In Jacob Rasmussen's paper [Khovanov homology and the slice genus](https://arxiv.org/pdf/math/0402131.pdf), he states as Corollary 3.6 that $s(\mathfrak s\_o)=s(\mathfrak s\_{\bar o})=s\_{min}(K)$, where $s$ is the $s$-grading and $\mathfrak s\_o,\mathfrak s\_{\bar o}$ are Lee's canonical generators for her homology theory $Kh'$. I don't really understand why this follows from the previous lemma. And in fact, I feel that at least one of $s(\mathfrak s\_o),s(\mathfrak s\_{\bar o})$ should be $s\_{max}$. After all, if any state $S$ can be written as $a\mathfrak s\_o+b\mathfrak s\_{\bar o}$, then we should have $s(S)\le\text{max}(s(\mathfrak s\_o),s(\mathfrak s\_{\bar o}))$. I contemplated that this could've been a typo in the corollary statement, but I still don't see why $s(\mathfrak s\_o)=s(\mathfrak s\_{\bar o})$, and I also don't see how this relates to the lemma.


Can someone explain the proof of this corollary, or at least explain why my "contradiction" to it is incorrect? Thanks.



Corollary in Rasmussen's paper about $s$-grading of Lee's canonical generators
Let $ C ( D) $ denote the Lee chain complex and $ Kh ' ( K ) $ its homology, $ q $ denote the grading on $ C ( D) $ (associated to the filtration) and $ s $ the induced grading on $ Kh ' ( K ) $.


Recall that $ \mathfrak{s}\_\overline{o} $ is obtained from $ \mathfrak{s}\_o $ by interchanging $ r = v\_+ + v\_- $ and $ g = v\_- - v\_+ $ on all circles of the oriented smoothing of $ D $. As $ q ( r ) = q ( g ) $ it follows that $ q ( \mathfrak{s}\_o ) = q ( \mathfrak{s}\_\overline{o} ) $ and that $ s ( \mathfrak{s}\_o ) = s ( \mathfrak{s}\_\overline{o} ) ~\text{mod}~ 4 $. By Lemma 3.5 $ Kh' ( K ) $ is supported in s-degrees $ 1 ~\text{mod}~ 4 $ and $ 3 ~\text{mod}~ 4 $, so that $ s ( \mathfrak{s}\_o ) = s ( \mathfrak{s}\_\overline{o} ) $ exactly.


The filtration on $ C ( D ) $ is upward: $ F^{i+1} C ( D ) \subset F^{i} C ( D ) $. As $ s ( \mathfrak{s}\_o ) = s ( \mathfrak{s}\_\overline{o} ) $ we must have $ s ( \mathfrak{s}\_o \pm \mathfrak{s}\_\overline{o} ) \geq s ( \mathfrak{s}\_o ) , s ( \mathfrak{s}\_\overline{o} ) $, so that $ s ( \mathfrak{s}\_o ) = s ( \mathfrak{s}\_\overline{o} ) = s\_{\text{min}} ( K ) $.


