The rational Khovanov homology of 3-strand pretzel links

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$MATH 201 - FINAL EXAM Friday January 24 2003, 8:30am Problem 1$

16 ANDREW MANION contradicting our assumption. □ In particular, if V and W have no exceptional pairs (i.e. E V and E W are identically zero), then there can be no nontrivial well-structured cancellation of V ⊕ W. One type of cancellation which occurs frequently is the following “standard cancellation:” suppose V and W are contained in the same two adjacent δ-gradings, with E V (i) > 0 and E W (i+1) > 0. Then one can single out two generators each from V and W in the configuration of Figure 7. Cancelling the generators in the middle q-grading produces a knight’s move, so the resulting space is still well-structured. Remark 3.7. Below, we will often see a set of cancellations containing several standard cancellations, and we will want to show no more cancellations can occur. We can argue as follows: suppose we did some standard cancellations on V ⊕ W to get X, and then did another cancellation. The additional cancellation would, in fact, be a cancellation of V ′ ⊕W ′ , a proper subspace of X obtained from V ⊕W by removing all four generators at each standard cancellation (rather than only those in the middle q-grading). The reason is that, after a standard cancellation, the two remaining generators under consideration are in the wrong δ-gradings to cancel (the generator of V is in the lower δ-grading, and the generator of W is in the higher one). Now, the spaces V ′ and W ′ are well-structured and have fewer exceptional pairs than V and W, so in many cases we will be able to apply Lemma 3.6 and derive a contradiction. We can also identify a situation in which we can conclude a standard cancellation occurred, given some information about X: Lemma 3.8. Assume V and W are contained in the same two adjacent δ-gradings. Suppose i is the lowest t-grading of W and E W (i) > 0. Assume also that E V (i − 1) > E X (i − 1). Then a standard cancellation must have occurred between t-gradings i − 1 and i. These two cancel Q Q Q Q Q Q Figure 7. A standard cancellation. Red generators are from V , blue generators are from W, and green generators belong to X after the cancellation.
THE RATIONAL KHOVANOV HOMOLOGY OF 3-STRAND PRETZEL LINKS 17 Proof. Let j be the lowest q-grading of W. As in the previous lemma, we have dimV i−2,j−4 − E V (i−2) = dimV i−1,j −E V (i−1) and dimX i−2,j−4 −E X (i−2) = dimX i−1,j −E X (i−1). We also know V i−2,j−4 = X i−2,j−4 since i is the lowest t-grading of W. Suppose no standard cancellation occurred; then dimX i−1,j = dimV i−1,j . After some substitutions, this equation becomes E X (i − 1) − E X (i − 2) = E V (i − 1) − E V (i − 2). But since X = V in t-gradings ≤ i − 2, we have E V (i − 2) = E X (i − 2) by Remark 3.5. Thus E X (i − 1) = E V (i − 1), a contradiction. □ By interchanging V with W and replacing i − 1 with i + 1, we get the following: Lemma 3.9. Suppose i is the highest t-grading of V and E V (i) > 0. Assume also that E W (i + 1) > E X (i + 1). Then a standard cancellation must have occurred between t-gradings i and i + 1. 3.2. Jones polynomial calculation. We need a Jones polynomial calculation to begin our inductive proofs. To fix notation, if L is a link, then V L (q 2 ) will denote the (normalized) Jones polynomial of L. (As usual, we will write everything in terms of the variable q, which squares to the standard argument of the Jones polynomial.) The unnormalized Jones polynomial V L (q 2 ) is defined to be (q + q −1 )V L (q 2 ). Lemma 3.10. Let l ≥ 2. The unnormalized Jones polynomial of the link P(−l,l, 0) LR is ∑l−3 V P(−l,l,0)LR (q 2 ) = (−1) l q 2l+1 + (−1) j+l+1 q 2l−2j−1 + (2q + 2q −1 ) j=1 ∑l−3 + (−1) i+1 q −2i−3 + (−1) l q −2l−1 . i=1 Remark 3.11. When l is odd, the orientation LR is forced by the choices we made earlier. When l is even, we will need to consider P(−l,l, 0) RL , whose Jones polynomial differs from that of P(−l,l, 0) LR by an overall factor of q. To pin down this factor, note that P(−l,l, 0) RL is (after flipping the diagram over) just P(−l,l, 0) LR with the orientation of the “outer” component reversed. The linking number of this component with the rest of the link in P(−l,l, 0) LR is −l/2. Hence P(−l,l, 0) RL picks up a factor of q −6(l/2) = q 3l . The resulting formula for the unnormalized Jones polynomial of the link P(−l,l, 0) RL is (2) ∑l−3 V P(−l,l,0)RL (q 2 ) = (−1) l q 5l+1 + (−1) j+l+1 q 5l−2j−1 + (2q 3l+1 + 2q 3l−1 ) j=1 ∑l−3 + (−1) i+1 q 3l−2i−3 + (−1) l q l−1 . i=1
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The rational Khovanov homology of 3-strand pretzel links

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