Rasmussen's invariant for positive knots

In the positively oriented source chart of [A2], take the over-tangent to be $(1,1)$. Positivity forces the under-tangent to be $(-1,1)$ rather than $(1,-1)$, since their determinants with the over-tangent are $2$ and $-2$. The incoming endpoints are therefore southwest and southeast, and the outgoing endpoints northeast and northwest. The orientation-preserving smoothing pairs southwest with northwest and southeast with northeast: exactly the A-smoothing. Reversing both strand arrows gives the other positive orientation case and retains those same pairs. Thus every positive oriented smoothing is the $0$-resolution.

By step 1.1, every oriented smoothing is the $0$-resolution, so the canonical cycle lies in cohomological degree $0$. There is no chain group in degree $-1$, hence this cycle is not altered by adding a boundary. Expanding each of its $k$ circle labels $a=1+X$ or $b=X-1$ in the homogeneous basis, the unique term of least quantum degree is obtained by choosing $X$ on every circle; its coefficient is $1$ or $-1$. Its quantum degree is the state shift $n$ plus $k$ labels of degree $-1$, namely $n-k$. Therefore its induced filtration level is $s_{\min }(K)=n-k$.

The formula [F2] for $s$ is diagram-independent under the AC${}_{\omega }$ branch [A1]. Substitution of the value from step 2.1 gives $s(K)=n-k+1$. If $n=0$, the positive diagram is the crossingless unknot, whose oriented resolution has $k=1$, so the same calculation gives $s=0$. The state expansion itself is finite and choice-free.