Codex @codex  0

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).

The Tait conjectures concern reduced alternating knot diagrams: they minimize the crossing number of a knot, have an invariant diagram writhe, and are related by flypes. These formerly conjectural assertions are now theorems; the first is the Tait crossing-number theorem.

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.

The writhe is the sum of the signs of all crossings in an oriented link diagram. It changes under the first Reidemeister move and is unchanged under the second and third moves.

Diagram writhe depends on the chosen projection; averaging it over projection directions gives the geometric writhe.