed from the all-A state, the first change lowers its circle count by one and the remaining $k-1$ changes increase it by at most $k-1$. The largest exponent of that state is at most $n+2s_A-6$, four below the all-A extreme $n+2s_A-2$, whose coefficient is $(-1{)}^{s_A-1}$. The all-B extreme is uniquely attained for the same reason. Consequently

For each alternating knot $K_i$, remove nugatory crossings from an alternating diagram. The equality just proved, together with the general breadth bound for any knot diagram, shows that its reduced crossing count is minimal: $c(K_i)=\mathrm{br}{V}_{K_i}$. Joining two knot diagrams by an orientation-compatible connected sum of knots gives $V_{K_1\#K_2}=V_{K_1}{V}_{K_2}$. Indeed the state circles joined along the sum arc become one, so the bracket state sums multiply, and writhe adds. Nonzero Laurent polynomials have additive breadth, because their extreme coefficients multiply. Hence every diagram of the sum has at least $c(K_1)+c(K_2)$ crossings, while joining the two minimal alternating diagrams achieves that count. Thus

Finally suppose that the two-component link is split. Its Jones polynomial satisfies $V_L=-(t^{1/2}+t^{-1/2})V_{K_1}{V}_{K_2}$, by the disjoint-union bracket rule, so $V_L(-1)=0$. If the given alternating diagram were connected, its Tait graph $G$ would be connected. Evaluate the bracket at $A=e^{\pi i/4}$, where $\delta =0$ and $t=-1$. Only one-circle states survive. These are exactly the spanning trees of $G$: a regular neighborhood of a chosen spanning subgraph has one boundary circle exactly when it is connected and has no cycle. All spanning trees use the same number of black-joining smoothings, $|V(G)|-1$. Since the crossing-to-checkerboard smoothing choice is uniform in an alternating diagram, all surviving state weights are the same nonzero power of $A$. Therefore $⟨D⟩$ is that power times the positive number of spanning trees, and $V_L(-1)\ne 0$, a contradiction.

The diagram is thus disconnected. Each of the two link components projects to a connected subset, so the two projection components can be separated by a boundary curve of a small regular neighborhood in the projection sphere. Hence a split two-component link represented by a reduced alternating diagram has a split diagram.

Write $\delta =-A^2-A^{-2}$. Normalize the reduced Kauffman bracket by $⟨◯⟩=1$ and $⟨D\bigsqcup ◯⟩=\delta ⟨D⟩$. At a positive braid crossing choose the smoothing rule $R_i=A1+A^{-1}{e}_i$, where $1$ is the two parallel strands and $e_i$ joins the adjacent top endpoints and adjacent bottom endpoints. At its mirror crossing the coefficients are interchanged. This convention fixes the two smoothing labels. Equivalently, for a nonempty diagram,where each bracket smoothing state chooses one smoothing at each crossing, $a,b$ count the choices, and $c$ counts the resulting circles. It is a Laurent polynomial in $A$ and is independent of the order in which crossings are resolved.

We prove the local invariance needed for the Jones polynomial. In the Temperley-Lieb diagram algebra, stacking the smoothing tangles gives $e_i^2=\delta e_i$, $e_i{e}_{i+1}{e}_i=e_i$ and its reflected version. These identities follow just by following the planar strands; only the first removes an extra circle. The inverse crossing is $R_i^{-1}=A^{-1}1+A{e}_i$, andThus the second Reidemeister move preserves the bracket. Expanding three crossings givesso the third Reidemeister move does too; inverse versions follow by multiplying by inverse crossings. A positive curl resolves to $A\delta +A^{-1}=-A^3$ times the straight strand, and a negative curl gives $A^{-1}\delta +A=-A^{-3}$. Thus the first move changes the bracket by the indicated factor.

Orient the diagram and let $w(D)$ be its writhe, the sum of the signed crossings. DefineThe first move changes $w$ by $+1$ or $-1$, exactly cancelling the bracket's curl factor. The other two moves leave writhe unchanged. Planar isotopy leaves both quantities unchanged. The Reidemeister theorem states that diagrams of ambient-isotopic tame oriented links are related by planar isotopy and these three local moves; therefore $V_D$ depends only on the oriented link.

This substitution really lies in $\mathbb{Z}[t^{1/2},t^{-1/2}]$. If there are $N$ crossings, each state exponent $a-b$ has parity $N$, and $w(D)$ also has parity $N$. Expanding each power of $\delta$ adds only even exponents, so every exponent of the normalized bracket has parity $N-3w(D)=0(\mathrm{mod}2)$. Every even power of $A$ is an integral power of $t^{1/2}$. The normalization gives $V_{\mathrm{unknot}}=1$.

Use the standard closed three-strand braid $B=({\sigma }_1{\sigma }_2^{-1}{)}^2$, a four-crossing diagram of the figure-eight knot. The diagram has two positive and two negative crossings, so $w(B)=0$.

We calculate its Kauffman bracket, making the inversion symmetry explicit rather than merely asserting amphichirality.

In the Temperley-Lieb diagram algebra on three strands put $u=e_1$, $v=e_2$. The relations are $u^2=\delta u$, $v^2=\delta v$, $uvu=u$, $vuv=v$. The bracket image of one two-crossing block isClosing planar diagrams gives the bracket trace on the three-strand Temperley-Lieb algebra:These values follow because the closures have respectively three, two and one circles. Expanding and reducing $H^2$ givesTaking its trace and substituting $\delta =-A^2-A^{-2}$ yieldsThere is no writhe correction, soThe equality is immediate from the five displayed coefficients. The same braid model supplies the meridians used in part (d).

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 Kauffman bracket skein relation to every crossing in the two-string tangle $D_1$. Each complete smoothing is a collection of closed circles together with one of the two crossingless pairings of the four boundary points. Rotation through $180^{\circ }$ about the indicated vertical axis preserves both crossingless pairings and the number of closed circles. Gluing the unchanged tangle $D_2$ to corresponding smoothings therefore gives equal terms, including equal loop factors, so $⟨D⟩=⟨D^{\prime }⟩$.

Choose an orientation of $L$, and transport the induced orientations of the four ends of $D_1$ through the rotation to orient $L^{\prime }$. Crossing signs inside the rotated tangle and outside it then have the same total, so the writhe of a link diagram satisfies $w(D)=w(D^{\prime })$. Applying the writhe normalization of the Jones polynomial gives

The relative skein module fixes finitely many framed endpoints on the boundary of a 3-manifold and uses properly embedded framed tangles with those endpoints. Local Kauffman bracket relations are imposed away from the boundary. Gluing compatible endpoint sets composes tangles; a rectangle cylinder with n endpoints on each horizontal side gives the Temperley-Lieb diagram algebra.

Use the Kauffman bracket assigning the empty diagram value $1$ and a disjoint circle factor $-A^2-A^{-2}$. Correct by the writhe of a link diagram and substitute $t=A^{-4}$. Relative to the Jones polynomial normalized to be $1$ on the unknot, this unreduced invariant has an additional factor $-(t^{1/2}+t^{-1/2})$. For a nonempty $l$-component link, its value at $t=1$ is $(-2{)}^l$.