The rational Khovanov homology of 3-strand pretzel links

THE RATIONAL KHOVANOV HOMOLOGY OF 3-STRAND PRETZEL LINKS 29
But we must have dimX l+n−1,3l+3n−1 = dimX l+n−2,3l+3n−5 since X has no exceptional
pair in t = l + n − 1 or t = l + n − 2. So both e 2 and e 3 must cancel.
Finally, if e 4 did not cancel, then we would have dimX l+n−1,3l+3n−3 = dimX l+n,3l+3n+1 +
1. But we need dim X l+n−1,3l+3n−3 = dimX l+n,3l+3n+1 , again because the only exceptional
pairs of X are in t = 0 and t = 2l. Hence e 4 must cancel, and we have proved
our formula when n is odd and less than l.
The next case for odd n is n = l + 1. The middle of Figure 12 is a reference here. A
similar argument implies that e 2 , e 3 , and e 4 all cancel, and a quick check verifies that
our results for the lower and upper summands of X agree with Definition 2.5(4) and
Definition 2.6(7).
If n is odd and n > l + 1, then the right side of Figure 12 depicts the situation. The
generator e 1 still cannot cancel. But logic very similar to before implies that e 4 and
a one-dimensional subspace of (e 2 ,e 3 ) must cancel, since X has no exceptional pairs
in t = l + n. Thus, the lower summand loses an exceptional pair (as predicted by
Definition 2.5), and the upper summand turns an exceptional pair into a knight’s move.
Finally, suppose n is even; luckily, this case is easier. First, if n < l, we want to
show that only e 4 cancels (using the above notation). Again, e 1 cannot cancel. If
some combination of e 2 and e 3 cancelled, then we would have dimX l+n−1,3l+3n−1 <
dimX l+n−2,3l+3n−5 , an impossibility since X has no exceptional pairs in t = l + n − 2.
So both e 2 and e 3 survive.
Since n < l, P(−l,l,n) is quasi-alternating by [4], so X l+n,3l+3n−3 = 0 for δ-grading
reasons, and e 4 has nowhere to live. The cancellation is responsible for the next (−1)
in the sequence d (n) , and e 2 and e 3 are responsible for the (2) following it.
On the other hand, if n > l, then X has two exceptional pairs in t = l + n but
V l+n = 0, so none of the e i can cancel. If n = l, the same argument holds: X must
have three exceptional pairs in t = l + n but dimV l+n is only two, so none of the e i can
cancel. Thus e 1 and (say) e 2 add an exceptional pair to the lower summand, and e 3 and
e 4 add an exceptional pair to the upper summand. These additions are precisely what
we were expecting, completing the inductive argument.
□
5.2. P(−l,l + 1,n) for even l. Now we will make a similar computation for P(−l,l +
1,n) which we will use as the base case in the induction to follow. As with odd l, we
do not need to start with n = 0. We can use n = l instead; since P(−l,l + 1,l) =
P(−l,l,l + 1), the previous section tells us the formula for Kh(P(−l,l + 1,l)).
Lemma 5.2. For n ≥ l+1, Kh(P(−l,l+1,n)) is given by the formula in Theorem 2.9.
Proof. Resolve the top crossing on the rightmost strand, a positive crossing. Our diagram
for P(−l,l + 1,n) has no negative crossings, while the unoriented resolution (a
diagram for the unknot) has l + n − 1 negative crossings. The oriented resolution is a
diagram for P(−l,l + 1,n − 1). Hence the skein sequence is
f
q 3l+3n−1 t l+n Kh(U) Kh(P(−l,l + 1,n)) qKh(P(−l,l + 1,n − 1))
f
,