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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.

True. In a connected sum of knots, a bracket smoothing state from each summand contributes a state of the joined diagram. Joining the two distinguished state circles reduces their total count by one, so the factors ${\delta }^{s-1}$ multiply. The smoothing weights also multiply and the writhe of a link diagram adds. Thus the Jones polynomial of a connected sum is

False as written for arbitrary links. The two-component unlink has a diagram with no crossings butThe correct connected-diagram statement does hold, in particular for every knot diagram. If $s_A,s_B$ count the all-$A$ and all-$B$ bracket smoothing state circles, changing $j$ smoothings increases the circle count by at most $j$. The maximum and minimum $A$ exponents therefore lie between $n+2s_A-2$ and $-n-2s_B+2$. This gives the Kauffman bracket breadth bound ${\mathrm{br}}_A⟨D⟩\le 2n+2s_A+2s_B-4$. For a connected projection, the Turaev surface has Euler characteristic $s_A+s_B-n\le 2$, so the bound is $4n$. Writhe of a link diagram normalization changes only a monomial, and $t=A^{-4}$ divides the breadth by four, giving $\mathrm{br}{V}_L\le n$.

For a diagram with $k$ projection components, sum the same Euler characteristic inequality over the components. Then $s_A+s_B\le n+2k$ and the Jones polynomial breadth bound for a disconnected diagram is $\mathrm{br}{V}_L\le n+k-1$. The extra $k-1$ is indispensable, as the unlink shows.

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.

Let $F$ be an oriented Seifert surface of genus $g$ for $K$, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is a torsion module. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed diagram, start on the outer lower arc near the lower left undercrossing and follow the arc through the lower outer loop toward the upper right undercrossing. Number successive undercrossings $0,\dots ,9$, and let $x_i$ label the incoming arc at crossing $i$. Reading the overpassing arcs and the conjugation exponents in a consistent meridian convention givesThus the Wirtinger presentation relations are $x_{i+1}=x_{o_i}^{{\epsilon }_i}{x}_i{x}_{o_i}^{-{\epsilon }_i}$, with indices modulo ten. Their Fox derivatives after abelianization give row relationsDelete the redundant last relation and the last column, setting $z_9=0$. This is a nine-generator presentation of the Alexander module. The cofactor determinant is $t^{-6}(t^4+t^3-3t^2+t+1)$; here is a smaller elimination check. Retain $z_1,z_4$. Seven relations solve the other generators asThe two remaining rows, at undercrossings two and seven, are the matrixIts determinant is $t^{-3}(t^4+t^3-3t^2+t+1)$, differing from the original cofactor only by a Laurent unit. ConsequentlyThe factor has value one at $t=1$ and is reciprocal, as required. Its Conway-normalized Alexander polynomial is $t^2+t-3+t^{-1}+t^{-2}$, and its knot determinant is three.

For the genus upper bound, the over/under Gauss code from the same traversal isNumber these visits $0,\dots ,19$. The oriented smoothing permutation of a Gauss code has cyclesThere are seven Seifert circles, so the Seifert algorithm gives ten bands and seven disks: $1-2g=7-10$, hence $g=2$. The Alexander breadth bound on Seifert genus supplies the matching lower bound from breadth four. Therefore $\boxed{g_s(K)=2}$.

Modulo two, the primitive Alexander polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root, and it is not $(t^2+t+1{)}^2=t^4+t^2+1$, the only possible product of irreducible monic quadratics over ${\mathbb{F}}_2$. It is therefore irreducible over ${\mathbb{F}}_2$, hence over $\mathbb{Q}$ and $\mathbb{Z}$ by Gauss lemma for polynomials. The full-degree irreducible Alexander polynomial implies a prime knot criterion now applies, since its breadth is exactly twice the genus. Thus the knot is prime. The full-degree condition is needed to rule out a nontrivial summand with unit Alexander polynomial.

To test reflection, calculate the Jones polynomial from the same crossing data. The diagram has nine positive crossings and one negative crossing, so its writhe of a link diagram is eight. Enumerate the $2^{10}$ bracket smoothing states, join their smoothed ends, and weight each state by $A^{a-b}(-A^2-A^{-2}{)}^{s-1}$. Correct by $(-A^3{)}^{-8}$ and put $t=A^{-4}$. This givesAs checks, $V_K(1)=1$ and $V_K(-1)=-3={\Delta }_K(-1)$. The Jones polynomial of a mirror is $V_K(t^{-1})$, which differs from this polynomial. Therefore the knot is not equivalent to its reflection. Reversing all crossing conventions reciprocates the displayed Jones polynomial and leaves the conclusion unchanged.

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

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 $(-2{)}^l$.

For an oriented closed curve, geometric writhe is the signed crossing number averaged over generic viewing directions. For an individual link diagram, writhe of a link diagram is its signed crossing sum. Both depend on geometry or projection rather than only the link isotopy class.