The HHH of the positive two-strand torus knot

Reduction by induction. Write the two outer copies of $y$ as $a,c$ and the middle copy in $B_1{\otimes }_R{B}_1$ as $b$. Its unshifted coefficient ring has $b^2=a^2=c^2$ and decomposes as $B_1\oplus B_1\{2\}$ on the middle basis $1,b$, as in [L2]. Tensor the displayed minimal complex for $n$ with $[R\{2\}\to B_1]$. For every existing $B_1\{r\}$ term, the new vertical map $B_1\{r+2\}\to B_1\{r\}{\otimes }_R{B}_1$ sends $x$ to $x(b+c)$; its component in the middle-$b$ summand is the identity. Cancel these blocks by [L7], starting from the highest cohomological degree and continuing through the bounded complex. The remaining terms are $R\{2n+2\}$ and $B_1\{2n\},B_1\{2n-2\},\dots ,B_1$, in degrees $-n-1,\dots ,0$. The base $n=1$ is the defined generator complex, so this gives the asserted terms for every $n$.

Compute the induced maps on $H{H}_0$. By [L3] the group $H{H}_0(B_1)$ is the quotient of $B_1$ by the commutator submodule is free on the class of $1\otimes 1$, with the class of $1\otimes y$ equal to $y$ times it. A bimodule map $\phi$ with $\phi (1\otimes 1)=1\otimes y-y\otimes 1$ therefore induces $0$ on $H{H}_0$, because $1\otimes y$ and $y\otimes 1=y(1\otimes 1)$ both lie in the class of $y(1\otimes 1)$, so their difference maps to $0$; while a map with $\phi (1\otimes 1)=1\otimes y+y\otimes 1$ induces multiplication by $2y$.

Compute the induced maps on $H{H}_1$. By [L3] the group $H{H}_1(B_1)$ is free on the class of $1\otimes y+y\otimes 1$; a bimodule map $\phi$ acts on this class by $\phi (1\otimes y+y\otimes 1)=\phi (1\otimes 1)\cdot y+y\phi (1\otimes 1)$. For $\phi (1\otimes 1)=1\otimes y-y\otimes 1$ this gives $(1\otimes y-y\otimes 1)y+y(1\otimes y-y\otimes 1)=(y^2\otimes 1-y\otimes y)+(y\otimes y-y^2\otimes 1)=0$, so the induced map is $0$; for $\phi (1\otimes 1)=1\otimes y+y\otimes 1$ the same computation with the signs reversed gives $2y$ times the generator, so the induced map is multiplication by $2y$.

The surviving differential. In the middle basis $1,b$, multiplication by $b\pm a$ has matrix $\left(\begin{array}{cc}\pm a& a^2\\ 1& \pm a\end{array}\right)$. Canceling the identity component of the column $(c,1{)}^T$ gives the projection $(q_0,q_1)\mapsto q_0-c{q}_1$, so on the remaining factor this multiplication becomes $\pm a-c$. Thus an old difference map changes to a sum map and an old sum map to a difference map, up to a unit sign, exactly when its position in the minimal complex advances by one. The new leading map is $a+c$ from the unit term, and the next map is $c-a$: their product is zero because $a^2=c^2$. Choosing signs of the surviving terms successively normalizes all unit signs to give the displayed $d_i$, alternating $c-a$ and $c+a$. This is the actual Schur-complement computation, and each canceled pair has the identity inverse as its homotopy by [L7]. The maps have degree zero between the displayed internal shifts.

The two complexes. By [L5] and [L2] the termwise complexes have the terms shown in the Example, and the differentials are those of steps 1.2 and 1.3 applied to the differential shapes $d_i$ of [L2]; the terms coming from the unit term $R\{2n\}$ contribute $H{H}_0(R,R)=R$ in degree $0$ and $H{H}_1(R,R)=R\{2\}$ in degree $1$ by [L4], which is the extra initial term of the $h=1$ complex. The leading $H{H}_1$ map is the identity in these bases: it sends the source Koszul symbol $\theta$ to $r{b}_1(1)\theta$, the target generator of [L3]. Hence the $h=0$ complex is $R\{2n\}\stackrel{2y}{\to }R\{2n-2\}\stackrel{0}{\to }R\{2n-4\}\stackrel{2y}{\to }\cdots$ and the $h=1$ complex begins $R\{2n+2\}\stackrel{1}{\to }R\{2n+2\}\stackrel{0}{\to }R\{2n\}\stackrel{2y}{\to }\cdots$, exactly as displayed.

Cohomology and actual Rouquier degrees. Number the minimal terms by $k=0,\dots ,n$; their cohomological degrees are $c=k-n$. In $h=0$, multiplication by $2y$ is injective, the intervening maps are zero, and for odd $n$ the last map is $2y$. The only cohomology is $R\{2n-2k\}\mathrm{/}(2y)\cong \mathbb{Q}\{2n-2k\}$ at odd $1\le k\le n$, giving the first list in the Example. In $h=1$ the initial identity pair cancels. For $n=1$ this is the entire complex and the second list is empty. For odd $n\ge 3$ the remaining maps alternate $0,2y$ with a final $2y$, so the only cohomology is $R\{2n-2k+4\}\mathrm{/}(2y)$ at odd $3\le k\le n$, giving the second list. Hence the class degrees are $(h,p,c)$ as displayed, with no unrecorded cohomological translation by $n$.

Rank count and comparison. The $h=0$ list has $(n+1)\mathrm{/}2$ classes and the $h=1$ list has $(n-1)\mathrm{/}2$ classes, each one-dimensional over $\mathbb{Q}$; the total rank of $HHH({\sigma }_1^n)$ is therefore $n$, in agreement with Khovanov's two-strand computation. For $n=2k+1$, the $h=0$ list has $k+1$ entries, the number of numerator monomials in the $A=0$ series in the cited two-strand example of Gorsky-Kivinen-Simental; that partial-sector count alone does not give the total rank. The pairs $(p,c)$ computed here are the internal and Rouquier degrees, and the identification with the full trigrading uses the dictionary of the comparison on the A page. The only use of AC is in step 3.1 through [L4].