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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.

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.

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.

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 $V_{\mathrm{unknot}}=1$.