Proof. When $K$ and $L$ do not have endpoints on the same boundary component of $\unicode[STIX]{x1D6F4}$, the generators for $\operatorname{end}(K\times L)$ are just given by $k\,\otimes \,\operatorname{id},\operatorname{id}\,\otimes \,l$ where $k\in \operatorname{End}(K)$ and $l\in \operatorname{End}(L)$. The differential on $\operatorname{end}(K\times L)$ is zero, and $k$ and $l$ commute. For example,

$$\begin{eqnarray}\displaystyle \operatorname{End}(L_{0}\times L_{3})=\mathbf{k}\langle \operatorname{id}\,\otimes \,x_{3},x_{1}\,\otimes \,\operatorname{id}\rangle =R/(x_{2}). & & \displaystyle \nonumber\end{eqnarray}$$

On the other hand, when both $K$ and $L$ have an end in the $i$th boundary component of $\unicode[STIX]{x1D6F4}$, we have that $\text{end}(K\times L)$ contains the vector subspace spanned by the elements

(3.3)$$\begin{eqnarray}\displaystyle (v_{i}u_{i})^{m}\,\otimes \,(u_{i}v_{i})^{n},\quad u_{i}(v_{i}u_{i})^{m}\,\otimes \,v_{i}(u_{i}v_{i})^{n}\quad \text{for }m,n\geqslant 0. & & \displaystyle\end{eqnarray}$$

To understand the algebra structure let us set

$$\begin{eqnarray}\displaystyle a_{i}=v_{i}u_{i}\,\otimes \,\operatorname{id}_{L},\quad b_{i}=\operatorname{id}_{K}\,\otimes \,u_{i}v_{i},\quad c_{i}=u_{i}\,\otimes \,v_{i}. & & \displaystyle \nonumber\end{eqnarray}$$

Then we have the relations

$$\begin{eqnarray}\displaystyle a_{i}b_{i}=b_{i}a_{i}=0,\quad a_{i}c_{i}=c_{i}b_{i},\quad b_{i}c_{i}=c_{i}a_{i}, & & \displaystyle \nonumber\end{eqnarray}$$

where the first relation comes from the product rule explained in Figure 4.

The quadratic algebra with these relations has the Gröbner bases

$$\begin{eqnarray}\displaystyle (c_{i}^{n},a_{i}^{m}c_{i}^{n},b_{i}^{m}c_{i}^{n})_{n\geqslant 0,m>0}, & & \displaystyle \nonumber\end{eqnarray}$$

consisting precisely of the elements (3.3). The differential is given by

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x2202}(a_{i})=-\unicode[STIX]{x2202}(b_{i})=c_{i},\quad \unicode[STIX]{x2202}(c_{i})=0 & & \displaystyle \nonumber\end{eqnarray}$$

and extended by the graded Leibniz rule, where we have $\deg (a_{i})=\deg (b_{i})=0$, $\deg (c_{i})=1$. It is easy to check that the relations are preserved. This also determines the signs. Furthermore, we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x2202}(c_{i}^{n})=0,\quad \unicode[STIX]{x2202}(a_{i}c_{i}^{n})=-\unicode[STIX]{x2202}(b_{i}c_{i}^{n})=c_{i}^{n+1},\quad \unicode[STIX]{x2202}(a_{i}^{m}c_{i}^{n})=-\unicode[STIX]{x2202}(b_{i}^{m}c_{i}^{n})=(a_{i}^{m-1}+b_{i}^{m-1})c_{i}^{n+1}, & & \displaystyle \nonumber\end{eqnarray}$$

where $m\geqslant 2$.

If $K$ and $L$ end at the $i$th boundary component of $\unicode[STIX]{x1D6F4}$, we let

$$\begin{eqnarray}\displaystyle x_{i}=a_{i}+b_{i}=v_{i}u_{i}\,\otimes \,\operatorname{id}_{L}+\operatorname{id}_{K}\,\otimes \,u_{i}v_{i}. & & \displaystyle \nonumber\end{eqnarray}$$

One can see by a straightforward calculation from the explicit description of the chain complex given above, that the contribution to the cohomology $\operatorname{End}(K\times L)$ from the $i$th boundary components comes from $x_{i}$ and its positive powers. This determines the cohomology. For example,

$$\begin{eqnarray}\displaystyle \operatorname{End}(L_{1}\times L_{2})=\mathbf{k}\langle u_{1}v_{1}\,\otimes \,\operatorname{id},v_{2}u_{2}\,\otimes \,\operatorname{id}+\operatorname{id}\,\otimes \,u_{2}v_{2},\operatorname{id}\,\otimes \,v_{3}u_{3}\rangle =R/(x_{1}x_{2}x_{3}). & & \displaystyle \nonumber\end{eqnarray}$$

The morphisms between different Lagrangians are calculated in the same way (see Theorem 3.2.5 below for a more general calculation). ◻

Proof. In terms of the generators $a_{i},b_{i},c_{i}$, our subcomplex is spanned by the elements

$$\begin{eqnarray}\displaystyle (a_{i}^{m}c_{i}^{n})_{m\geqslant 0,n>0},\quad (b_{i}^{m}c_{i}^{n})_{m>0,n\geqslant 0}. & & \displaystyle \nonumber\end{eqnarray}$$

This complex splits into a direct sum of subcomplexes with fixed total degree (given by $m+n$). The subcomplex $C^{\bullet }$ corresponding to the degree $n>0$ has terms

$$\begin{eqnarray}\displaystyle & C^{0}=\langle b_{i}^{n}\rangle ,\quad C^{1}=\langle a_{i}^{n-1}c_{i},b_{i}^{n-1}c_{i}\rangle ,\quad C^{2}=\langle a_{i}^{n-2}c_{i}^{2},b_{i}^{n-2}c_{i}^{2}\rangle ,\ldots , & \displaystyle \nonumber\\ \displaystyle & C^{n-1}=\langle a_{i}c_{i}^{n-1},b_{i}c_{i}^{n-1}\rangle ,\quad C^{n}=\langle c_{i}^{n}\rangle . & \displaystyle \nonumber\end{eqnarray}$$

Now we see that for $m\in [1,n-1]$, we have

$$\begin{eqnarray}\displaystyle \operatorname{ker}(d:C^{m}\rightarrow C^{m+1})=\langle (a_{i}^{n-m}+b_{i}^{n-m})c_{i}^{m}\rangle =\operatorname{im}(d:C^{m-1}\rightarrow C^{m}), & & \displaystyle \nonumber\end{eqnarray}$$

while $\operatorname{ker}(d:C^{0}\rightarrow C^{1})=0$ and $\operatorname{im}(d:C^{n-1}\rightarrow C^{n})=C^{n}$.◻

Proof. (i) Note that for every $i$ we have $g(S\,\cap \,[0,i])\subset S^{\prime }\,\cap \,[0,i+1]$ and $g^{-1}(S^{\prime }\,\cap \,[0,i])\subset S\,\cap \,[0,i+1]$. Now, since $t_{i}\in T$ and $t_{i}+1\not \in T$, we have either $t_{i}+1\in S\setminus S^{\prime }$ or $t_{i}+1\in S^{\prime }\setminus S$. In the former case we have

$$\begin{eqnarray}\displaystyle g(S\cap [0,t_{i}])\subset S^{\prime }\cap [0,t_{i}+1]=S^{\prime }\cap [0,t_{i}], & & \displaystyle \nonumber\end{eqnarray}$$

while in the latter case we have

$$\begin{eqnarray}\displaystyle g^{-1}(S^{\prime }\cap [0,t_{i}])\subset S\cap [0,t_{i}+1]=S\cap [0,t_{i}]. & & \displaystyle \nonumber\end{eqnarray}$$

Thus, we get $g(S\cap [0,t_{i}])=S^{\prime }\cap [0,t_{i}]$.

Next, we have $s_{i}-1\not \in T$ and $s_{i}\in T$, so either $s_{i}\in S\setminus S^{\prime }$ or $s_{i}\in S^{\prime }\setminus S$. In the former case we have

$$\begin{eqnarray}\displaystyle g(S\cap [s_{i}+1,k])\subset S^{\prime }\cap [s_{i},k]=S^{\prime }\cap [s_{i}+1,k], & & \displaystyle \nonumber\end{eqnarray}$$

while in the latter case we have

$$\begin{eqnarray}\displaystyle g^{-1}(S^{\prime }\cap [s_{i}+1,k])\subset S\cap [s_{i},k]=S\cap [s_{i}+1,k]. & & \displaystyle \nonumber\end{eqnarray}$$

Since $\#(S\cap [s_{i}+1,k])=\#(S^{\prime }\cap [s_{i}+1,k])$, we deduce that $g(S\cap [s_{i}+1,k])=S^{\prime }\cap [s_{i}+1,k]$. This implies our first assertion.

Assume that $s_{i}+1\leqslant t_{i}$. Since both $s_{i}$ and $s_{i}+1$ are in $T$, either the element $s_{i}+1$ belongs to $S\cap S^{\prime }$, or it does not belong to both $S$ and $S^{\prime }$. Proceeding in the same way, we see that $S\cap [s_{i}+1,t_{i}]=S^{\prime }\cap [s_{i}+1,t_{i}]$.

Next, let us consider the interval $[t_{i}+1,s_{i+1}]$. For every $i$ in this interval, except for the right end, we have $i\not \in T$. Assume first that $t_{i}+1\in S$. Then, since $t_{i}\in T$ and $t_{i}+1\not \in T$, we should have $t_{i}+1\not \in S^{\prime }$, so $g(t_{i}+1)=t_{i}+2\in S^{\prime }$. If $t_{i}+2<s_{i+1}$ we can continue in the same way: since $\#(S\cap [0,t_{i}+1])=\#(S^{\prime }\cap [0,t_{i}+1])+1$ and $t_{i}+2\in S^{\prime }\setminus T$, we should have $t_{i}+2\in S$, and so $g(t_{i}+2)=t_{i}+3$, and so on. In the case $t_{i}+1\not \in S$, we should have $t_{i}+1\in S^{\prime }$, so we can apply the same argument with $S$ and $S^{\prime }$ swapped and $g$ replaced by $g^{-1}$.

(ii) We use induction on the cardinality of $S$. Let $i_{0}$ be the minimal element of $S$. Then we have either $g(i_{0})=i_{0}$ or $g(i_{0})=i_{0}+1$. In the former case we have $g(S\setminus \{i_{0}\})=S\setminus \{i_{0}\}$, so we can apply the induction assumption to $S\setminus \{i_{0}\}$. Now let us assume that $g(i_{0})=i_{0}+1$. Then $g^{-1}(i_{0})\neq i_{0}$, so we should have $g^{-1}(i_{0})=i_{0}+1$. Thus, in this case $g$ swaps $i_{0}$ and $i_{0}+1$. Now we can replace $S$ by $S\setminus \{i_{0},i_{0}+1\}$ and apply the induction assumption.◻

Proof of Proposition 3.2.1.

By Lemma 3.2.2(i), we can partition $[0,k]$ into subintervals of two kinds: those for which one of the conditions (2) or (3) of Proposition 3.2.1 holds, and subintervals $I$ such that $S\cap I=S^{\prime }\cap I$ and $g(S\cap I)=S^{\prime }\cap I$. It remains to apply Lemma 3.2.2(ii) to all subintervals of the second kind.◻

Proof. (i) Recall that a grading structure on $M_{n,k}$ is given by a fiberwise universal cover of the Lagrangian Grassmannian of the tangent bundle [Reference SeidelSei00]. Such a cover exists since $2c_{1}(M_{n,k})=0$, and the set of possible grading structures up to homotopy is a torsor over $H^{1}(M_{n,k},\mathbb{Z})\simeq \mathbb{Z}^{k}$. As all our Lagrangians $L_{S}$ are contractible, they can be graded (uniquely up to a shift by $\mathbb{Z}$). Changing the homotopy class of the grading by some cohomology class $c\in H^{1}(M_{n,k};\mathbb{Z})\simeq H^{1}(M_{n,k},L;\mathbb{Z})$ changes the degree of a Reeb chord $x\in \operatorname{end}(L_{S})$ by $\langle c,[x]\rangle$, where $[x]\in H_{1}(M_{n,k},L_{S};\mathbb{Z})$ (see [Reference Abouzaid and SeidelAS10, § 9a]).

Now we use the fact that away from the big diagonal $\unicode[STIX]{x1D6E5}\subset M_{n,k}$ the Reeb flow is just the product of the Reeb flows on the surface. In addition, we observe that the homotopy class of any Reeb loop of the form $(pt.\times \cdots \times pt.\times \unicode[STIX]{x1D6FE}_{i}\times pt.\times \cdots \times pt.)$ in $M_{n,k}\setminus \unicode[STIX]{x1D6E5}$, coming from the loop $\unicode[STIX]{x1D6FE}_{i}=u_{i}v_{i}$ in $\unicode[STIX]{x1D6F4}$, for fixed $i$ does not depend on the choice of points in the other components. It follows that the degree of the generator of $\operatorname{end}(L_{S})$ corresponding to such a loop is independent of $S$ (it depends only on $i$). Hence, the fact that grading structures on $M_{n,k}$ form a torsor over $H^{1}(M_{n,k},\mathbb{Z})$ implies that, for a given $(d_{1},d_{2},\ldots ,d_{k})\in \mathbb{Z}^{k}$, there exists a unique grading structure such that $|x_{i}|=d_{i}$.

Let us first calculate $\operatorname{End}(L_{S})$ in the case $S=[i,j]$, where $i>0$, $j<k$. Let us set

$$\begin{eqnarray}\displaystyle E[i,j]:=\operatorname{end}(L_{[i,j]}). & & \displaystyle \nonumber\end{eqnarray}$$

First, we claim that there is a natural quasi-isomorphism of dg-algebras

(3.6)$$\begin{eqnarray}\displaystyle k[x_{i},\ldots ,x_{j+1}]/(x_{i}\ldots x_{j+1})\rightarrow E[i,j], & & \displaystyle\end{eqnarray}$$

such that

$$\begin{eqnarray}\displaystyle & x_{i}\mapsto (u_{i}v_{i})\,\otimes \,\operatorname{id}\,\otimes \;\cdots \,, & \displaystyle \nonumber\\ \displaystyle & x_{s}\mapsto \cdots \,\otimes \,\operatorname{id}\,\otimes \,(v_{s}u_{s})\,\otimes \,\operatorname{id}\,\otimes \,\cdots +\cdots \,\otimes \,\operatorname{id}\,\otimes \,(u_{s}v_{s})\,\otimes \,\operatorname{id}\,\otimes \,\cdots \quad \text{for }i<s\leqslant j, & \displaystyle \nonumber\\ \displaystyle & x_{j+1}\mapsto \cdots \,\otimes \,\operatorname{id}\,\otimes \,(v_{j+1}u_{j+1}). & \displaystyle \nonumber\end{eqnarray}$$

Here we view the source as a complex with zero differential. It is easy to check that the morphisms corresponding to $x_{s}$ are closed, pairwise commute and their product is zero, so the map (3.6) is well defined.

Next, we want to prove that (3.6) is a quasi-isomorphism. We proceed by induction on $j-i$. In the base case $j=i$ this is easy to see. For the induction step we start with a decomposition into a direct sum of subcomplexes,

$$\begin{eqnarray}\displaystyle E[i,j]=\bigoplus _{n\geqslant 0}E[i,j](n). & & \displaystyle \nonumber\end{eqnarray}$$

Here, for $n>0$, $E[i,j](n)$ is spanned by elements of the form $(u_{i}v_{i})^{n}\,\otimes \,\cdots \,$, while $E[i,j](0)$ is spanned by the remaining basis elements.

It is easy to see that, for $n>0$, $E[i,j](n)$ is isomorphic (by multiplication with $x_{i}^{n}$) to $E[i+1,j]$, so we know its cohomology from the induction assumption. More precisely, we know that the map

$$\begin{eqnarray}\displaystyle x_{i}^{n}\cdot k[x_{i+1},\ldots ,x_{j+1}]/(x_{i+1}\ldots x_{j+1})\rightarrow E[i,j](n) & & \displaystyle \nonumber\end{eqnarray}$$

is a quasi-isomorphism.

It remains to deal with $E[i,j](0)$. Namely, we want to prove that the natural map

$$\begin{eqnarray}\displaystyle k[x_{i+1},\ldots ,x_{j+1}]\rightarrow E[i,j](0) & & \displaystyle \nonumber\end{eqnarray}$$

is a quasi-isomorphism.

Let us define subcomplexes $C(0)\subset C(1)\subset E[i,j](0)$ as follows. We define $C(0)$ to be the span of all elements of the form

$$\begin{eqnarray}\displaystyle & ((v_{i+1}u_{i+1})^{m}\,\otimes \,(u_{i+1}v_{i+1})^{n}\,\otimes \,\operatorname{id}_{L_{i+2}}\,\otimes \,\cdots \,)_{m\geqslant 0,n>0}, & \displaystyle \nonumber\\ \displaystyle & (u_{i+1}(v_{i+1}u_{i+1})^{m}\,\otimes \,v_{i+1}(u_{i+1}v_{i+1})^{n}\,\otimes \,\operatorname{id}_{L_{i+2}}\,\otimes \,\cdots \,)_{m\geqslant 0,n\geqslant 0}. & \displaystyle \nonumber\end{eqnarray}$$

By Lemma 3.1.2, the complex $C(0)$ is acyclic.

Next, we define $C(1)$ to be spanned by $C(0)$ and by all elements of the form $\text{id}_{L_{i}}\,\otimes \,\cdots \,$. We have an isomorphism

$$\begin{eqnarray}\displaystyle E[i+1,j](0)\overset{{\sim}}{\longrightarrow }C(1)/C(0) & & \displaystyle \nonumber\end{eqnarray}$$

induced by the natural embedding of $E[i+1,j](0)$ into $C(1)$. Thus, by the induction assumption, the natural map

$$\begin{eqnarray}\displaystyle k[x_{i+2},\ldots ,x_{j+1}]\rightarrow C(1)/C(0) & & \displaystyle \nonumber\end{eqnarray}$$

is a quasi-isomorphism.

Finally, the quotient $E[i,j](0)/C(1)$ splits into a direct sum of subcomplexes $E(n)$ numbered by $n>0$, where $E(n)$ is spanned by elements of the form $(v_{i+1}u_{i+1})^{n}\,\otimes \,\cdots \,$ (modulo $C(1)$). Note that we have

$$\begin{eqnarray}\displaystyle x_{i+1}^{n}\equiv (v_{i+1}u_{i+1})^{n}\,\otimes \,\operatorname{id}\,\otimes \,\cdots \,\operatorname{mod}C(1), & & \displaystyle \nonumber\end{eqnarray}$$

and it is easy to see that the multiplication by $x_{i+1}^{n}$ gives an isomorphism of complexes,

$$\begin{eqnarray}\displaystyle E[i+1,j](0)\overset{{\sim}}{\longrightarrow }E(n). & & \displaystyle \nonumber\end{eqnarray}$$

Therefore, we deduce that the natural map

$$\begin{eqnarray}\displaystyle x_{i+1}^{n}k[x_{i+2},\ldots ,x_{j+1}]\rightarrow E(n) & & \displaystyle \nonumber\end{eqnarray}$$

is a quasi-isomorphism for every $n>0$.

Thus, we have a morphism of exact triples of complexes

in which the left and right vertical arrows are quasi-isomorphisms, hence the middle one is also a quasi-isomorphism.

Next, we can slightly modify the above argument for the cases $i=0$ and $j<k$ (respectively, $i>0$ and $j=k$). In these cases we still claim that the natural maps

$$\begin{eqnarray}\displaystyle k[x_{1},\ldots ,x_{j+1}]\rightarrow E[0,j],\quad k[x_{i},\ldots ,x_{k}]\rightarrow E[i,k] & & \displaystyle \nonumber\end{eqnarray}$$

are quasi-isomorphisms. Namely, in the case of $E[0,j]$ we skip the first step and just deal with this complex exactly as with the complexes $E[i,j](0)$, considering the subcomplexes analogous to $C(0)$ and $C(1)$. In the case of $E[i,k]$, we repeat exactly the same steps as for $E[i,j]$, considering first the subcomplexes $E[i,k](n)$ and using the induction assumption to see that the maps

$$\begin{eqnarray}\displaystyle x_{i}^{n}\cdot k[x_{i+1},\ldots ,x_{k}]\rightarrow E[i,k](n) & & \displaystyle \nonumber\end{eqnarray}$$

are quasi-isomorphisms.

Now for $S=[i_{1},j_{1}]\sqcup [i_{2},j_{2}]\sqcup \cdots \sqcup [i_{r},j_{r}]$ with $j_{s}+1<i_{s+1}$, we have a natural identification

$$\begin{eqnarray}\displaystyle \operatorname{end}(L_{S})\simeq \operatorname{end}(L_{[i_{1},j_{1}]})\,\otimes \,\cdots \,\otimes \,\operatorname{end}(L_{[i_{r},j_{r}]}), & & \displaystyle \nonumber\end{eqnarray}$$

and so we have a similar decomposition of $\operatorname{End}(L_{S})$.

(ii) Now let us compute $\operatorname{Hom}(L_{S},L_{S^{\prime }})$ for a pair of subsets $S,S^{\prime }\subset [0,k]$ of size $n$. If $S$ and $S^{\prime }$ are not close then we have $\operatorname{hom}(L_{S},L_{S^{\prime }})=0$. Otherwise, we have a decomposition

(3.7)$$\begin{eqnarray}\displaystyle \operatorname{hom}(L_{S},L_{S^{\prime }})\simeq \operatorname{end}(L_{([0,k]\setminus \sqcup _{a}I_{a}\sqcup \sqcup _{b}J_{b})\cap S})\,\otimes \,\bigotimes _{a=1}^{r}\operatorname{hom}(L_{I_{a}\cap S},L_{I_{a}\cap S^{\prime }})\,\otimes \,\bigotimes _{b=1}^{s}\operatorname{hom}(L_{J_{b}\cap S},L_{J_{b}\cap S^{\prime }}), & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where we use the subintervals $(I_{a})$, $(J_{b})$ from Corollary 3.2.4.

It remains to identify the complexes $\operatorname{hom}(L_{I_{a}\cap S},L_{I_{a}\cap S^{\prime }})$ and $\operatorname{hom}(L_{J_{b}\cap S},L_{J_{b}\cap S^{\prime }})$. First, let $I=I_{a}=[i,j]$, so that $I\cap S=[i,j-1]$ and $I\cap S^{\prime }=[i+1,j]$. We have

$$\begin{eqnarray}\displaystyle \operatorname{hom}(L_{[i,j-1]},L_{[i+1,j]})=\bigotimes _{p\in [i,j-1]}\operatorname{hom}(L_{p},L_{p+1}), & & \displaystyle \nonumber\end{eqnarray}$$

with the basis given by

$$\begin{eqnarray}\displaystyle u_{i+1}(v_{i+1}u_{i+1})^{m_{1}}\,\otimes \,\cdots \,\otimes \,u_{j}(v_{j}u_{j})^{m_{j-i}},\quad m_{1}\geqslant 0,\ldots ,m_{j-i}\geqslant 0. & & \displaystyle \nonumber\end{eqnarray}$$

It is easy to check that we have

$$\begin{eqnarray}\displaystyle u_{i+1}(v_{i+1}u_{i+1})^{m_{1}}\,\otimes \,\cdots \,\otimes \,u_{j}(v_{j}u_{j})^{m_{j-i}}=(x_{i+1}^{m_{1}}\ldots x_{j}^{m_{j-i}})u_{[i,j]}=u_{[i,j]}(x_{i+1}^{m_{1}}\ldots x_{j}^{m_{j-i}}), & & \displaystyle \nonumber\end{eqnarray}$$

where we view the monomial in $x_{1},\ldots ,x_{k}$ as an endomorphism of $L_{[i,j-1]}$ or of $L_{[i+1,j]}$. In particular, the differential on $\operatorname{hom}(L_{[i,j-1]},L_{[i+1,j]})$ is zero.

Similarly, we see that for $J=J_{b}=[i,j]$, the space $\operatorname{hom}(L_{[i+1,j]},L_{[i,j-1]})$ is spanned by

$$\begin{eqnarray}\displaystyle (x_{i+1}^{m_{1}}\ldots x_{j}^{m_{j-i}})v_{[i,j]}=v_{[i,j]}(x_{i+1}^{m_{1}}\ldots x_{j}^{m_{j-i}}), & & \displaystyle \nonumber\end{eqnarray}$$

and has zero differential.

(iii) Formula (3.5) can be checked directly. To compose arbitrary elements in $\operatorname{Hom}(L_{S},L_{S^{\prime }})$ and $\operatorname{Hom}(L_{S^{\prime }},L_{S^{\prime \prime }})$ we can use (ii) to write them in the form $f_{S,S^{\prime }}p(x)$ and $q(x)f_{S^{\prime },S^{\prime \prime }}$, with $p(x)$ and $q(x)$ some polynomials in $x_{1},\ldots ,x_{k}$, and then use (3.5).

(iv) The first assertion is clear from our previous computations: the subalgebra in question is generated by $f_{S,S^{\prime }}$ and by $x_{i}\in \text{end}(L_{S})$.

By (i), we know that for each choice $\deg (x_{i})=d_{i}\in \mathbb{Z}$, there exists a $\mathbb{Z}$-grading on ${\mathcal{A}}^{\circ \circ }$ coming from a grading structure on $M_{n,k}$. Thus, it is enough to prove uniqueness of $(d_{S,S^{\prime }})$ up to adding $d_{S^{\prime }}-d_{S}$ (for given $d_{i}\in D$). The numbers $d_{S,S^{\prime }}$ are constrained by equations (3.5), corresponding to triples $(S,S^{\prime },S^{\prime \prime })$ of pairwise close subsets. Thus, a difference between two systems $(d_{S,S^{\prime }})$ is a $1$-cocycle for the following simplicial complex $X$. The set of vertices, $X_{0}$, is the set of all subsets in $[0,k]$ of size $n$. The set of edges, $X_{1}$, is given by pairs $(S,S^{\prime })$ of subsets which are close, and the set of $2$-simplexes, $X_{2}$, is the set of triples $(S,S^{\prime },S^{\prime \prime })$ of pairwise close subsets. Our uniqueness statement would follow from the vanishing of $H^{1}(X,D)$. We claim that in fact the geometric realization $|X|_{\operatorname{real}}$ is simply connected. The proof is by induction on $k$. The case $k=1$ is straightforward (for $n=0$ we get a point and for $n=1$ we get a segment). Another special case we have to include to make our proof work is $n=k+1$, when $X$ reduces to a point.

Let us write $X=X(n,k)$. We have natural embeddings as simplicial subcomplexes,

$$\begin{eqnarray}\displaystyle & i_{k}:X(n,k-1){\hookrightarrow}X(n,k):S\subset [0,k-1]\mapsto S, & \displaystyle \nonumber\\ \displaystyle & j_{k}:X(n-1,k-1){\hookrightarrow}X(n,k):S\subset [0,k-1]\mapsto S\sqcup \{k\}, & \displaystyle \nonumber\end{eqnarray}$$

so that every vertex of $X(n,k)$ is either in the image of $i_{k}$ or in the image of $j_{k}$.

Let us in addition consider the subcomplex $Y\subset X(n,k)$ spanned by all vertices of the form $S\sqcup \{k-1\}$ and $S\sqcup \{k\}$, with $S\subset [0,k-2]$, $|S|=n-1$. Note that whenever $S$ and $S^{\prime }$ are close subsets of $[0,k-2]$, all four subsets

$$\begin{eqnarray}\displaystyle S\sqcup \{k-1\},\quad S\sqcup \{k\},\quad S^{\prime }\sqcup \{k-1\},\quad S^{\prime }\sqcup \{k\} & & \displaystyle \nonumber\end{eqnarray}$$

are pairwise close. This implies that $|Y|_{\operatorname{real}}$ can be identified with the $2$-skeleton of $|X(n-1,k-2)|_{\operatorname{real}}\times I$, where $I$ is a segment. In particular, by the induction assumption, $|Y|_{\operatorname{real}}$ is simply connected.

Now assume that we have an edge $e$ between a vertex $S$ in $i_{k}(X(n,k-1))$, where $S\subset [0,k-1]$, and a vertex $S^{\prime }\sqcup \{k\}$ in $j_{k}(X(n-1,k-1))$. This means that $S$ and $S^{\prime }\sqcup \{k\}$ are close, which can happen only if $S=T\sqcup \{k-1\}$, with $T$ and $S^{\prime }$ close. In this case the triangle

$$\begin{eqnarray}\displaystyle (S=T\sqcup \{k-1\},T\sqcup \{k\},S^{\prime }\sqcup \{k\}) & & \displaystyle \nonumber\end{eqnarray}$$

gives a homotopy between $e$ and a segment $(T\sqcup \{k-1\},T\sqcup \{k\})$ in $Y$, followed by a segment in $j_{k}(X(n-1,k-1))$. Since $i_{k}(X(n,k-1))$, $j_{k}(X(n-1,k-1))$ and $Y$ are simply connected, this implies that $X(n,k)$ is also simply connected.◻

Proof. This is obtained by repeatedly applying the exact triangle given by (2.5). More precisely, first, as in Figure 8, we slide $L_{j_{k+1}}$ along the components of $X$ until one of its legs is next to $L_{j_{k}}$. This gives an isomorphism

$$\begin{eqnarray}\displaystyle L_{j_{k+1}}\times X\simeq \tilde{L}_{j_{k+1}}\times X. & & \displaystyle \nonumber\end{eqnarray}$$

Next, let us consider

$$\begin{eqnarray}\displaystyle C:=\text{cone}(\tilde{L}_{j_{k+1}}\times X\rightarrow L_{j_{k}}\times X). & & \displaystyle \nonumber\end{eqnarray}$$

Because of the above isomorphism we also have an exact triangle

$$\begin{eqnarray}\displaystyle L_{j_{k+1}}\times X\rightarrow L_{j_{k}}\times X\rightarrow C\rightarrow \cdots & & \displaystyle \nonumber\end{eqnarray}$$

which is shown in Figure 9.

Thus, applying this kind of triangle repeatedly, we can express $\tilde{T}_{2}\times X$, shown in Figure 10, as

$$\begin{eqnarray}\displaystyle \tilde{T}_{2}\times X\simeq \{L_{j_{k-n+2}}\times X\rightarrow L_{j_{k-n+1}}\times X\rightarrow \cdots \rightarrow L_{j_{1}}\times X\}, & & \displaystyle \nonumber\end{eqnarray}$$

where $\tilde{T}_{2}$ has the property that all the objects $L_{0},\ldots ,L_{i}$ to the left of the left leg of $\tilde{T}_{2}$ are in $X$ and all the objects $L_{j},\ldots ,L_{k}$ to the right of the right leg of $\tilde{T}_{2}$ are in $X$. Therefore, sliding $\tilde{T}_{2}$ over these Lagrangians, we exhibit a Hamiltonian isotopy between $\tilde{T}_{2}\times X$ and $T_{2}\times X$, which completes the proof of the proposition for $T_{2}\times X$. The case of $T_{1}\times X$ is considered similarly.◻

Proof. (i) For any ${\mathcal{B}}^{\circ \circ }$-module $M$, we have

$$\begin{eqnarray}\displaystyle \operatorname{Hom}_{B^{\circ \circ }}(P_{[1,i]}^{\circ \circ },M)\simeq M_{[1,i]}=r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}(M)_{[1,i]}\simeq \operatorname{Hom}_{{\mathcal{B}}^{\circ }}(P_{[1,i]}^{\circ },r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}(M)), & & \displaystyle \nonumber\end{eqnarray}$$

which gives an isomorphism $i_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}(P_{[1,i]}^{\circ })\simeq P_{[1,i]}^{\circ \circ }$. The proof of the isomorphism $i_{R}^{{\mathcal{B}}^{\circ }}(R)\simeq P$ is similar.

This implies that our two adjunction maps are isomorphisms on all projective modules, hence on the perfect derived subcategories. It is also easy to see that these maps are isomorphisms on the abelian categories (since the induction functors on the abelian categories can be explicitly described).

(ii), (iii). The proofs are straightforward and similar to (i). ◻

Proof. The surjectivity of the map $R/(x_{[1,i]})\rightarrow R/(x_{[1,i-1]})$ implies that the map $\unicode[STIX]{x1D6FC}\{[1,i-1],[i]\}$ is an embedding. Furthermore, it is easy to see that

$$\begin{eqnarray}\displaystyle (\overline{P}_{i})_{[1,j]}=\left\{\begin{array}{@{}ll@{}}R/(x_{i}),\quad & j\geqslant i,\\ 0,\quad & j<i.\end{array}\right. & & \displaystyle \nonumber\end{eqnarray}$$

In particular, $\operatorname{Ext}^{\ast }(P_{[1,j]},\overline{P}_{i})=0$ for $j<i$ and $\operatorname{Ext}^{\ast }(P_{i},\overline{P}_{i})=R/(x_{i})$. This immediately implies the required semiorthogonalities and equality (4.3).

Now let us consider the restriction and induction functors

$$\begin{eqnarray}\displaystyle r_{{\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}]}^{{\mathcal{B}}^{\circ }}:\operatorname{Perf}({\mathcal{B}}^{\circ })\rightarrow D^{b}({\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}]),\quad i_{{\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}]}^{{\mathcal{B}}^{\circ }}:D^{b}({\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}])\rightarrow D^{b}({\mathcal{B}}^{\circ }). & & \displaystyle \nonumber\end{eqnarray}$$

We have $r_{{\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}]}^{{\mathcal{B}}^{\circ }}\circ i_{{\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}]}^{{\mathcal{B}}^{\circ }}=\operatorname{Id}$, so the functor $i_{{\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}]}^{{\mathcal{B}}^{\circ }}$ is fully faithful. Since for $i\leqslant k-1$ one has

(4.4)$$\begin{eqnarray}\displaystyle i_{{\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}]}^{{\mathcal{B}}^{\circ }}(P_{[1,i]}[x_{k}])=P_{[1,i]}, & & \displaystyle\end{eqnarray}$$

we deduce that this induction functor sends perfect complexes to perfect complexes and

$$\begin{eqnarray}\displaystyle i_{{\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}]}^{{\mathcal{B}}^{\circ }}(\overline{P}_{i}[x_{k}])\simeq \overline{P}_{i}, & & \displaystyle \nonumber\end{eqnarray}$$

for $i\leqslant k-1$. Using the induction on $k$, we see that all that remains is to prove that there is a semiorthogonal decomposition

(4.5)$$\begin{eqnarray}\displaystyle \operatorname{Perf}({\mathcal{B}}^{\circ })=\langle \langle \overline{P}_{k}\rangle ,i_{{\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}]}^{{\mathcal{B}}^{\circ }}\operatorname{Perf}({\mathcal{B}}_{[1,k-1]}^{\circ }[x_{k}])\rangle . & & \displaystyle\end{eqnarray}$$

Since we already proved the semiorthogonality, by [Reference Bondal, Larsen and LuntsBLL04, Lemma 1.20], it is enough to check that the two subcategories on the right generate $\operatorname{Perf}({\mathcal{B}}^{\circ })$. In view of (4.4), all that remains is to see that they generate the module $P_{[1,k]}$. But this follows from the exact sequence defining $\overline{P}_{k}$.◻

Proof. We just need to show that the kernel of the morphism $P_{I\sqcup J}\rightarrow P_{J}$ is isomorphic to $P_{I}$. But this immediately follows from the exact sequence

$$\begin{eqnarray}\displaystyle 0\rightarrow R/(x_{J})\overset{x_{I}}{\longrightarrow }R/(x_{I\sqcup J})\rightarrow R/(x_{I})\rightarrow 0.\Box & & \displaystyle \nonumber\end{eqnarray}$$

Proof. We need to compute the cokernel $C$ of the map $\unicode[STIX]{x1D6FD}\{[j-1],[j,m]\}:P_{[j-1,m]}\rightarrow P_{[j,m]}$. We have

$$\begin{eqnarray}\displaystyle (P_{[j,m]})_{I}=\operatorname{Hom}(R/(x_{[j,m]}),R/(x_{I}))=(x_{I\setminus [j,m]})/(x_{I}),\quad (P_{[j-1,m]})_{I}=(x_{I\setminus [j-1,m]})/(x_{I}), & & \displaystyle \nonumber\end{eqnarray}$$

and our map is given by multiplication by $x_{j-1}$.

Assume first that $j-1\not \in I$. Then $I\setminus [j,m]=I\setminus [j-1,m]$, so we get

$$\begin{eqnarray}\displaystyle C_{I}=(P_{[j,m]})_{I}\,\otimes \,R/(x_{i}). & & \displaystyle \nonumber\end{eqnarray}$$

Note that in this case we have either $I\subset [1,j-1]$ or $I\subset [j,k]$, and in the former case, $(P_{[j,m]})_{I}=0$, so $C_{I}=0$.

On the other hand, if $j-1\in I$ then, setting $I^{\prime }=I\setminus [j-1,m]$, we have

$$\begin{eqnarray}\displaystyle (P_{[j,m]})_{I}=(x_{j-1}x_{I^{\prime }})/(x_{I}),\quad (P_{[j-1,m]})_{I}=x_{I^{\prime }}/(x_{I}), & & \displaystyle \nonumber\end{eqnarray}$$

so in this case the map $(P_{[j-1,m]})_{I}\overset{x_{j-1}}{\longrightarrow }(P_{[j,m]})_{I}$ is surjective, so $C_{I}=0$.

This easily leads to the identification $C\simeq i_{!}\{j\}(P_{[j,m]}[x_{1},\ldots ,x_{j-2}])$.◻

Proof. Recall that by Lemma 4.1.1(i), the functor $i_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}:\operatorname{Perf}({\mathcal{B}}^{\circ })\rightarrow \operatorname{Perf}({\mathcal{B}}^{\circ \circ })$ is fully faithful. Thus, the functor $\unicode[STIX]{x1D6F7}_{1}$ is fully faithful.

To check that the functor $\unicode[STIX]{x1D6F7}_{j}$, for $j\geqslant 2$, is fully faithful and takes values in $\operatorname{Perf}({\mathcal{B}}^{\circ \circ })$, we observe that by Lemma 4.1.5 together with Lemma 4.1.1(i), we have

(4.7)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}_{j}(P_{[j,m]}^{\circ }[x_{1},\ldots ,x_{j-2}])=M\{[j-1],[j,m]\}. & & \displaystyle\end{eqnarray}$$

By definition, the latter module is perfect. Also, for any ${\mathcal{B}}_{[j,k]}^{\circ \circ }[x_{1},\ldots ,x_{j-2}]$-module $M$ one has $\operatorname{Hom}(P_{[j-1,m]},i_{!}\{j\}M)=0$ for any $m\geqslant j-1$. Hence, using (4.7) and the defining exact sequence for $M\{[j-1],[j,m]\}$, we get an identification

$$\begin{eqnarray}\displaystyle & & \displaystyle \operatorname{Hom}(\unicode[STIX]{x1D6F7}_{j}(P_{[j,m]}^{\circ }[x_{1},\ldots ,x_{j-2}]),i_{!}\{j\}M)\simeq \operatorname{Hom}(P_{[j,m]},i_{!}\{j\}M)\simeq (i_{!}\{j\}M)_{[j,m]}\nonumber\\ \displaystyle & & \displaystyle \quad =M_{[j,m]}\simeq \operatorname{Hom}(P_{[j,m]}[x_{1},\ldots ,x_{j-2}],M).\nonumber\end{eqnarray}$$

Applying this to $M=i_{{\mathcal{B}}_{[j,k]}^{\circ }}^{{\mathcal{B}}_{[j,k]}^{\circ \circ }}(P_{[j,m^{\prime }]}^{\circ }[x_{1},\ldots ,x_{j-2}])$, we derive that $\unicode[STIX]{x1D6F7}_{j}$ is fully faithful.

For $j\leqslant k$, let us denote by ${\mathcal{P}}_{j}\subset \operatorname{Perf}({\mathcal{B}}^{\circ \circ })$ the subcategory generated by the projective modules $P_{[j^{\prime },m]}$ with $j^{\prime }\leqslant j$, so that ${\mathcal{P}}_{1}=\unicode[STIX]{x1D6F7}_{1}\operatorname{Perf}({\mathcal{B}}_{[1,k]}^{\circ })$. Let us prove by induction on $j$ that we have a semiorthogonal decomposition

(4.8)$$\begin{eqnarray}\displaystyle {\mathcal{P}}_{j}=\langle \operatorname{im}\unicode[STIX]{x1D6F7}_{j},{\mathcal{P}}_{j-1}\rangle & & \displaystyle\end{eqnarray}$$

(for $j=k$ this will give the theorem). Note that (4.7) and the defining exact sequence for $M\{[j-1],[j,m]\}$ together imply that $\operatorname{im}\unicode[STIX]{x1D6F7}_{j}\subset {\mathcal{P}}_{j}$. The semiorthogonality immediately follows from the fact that the image of $i_{!}\{j\}$ (and hence of $\unicode[STIX]{x1D6F7}_{j}$) is right orthogonal to ${\mathcal{P}}_{j-1}$. Finally, the defining exact sequence for $M\{[j-1],[j,m]\}$ shows that $P_{[j,m]}$ belongs to the subcategory generated by $\operatorname{im}\unicode[STIX]{x1D6F7}_{j}$ and ${\mathcal{P}}_{j-1}$. Hence, our assertion follows from [Reference Bondal, Larsen and LuntsBLL04, Lemma 1.20].◻

Proof. We use the fact that the diagonal bimodule is perfect, so it is a homotopy direct summand of a bounded complex of free $B\otimes _{\mathbf{k}}B^{\text{op}}$-modules of finite rank. Applying the corresponding functors $D(B)\rightarrow D(B)$ to $M$, we get that $M$ is a homotopy direct summand in a bounded complex of $B$-modules whose terms are finite direct sums of $B\otimes _{\mathbf{k}}M$. By assumption, these terms have finite projective dimension over $B$, so $M$ also has finite projective dimension over $B$. Since $M$ is finitely generated, this implies that it is perfect.◻

Proof. To apply Lemma 4.2.3 we need to check that the $B_{1}-B_{2}$-bimodule $\hom (M_{1},M_{2})$ is perfect. Since $B_{2}$ is smooth, it is enough to check that it is perfect as a $B_{1}$-module (see [Reference Toën and VaquiéToë07, Lemma 2.8.2]). By Lemma 4.2.1, this follows from our assumptions.◻

Proof of Proposition 4.2.2.

To prove that ${\mathcal{B}}^{\circ }$ is smooth we apply Lemma 4.2.4 iteratively to semiorthogonal decompositions (4.5). Since the algebras $R/(x_{i})$ are homologically smooth, it suffices to check that the $R$-modules $\operatorname{Ext}^{\ast }(\overline{P}_{k},P_{[1,i]})$, for $i<k$, are finitely generated and are free as $\mathbf{k}$-modules. Computing these using the projective resolution of $\overline{P}_{k}$, we realize them as cohomology of the complex (in degrees $[0,1]$)

$$\begin{eqnarray}\displaystyle R/(x_{[1,i]})\xrightarrow[{}]{x_{[i+1,k]}}R/x_{[1,i]}. & & \displaystyle \nonumber\end{eqnarray}$$

Thus, we only have non-trivial $\operatorname{Ext}^{1}$ isomorphic to $R/(x_{[1,i]},x_{[i+1,k]})$, and the assertion follows.

Similarly, to prove that ${\mathcal{B}}^{\circ \circ }$ is smooth we apply Lemma 4.2.4 iteratively to semiorthogonal decompositions (4.8). It suffices to prove that the $R$-modules $\operatorname{Ext}^{\ast }(M\{[j-1],[j,m]\},P_{I})$, where $I=[j^{\prime },m^{\prime }]$ with $j^{\prime }<j$, are finitely generated and are free as $\mathbf{k}$-modules. Using the defining projective resolution of $M\{[j-1],[j,m]\}$, we get the complex (in degrees $[0,2]$) that computes this $\operatorname{Ext}^{\ast }$. If $m^{\prime }<j-1$ then this complex is zero. For $m^{\prime }\geqslant j-1$ it has the form

$$\begin{eqnarray}\displaystyle R/(x_{[j,m]\cap I})\overset{x_{j-1}}{\longrightarrow }R/(x_{[j-1,m]\cap I})\xrightarrow[{}]{x_{[j,m]\setminus I}}R/(x_{j-1}), & & \displaystyle \nonumber\end{eqnarray}$$

Thus, only $\operatorname{Ext}^{2}$ is non-trivial and it is isomorphic to $R/(x_{j-1},x_{[j,m]\setminus I})$, which is free over $\mathbf{k}$.◻

Proof. (i) The first assertion follows immediately from the semiorthogonal decomposition of Theorem 4.1.6 together with the adjunction of the induction and restriction functors $(i_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }},r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }})$.

To check that $M\{[i,j],[j+1,m]\}$ belongs to $\operatorname{ker}(r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }})$, we need to prove the surjectivity of the map

$$\begin{eqnarray}\displaystyle (P_{[i,m]})_{[1,r]}\rightarrow (P_{[j+1,m]})_{[1,r]} & & \displaystyle \nonumber\end{eqnarray}$$

induced by $x_{[i,j]}$. In other words, we need to check that the map

$$\begin{eqnarray}\displaystyle (x_{[1,r]\setminus [i,m]})/(x_{[1,r]})\overset{x_{[i,j]}}{\longrightarrow }(x_{[1,r]\setminus [j+1,m]})/(x_{[1,r]}) & & \displaystyle \nonumber\end{eqnarray}$$

is surjective for every $m$. Indeed, if $j\geqslant r$ then the target is zero. Otherwise $j<r$ and the assertion follows from the equality $x_{[1,r]\setminus [j+1,m]}=x_{[i,j]}x_{[1,r]\setminus [i,m]}$.

By Theorem 4.1.6, the subcategory $\operatorname{ker}(r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }})$ is generated by the images of $\unicode[STIX]{x1D6F7}_{2},\ldots ,\unicode[STIX]{x1D6F7}_{n}$. Hence, it is generated by the images of projective modules under $\unicode[STIX]{x1D6F7}_{2},\ldots ,\unicode[STIX]{x1D6F7}_{n}$. Since these images are given by formula (4.7), the last assertion follows.

(ii) The restriction functor between abelian categories

$$\begin{eqnarray}\displaystyle r_{R}^{{\mathcal{B}}^{\circ \circ }}:{\mathcal{B}}^{\circ \circ }-\operatorname{mod}\rightarrow R-\operatorname{mod} & & \displaystyle \nonumber\end{eqnarray}$$

is exact. Furthermore, by Lemma 4.1.1(i) the induction functor $i_{R}^{{\mathcal{B}}^{\circ \circ }}$ provides a cosection, so the category $R-\operatorname{mod}$ is a colocalization of ${\mathcal{B}}^{\circ \circ }-\operatorname{mod}$, which gives an equivalence

$$\begin{eqnarray}\displaystyle R-\operatorname{mod}\simeq {\mathcal{B}}^{\circ \circ }-\operatorname{mod}/(\operatorname{ker}(r_{R}^{{\mathcal{B}}^{\circ \circ }})\cap {\mathcal{B}}^{\circ \circ }-\operatorname{mod}). & & \displaystyle \nonumber\end{eqnarray}$$

The similar equivalence with derived categories follows from the work [Reference MiyachiMiy91].

As in (i), the assertion that $M\{I,J\}$ belongs to $\operatorname{ker}(r_{R}^{{\mathcal{B}}^{\circ \circ }})$ amounts to the surjectivity of the map

$$\begin{eqnarray}\displaystyle (x_{[1,k]\setminus (I\sqcup J)})\overset{x_{I}}{\longrightarrow }(x_{[1,k]\setminus J}), & & \displaystyle \nonumber\end{eqnarray}$$

which follows from the equality $x_{[1,k]\setminus J}=x_{I}x_{[1,k]\setminus (I\sqcup J)}$.

Let ${\mathcal{C}}\subset {\mathcal{B}}^{\circ }-\operatorname{mod}$ denote the kernel of $r_{R}^{{\mathcal{B}}^{\circ }}$, that is, the full subcategory consisting of ${\mathcal{B}}^{\circ }$-modules $M$ such that $M_{[1,k]}=0$. It is enough to prove that the modules $r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}(M\{[j],[i,j-1]\})$ generate ${\mathcal{C}}$ as an abelian category. We will use induction on $k$. In the case $k=1$, we have ${\mathcal{C}}=0$ so there is nothing to prove.

Let ${\mathcal{D}}\subset {\mathcal{B}}^{\circ }-\operatorname{mod}$ denote the full subcategory of ${\mathcal{B}}^{\circ }$-modules $M$ such that $M_{1}=0$. Let us set $\overline{{\mathcal{B}}}^{\circ \circ }={\mathcal{B}}_{[2,k]}^{\circ \circ }[x_{1}]$, $\overline{{\mathcal{B}}}^{\circ }={\mathcal{B}}_{[2,k]}^{\circ }[x_{1}]$. Note that the restriction functor $r_{\overline{{\mathcal{B}}}^{\circ }}^{{\mathcal{B}}^{\circ }}:{\mathcal{B}}^{\circ }-\operatorname{mod}\rightarrow \overline{{\mathcal{B}}}^{\circ }-\operatorname{mod}$ induces an equivalence

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}:{\mathcal{D}}\simeq \overline{{\mathcal{B}}}^{\circ }-\text{mod}. & & \displaystyle \nonumber\end{eqnarray}$$

Furthermore, as we have seen in Lemma 4.1.1(ii), we have $r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}i_{\overline{{\mathcal{B}}}^{\circ \circ }}^{{\mathcal{B}}^{\circ \circ }}(M)\in {\mathcal{D}}$ and

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}(r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}i_{\overline{{\mathcal{B}}}^{\circ \circ }}^{{\mathcal{B}}^{\circ \circ }}(M))=r_{\overline{{\mathcal{B}}}^{\circ }}^{\overline{{\mathcal{B}}}^{\circ \circ }}(M). & & \displaystyle \nonumber\end{eqnarray}$$

Let ${\mathcal{C}}^{\prime }\subset \overline{{\mathcal{B}}}^{\circ }-\operatorname{mod}$ denote the kernel of $r_{R[x_{1}]}^{\overline{{\mathcal{B}}}^{\circ }}$. By induction assumption, ${\mathcal{C}}^{\prime }$ is generated by the modules $r_{\overline{{\mathcal{B}}}^{\circ }}^{\overline{{\mathcal{B}}}^{\circ \circ }}(\overline{M}\{[j],[i,j-1]\})$, where $2\leqslant i<j\leqslant k$, and $\overline{M}\{[j],[i,j-1]\}$ are $\overline{{\mathcal{B}}}^{\circ \circ }$-modules defined similarly to $M\{[j],[i,j-1]\}$. We have $i_{\overline{{\mathcal{B}}}^{\circ \circ }}^{{\mathcal{B}}^{\circ \circ }}\overline{M}\{[j],[i,j-1]\}=M\{[j],[i,j-1]\}$ for $2\leqslant i<j\leqslant k$, so

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}^{-1}(r_{\overline{{\mathcal{B}}}^{\circ }}^{\overline{{\mathcal{B}}}^{\circ \circ }}(\overline{M}\{[j],[i,j-1]\}))=r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}i_{\overline{{\mathcal{B}}}^{\circ \circ }}^{{\mathcal{B}}^{\circ \circ }}(\overline{M}\{[j],[i,j-1]\})=r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}(M\{[j],[i,j-1]\}). & & \displaystyle \nonumber\end{eqnarray}$$

Thus, the subcategory $\unicode[STIX]{x1D6F7}^{-1}({\mathcal{C}}^{\prime })={\mathcal{C}}\cap {\mathcal{D}}$ belongs to the subcategory generated by the modules $r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}(M\{[j],[i,j-1]\})$, with $2\leqslant i<j\leqslant k$.

It remains to check that ${\mathcal{C}}$ is generated by ${\mathcal{C}}\cap {\mathcal{D}}$ and by the modules $r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}(M\{[j],[1,j-1]\})$. Note that ${\mathcal{C}}\cap {\mathcal{D}}$ is precisely the kernel of the functor

$$\begin{eqnarray}\displaystyle r:{\mathcal{C}}\rightarrow R/(x_{1},x_{2}\ldots x_{k})-\operatorname{mod}:M\mapsto M_{1}. & & \displaystyle \nonumber\end{eqnarray}$$

Now for $2\leqslant j\leqslant k$, let us define the functor

$$\begin{eqnarray}\displaystyle i_{\ast }(j):R/(x_{1},x_{j})\rightarrow {\mathcal{C}} & & \displaystyle \nonumber\end{eqnarray}$$

sending $N$ to the ${\mathcal{B}}^{\circ }$-module $M$ such that

$$\begin{eqnarray}\displaystyle M_{[1,m]}=\left\{\begin{array}{@{}ll@{}}N\quad & m<j,\\ 0\quad & m\geqslant j,\end{array}\right. & & \displaystyle \nonumber\end{eqnarray}$$

with the maps $M_{[1,m-1]}\rightarrow M_{[1,m]}$ (respectively, $M_{[1,m]}\rightarrow M_{[1,m-1]}$) being the identity maps (respectively, multiplication by $x_{m}$) for $m<j$.

For $2\leqslant j\leqslant k$, let ${\mathcal{C}}_{j}\subset {\mathcal{C}}$ denote the full subcategory of $M\in {\mathcal{C}}$ such that $x_{j}M_{1}=0$. Since for every $M\in {\mathcal{C}}$ one has $x_{2}\ldots x_{k}M_{1}=0$, it is easy to see that the subcategories ${\mathcal{C}}_{j}$ generate ${\mathcal{C}}$ as an abelian category. Now for every $M\in {\mathcal{C}}_{j}$, there is a natural morphism

$$\begin{eqnarray}\displaystyle i_{\ast }(j)(M_{1})\rightarrow M & & \displaystyle \nonumber\end{eqnarray}$$

inducing the identity map $i_{\ast }(j)(M_{1})_{1}\rightarrow M_{1}$. Hence, the kernel and the cokernel of the morphism $i_{\ast }(j)(M_{1})\rightarrow M$ are in ${\mathcal{C}}\cap {\mathcal{D}}$. It follows that the images of the functors $i_{\ast }(j)$ together with ${\mathcal{C}}\cap {\mathcal{D}}$ generate ${\mathcal{C}}$.

It remains to observe that for each $j\geqslant 2$, there is an exact sequence

$$\begin{eqnarray}\displaystyle 0\rightarrow i_{\ast }(j)(R/(x_{1},x_{j}))\rightarrow r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}M\{[j],[1,j-1]\}\rightarrow r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}M\{[j],[2,j-1]\}\rightarrow 0 & & \displaystyle \nonumber\end{eqnarray}$$

(where for $j=2$ the last term is zero). This implies that the image of $i_{\ast }(j)$ is contained in the subcategory generated by $r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}M\{[j],[1,j-1]\}$ and $r_{{\mathcal{B}}^{\circ }}^{{\mathcal{B}}^{\circ \circ }}M\{[j],[2,j-1]\}$.◻

Proof. We have

$$\begin{eqnarray}\displaystyle \operatorname{Hom}_{{\mathcal{B}}^{\circ \circ }}(P_{I},P_{J})\simeq \operatorname{Hom}_{R}(R/(x_{J}),R/(x_{I}))\simeq (x_{I\setminus J})/(x_{I})=x_{I\setminus J}\cdot R/(x_{I\cap J}). & & \displaystyle \nonumber\end{eqnarray}$$

It is easy to see that mapping $\operatorname{Hom}_{{\mathcal{B}}^{\circ \circ }}(P_{I},P_{J})$ to $\operatorname{Hom}_{{\mathcal{B}}^{\circ \circ }}(P_{J},P_{I})$ using these identifications gives an isomorphism of the algebra ${\mathcal{B}}^{\circ \circ }$ with its opposite algebra (see below for the computation of compositions).

On the other hand, for $S=[0,k]\setminus \{a,b\}$ and $S^{\prime }=[0,k]\setminus \{a^{\prime },b^{\prime }\}$ we have the following four possibilities, and we can calculate the subintervals $I_{\bullet }$, $J_{\bullet }$ of Corollary 3.2.4 in each of them.

Case 1: $a<a^{\prime }$, $b<b^{\prime }$. If $a^{\prime }\geqslant b$ then $S$ and $S^{\prime }$ are not close, so we can assume that $a^{\prime }<b$. In this case we have $J_{1}=[a,a^{\prime }]$ and $J_{2}=[b,b^{\prime }]$, so, using Theorem 3.2.5(ii), we get

$$\begin{eqnarray}\displaystyle {\mathcal{A}}(S,S^{\prime }) & = & \displaystyle \mathbf{k}[x_{1},\ldots ,x_{a}]\,\otimes \,\mathbf{k}[x_{a+1},\ldots ,x_{a^{\prime }}]\,\otimes \,\mathbf{k}[x_{a^{\prime }+1},\ldots ,x_{b}]/(x_{[a^{\prime }+1,b]})\nonumber\\ \displaystyle & & \displaystyle \,\otimes \,\,\mathbf{k}[x_{b+1},\ldots ,x_{b}^{\prime }]\,\otimes \,\mathbf{k}[x_{b^{\prime }+1},\ldots ,x_{k}]\cdot f_{S,S^{\prime }}=R/(x_{[a^{\prime }+1,x_{b}]})\cdot f_{S,S^{\prime }}.\nonumber\end{eqnarray}$$

Case 2: $a<a^{\prime }$, $b\geqslant b^{\prime }$. In this case we have $J_{1}=[a,a^{\prime }]$ and $I_{1}=[b^{\prime },b]$ ($I_{1}$ should be omitted if $b=b^{\prime }$), so by Theorem 3.2.5(ii) we get

$$\begin{eqnarray}\displaystyle {\mathcal{A}}(S,S^{\prime }) & = & \displaystyle \mathbf{k}[x_{1},\ldots ,x_{a}]\,\otimes \,\mathbf{k}[x_{a+1},\ldots ,x_{a^{\prime }}]\,\otimes \,\mathbf{k}[x_{a^{\prime }+1},\ldots ,x_{b^{\prime }}]/(x_{[a^{\prime }+1,b^{\prime }]})\nonumber\\ \displaystyle & & \displaystyle \,\otimes \,\,\mathbf{k}[x_{b^{\prime }+1},\ldots ,x_{b}]\,\otimes \,\mathbf{k}[x_{b+1},\ldots ,x_{k}]\cdot f_{S,S^{\prime }}=R/(x_{[a^{\prime }+1,x_{b^{\prime }}]})\cdot f_{S,S^{\prime }}.\nonumber\end{eqnarray}$$

Case 3: $a\geqslant a^{\prime }$, $b<b^{\prime }$. In this case we have $I_{1}=[a^{\prime },a]$ and $J_{1}=[b,b^{\prime }]$ ($I_{1}$ should be omitted if $a=a^{\prime }$), so by Theorem 3.2.5(ii) we get

$$\begin{eqnarray}\displaystyle {\mathcal{A}}(S,S^{\prime }) & = & \displaystyle \mathbf{k}[x_{1},\ldots ,x_{a^{\prime }}]\,\otimes \,\mathbf{k}[x_{a^{\prime }+1},\ldots ,x_{a}]\,\otimes \,\mathbf{k}[x_{a+1},\ldots ,x_{b}]/(x_{[a+1,b]})\nonumber\\ \displaystyle & & \displaystyle \,\otimes \,\,\mathbf{k}[x_{b+1},\ldots ,x_{b^{\prime }}]\,\otimes \,\mathbf{k}[x_{b^{\prime }+1},\ldots ,x_{k}]\cdot f_{S,S^{\prime }}=R/(x_{[a+1,x_{b}]})\cdot f_{S,S^{\prime }}.\nonumber\end{eqnarray}$$

Case 4: $a\geqslant a^{\prime }$, $b\geqslant b^{\prime }$. If $a\geqslant b^{\prime }$ then $S$ and $S^{\prime }$ are not close, so we can assume that $a<b^{\prime }$. In this case we have $I_{1}=[a^{\prime },a]$ and $I_{2}=[b^{\prime },b]$ (where $I_{1}$ is omitted if $a=a^{\prime }$ and $I_{2}$ is omitted if $b=b^{\prime }$). By Theorem 3.2.5(ii) we get

$$\begin{eqnarray}\displaystyle {\mathcal{A}}(S,S^{\prime }) & = & \displaystyle \mathbf{k}[x_{1},\ldots ,x_{a^{\prime }}]\,\otimes \,\mathbf{k}[x_{a^{\prime }+1},\ldots ,x_{a}]\,\otimes \,\mathbf{k}[x_{a+1},\ldots ,x_{b^{\prime }}]/(x_{[a+1,b^{\prime }]})\nonumber\\ \displaystyle & & \displaystyle \,\otimes \,\,\mathbf{k}[x_{b^{\prime }+1},\ldots ,x_{b}]\,\otimes \,\mathbf{k}[x_{b+1},\ldots ,x_{k}]\cdot f_{S,S^{\prime }}=R/(x_{[a+1,x_{b^{\prime }}]})\cdot f_{S,S^{\prime }}.\nonumber\end{eqnarray}$$

In all of these cases we deduce that

$$\begin{eqnarray}\displaystyle {\mathcal{A}}(S,S^{\prime })\simeq R/(x_{I\cap J})\cdot f_{S,S^{\prime }}, & & \displaystyle \nonumber\end{eqnarray}$$

where $I=[a+1,b]$ and $J=[a^{\prime }+1,b^{\prime }]$. So we get an identification of $R$-modules

$$\begin{eqnarray}\displaystyle {\mathcal{A}}(S,S^{\prime })\simeq \operatorname{Hom}(P_{I},P_{J}) & & \displaystyle \nonumber\end{eqnarray}$$

sending $f_{S,S^{\prime }}$ to $x_{I\setminus J}$.

To check the compatibility with the composition, we note that for three intervals $I,J,K\subset [1,k]$ one has

$$\begin{eqnarray}\displaystyle x_{I\setminus J}\cdot x_{J\setminus K}=x_{I\setminus K}\cdot \mathop{\prod }_{i\in (J\setminus (I\cup K))\cup (I\cap K\setminus J)}x_{i}. & & \displaystyle \nonumber\end{eqnarray}$$

Thus, the assertion follows from Theorem 3.2.5(iii) and the equality

(5.2)$$\begin{eqnarray}\displaystyle \{i:~\mid [i-1,i]\subset T(S,S^{\prime },S^{\prime \prime })\}=(J\setminus (I\cup K))\cup (I\cap K\setminus J), & & \displaystyle\end{eqnarray}$$

where $L_{S}$, $L_{S^{\prime }}$ and $L_{S^{\prime \prime }}$ correspond to $I$, $J$ and $K$ under (5.1). The proof of (5.2) is a straightforward but tedious check, so we will consider only one of the cases. Let

$$\begin{eqnarray}\displaystyle & \displaystyle S=[0,k]\setminus \{a,b\},\quad S^{\prime }=[0,k]\setminus \{a^{\prime },b^{\prime }\},\quad S^{\prime \prime }=[0,k]\setminus \{a^{\prime \prime },b^{\prime \prime }\}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle I=[a+1,b],\quad J=[a^{\prime }+1,b^{\prime }],\quad K=[a^{\prime \prime }+1,b^{\prime \prime }]. & \displaystyle \nonumber\end{eqnarray}$$

Assume that $a<a^{\prime }$, $b<b^{\prime }$, $a^{\prime \prime }<a^{\prime }$, $b^{\prime \prime }<b^{\prime }$. Then we have $J_{1}=[a,a^{\prime }]$, $J_{2}=[b,b^{\prime }]$, $I_{1}^{\prime }=[a^{\prime \prime },a^{\prime }]$ and $I_{2}^{\prime }=[b^{\prime \prime },b^{\prime }]$, hence

$$\begin{eqnarray}\displaystyle T(S,S^{\prime },S^{\prime \prime })=[\max (a,a^{\prime \prime }),a^{\prime }]\cup [\max (b,b^{\prime \prime }),b^{\prime }]. & & \displaystyle \nonumber\end{eqnarray}$$

On the other hand,

$$\begin{eqnarray}\displaystyle I\cap K\setminus J=[\max (a+1,a^{\prime \prime }+1),a^{\prime }],\quad J\setminus (I\cup K)=[\max (b,b^{\prime \prime })+1,b^{\prime }], & & \displaystyle \nonumber\end{eqnarray}$$

so equality (5.2) follows in this case. The other cases are considered similarly. ◻

Proof. Note that by Theorem 3.2.5(iv), shifting the graded structures on each $L_{S}$ if needed, we can achieve that the algebra ${\mathcal{A}}^{\circ \circ }$ is concentrated in degree $0$. Since the Lagrangians $(L_{S})$ generate the wrapped Fukaya category, the first equivalence is an immediate consequence of the isomorphism of algebras proved in Lemma 5.1.1.