Corollary in Rasmussen's paper about $s$-grading of Lee's canonical generators In Jacob Rasmussen's paper Khovanov homology and the slice genus, 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.
 A: 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 ) $.
