Rasmussen's invariant changes sign under mirror image

On $A_L=\mathbb{Q}[X]\mathrm{/}(X^2-1)$ use $⟨u,v⟩={ϵ}_L(uv)$, whose matrix in the basis $1,X$ is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Thus $1,X$ are dual labels of opposite quantum degrees. The displayed Frobenius maps give $⟨m(a,b),c⟩=⟨a\otimes b,\mathrm{\Delta }(c)⟩$ on the eight basis triples, and the analogous split identity follows by transposition. Number the crossings identically on $D$ and its mirror. For a mirror vertex $w$, its corresponding original vertex is $v=1-w$. The local smoothing chart interchanges A and B under reflection, so their resolved circles agree. On the dual complex put $(C^{\vee }{)}^i={\mathrm{Hom}}_{\mathbb{Q}}(C^{-i},\mathbb{Q})$ and $d^{\vee }=(d{)}^{\ast }$; this convention squares to zero directly. For the mirror edge changing coordinate $j$, the original reversed edge has sign $(-1{)}^{{\sum }_{k<j}{v}_k}$, while its mirror sign is $(-1{)}^{{\sum }_{k<j}{w}_k}=(-1{)}^{j-1}(-1{)}^{{\sum }_{k<j}{v}_k}$. Multiply the tensor-pairing identification on mirror vertex $w$ by $\sigma (w)=(-1{)}^{{\sum }_j(j-1)w_j}$. Since $\sigma (w+e_j)\mathrm{/}\sigma (w)=(-1{)}^{j-1}$, this makes every edge commute with the transposed original edge. It is therefore an actual cochain isomorphism, including cube signs. With $n=n_{+}+n_{-}$, complementation gives $i^{\mathrm{\prime }}=-i$ and state shift $\mathrm{\mid }w\mathrm{\mid }+n_{+}^{\mathrm{\prime }}-2n_{-}^{\mathrm{\prime }}=-\mathrm{\mid }v\mathrm{\mid }-n_{+}+2n_{-}$, the negative original shift. Dual tensor labels have opposite label degrees as well, giving $q^{\mathrm{\prime }}=-q$. Reversing the knot orientation leaves crossing signs and resolutions unchanged, so the same isomorphism applies to $K^{†}$.

For a finite filtered cochain complex $C$, the just-described dual quantum filtration is $F^p{C}^{\vee }=\mathrm{Ann}(F^{1-p}C)$ in complementary cohomological degrees. Evaluation induces a perfect pairing $H(C^{\vee })\times H(C)\to \mathbb{Q}$: cocycles are exactly the functionals vanishing on boundaries, and coboundaries are exactly those vanishing on cycles, by finite elimination. Its induced homology filtration is exactly the annihilator filtration. Indeed one direction follows by evaluating representatives. Conversely a homology functional vanishing on classes represented by cycles in $F^{1-p}C$ defines a functional on the sum of the cycle space and $F^{1-p}C$, using its given values on cycles and zero on the latter subspace. The intersection condition makes this well-defined. Extend it across a finite vector-space complement; it is a cocycle because it vanishes on boundaries and represents the required dual class in $F^p$. Thus the finite homology graded quotients occur in opposite degrees under duality. Applied to the two Lee lines, $s_{\max }(K^{†})=-s_{\min }(K)$ and $s_{\min }(K^{†})=-s_{\max }(K)$. By [A1] and [F2], $s_{\max }(K)=s(K)+1$ and $s_{\min }(K)=s(K)-1$, so $s(K^{†})=s_{\max }(K^{†})-1=-s(K)$. This finite basis argument adds no choice assumption.