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Here $n$ is the crossing count and $k$ is the number of connected components of the planar link diagram, not necessarily the number of link components. The all-$A$ and all-$B$ state counts satisfy $s_A+s_B\le n+2k$, by summing the Turaev surface Euler characteristic bound over projection components. The Kauffman bracket breadth bound and $t=A^{-4}$ give the displayed bound. It is sharp for an unlink diagram with $n=0$, whose Jones polynomial has breadth of a Laurent polynomial $k-1$.

A state with $k$ B smoothings has at most $s_A+k$ circles, so its largest A exponent is at most $c(D)+2s_A-2$. The dual argument bounds the smallest exponent below by $-c(D)-2s_B+2$. For a connected link diagram, the Turaev surface inequality bounds the resulting breadth by $4c(D)$. For a connected reduced alternating diagram, adequacy prevents cancellation at both extremes and planar Euler characteristic gives equality. A disconnected projection needs the additional term in the Jones polynomial breadth bound for a disconnected diagram.

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, 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.