The reduced Khovanov homology of a trefoil has rank three

Let $r$ be the number of one-resolutions. The all-through closure has two circles; for $r\ge 1$, the first cap-cup gives one circle and each additional cap-cup contributes one closed circle. Thus the eight state circle counts in lexicographic order are $2,1,1,2,1,2,2,3$. Removing the marked circle gives reduced state dimensions $2,1,1,2,1,2,2,4$, so the cube cochain dimensions are $2,3,6,4$.

Order states within each cohomological degree lexicographically and order free circle factors by their first encounter in a fixed diagram scan, with $1$ before $X$. The quotient by the marked-circle action in [F1], the Frobenius maps in [F2], and the edge procedure in [F4] give $d^0=\left(\begin{array}{cc}1& 0\\ 1& 0\\ 1& 0\end{array}\right)$, $d^1=\left(\begin{array}{ccc}0& 0& 0\\ 1& 1& 0\\ 0& 0& 0\\ 1& 0& 1\\ 0& 0& 0\\ 0& 1& 1\end{array}\right)$, and $d^2=\left(\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 1& 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 1& 0\\ 0& 1& 0& 1& 0& 1\end{array}\right)$. In $d^0$ each merge sends $1\mapsto 1$ and $X\mapsto 0$. In $d^1$ each target gets two marked-circle splits $1\mapsto X$. For $d^2$, on target basis $(1\otimes 1,1\otimes X,X\otimes 1,X\otimes X)$ the three source-state maps are $011:(1\mapsto 1\otimes X,X\mapsto X\otimes X)$, $101:(1\mapsto 1\otimes X+X\otimes 1,X\mapsto X\otimes X)$, and $110:(1\mapsto X\otimes 1,X\mapsto X\otimes X)$. These are exactly the marked quotient split maps and the ordinary Frobenius split map, in the stated basis order.

The matrix $d^0$ has rank $1$. In $d^1$, the first two columns are independent and their sum is the third, so $\mathrm{rank}{d}^1=2$. The last matrix has three independent nonzero rows, so $\mathrm{rank}{d}^2=3$. By [F1] the quotient is a cochain complex, and [F3] gives its finite support; hence its cohomology dimensions are $2-1=1$, $3-1-2=0$, $6-2-3=1$, and $4-3=1$. The total rank is therefore $3$, and grading shifts preserve total dimension.