Hochschild Homology and Triply-Graded Link Homology — Examples

Compute the differential in the displayed basis. [L1, L2, given, algebra] In the left $R$-basis $\{1\otimes 1,1\otimes y\}$ one has $y\otimes 1=y(1\otimes 1)$ and $1\otimes y$ as the second basis vector, so $\delta (1\otimes 1)=y\otimes 1-1\otimes y$ has coordinates $(y,-1)$. Using the balancing relation $1\otimes y^2=y^2\otimes 1=y^2(1\otimes 1)$ and $y\otimes y=y(1\otimes y)$, $\delta (1\otimes y)=y\otimes y-1\otimes y^2$ has coordinates $(-y^2,y)$. The matrix is therefore $\left(\begin{array}{cc}y& -y^2\\ -1& y\end{array}\right)$, with columns the images of the two basis vectors, and both columns are homogeneous of degree $2$ since $y$ has degree $2$.

Compute kernel and cokernel. [step 1.1, algebra] An element $a(1\otimes 1)+b(1\otimes y)$ lies in $\ker \delta$ exactly when $ay-b{y}^2=0$ and $-a+by=0$; since $R$ is a domain the second equation gives $a=by$, and the first is then automatic. Hence $\ker \delta =\mathbb{Q}[y]\cdot (y\otimes 1+1\otimes y)$, free of rank one with generator of degree $2$. The first column $(y,-1)$ generates the image, because the second column $(-y^2,y)$ equals $-y$ times the first; the cokernel is therefore free of rank one and the class of $1\otimes 1$ is a generator of degree $0$, so $\mathrm{coker}\delta \cong R$ with the class of $1\otimes 1$ as basis element.

Apply the diagonal theorem and read the Hochschild groups. [L1, L2, L3, L5, step 2.1, algebra] By [L3] applied to $M=B_1$, $H{H}_0(R,B_1)\cong H_0=\mathrm{coker}\delta \cong R$ in degree $0$, and $H{H}_1(R,B_1)\cong H_1=\ker (\delta :B_1\{2\}\to B_1)$. The kernel of $\delta$ has its generator in degree $2$ inside $B_1$; the shifted term $B_1\{2\}$ represents the coefficient together with its single Koszul symbol $\theta$, so its degree is $2+2=4$. Thus the class $[(y\otimes 1+1\otimes y)\theta ]$ has total internal degree $4$ and $H_1\cong R\{4\}$. For $h\ge 2$ the complex has no terms, so $H{H}_h=0$. AC is used only here, in [L3].

Record the normalization. [L1, L4, step 3.1] The generator of $H{H}_1$ is the class of $r{b}_1(1)$ times the symbol, so it is the degree-$2$ element $r{b}_1(1)$ shifted by the degree-$2$ symbol; the resulting degree is $4$, and no other normalization is asserted. The two other boundary cases are immediate: for $h=0$ there is no symbol and the degree is the degree of the class of $1\otimes 1$, namely $0$; the complex is empty above $h=1$.

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].

Compute $HHH({\sigma }_{\ast })$. By [L1] the complex $F({\sigma }_{\ast })$ is $\mathbb{Q}$ in cohomological degree $0$; by [L2] the termwise complex in Hochschild degree $0$ is the single space $H{H}_0(\mathbb{Q},\mathbb{Q})=\mathbb{Q}\mathrm{/}⟨rm-mr⟩=\mathbb{Q}$, because $\mathbb{Q}$ is commutative, and the space $\mathbb{Q}$ has internal degree $0$. For $h\ge 1$ the Hochschild chain groups are $C_h(\mathbb{Q},\mathbb{Q})=\mathbb{Q}$ and the alternating boundary acts on the one-dimensional space as the sum ${\sum }_{i=0}^h(-1{)}^i$, which is $1$ in characteristic zero for $h$ even and $0$ for $h$ odd; hence each boundary is either zero or an isomorphism and $H{H}_h(\mathbb{Q},\mathbb{Q})=0$ for all $h\ge 1$. Thus the termwise complex is $\mathbb{Q}$ in cohomological degree $0$ and internal degree $0$, so $HHH({\sigma }_{\ast })=\mathbb{Q}$ in $(h,p,c)=(0,0,0)$.

The Khovanov-Rozansky side and the correction. By [L3] the reduced unknot class of the one-strand diagram sits in the raw tridegree $(-1,1,0)$. The global correction $(k,l)\mapsto (k+1,l-1)$ moves it to $(k+1,l-1,j)=(0,0,0)$, and the dictionary of [L4] maps the HHH class $(h,p,c)=(0,0,0)$ to $(k,l,j)=(0,0,0)$: indeed $k=-h=0$, $l=p-h=0$ and $j=c=0$. Thus the two one-strand classes agree in the corrected trigrading.

Fix the global constant. The dictionary is determined up to the global shift that carries the one-strand class of one theory to the class of the other; step 2.1 computes that shift to be exactly the correction $(k,l)\mapsto (k+1,l-1)$ used in the dictionary, so no further constant is available: any other correction would move the class $(0,0,0)$ away from itself and contradict the identification of the two one-dimensional classes. Caveat: the unreduced theory keeps the tower $\mathbb{Q}[x]\{-1,1\}$ and is not one-dimensional, so the normalization is a statement about the reduced theories.

Compute $H{H}_{\bullet }(R,R)$. By [L1] the diagonal Koszul complex of $R$ over $R$ has terms $R\{2\}$ in degree one and $R$ in degree zero, and the differential is multiplication by $u=x^L-x^R$, which acts on the regular bimodule $R$ as $xm-mx=0$. Hence the complex is $0\to R\{2\}\stackrel{0}{\to }R\to 0$ and [L2] gives $H{H}_0(R,R)=R$ with generator in internal degree $0$, $H{H}_1(R,R)=R\{2\}$ with generator in internal degree $2$ (the Koszul symbol contributes the degree), and $H{H}_h(R,R)=0$ for $h\ge 2$, since the complex has no terms above degree one. AC is used only here, in [L2].

The termwise complex has four free $R$-generators. By [L3] the termwise complex in Hochschild degree $h$ has terms $H{H}_h(R,F^c)$ for $c=0,1$; by step 1.1 these are nonzero exactly for $h\in \{0,1\}$, where they are copies of $R\{2h\}$, with generator degree $2h$, and the induced differentials vanish because $d_F=0$. The page $E_1$ therefore has the four displayed free generators, and every higher differential $d_r\text{\hspace{0.17em}⁣}:E_r^{c,-h}\to E_r^{c+r,-h-r+1}$ vanishes for bidegree reasons: for $r\ge 2$ the target column $c+r\ge 2$ is zero since $F$ is concentrated in degrees $0,1$. Hence $E_2=E_{\mathrm{\infty }}=E_1$ and the termwise output is the trigraded object with the four labels $(c,h,p)$ displayed in the Example.

The total hyperhomology. By [L4] the total complex has $T^n=C_{-n}(R,F^0)\oplus C_{1-n}(R,F^1)$ for $n\le 1$ and $T^n=0$ for $n\ge 2$, with $D=+b$ on the $F^0$-summand and $D=-b$ on the $F^1$-summand because $d_F=0$; hence $T^{\bullet }$ is the direct sum of the two column complexes $C_{\bullet }(R,F^0)$ and $C_{\bullet }(R,F^1)[-1]$ in which the differential is $\pm b$ up to the index. Taking homology and using $H{H}_j(R,R)=0$ for $j\ge 2$ from step 1.1 gives $H^{-1}=H{H}_1(R,F^0)=R\{2\}$, $H^0=H{H}_0(R,F^0)\oplus H{H}_1(R,F^1)=R\oplus R\{2\}$, $H^1=H{H}_0(R,F^1)=R$, and $H^n=H{H}_{-n}(R,F^0)\oplus H{H}_{1-n}(R,F^1)=0$ for $n\le -2$ and $n\ge 2$; this is the displayed computation of the Example.

Compare the two outputs and identify the filtration. By [L5] the page $E_{\mathrm{\infty }}^{i,-j}$ contributes to $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},i-j}$ as ${\mathrm{gr}}^i$, so the four classes of step 2.1 regroup by the total degree $n=c-h$ as follows: $(c,h)=(0,1)$ gives ${\mathrm{gr}}^0$ of $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},-1}$, the pairs $(0,0)$ and $(1,1)$ give ${\mathrm{gr}}^0$ and ${\mathrm{gr}}^1$ of $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},0}$, and $(1,0)$ gives ${\mathrm{gr}}^1$ of $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},1}$. This matches step 2.2, where the two summands of $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},0}=R\oplus R\{2\}$ are exactly the two filtered pieces $H{H}_0(R,F^0)$ and $H{H}_1(R,F^1)$: the declared hyperhomology grading uses $(n,p)$, while in this particular example the vanishing outer differential supplies a canonical additional column decomposition. The termwise theory of step 2.1 keeps $c$ and $h$ separately inside the trigrading $(c,h,p)$, while the total theory of step 2.2 is declared graded by $n=c-h$ and $p$, with the indicated split filtration in this example; this is the asserted difference of grading outputs, and no nonzero higher differential occurs by step 2.1.