Let X be a (2 + 1)-dimensional globally hyperbolic spacetime with a Cauchy surface Σ whose universal cover is homeomorphic to R2. We provide empirical evidence suggesting that the Jones polynomial detects causality in X. We introduce a new invariant of certain tangles related to the Conway polynomial and prove that the Conway polynomial does not detect the connected sum of two Hopf links among relevant three-component links, which suggests that the Conway polynomial does not detect causality in the scenario described.

Let X be a (2 + 1)-dimensional globally hyperbolic spacetime with Cauchy surface Σ homeomorphic to R2, and let N be the set of future-directed null geodesics in X. The set N can be identified with the spherical cotangent bundle ST* Σ of Σ, which in this case is homeomorphic to a solid torus S1×R2. The sky of xX, denoted SxN, is the set of all future-directed null geodesics through x. The sky Sx is homeomorphic to a circle and viewed as a subset of the solid torus S1×R2, and Sx is isotopic to S1 × {0}. For more explanation, see Ref. 1 or 2.

Chernov and Nemirovski proved the Low conjecture,3 which says that as long as Σ is not a closed two-manifold, two events x, yX are causally related if and only if their skies SxSy are linked. In our context (Σ homeomorphic to R2), linked means either SxSy or SxSy is not isotopic in N to S1 × {a} ⊔ S1 × {b} for a,bR2. Nemirovski and Chernov actually proved more; they showed that the relationship between linking and causality holds as long as Σ is not homeomorphic to S2 or RP2. They also proved the Legendrian Low conjecture, which is analogous to the Low conjecture but for higher-dimensional spacetimes, where topological linking is replaced by Legendrian linking.3 

A natural question is whether linking of Sx and Sy, and thus causality, could be detected by various link invariants. Natário and Tod2 provided a large family of pairs of skies corresponding to causally related events such that for each pair, the associated link has a nontrivial Kauffman polynomial. The Kauffman polynomial is related to the Jones polynomial V(L) by a change of variables, but there is another invariant that contains strictly more information than both. Specifically, Khovanov homology provides a “categorification” of the Jones polynomial.4 In fact, Khovanov homology detects causality in X.1 Another common link polynomial is the Alexander–Conway polynomial (also called the Conway polynomial) ∇(L), which is categorified by link Floer homology.5 It is known that link Floer homology detects braid closures (see Ref. 6, Corollary 2) and detects the trivial braid amongst braid closures (see Ref. 7, Theorem 1). In combination with these results, the work of Chernov, Martin, and Petkova implies that link Floer homology also detects causality in this setting.1 The related knot polynomials, the Jones polynomial and the Conway polynomial, are strictly weaker link invariants than their respective categorifications. On the other hand, Ref. 2 indicates that these polynomials still might detect causality.

By a remark in Ref. 1, if an invariant can detect causality in X as described, then causality can be detected in any (2 + 1)-dimensional globally hyperbolic spacetime X′ whose Cauchy surface Σ′ is not homeomorphic to S2 or RP2. The reason is that the universal cover X̃ of such a spacetime is also a globally hyperbolic spacetime with Cauchy surface Σ̃ that is a universal cover of Σ′. By lifting of paths, causal relationships between points in X′ are equivalent to causal relationships between points in X̃. If Σ′ is not homeomorphic to S2 or RP2, then the universal cover Σ̃ is homeomorphic to R2. As a result, it will be sufficient for our purposes to consider the case described above, where Σ is homeomorphic to R2.

In this paper, we show that the Conway polynomial does not detect causality (linking) in the given setting, and we conjecture that the Jones polynomial does detect causality. At first glance, this hypothesis might seem strange: there are infinitely many nontrivial k-component links indistinguishable from unlinks via their Jones polynomials (see Ref. 8, Corollary 3.3.1). There are links with similarly trivial Conway polynomials, such as L10n32 and L10n599. However, there are topological restrictions on the links being considered. Components of links given by skies of events exist in a solid torus, and each sky must be isotopic to a longitude of the solid torus S1×R2. While this solid torus has a natural embedding into R3 as the neighborhood of an unknotted circle, there are links that are nontrivial in S1×R2 yet become unlinked when embedded into R3 via this embedding. In Sec. III, we exhibit an infinite family of such links that have a nontrivial Jones polynomial. In a sense, these are the simplest type of links one could consider, which satisfy the constraints.

Proposition 1.1.

There are infinitely many nontrivial two-component links inS1×R2that are unlinked when embedded intoR3but whose linking inS1×Rcan be detected by the Jonespolynomial.

Following the example of Eliahou, Kauffman, and Thistlethwaite,8 in Sec. IV, we define for each four-ended tangle T with a specific orientation (which we will call a left-right oriented tangle) an invariant Con(T) related to the Conway polynomial. This invariant Con(T) is an element of the free Z[z]-module Z[z]2 and can be used to calculate the Jones and Conway polynomials of links depending on a tangle.

Theorem 1.2.

LetLbe a diagrammatic operator, which takes a single left-right oriented tangleTand yields an oriented linkL(T). Then, this operator induces aZ[z]-module homomorphismφL:Z[z]2Z[z]such that ∇(L(T)) = φL(Con(T)).

We combine this theorem and results of Kauffman10 to produce an explicit example indicating that the Conway polynomial does not detect causality in the given setting.

After examining the Jones polynomial in more detail, we then make the following conjecture:

Conjecture 1.3.

LetXbe a (2 + 1)-dimensional globally hyperbolic spacetime with Cauchy surface Σ not homeomorphic toS2orRP2. Then, the Jones polynomial detects causality between any two events inX.


The numerical data for this paper were found by Jacob Swenberg during his undergraduate research project conducted under the guidance of Samantha Allen, Vladimir Chernov, and Ina Petkova. The problem considered in this paper was proposed by Vladimir Chernov and Ina Petkova.

In this section, we give definitions and useful results related to the Conway polynomial, the Jones polynomial, and tangles.

Let L1 be a link diagram with a distinguished crossing of the form . Denote by L0 and L the resulting link diagrams arising from performing smoothing changes at the distinguished crossing, as in Fig. 1.

Definition 2.1.
The Kauffman bracket of a linkL,L, is the Laurent polynomial in a variableAdefined by

The Kauffman bracket is not a link invariant; one must adjust for the writhe of the link diagram.

Definition 2.2.
The Kauffman polynomial of a linkL,K(L), is defined to be
wherew(L) is the writhe of the diagram. The Jones polynomialV(L) can be obtained by substitutingt−1/4forAin the Kauffman polynomial (Ref.11 , Theorem 2.8).

Theorem 2.3

(Ref.11 , Theorem 2.6).The Kauffman polynomial (and, therefore, the Jones polynomial) is a link invariant.

For an oriented link diagram, let L+, L, L0 be the resulting link diagrams arising from crossing and smoothing changes on a local region of a specified crossing of the diagram, as in Fig. 2.

Definition 2.4.
The Conway polynomial ofL, ∇(L), is defined by the following skein relations:

The Conway polynomial of a link gives the Alexander polynomial via a change of variables. Thus, the Conway polynomial is a link invariant.10 

We will be interested in computing the Jones and Conway polynomials of links containing specific tangles.

Definition 2.5.

A (two-string) tangleTis a three-dimensional ball with two strings properly embedded in it, that is, the endpoints of the strings are fixed on the boundary of a ball.

Note that in some situations, we may also allow for additional (knotted and linked) components inside the ball.

Choosing the fixed points on the boundary of the ball to be along the equator, one can arrange for the tangle to be in general position with respect to the disk bounded by the equator. This allows for a diagrammatic representation of such a tangle (see Fig. 3). It will be useful to denote certain tangles by integers (see Fig. 4).

Given tangles T and U, we denote by

  • T the reflection of T with respect to the equatorial disk,

  • Tρ the reflection of T across the NW–SE axis (see Fig. 5),

  • TN the numerator closure of T,

  • TD the denominator closure of T,

  • T + U the tangle sum of T with U, and

  • T * U the tangle “vertical” sum of T with U.

See Fig. 6 for schematic pictures of the latter four operations.

We denote by LT a link diagram that, in a disk intersecting the link diagram at four points, contains the tangle diagram T. For example, TN and TD are such link diagrams. If LT is such a link diagram with T contained in some disk D, then performing local moves (such as crossing changes and smoothings) in T results in a new tangle diagram T′ ⊂ D and a new link diagram LT, which is unchanged outside of D. With this notation, the bracket polynomial ⟨LT⟩ can be formally expanded to f(T)⟨L0⟩ + g(T)⟨L⟩, where f(T) and g(T) are Laurent polynomials in Z[A,A1], which depend only on the tangle diagram T.

Definition 2.6
(Ref. 8). LetTbe a tangle. Define the bracket vector ofTto be

With the notation from the previous paragraph, the definition of br(T) gives


Proposition 2.7
(Ref. 8). LetTandUbe tangles. Then, we have the formula
whereδ = −A2A−2.

In Sec. IV, we define a similar invariant for computing the Conway polynomial.

A two-component link (N, K1K2) in the solid torus N can equivalently be thought of as a three-component link (S3, K1K2 ⊔ μ) in S3, where the additional component μ is a meridian of N.1 In this setting, where we are concerned with skies of causally related events, the meridian μ may be oriented in either direction, while K1 and K2 must be oriented coherently with a longitude λ. In addition, we require that each of K1 and K2 be independently isotopic to λ so that the linking numbers lk(K1μ), lk(K2μ) are either both 1 or both −1. We call such three-component links two-sky-like, and all links we will consider will be of this form.


While all links arising from pairs of skies in our setting are two-sky-like, it is possible that there are two-sky-like links that do not arise as pairs of skies.

We consider the family of link diagrams C(T) depicted in Fig. 7(a), where T can be any tangle. When the tangle T has a diagram that can be appropriately oriented, we also will consider C(T) with two specific orientations: C+(T) and C(T) as in Figs. 7(b) and 7(c). This is a natural family to consider since the simplest examples of two-sky-like links have geometric intersection number two with the meridian. Note that if C(T) is two-sky-like, then the two-component link where μ is ignored is the numerator closure of T.

Let H be the connected sum of two Hopf links, i.e., C(0). With the orientations as in C+(0) and C(0), H is two-sky-like. In fact, H is a pair of skies in the following sense. Let X is a (2 + 1)-dimensional globally hyperbolic spacetime with Cauchy surface homeomorphic to R2. The solid torus N is the space of future-directed null geodesics (light rays) in X and is homeomorphic to a solid torus. Then, H corresponds to an unlink of two longitudinal loops in the solid torus N, which corresponds to the skies of a pair of causality unrelated points in X.

Using Definition 2.1, we calculate the bracket vector of C(T).

Proposition 3.1.


Apply Definition 2.1 iteratively to each crossing in the diagram of C(T).□

Before we apply Proposition 3.1 to show Theorem 3.6, we lay out some facts about the Kauffman bracket.

Lemma 3.2.
LetLbe a link diagram, and letLt+andLtbe the result of adding a positive kink and a negative kink, respectively, toLvia a type I Reidemeister move. Then
  1. Lt+⟩ = −A3L.

  2. Lt⟩ = −A−3L.

  3. For a positive integerk,

  1. For a negative integerk,

  1. IfLis another link diagram, then


To see (1) and (2), apply Definition 2.1 to Lt+ and Lt, respectively (see, for example, Ref. 11, Proposition 2.5). To prove (3), we apply Definition 2.1 at any crossing of kN. Combining this with (1) and (2), we obtain the recursive relation
Part (3) follows by induction on k. Part (4) follows immediately from the fact that the Kauffman bracket of a mirror diagram is obtained by substituting A−1 for A (Ref. 11, Proposition 2.7). Finally, note that that V(L#L′) = V(L)V(L′) and w(L#L′) = w(L) + w(L′) for any appropriate orientation on L and L′. Then, (5) follows easily from Definition 2.2.□

Definition 3.3.

A regular isotopy between link diagrams is a series of type II or type III Reidemeister moves. A balanced isotopy between link diagrams is a series of type II Reidemeister moves, type III Reidemeister moves, or pairs of type I Reidemeister moves that preserve the writhe of the diagram for any orientation, e.g., a positive kink and a negative kink. Two link diagrams are said to be balanced-isotopic if they are related by a balanced isotopy.

Corollary 3.4.

The bracket polynomial is an invariant of regular-isotopic and balanced-isotopic link diagrams.


This is a direct consequence of Definition 2.2 and Lemma 3.2.□

We now consider the family of two-sky-like links U(n, m) = C(TU(n, m)) (see Fig. 8), where TU(n, m) is as in Fig. 9 for integers n and m.

Definition 3.5.

Lety(x) be a nonzero Laurent polynomial. Denote byh(y) the highest exponent ofxand(y) the lowest exponent ofxiny(x). The span of a nonzero Laurent polynomialy(x), denoted span(y(x)), is defined to beh(y) − (y).

Clearly, span(y1y2) = span(y1) + span(y2). We will use this fact in the next theorem.

Theorem 3.6.

For (n, m) ≠ (0, 0),V(U(n, m)) ≠ V(H) for any orientation ofU(n, m) andH.

From Definition 2.2, we have that for any link diagram L,
The Jones polynomial of the Hopf link is − t5/2t1/2 for some orientation,9 so V(H) = t5 + 2t3 + t for some orientation. It follows that
Thus, it suffices to show that span(⟨U(n, m)⟩) ≠ 16 for (n, m) ≠ (0, 0).

The mirror image of U(n, m) is balanced-isotopic to U(−n − 1, −m + 1), and by flipping the link across the horizontal axis, we see that U(n, m) is isotopic to U(m − 1, n + 1) for all integers n, m. The composition of flipping and mirroring takes U(n, m) to U(−m, −n). Thus, it is sufficient to consider n ≥ 0 and m ≥ −n.

First, we consider U(n, −n) for n positive. Note that TU(n,−n)N is regularly isotopic to a split union of two unknots. By Proposition 3.1, we have
Applying Theorem 5.7 from Ref. 2 and Lemma 3.2 from this paper, we see that
Therefore, we have
and so span(⟨U(n, −n)⟩) = 4n + 12 − (−4n − 12) ≥ 24 > 16.
Next, we consider U(n, −m), with 0 ≤ m < n. In this case, TU(n, −m)N is no longer a trivial link. However, observe that after performing the regular isotopy shown in Fig. 10, we obtain a link that is balanced isotopic to C((n + m)ρ). Therefore, we can apply Proposition 3.1. Using this and Lemma 3.2, we see that
Applying Theorem 5.7 from Ref. 2 and Lemma 3.2 from this paper, we calculate that
(Here, the empty sum when m = 0 is taken to be 0.) This allows for a full calculation of ⟨U(n, −m)⟩, as Proposition 3.1 gives
After checking that U(1, 0) does not have a trivial Jones polynomial, we consider the few remaining cases. We omit many of the computations as they are very similar to the computation of span(⟨U(n, −n)⟩) above.
  • If m = 0, and n > 1, we compute that
    Note that in this case, the highest terms in the two summands cancel.
  • If nm = 1, then n = m + 1 and −m < 0. In this case, there is cancellation in the sum in the formula for ⟨TU(n, −m)N⟩. We see that

  • If nm > 1 and −m < 0, we compute that

Finally, we consider the case of U(n, m) with n ≥ 0 and m > 0. We can always arrange that n < m for if nm, we can instead consider U(m − 1, n + 1) [obtained by flipping U(n, m) over] and we have m − 1 < n + 1. Because U(0, 1) is isotopic to the link L8n3 in Ref. 9, which has the Jones polynomial not equal to that of H, we can further restrict to m ≥ 2. Now,
First, we consider the case where n = 0. Then, we have
h(TU(0,m)N)=3m+6 and (TU(0,m)N)=m6,
h(TU(0,m)D)=3m7 and (TU(0,m)D)=m11,
and so
Now, we consider the last remaining case, m > n > 0,
h(TU(n,m)N)=3(n+m)+6 and (TU(n,m)N)=(n+m)6,
h(TU(n,m)D)=3(n+m)+1 and (TU(n,m)D)=(n+m)11,
and so
Thus, U(n, m) does not give the trivial Jones polynomial for any integers (n, m) ≠ (0, 0).□

Proposition 1.1 follows from Theorem 3.6. Namely, consider U(n, −n). The corresponding two-component link TU(n,−n)N is evidently an unlink. On the other hand, U(n, −n) and the connected sum of two Hopf links H have different Jones polynomials for any orientation of the components by Theorem 3.6. Natário and Tod2 had already found an infinite family of links with a nontrivial Jones polynomial, but did not consider links in the solid torus that were unlinked when embedded into R3.

Can the Jones polynomial detect causality in the given setting? In other words, are all two-sky-like links distinguishable from the connected sum of two Hopf links H by their Jones polynomials? To answer this question, it is necessary to ask if any non-trivial two-sky-like link has the same bracket polynomial as H = C(0) up to multiplication by a power of A. We first check LinkInfo9 and find no three-component links up to 11 crossings with the same Jones polynomial as the connected sum of two Hopf links. Next, we try to find properties of a link with this Jones polynomial. For simplicity, we still consider the family C(T). One can calculate, using Definition 2.1 that,


Then, by Definition 2.6, for any tangle T,


If, for some orientation of C(T), V(C(T)) = V(C(0)), then C(T)=(A3)nC(0) for some integer n. One way we could hope to get this is by finding T such that br(T)=(A3)nbr(0)=[(A3)n0]t. However, we have the following theorem:

Theorem 3.7

(Ref.12 , Theorem 13). There exists a nontrivial knot with Jones polynomial 1 if and only if there exists a tangleT ≠ 0 such thatbr(T)=[rAn0]tfor some integersr,nZ.

It remains an open problem whether there exists a nontrivial knot with trivial Jones polynomial.13 We also do not know that C(T)=(A3)nC(0) necessarily implies br(T)=(A3)nbr(0). There are certainly many vectors VZ[A,A1]2 that satisfy


but we do not know if these vectors arise as bracket vectors of tangles. We pose this as an open question.

Question 3.8.
imply br(T) = br(0)?

If Question 3.8 is answered in the positive, then whether or not the Jones polynomial detects causality in our setting is tied to whether or not the Jones polynomial detects the unknot by Theorem 3.7. If Question 3.8 is answered in the negative, then there must exist some tangle T that provides a counterexample, and this tangle will necessarily be nontrivial but not necessarily two-sky-like. Our next step would be to check if it is two-sky-like and then whether it appears as a pair of skies. Since there is no clear answer either way, we pose the following conjecture:

Conjecture 3.9

(Conjecture 1.1). LetXbe a (2 + 1)-dimensional globally hyperbolic spacetime with Cauchy surface Σ not homeomorphic toS2orRP2. Then, the Jones polynomial detects causality between any two events inX.


Not all two-sky-like links are of the form C(T). For a two-sky-like link to be C(T) for some tangle T, it is necessary and sufficient that the corresponding two-component link in the solid torus has some meridional disk that meets each component exactly once. An example of a two-sky-like link that does not seem to satisfy this condition is given in Fig. 11.

We now investigate the Conway polynomial and causality by noting some tangle invariants. Of particular importance is the way that tangle invariants are related to certain link invariants. We first distinguish two classes of tangles.

Definition 4.1.

An oriented tangleTis left-right oriented if its ends are oriented as inFig. 12(a),. IfTis oriented as inFig. 12(b)or (c), we say thatTis diagonally oriented.

Note that while a diagonally oriented tangle can have consistently oriented numerator and denominator closures, a left-right oriented tangle only has a consistently oriented numerator closure. It is also important to note that the tangles 0 and can be diagonally oriented, but only 0 can be left-right oriented.

For diagonally oriented tangles, Kauffman10 gave a definition for an invariant, which is well-behaved under tangle addition.

Definition 4.2

(Ref. 10). Given a diagonally oriented tangleT, let the fraction of the tangle be denotedF(T) = ∇(TN)/∇(TD).

We will usually not consider a/b equivalent to ac/bc so that the numerator and the denominator of the fraction are well-defined quantities.

Theorem 4.3
(Ref. 10, originally due to Conway). Given tanglesTandUthat are diagonally oriented such thatT + Uis coherently oriented,F(T + U) = F(T) + F(U). Specifically,

Recall that we denote by LT a link diagram that, inside some disk intersecting the link diagram in four points, contains a tangle diagram T. When LT is oriented, then T is also oriented and the resulting orientation may be left-right or diagonal as in Definition 4.1. The following fact will be useful:

Theorem 4.4.
LetLTbe an oriented link diagram, which contains a left-right oriented tangle diagramT. Then, Definition 2.4 can be used to calculate ∇(LT) as a linear combination
for somep(T),q(T)Z[z]depending only on the tangle diagramT, where 0 and 1 denote the corresponding tangles.
Similarly, ifLTis an oriented link diagram, which contains a diagonally oriented tangle diagramT, then ∇(LT) can be calculated as
for someP(T),Q(T)Z[z]depending only on the tangle diagramT.


We first assume that all tangles are left-right oriented unless otherwise stated. We induct on the number of crossings n of a tangle T. For n = 0, T must be the (split) union of tangle 0 and possibly some finite number of simple loops. If the number of loops is nonzero, ∇(LT) = 0 since the Conway polynomial of a split link is 0. Otherwise, ∇(LT) = ∇(L0).

Let T be a left-right oriented tangle with n = k crossings. Suppose that ∇(LT) can be formally expanded as a linear combination of ∇(L0) and ∇(L1) for some k ≥ 1. We show that if T has n = k + 1 crossings, ∇(LT) can be formally expanded as a linear combination of ∇(L0) and ∇(L1). Let strand 1 be the strand starting at the NW corner, and let strand 2 be the strand starting at the SW corner. We induct on N = u1(T) + o2(T), where u1(T) is the number of undercrossings of strand 1 not with itself and o2(T) is the number of overcrossings of strand 2 not with itself.

Suppose N = 0. Then, strand 1 passes over all other components, and strand 2 passes under all other components. If there is another connected component, it must be split from both strands, and ∇(LT) = 0. Otherwise, the diagram involving LT is isotopic to a connected sum, either L0#K1#K2 or L1#K1#K2 for some knots K1 and K2, depending on whether strand 1 ends in the NE or SE corner. Specifically, K1 comes from strand 1, and K2 comes from strand 2. Then, ∇(LT) = ∇(K1)∇(K2)∇(L0) or ∇(LT) = ∇(K1)∇(K2)∇(L1).

Now suppose N = m for n = k + 1 crossings. Either strand 1 has an undercrossing not with itself or strand 2 has an overcrossing not with itself. In either case, let T0 be T with this crossing smoothed (with the orientation), and let T× be T with this crossing switched. We know that (LT0) is a linear combination of ∇(L0) and ∇(L1) by the induction hypothesis on n, the number of crossings of T. Since T× has u1(T×) + o2(T×) equal to either m − 1 or m − 2, ∇(T×) is a linear combination of ∇(L0) and ∇(L1) by the induction hypothesis on N. Then,
is also a linear combination of ∇(L0) and ∇(L1). By induction, the result holds for all left-right oriented tangles T.

The proof for diagonally oriented tangles is identical except that strands 1 and 2 start at diagonally opposite corners, and ∇(LT) is reduced to a linear combination of ∇(L0) and ∇(L).□

The next lemma motivates the definition of Con(T).

Lemma 4.5.
LetLTbe an oriented link diagram, which contains a diagonally oriented tangle diagramT. Then,

Note that 0N and D are split links and thus have Conway polynomial 0. By Theorem 4.4, for any oriented link diagram LT containing the diagonally oriented tangle T, the Conway polynomial ∇(LT) can be computed as a sum P(T)∇(L0) + Q(T)∇(L), where P(T),Q(T)Z[z] only depend on T. Recall that TN and TD are links containing T. If T is diagonally oriented, then there is an induced orientation on TN and TD. We have
as desired.□

This lemma and the bracket vector definition in Ref. 8 indicate that it is useful to reduce a tangle diagram into a formal sum via a skein relation.

Definition 4.6.
LetLTbe an oriented link diagram, which contains a left-right oriented tangle diagramT. Then, ∇(LT) can be computed asp(T)∇(L0) + q(T)∇(L1) for some polynomialsp(T),q(T)Z[z]by Theorem 4.4. Define the Conway vector ofTto be

Proposition 4.7.

LetTbe a left-right oriented tangle. Then, Con(T) is a well-defined invariant of isotopy classes of tangles. Explicitly, Con(T) = (∇((T − 1)N), ∇(TN)).

We reduce TN and (T − 1)N using Theorem 4.4 since these are both oriented tangle diagrams containing T as a left-right oriented tangle diagram,
Then, if T is changed by an isotopy, ∇(TN) and ∇((T − 1)N) remain constant. Thus, Con(T) is defined for isotopy classes of tangles.□

Let L be a diagrammatic operator, which takes a single left-right oriented tangle T and yields an oriented link L(T). In particular, L(T) is an oriented link containing the tangle T in some ball. Applying Definition 4.6, we get that


This proves Theorem 1.2 as this defines a Z[z]-module homomorphism φL:Z[z]2Z[z] with


such that φL(Con(T)) = ∇(L(T)).

The Conway vector gives formulas for various combinations of tangles. These formulas mirror those from Ref. 8, Proposition 2.2.

Proposition 4.8.
LetTbe a left-right oriented tangle.
  1. LetUbe a left-right oriented tangle. Then,

  1. LetWbe a diagonally oriented tangle such thatT * Wis a left-right oriented tangle. Then,


First, we use Definition 2.4 to see that if L2 is any link diagram containing the integer tangle 2, then
Note that if T and U are left-right oriented tangles, then so is T + U. Consider an oriented link diagram LT+U containing T + U in some disk intersecting the link diagram in four points. Reducing T and U, in turn, using the skein relation,
This proves the first statement.
To prove the second statement, we first note that ((T * 0) − 1)N is isotopic to TN, T * = T, and (T * 0)N = TNO is a split link, which thus has Conway polynomial 0. Recall that the Conway polynomial is a link invariant and therefore does not change under isotopy. We can treat diagrams of ((T * W) − 1)N and (T * W)N as oriented link diagrams containing a diagonally oriented tangle W. Applying Lemma 4.5, Definition 4.6, and Proposition 4.7, we see that

In this section, we find an infinite family of two-sky-like links with a Conway polynomial indistinguishable from H, the connected sum of two Hopf links. We show using the Jones polynomial that all of the links in this family are distinct from each other and distinct from H. More specifically, we will show that


for infinitely many left-right oriented tangles T. First, we identify a necessary and sufficient condition for this equation to be satisfied.

Lemma 5.1.

Equation (1)is satisfied by a tangleTif and only if Con(T) = (1, 0).

We note that C+(T) and C(T) are oriented links containing a left-right oriented tangle T. Using the Definition 4.6 of Con(T) and the fact that C+(1) is L4a1{1} and C(1) is L4a1{0} in Ref. 9, we have
If Con(T) = (1, 0), then Eq. (1) is satisfied. Conversely, suppose Eq. (1) is satisfied. Since ∇(C+(T)) = ∇(C(T)), we must have q(T) = 0, and it follows that p(T) = 1.□

We now construct solutions T to Eq. (1). In Ref. 10, Kauffman identifies three diagonally oriented tangles TA = 2ρ, TB = −6, and TC with fractions F(TA) = 1/z, F(TB) = 3z/1, and F(TC) = −3z/1. We refer the reader to Sec. II for definitions of Tρ and integer tangles. For a diagram of TC, see Fig. 13. Using Theorem 4.3,


One can check that with the same orientation, F(−TA) = 1/(−z). Then, a similar calculation gives


Let T+ be TA + TB + TC rotated 90° clockwise and T be −TA + TB + TC rotated 90° clockwise. Since the rotation swaps numerator and denominator closures, F(T+) = z/1 and F(T) = −z/1. Let T0 = T+ + T. Then, T0 is a diagonally oriented tangle, which has


In other words, (T0N)=0 and (T0D)=1. Let T0(n) denote an n-fold sum T0 + ⋯ + T0. Then, F(T0(n)) = 0/1 for all positive integers n by repeated application of Theorem 4.3. If tangle 1 is given a left-right orientation, then 1 * T0(n) is left-right oriented. The link C(1 * T0(1)) is shown in Fig. 14. The links C+(1 * T0(n)) and C(1 * T0(n)) are visibly two-sky-like for all nZ+.

Theorem 5.2.

For any positive integern,Eq. (1) is satisfied by 1 * T0(n), and the (unoriented) linkC(1 * T0(n)) is not isotopic to the (unoriented) connected sum of two Hopf linksH. The linksC(1 * T0(n)) are distinct as unoriented links. Thus, there are infinitely many two-sky-like links indistinguishable fromHby the Conway polynomial.

We show that the span of the bracket polynomial distinguishes all of the links C(1 * T0(n)), nZ+, from each other and from C(0). With the help of KnotTheory’ in Mathematica,14 we compute T0N and T0D. By Definition 2.6, for any tangle T,
where δ = −A2A−2. Similarly,
and so
For ease of notation, let Poly(y1;y2)Z[A,A1] be the set of Laurent polynomials with the highest degree monomial y1 and lowest degree monomial y2. Using the above formula, we calculate that
By Proposition 2.7, for nZ+,
Assume for some n that
which is true for n = 1. Then,
and so
By induction, Eq. (2) holds for all nZ+. From Definition 2.6,
We calculate that ⟨C(1 * 0)⟩ = −A11 − 2A3A−5 and ⟨C(1 *)⟩ = A9 + AA−3 + A−7. Hence, for nZ+,
and so
Then, the Jones polynomial of C(1 * T0(n)) is distinct for distinct nZ+ and not equal to the Jones polynomial of C(0) for any orientation of the components. It follows that the links C(1 * T0(n)) are distinct from each other and distinct from C(0).
It remains to show that 1 * T0(n) satisfies Eq. (1). By Lemma 5.1, it is sufficient to show that Con(1 * T0(n)) = (1, 0). Recall that ∇(T0(n)N) = 0 and ∇(T0(n)D) = 1, and observe that Con(1) = [0 1]t. Using Proposition 4.8,
This gives the desired result.□


We have not proved conclusively that the Conway polynomial does not detect causality in the given setting. The link C(1 * T0(n)) might not correspond to the skies of a pair of causally related events. However, C(1 * T0(n)) satisfies the basic topological restrictions of a two-sky-like link, giving strong evidence that the Conway polynomial does not detect causality in the given setting.


though it is not directly relevant to this setting, the Conway vector cannot distinguish the tangle T0(n) from tangle 0 even when the orientation of one of the strands is reversed. Let TA = (−2)ρ, left-right oriented, and let U0 be TB + TC rotated 90° clockwise and oriented as above. Note that T0′ ≔ (TA′ * U0) + ((−TA′) * U0) is T0 but left-right oriented. The above calculation gives that (U0N)=1 and ∇(U0) = 0, so Proposition 4.8 gives Con(TA′ * U0) = Con(TA′) and Con((−TA′) * U0) = Con(−TA′). Then, Con(T0′) = Con(TA′ + (−TA′)) = Con(0) since Proposition 4.8 gives the Conway vector of a sum of tangles in terms of the Conway vectors of the tangle summands. Thus, T0 is indistinguishable from tangle 0 via the Conway polynomial no matter how it is oriented. A similar calculation to the above will show that C+(1 * T0(n)) and C(1 * T0(n)), when T0 is left-right oriented, have the same Conway polynomial as C+(0) and C(0) with the longitudinal components oriented oppositely relative to the meridional component. Thus, the Conway polynomial cannot distinguish C(1 * T0(n)) from the connected sum of two Hopf links H.

The authors would like to thank Vladimir Chernov and Ina Petkova for their guidance throughout the project. The authors would also like to thank Gage Martin for helpful comments. S.A. would like to thank Charles Livingston for a helpful conversation. J.H.S. would like to thank Vanessa Pinney for her insight on enumeration of links. J.H.S. received support from NSF Grant No. DMS-1711100.

Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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