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The Conway polynomial is normalized by ${\nabla }_U=1$ for the unknot, zero for a split unlink with more than one component, and the skein relation ${\nabla }_{L_{+}}-{\nabla }_{L_{-}}=z{\nabla }_{L_0}$. For a knot, the symmetric Alexander polynomial with value one at $t=1$ is ${\Delta }_K(t)={\nabla }_K(t^{1/2}-t^{-1/2})$. This fixes the Laurent-unit ambiguity of the Seifert-matrix presentation of the Alexander module.

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 $k-1$.

For a nonempty link with $\ell$ components, the Jones polynomial skein relation at $t=1$ makes a crossing change preserve the value. Crossing changes turn the link into an unlink with the same number of components. Repeated Jones polynomial of a split union gives the formula. In particular the Jones polynomial is never the zero polynomial.

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.

False. The zero polynomial is already a counterexample. The Jones polynomial evaluation at one of a nonempty link with $\ell$ components isTo prove this, specialize the Jones polynomial skein relation at $t=1$: switching a crossing preserves the value. Switch crossings until the diagram is an unlink without changing its component count. The Kauffman bracket loop rule, equivalently the Jones polynomial of a split union, gives the displayed evaluation. Hence arbitrary Laurent polynomials cannot occur.

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.

True with the stated Conway normalization. Fix $t^{1/2}=i$ at $t=-1$. The Jones polynomial skein relation becomesThe Conway polynomial satisfies ${\nabla }_{L_{+}}(z)-{\nabla }_{L_{-}}(z)=z{\nabla }_{L_0}(z)$. Put $W_L=(-1{)}^{\ell (L)-1}{\nabla }_L(2i)$. An oriented smoothing changes the number of components by one, so these quantities satisfy precisely $W_{L_{+}}-W_{L_{-}}=-2i{W}_{L_0}$. Both $W$ and $V(-1)$ are one on the unknot and zero on every multi-component unlink: for $V$, the split-union factor vanishes at this substitution.

These skein relations and unlink values determine the invariants uniquely. Switching a first undesired crossing gives a diagram closer to a descending diagram, while the smoothing term has fewer crossings; induction first on crossing count and then on undesired crossings reduces everything to unlinks. Thus $W_L=V_L(-1)$. For a knot there is one component and the Conway-normalized Alexander polynomial givesThis is the signed Jones evaluation at minus one. Replacing ${\Delta }_K$ by an arbitrary Laurent unit multiple would destroy the signed conclusion.

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.

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.

Let $H_{+}$ be the positively oriented Hopf link obtained by closing a two-strand braid with two positive crossings. Switching one crossing gives the two-component unlink; the oriented smoothing gives the unknot. The printed skein relation givesRepresent the right-handed trefoil knot by the closure of the two-strand braid with three positive crossings. Switching one crossing gives the unknot, while its oriented smoothing gives $H_{+}$. Consequentlyso the two Jones polynomials areThe second expression follows from the mirror of a link identity. They are unequal Laurent polynomials, so invariance under link isotopy proves that the right-handed and left-handed trefoils are not isotopic.

In either diagram of the Stevedore knot, the evident ribbon band can be cut to leave a two-component unlink. Cap those components by disjoint disks in $B^4$ and restore the band. This constructs a ribbon disk, hence proves that the Stevedore knot is slice.

The two displayed Stevedore diagrams give two distinct ribbon-band presentations. Use one below the equator and the reverse of the other above it. In banded-unlink diagram language, draw their common unlink and include the two ribbon bands, one from each presentation. Reading the movie from bottom to top gives the indicated birth level, the two saddle bands, and the death level. The resulting surface is knotted: the two-band movie is the standard presentation obtained by gluing the two Stevedore ribbon disks, and a van Kampen calculation retains a noncyclic quotient coming from the trefoil group, whereas the complement of the unknotted model has cyclic fundamental group.

There is a terminology issue in the question. If “2-knot” means an embedded $S^2$, as it normally does, one minimum, two index-one saddles, and one maximum have Euler characteristicso the resulting orientable surface knot is a torus, not a two-sphere. The construction above answers the question under the broader usage in which “2-knot” means a connected knotted surface. Under the standard narrow definition, the requested critical-point data are impossible.

Use the Conway-normalized Alexander polynomial. Choosing $t^{1/2}=i$ in the Jones polynomial skein relation gives $V_{L_{+}}(-1)-V_{L_{-}}(-1)=-2i{V}_{L_0}(-1)$. The quantities $(-1{)}^{\ell -1}{\nabla }_L(2i)$ obey the same relation, because smoothing changes the component count by one. Their unlink values agree, so crossing-change induction proves equality for every link. For a knot, the sign is one and ${\nabla }_K(2i)={\Delta }_K(-1)$. This proves the signed identity, not only equality of knot determinants.