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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.

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 $k-1$.

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.

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.

Apply an orientation-reversing homeomorphism of $S^3$ to a link and transport its component orientations. Reflection through the projection plane produces a link diagram by switching every crossing.

For tame or smooth embeddings $f_0,f_1:S^1\hookrightarrow S^3$, ambient isotopy means there is a continuous family $F_t:S^3\to S^3$, $0\le t\le 1$, of homeomorphisms, with $F_0=\mathrm{id}$ and $F_1\circ f_0=f_1$. Equivalently the map $(x,t)\mapsto (F_t(x),t)$ is a homeomorphism of $S^3\times [0,1]$ preserving the time coordinate. In the smooth category one uses diffeomorphisms and a smooth isotopy. If only unparametrized images are compared, the last condition is $F_1(f_0(S^1))=f_1(S^1)$; for oriented knots it must also preserve the strand orientation.

A link is an embedding of a finite disjoint union of circles. Choose a point of $S^3$ outside it and identify the complement of that point with ${\mathbb{R}}^3$. A generic planar projection gives an immersion with only finitely many transverse double points, no triple points and no singularities of the projected strands. A link diagram is this planar immersed union together with the over/under information at each double point, recording which strand has the larger height. An oriented link diagram also records each component's orientation. The crossing data are indispensable: the planar immersion alone does not specify the link.

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.

The drawn component $L_1$ is an unknot. Its exterior is a solid torus $V$, in which $L_2$ is a pattern of a satellite knot. A meridian of a solid torus of $V$ is ${\ell }_1$, whereas its longitudinal direction is $m_1$. Thus the prescribed gluing sends these directions to $m$ and $\ell$, respectively. This identifies $V$ with a tubular neighborhood of $K$ with its zero Seifert framing.

Removing the pattern before making this identification givesEquivalently, meridionally filling the remaining boundary first removes $L_2$ from the construction and leaves $E(K)\cup V=S^3$; its filling core is the required satellite knot $C(K)$.

In the original link diagram, the two parallel passages through a meridional disk of $V$ run in the same direction; closing them uses a single interchange. The winding number of a satellite pattern is therefore two. Untwisting the surrounding disk shows that $P(U)$ is the unknot (the pattern is the $(2,1)$ cable knot, up to the harmless sign of its single interchange). In particular ${\Delta }_{P(U)}=1$. The Satellite formula for the Alexander polynomial now gives the concise answerHere $\doteq$ allows multiplication by $\pm t^j$. Reversing the winding orientation replaces $t^2$ by $t^{-2}$, giving the same Alexander polynomial of a knot up to such a unit.

Interpret the paired $A,B,C$ disks as the three one-handles of a genus-three handlebody. Attach two-handles on the two displayed $\beta$ curves. Orient the three disk-crossing generators as $a,b,c$. Reading the signed crossings, starting at the upper-left portion of each attaching curve and changing the start point when needed, givesThe first says $abc=bca$ and the second $abc=cab$. Thus the fundamental group presentation isHere $[u,v]=uv{u}^{-1}{v}^{-1}$. Cyclically changing the starting point, reversing an attaching curve, or changing generator orientations gives equivalent presentations. With $h=abc$, these relations make $h$ central; eliminating $c=b^{-1}{a}^{-1}h$ givesFor the topological identification, thicken the diagram's two nested bands and identify the three paired disk mouths. The complement of those bands is the product region of a pair of pants with a circle; its two compressing curves are exactly $[a,bc]$ and $[ab,c]$. Equivalently, the standard cell decomposition of this product has three one-handles and the two commuting two-handle attachments shown. Thus this is a generalized Heegaard diagram of $P\times S^1$.

The Hopf fibration of $S^3$ has three disjoint regular fibers whose removal leaves $P\times S^1$. Its three fibers are the components of the torus link $T_{3,3}$, as is also apparent from the full three-strand twist in the later link diagram. This identifies $Y$ with that link exterior, using the diagram and product structure rather than just its fundamental group. In particular it is a link exterior in the three-sphere.

Orient the two components coherently through the twist region and assign meridian variables $x,y$. The link diagram is the torus link $T_{2,2n}$. It is obtained from the three-component torus link of part 2 by $-1/(n-1)$ rational Dehn surgery on the third component: removing its meridional disk adds $n-1$ full twists to the original single full twist. For $n=1$, this just means meridionally deleting the third component.

Write $z$ for the removed component's meridian. Its longitude is homologous to ${\mu }_1+{\mu }_2$, so the filling imposes $z=(xy{)}^{n-1}$. Under this substitution, the polynomial of part 2 becomes $(xy{)}^n-1$. The filling core is homologous, up to sign, to ${\mu }_1+{\mu }_2$. The Turaev-torsion Dehn-filling formula therefore removes the factor $xy-1$, givingFor positive $n$ this is the Laurent polynomial $1+xy+\cdots +(xy{)}^{n-1}$. For $n=1$ it is one, as for a Hopf link; for $n=0$ it is zero, as for the two-component unlink. For negative $n$ the displayed quotient is still a Laurent polynomial and agrees with the mirrored positive-twist answer up to a unit. These checks also fix the twist count: the exponent is $n$, rather than $n+1$.

Apply the given skein relation to a positive and a negative kink in a diagram of the unknot. Both crossed link diagrams represent the unknot, whereas the oriented smoothing represents the two-component unlink. ThusFactor the left-hand side and cancel the nonzero factor in the Laurent polynomial ring:The positive sign follows from the skein relation as printed in this paper. It differs from the frequently used Jones polynomial convention whose two-component unlink has value $-(q^{1/2}+q^{-1/2})$. With the printed convention, an $r$-component link has invariant $(-1{)}^{r-1}$ times that convention's Jones polynomial.

Insert a kink on any component of an oriented link diagram for $L$. Switching the kink's crossing gives another link diagram of $L$; its oriented smoothing separates off a small unknot in a ball disjoint from the rest of $L$. The skein relation therefore givesCancel the nonzero factor $q^{1/2}-q^{-1/2}$ in the Laurent polynomial ring to obtain

A skein relation relates the invariants of link diagrams agreeing outside a disk and differing by a positive crossing, a negative crossing, and a smoothing within it. Its signs and normalization determine the particular invariant convention.

Choose a checkerboard coloring of the complementary regions of a link diagram. The Tait graph has one vertex for each black region and one edge through each crossing joining its adjacent black regions. The other coloring gives its planar dual. For a connected alternating link diagram, all-A and all-B smoothing circles correspond to the regions of its two checkerboard colors; one-circle smoothing states correspond to spanning trees.

Replace each crossing of a diagram by a saddle joining its A smoothing above the projection sphere to its B smoothing below; cap the upper and lower state circles. For a connected classical link diagram this constructs a connected closed orientable surface. Its Euler characteristic is $s_A+s_B-c(D)$, so $s_A+s_B\le c(D)+2$. For an alternating diagram the surface has genus zero.

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.