The rational Khovanov homology of 3-strand pretzel links

24 ANDREW MANION
t = −1 0 1 n−l−1n−l n−l+1
Q
Q f 1
QQ f 2
Q
Q
Q
e 1
e 2
q = n−l+1
q = n−l−1
q = n−l−3
Q
Figure 10. The case m = l + 1 in Theorem 4.3. Red denotes V and
blue denotes W.
of U with a knight’s move; in effect, U becomes V [a] (up to a shift), where a is the
sequence a m−1−l ·( m−1−l ) n−m+1 ·a
2 m−1−l . Adding in the rest of W amounts to adding the
sequence (1, 0) (n−l)/2 , coordinate-wise, to the right side of a, where the first 1 carries an
exceptional pair. (“To the right side” means the addition is done such that the final 0
of (1, 0) (n−l)/2 lines up with the last nonzero index of a.) This addition replaces a m−l−1
with c m−l−1 and ( m−1−l ) n−m+1 with ( m−l+1 , m−l−1 ) (n−m+1)/2 . The exceptional pair is
2 2 2
on the first instance of m−l+1 , giving us the upper summand we want as specified in
2
Definition 2.6(3). We noted before that the summands are based at the right points, so
we have proved our formula.
Now assume m is odd. The diagram for P(−l,m,n) is oriented RR and has m + n
negative crossings, while the diagram for the unoriented resolution P(−l,m − 1,n) is
oriented LL and has l + m − 1 negative crossings. Thus ǫ = l − n − 1. The oriented
resolution is −T n−l,2 , the negatively oriented right-handed (n − l, 2) torus link. The