Let us study the face ring of $P$ and use Theorem 2.16. We denote by
$v_{1},...,v_{6}$ the variables of $[R(P),d]$ coming from the edges of
$6-$gon. Denote the corresponding exterior variables of $[R(P),d]$ by
$y_{1},\ldots,y_{6}$ (see Figure 4).

Figure 4: 6-gon

Since $v_{1}\cap v_{5}=\emptyset$, we see that $d(y_{1}v_{5})=v_{1}v_{5}=0$.
Therefore, $y_{1}v_{5}$ represent nontrivial element. Analogously,

$\begin{gathered}y_{1}v_{5},\;\;y_{1}v_{3},\;\;y_{1}v_{2},\;\;y_{4}v_{3},\;\;y_{4}v_{2},\;\;y_{4}v_{6},\\\
y_{5}v_{2},\;\;y_{5}v_{6},\;\;y_{3}v_{6}\quad\in
H^{3}(\widetilde{\mathcal{Z}}_{P},\mathbb{Z})\end{gathered}$

represent nontrivial elements in
$H^{3}(\widetilde{\mathcal{Z}}_{P},\mathbb{Z})$. In the same way

$\begin{gathered}y_{1}y_{4}v_{3},\;\;y_{1}y_{4}v_{2},\;\;y_{1}y_{5}v_{2},\;\;y_{4}y_{5}v_{2},\\\
y_{4}y_{5}v_{6},\;\;y_{4}y_{3}v_{6},\;\;y_{5}y_{3}v_{6},\;\;y_{3}y_{2}v_{1}\in
H^{4}(\widetilde{\mathcal{Z}}_{P},\mathbb{Z})\end{gathered}$

represent nontrivial elements in
$H^{4}(\widetilde{\mathcal{Z}}_{P},\mathbb{Z})$. For example, since $v_{1}\cap
v_{3}=v_{3}\cap v_{4}=\emptyset$, we have
$d(y_{1}y_{4}v_{3})=y_{4}v_{1}v_{3}-y_{1}v_{4}v_{3}=0$.

Now, let us study the face ring of $Q=P^{(k,k,k,k,k,k)}$. We denote the new
variables corresponding to the new facets by $v_{i,1},\ldots,v_{i,k}$. New
exterior variables are denoted by $y_{i,1},\ldots,y_{i,k}$. Theorem 2.22 says
that

$d(y_{1,1}v_{1,2}\ldots v_{1,k}v_{5,1}\ldots v_{5,k})=v_{1,1}\ldots
v_{1,k}v_{5,1}\ldots v_{5,k}=0$

because $v_{1,1}\cap\ldots\cap v_{1,2k}\cap v_{3,1}\ldots v_{3,2k}=\emptyset$.
Analogously, we see that

$\begin{gathered}y_{1,1}v_{1,2}\ldots v_{1,k}v_{5,1}\ldots
v_{5,k},\;\;y_{1,1}v_{1,2}\ldots v_{1,k}v_{3,1}\ldots
v_{3,k},\;\;y_{1,1}v_{1,2}\ldots v_{1,k}v_{2,1}\ldots v_{2,k},\\\
y_{4,1}v_{4,2}\ldots v_{4,k}v_{3,1}\ldots v_{3,k},\;\;y_{4,1}v_{4,2}\ldots
v_{4,k}v_{2,1}\ldots v_{2,k},\;\;y_{4,1}v_{4,2}\ldots v_{4,k}v_{6,1}\ldots
v_{6,k},\\\ y_{5,1}v_{5,2}\ldots v_{5,k}v_{2,1}\ldots
v_{2,k},\;\;y_{5,1}v_{5,2}\ldots v_{5,k}v_{6,1}\ldots
v_{6,k},\;\;y_{3,1}v_{3,2}\ldots v_{3,2k}v_{6,1}\ldots v_{6,k}\\\
\in\;\;H^{4k-1}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z})\end{gathered}$

So, we have

$\begin{gathered}h(s^{3}_{i})\in
H^{4k-1}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}).\end{gathered}$

Arguing as before, we show that

$\begin{gathered}h(s^{4}_{j})\in
H^{6k-2}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}).\end{gathered}$

As a result, we obtain that rings

$H^{*}(\widetilde{\mathcal{Z}}_{Q},\;\mathbb{Z}),\quad
H^{*}(\\#9(S^{4k-1}\times S^{8k-3})\\#8(S^{6k-2}\times
S^{6k-2}),\;\mathbb{Z})$

are isomorphic as ungraded rings. Since there is no torsion, we see that

$H^{*}(\widetilde{\mathcal{Z}}_{Q},\;\mathbb{Z}_{2}),\quad
H^{*}(\\#9(S^{4k-1}\times S^{8k-3})\\#8(S^{6k-2}\times
S^{6k-2}),\;\mathbb{Z}_{2})$

are isomorphic as ungraded rings. ∎

As we already noticed

$\widetilde{\mathcal{Z}}_{Q}=\widetilde{\mathcal{R}}^{(2,\ldots,2)}_{Q},$

i.e. $\widetilde{\mathcal{R}}_{P}^{(2k,\ldots,2k)}$ is associated to the
polytope $P^{(2k,\ldots,2k)}=Q^{(2,\ldots,2)}$. We see that
$\widetilde{\mathcal{R}}_{Q}^{(2,\ldots,2)}$ is given by the system

$\begin{gathered}\widetilde{\mathcal{R}}_{Q}^{(2,\ldots,2)}=\left\\{\begin{array}[]{l}|\overrightarrow{u}_{2}|^{2}+|\overrightarrow{u}_{3}|^{2}=2\\\
|\overrightarrow{u}_{2}|^{2}+|\overrightarrow{u}_{4}|^{2}=2\\\
|\overrightarrow{u}_{1}|^{2}+|\overrightarrow{u_{2}}|^{2}+|\overrightarrow{z}_{5}|^{2}=3\\\
|\overrightarrow{u}_{1}|^{2}+|\overrightarrow{u}_{2}|^{2}-|\overrightarrow{u}_{6}|^{2}=1\end{array}\right.\\\
\overrightarrow{u}_{r}=(u_{r,1},\ldots,u_{r,2k})\in\mathbb{R}^{2k}\\\
|\overrightarrow{u}_{r}|^{2}=u_{r,1}^{2}+\ldots+u_{r,2k}^{2}.\end{gathered}$
(5.18)

Let us write the system above in the following form:

$\begin{gathered}\widetilde{\mathcal{Z}}_{Q}=\widetilde{R}_{Q}^{(2,\ldots,2)}=\left\\{\begin{array}[]{l}|\overrightarrow{u}_{1}|^{2}+|\overrightarrow{u}_{2}|^{2}+|\overrightarrow{u}_{3}|^{2}+|\overrightarrow{u}_{4}|^{2}+|\overrightarrow{u}_{5}|^{2}+|\overrightarrow{u}_{6}|^{2}=6\\\
|\overrightarrow{u}_{2}|^{2}+|\overrightarrow{u}_{4}|^{2}=2\\\
|\overrightarrow{u}_{1}|^{2}+|\overrightarrow{u_{2}}|^{2}+|\overrightarrow{z}_{5}|^{2}=3\\\
|\overrightarrow{u}_{1}|^{2}+|\overrightarrow{u}_{2}|^{2}-|\overrightarrow{u}_{6}|^{2}=1\end{array}\right.\end{gathered}$
(5.19)

We see that system $(\ref{widenarrow3})$ can be written in the form
$(\ref{eqmain})$. From Section 4.2 we know that there exists a Lagrangian
$L\subset\mathbb{C}P^{12k-1}$ associated to system 5.19 (or equivalently to
the polytope $P^{(2k,\ldots,2k)}$). From Lemma 5.6 we get that

$L=\mathcal{R}_{Q}^{(2,\ldots,2)}\times
T^{3},\quad\;\;\mathcal{R}_{Q}^{(2,\ldots,2)}=\widetilde{\mathcal{R}}_{Q}^{(2,\ldots,2)}/\mathbb{Z}_{2}.$

######  Lemma 5.17.

The Lagrangian $L$ is embedded and monotone. The minimal Maslov number $N_{L}$
is equal to $2k$.

###### Proof.

The polytope $P$ is Delzant and Lemma 5.8 says that
$P^{(2k,\ldots,2k)}=Q^{(2,\ldots,2)}$ is Delzant. Moreover, the conditions of
Lemma 4.4 are satisfied for $C=2k$. Therefore $P^{(2k,\ldots,2k)}$ is Fano. It
follows from Theorem 4.12 that $L$ is embedded and monotone. From Lemma 5.5
and Theorem 5.1 we get that $N_{L}=2k$. ∎

Dimension of the polytope $P^{(2k,\ldots,2k)}$ equals $12k$ and dimension of
$Q=P^{(k,\ldots,k)}$ is equal to 6k. It follows from Lemma 1.2 that $L$ is
spin.

Let $G$ be a ring in which $2$ is invertible. Lemma 1.3 says that
$QH(L,G[T,T^{-1}])=0$.

Let us prove that $QH(L,\mathbb{Z}_{2}[T,T^{-1}])=0$. Recall that

$m(Q)=min\\{\ell\in\mathbb{N}\;|\;v_{i_{1}}\cap\ldots\cap
v_{i_{\ell}}=\emptyset\\}.$

We see that $m(P)=2$. It follows from Theorem 2.22 that

$m(Q)=m(P^{(k,\ldots,k)})=2k.$

For example, $v_{1,1}\cap\ldots\cap v_{1,k}\cap v_{3,1}\cap\ldots\cap
v_{3,k}=\emptyset$. By Lemma 5.16,
$H^{*}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}_{2})$ is isomorphic to

$H^{*}(\\#9(S^{4k-1}\times S^{8k-3})\\#8(S^{6k-2}\times
S^{6k-2}),\;\mathbb{Z}_{2}).$

We have the Cartan-Leray spectral sequence

$\begin{gathered}E_{2}^{p,q}=H^{p}(\mathbb{R}P^{\infty},H^{q}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}_{2}))\Rightarrow
H^{*}(\mathcal{Z}_{Q},\mathbb{Z}_{2}),\\\
\mathcal{Z}_{Q}=\widetilde{\mathcal{Z}}_{Q}/\mathbb{Z}_{2}=\widetilde{\mathcal{R}}_{Q}^{(2,\ldots,2)}/\mathbb{Z}_{2}.\end{gathered}$

The second page of the spectral sequence is shown in figure 5.

Figure 5: $E_{2}$ page

Let us denote

The nonzero element of $E^{0,12k-4}_{2}$ by $h$;
Generators of $E^{0,8k-3}_{2}$ by $s^{8k-3}_{i}$;
Generators of $E^{0,6k-1}_{2}$ by $s^{6k-2}_{j}$ and $\hat{s}^{6k-2}_{j}$,
where $s_{j}^{6k-2}$ is the Poincare dual to $\widehat{s}_{j}^{6k-2}$;
Generators of $E^{0,4k-1}_{2}$ by $s^{4k-1}_{i}$;
Generator of $H^{*}(\mathbb{R}P^{\infty},\mathbb{Z}_{2})$ by $\alpha$;
Assume that $s^{4k-1}_{i}$ is Poicare dual to $s^{8k-3}_{i}$.

The first nontrivial differential is $\partial_{2k}$ and $E_{2}=E_{2k}$. We
have

$\begin{gathered}\partial_{2k}(s_{i}^{8k-3})=\alpha^{2k}(\sum\limits_{r=1}^{8}a_{i,r}s_{r}^{6k-2}+\sum\limits_{r=1}^{8}b_{i,r}\widehat{s}_{r}^{6k-2}),\\\
\partial_{2k}(s_{j}^{6k-2})=\alpha^{2k}\sum\limits_{r=1}^{9}c_{j,r}s_{r}^{4k-1},\quad\partial_{2k-1}(\widehat{s}_{j}^{6k-2})=\alpha^{2k}\sum\limits_{r=1}^{9}\widehat{c}_{j,r}s_{r}^{4k-1},\\\
a_{i,r},\;b_{i,r},\;c_{j,r},\;\widehat{c}_{j,r}\;\in\mathbb{Z}_{2}.\end{gathered}$

In other words,

$\begin{gathered}\partial_{2k}\begin{pmatrix}s^{8k-3}_{1}\\\ .\\\ .\\\ .\\\
s^{8k-3}_{9}\end{pmatrix}=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}\alpha^{2k}s^{6k-2}_{1}\\\
.\\\ .\\\ .\\\ \alpha^{2k}s^{6k-2}_{8}\\\ \alpha^{2k}\widehat{s}^{6k-2}_{1}\\\
.\\\ .\\\ .\\\ \alpha^{2k}\widehat{s}^{6k-2}_{8}\end{pmatrix}\end{gathered}$
$\begin{gathered}\partial_{2k}\begin{pmatrix}s^{6k-2}_{1}\\\ .\\\ .\\\ .\\\
s^{6k-2}_{8}\end{pmatrix}=C\begin{pmatrix}\alpha^{2k}s^{4k-1}_{1}\\\ .\\\ .\\\
.\\\
\alpha^{2k}s^{4k-1}_{9}\end{pmatrix},\quad\;\;\partial_{2k}\begin{pmatrix}\widehat{s}^{6k-2}_{1}\\\
.\\\ .\\\ .\\\
\widehat{s}^{6k-2}_{8}\end{pmatrix}=\widehat{C}\begin{pmatrix}\alpha^{2k}s^{4k-1}_{1}\\\
.\\\ .\\\ .\\\ \alpha^{2k}s^{4k-1}_{9}\end{pmatrix},\end{gathered}$

where $A=(a_{i,r})$, $B=(b_{i,r})$, $C=(c_{j,r})$,
$\widehat{C}=(\widehat{c}_{j,r})$ are matrices. Since
$s^{8k-3}_{i}s^{6k-2}_{j}=0$, we have

$\begin{gathered}0=\partial_{2k}(s^{8k-3}_{i}s^{6k-2}_{j})=(b_{i,j}+c_{j,i})\alpha^{2k}h,\\\
0=\partial_{2k}(s^{8k-3}_{i}\widehat{s}^{6k-2}_{j})=(a_{i,j}+\widehat{c}_{j,i})\alpha^{2k}h,\\\
\Rightarrow b_{i,j}=c_{j,i},\quad a_{i,j}=\widehat{c}_{j,i},\end{gathered}$

This means that $A=\widehat{C}^{T}$, $B=C^{T}$, where $T$ stands for the
transposition. Therefore,

$dim(\partial_{2k}(E_{2k}^{0,8k-3}))=dim(\partial_{2k}(E_{2k}^{0,6k-2})).$
(5.20)

Also, since $dim(\partial_{2k}(E_{2k}^{0,8k-3}))\subset
ker(\partial_{2k}(E_{2k}^{2k,6k-2}))$ and $dim(E_{2k}^{2k,6k-2})=16$, we get
that

$dim(\partial_{2k}(E_{2k}^{0,6k-2}))\leqslant 8.$

Let us note that $E_{2k+1}^{*,6k-2}=E_{6k-1}^{*,6k-2}$ and
$E_{6k}^{*,6k-2}=E_{\infty}^{*,6k-2}$. We see that only $\partial_{6k-1}$ may
affect the $(6k-2)$th row of $E_{6k-1}$. The manifold $\mathcal{Z}_{Q}$ is
finite-dimensional and for any $r>12k-4$ we must have $E_{\infty}^{r,6k-2}=0$.
Since, $dim(E_{6k-1}^{0,12k-4}),dim(E_{6k-1}^{6k-1,0})\leqslant 1$, we get
that $dim(E_{6k-1}^{r,6k-2})\leqslant 2$ for all $r\geqslant 2k$. Hence,

$dim(\partial_{2k}(E_{2k}^{0,6k-2}))\geqslant 7.$

We have two cases:

1. Assume that $dim(\partial_{2k}(E_{2k}^{0,8k-3}))=7$. The next nontrivial differential may be $\partial_{4k-1}$ and $E_{2k+1}=E_{4k-1}$. There exist elements $x_{1},x_{2}\in E_{4k-1}^{0,4k-1}$, $y_{1},y_{2}\in E_{4k-1}^{0,8k-3}$, and $t_{1},t_{2}\in E_{4k-1}^{0,6k-2}$ such that for any $r\geqslant 2k$ and $m\geqslant 0$ we have $E_{4k-1}^{r,4k-1}=\mathbb{Z}_{2}\langle\alpha^{r}x_{1},\alpha^{r}x_{2}\rangle$, $E_{4k-1}^{r,6k-2}=\mathbb{Z}_{2}\langle\alpha^{r}t_{1},\alpha^{r}t_{2}\rangle$, and $E_{4k-1}^{m,8k-3}=\mathbb{Z}_{2}\langle\alpha^{r}y_{1},\alpha^{r}y_{2}\rangle$. Note that $\partial_{4k-1}$ can not change $E_{4k-1}^{*,6k-2}$. Since the row $E_{\infty}^{*,6k-2}$ is finite ($\mathcal{Z}_{Q}$ is finite-dimensional), we see that the differential $\partial_{6k-1}$ must be nonzero on $E_{6k-1}^{*,12k-4}$ and $E_{6k-1}^{*,6k-2}$ (in particular $E_{6k-1}^{0,12k-4}\neq 0$). Hence, rows $E_{\infty}^{*,4k-1}$, $E_{\infty}^{*,8k-3}$ are finite if and only if the map $\partial_{4k-1}:E_{4k-1}^{0,8k-3}\rightarrow E_{4k-1}^{4k-1,4k-1}$ is isomorphism. Therefore, $E_{4k}^{4k-1,4k-1}=0$. Let us consider $E_{4k}$ and let $x$ be an arbitrary element of $E_{4k}^{0,4k-1}$. If $\partial_{4k}(x)=\alpha^{4k}$, then $\partial_{4k}(\alpha^{4k-1}x)=\alpha^{8k-1}\neq 0$. On the other hand, $\alpha^{4k-1}x=0$ because it belongs to $E_{4k}^{4k-1,4k-1}=0$. This contradiction shows that

$dim(\partial_{2k}(E_{2k}^{0,8k-3}))=7\;\Rightarrow\;\partial_{4k}(E_{4k-1}^{0,4k-1})=0.$

2. Assume that $dim(\partial_{2k}(E_{2k}^{0,8k-3}))=8$. Then, $E_{\infty}^{r,6k-2}=0$ for any $r\geqslant 2k$. As in the previous case, the next nontrivial differential may be $\partial_{4k-1}$ and $E_{2k+1}=E_{4k-1}$. There exist elements $x\in E_{4k-1}^{0,4k-1}$ and $y\in E_{4k-1}^{0,8k-3}$ such that $E_{4k-1}^{r,4k-1}=\mathbb{Z}_{2}\langle\alpha^{r}x\rangle$, $E_{4k-1}^{m,8k-3}=\mathbb{Z}_{2}\langle\alpha^{m}y\rangle$ for any $r\geqslant 2k$ and $m\geqslant 0$.

Suppose that $\partial_{4k-1}(y)=\alpha^{4k-1}x$. Then, $E_{4k}^{r,4k-1}=0$
for all $r\geqslant 4k-1$. Let $w$ be an element of $E_{4k}^{0,4k-1}$ such
that $\partial_{4k}(w)=\alpha^{4k}$. We know that $\alpha^{4k-1}w=0$ and
$0=\partial_{4k}(\alpha^{4k-1}w)=\alpha^{8k-1}\neq 0$. This yields that

$dim(\partial_{2k}(E_{2k-1}^{0,8k-3}))=8,\;\;\;\partial_{4k-1}(y)=\alpha^{4k-1}x\;\Rightarrow\;\partial_{4k}(E_{4k-1}^{0,4k-1})=0.$

Suppose that $\partial_{4k-1}(y)=0$. We see that
$E_{4k}^{m,8k-3}=\mathbb{Z}_{2}\langle\alpha^{m}y\rangle$ for any $m\geqslant
0$. Since $E_{4k}^{0,4k-1}=E_{2}^{0,4k-1}$, we have an element $x\in
E_{4k}^{0,4k-1}$ such that $xy=0$. Then,

$0=\partial_{4k}(xy)=y\partial_{4k}(x)\;\Rightarrow\;\partial_{4k}(x)\neq\alpha^{4k}.$

We get the following:

$dim(\partial_{2k}(E_{2k}^{0,8k-3}))=8,\;\;\;\partial_{4k-1}(y)=0\;\Rightarrow\;\partial_{4k-1}(E_{4k-1}^{0,4k-1})=0.$

As a result, we see that $\partial_{4k}(E_{4k-1}^{0,4k-1})=0$. It follows from
Theorem 1.3 that $QH(L,\mathbb{Z}_{2}[T,T^{-1}])=0$.

###  5.7 Proof of Theorem 1.6

Let us consider three dimensional cube $I^{3}$ defined by

$I^{3}=\left\\{\begin{array}[]{l}x_{1}+1\geqslant 0\;\;\;x_{2}+1\geqslant
0\;\;\;x_{3}+1\geqslant 0\\\ -x_{1}+1\geqslant 0\;\;\;-x_{2}+1\geqslant
0\;\;\;-x_{3}+1\geqslant 0\\\ \end{array}\right.$

Let us cut off the vertices $(1,1,1)$, $(-1,-1,-1)$ and consider the polytope
$P$ defined by inequalities (see Figure 6)

$P=\left\\{\begin{array}[]{l}x_{1}+1\geqslant 0\;\;\;x_{2}+1\geqslant
0\;\;\;x_{3}+1\geqslant 0\\\ -x_{1}+1\geqslant 0\;\;\;-x_{2}+1\geqslant
0\;\;\;-x_{3}+1\geqslant 0\\\ -x_{1}-x_{2}-x_{3}+2\geqslant 0\quad
x_{1}+x_{2}+x_{3}+2\geqslant 0\end{array}\right.$

The complex moment-angle manifold
$\widetilde{\mathcal{Z}}_{P}\subset\mathbb{C}^{8}$ associated to $P$ is given
by a system

$\widetilde{\mathcal{Z}}_{P}=\left\\{\begin{array}[]{l}|z_{1}|^{2}+|z_{4}|^{2}=2\\\
|z_{2}|^{2}+|z_{5}|^{2}=2\\\ |z_{3}|^{2}+|z_{6}|^{2}=2\\\
|z_{1}|^{2}+|z_{2}|^{2}+|z_{3}|^{2}+|z_{7}|^{2}=5\\\
|z_{1}|^{2}+|z_{2}|^{2}+|z_{3}|^{2}-|z_{8}|^{2}=1\end{array}\right.$

Let us apply the multiwedge operation defined in Section 2.3 and consider

$Q=P^{(k,\ldots,k,2k,2k)}$

We have the associated complex moment angle manifold
$\widetilde{\mathcal{Z}}_{Q}$. We know that

$\widetilde{\mathcal{Z}}_{Q}=\widetilde{\mathcal{R}}_{Q}^{(2,\ldots,2)}$

and $\widetilde{\mathcal{R}}_{Q}^{(2,\ldots,2)}\subset\mathbb{R}^{20k}$ is
associated to the polytope $Q^{(2,\ldots,2)}=P^{(2k,\ldots,2k,4k,4k)}$. We
repeat $j_{i}$ times all variables (as in the proof of Theorem 1.5). By
$\overrightarrow{u}_{i}$ denote the vector $(u_{i,1},\ldots,u_{i,j_{i}})$ and
by $|\overrightarrow{u}_{i}|^{2}=u_{i,1}^{2}+\ldots+u_{i,j_{i}}^{2}$ the
squared length of the vector. We obtain that the real moment-angle manifold
$\widetilde{\mathcal{R}}_{Q}^{(2,\ldots,2)}\subset\mathbb{R}^{20k}$ is defined
by

$\begin{gathered}\widetilde{\mathcal{R}}^{(2,\ldots,2)}_{Q}=\left\\{\begin{array}[]{l}|\overrightarrow{u}_{1}|^{2}+|\overrightarrow{u}_{4}|^{2}=2\\\
|\overrightarrow{u}_{2}|^{2}+|\overrightarrow{u}_{5}|^{2}=2\\\
|\overrightarrow{u}_{3}|^{2}+|\overrightarrow{u}_{6}|^{2}=2\\\
|\overrightarrow{u}_{1}|^{2}+|\overrightarrow{u}_{2}|^{2}+|\overrightarrow{u}_{3}|^{2}+|\overrightarrow{u}_{7}|^{2}=5\\\
|\overrightarrow{u}_{1}|^{2}+|\overrightarrow{u}_{2}|^{2}+|\overrightarrow{u}_{3}|^{2}-|\overrightarrow{u}_{8}|^{2}=1\end{array}\right.\\\
j_{1}=\ldots=j_{6}=2k,j_{7}=j_{8}=4k\end{gathered}$ (5.21)

Since $P$ is Dlzant, we get that $Q^{(2,\ldots,2)}$ is Delzant (see Lemma
5.8). The conditions of Lemma 4.4 are satisfied (for $C=2k$) and we get that
$Q^{(2,\ldots,2)}$ is Fano. Therefore, there is embedded monotone Lagrangian
$L\subset\mathbb{C}P^{20k-1}$ associated to $Q^{(2,\ldots,2)}$ (see Theorem
$\ref{monotonecondition})$. It follows from Lemma 5.6 that

$L=\widetilde{\mathcal{R}}_{Q}^{(2,\ldots,2)}/\mathbb{Z}_{2}\times
T^{4}=\widetilde{\mathcal{Z}}_{Q}/\mathbb{Z}_{2}\times T^{4}.$

From Lemma 5.5 we have

$N_{L}=gcd(4k,10k,2k)=2k.$

Then, Theorem 2.21 says that $H^{*}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z})$
and $H^{*}(\widetilde{\mathcal{Z}}_{P},\mathbb{Z})$ are isomorphic as ungraded
rings. Moreover, by Theorem 2.18 we have

$\widetilde{\mathcal{Z}}_{P}=\mathcal{G}(\mathcal{G}((S^{3}\times S^{3}\times
S^{3}))\\#7(S^{3}\times S^{8})\\#12(S^{4}\times S^{7})\\#10(S^{5}\times
S^{6}).$ (5.22)

Let us prove that $QH(L,\mathbb{Z}_{2}[T,T^{-1}])=0$.

######  Lemma 5.18.

We have:
$H^{3}(\mathcal{G}(\mathcal{G}(S^{3}\times S^{3}\times
S^{3})),\mathbb{Z}_{2})=\mathbb{Z}_{2}^{3}$;
$H^{4}(\mathcal{G}(\mathcal{G}(S^{3}\times S^{3}\times
S^{3})),\mathbb{Z}_{2})=\mathbb{Z}^{6}_{2}$;
$H^{5}(\mathcal{G}(\mathcal{G}(S^{3}\times S^{3}\times
S^{3})),\mathbb{Z}_{2})=\mathbb{Z}_{2}^{3}$.

###### Proof.

By $(S^{3}\times S^{3}\times S^{3})_{-}$ we denote $S^{3}\times S^{3}\times
S^{3}$ minus a small $9-$dimensional disk. From Lefschetz duality

$H^{r}((S^{3}\times S^{3}\times S^{3})_{-}\times D^{2},\mathcal{G}(S^{3}\times
S^{3}\times S^{3}),\mathbb{Z}_{2})\approx H_{11-r}((S^{3}\times S^{3}\times
S^{3})_{-}\times D^{2},\mathbb{Z}_{2})$

and from the exact sequence of the pair $((S^{3}\times S^{3}\times
S^{3})_{-}\times D^{2},\;\;\mathcal{G}(S^{3}\times S^{3}\times S^{3}))$ we get

$\begin{gathered}H_{11-r}((S^{3}\times S^{3}\times S^{3})_{-}\times
D^{2},\mathbb{Z}_{2})\rightarrow\\\ H^{r}((S^{3}\times S^{3}\times
S^{3})_{-}\times D^{2},\mathbb{Z}_{2})\rightarrow
H^{r}(\mathcal{G}(S^{3}\times S^{3}\times S^{3}),\mathbb{Z}_{2})\rightarrow\\\
H_{10-r}((S^{3}\times S^{3}\times S^{3})_{-}\times
D^{2},\mathbb{Z}_{2})\rightarrow H^{r+1}((S^{3}\times S^{3}\times
S^{3})_{-}\times D^{2},\mathbb{Z}_{2})\rightarrow\\\ \end{gathered}$

and

$\begin{gathered}0\rightarrow H^{3}((S^{3}\times S^{3}\times S^{3})_{-}\times
D^{2},\mathbb{Z}_{2})\rightarrow H^{3}(\mathcal{G}(S^{3}\times S^{3}\times
S^{3}),\mathbb{Z}_{2})\rightarrow 0\\\ 0\rightarrow H^{6}((S^{3}\times
S^{3}\times S^{3})_{-}\times D^{2},\mathbb{Z}_{2})\rightarrow
H^{6}(\mathcal{G}(S^{3}\times S^{3}\times S^{3}),\mathbb{Z}_{2})\rightarrow
0\\\ 0\rightarrow H^{4}(\mathcal{G}(S^{3}\times S^{3}\times
S^{3}),\mathbb{Z}_{2})\rightarrow H_{6}((S^{3}\times S^{3}\times
S^{3})_{-}\times D^{2},\mathbb{Z}_{2})\rightarrow 0\\\ 0\rightarrow
H^{7}(\mathcal{G}(S^{3}\times S^{3}\times S^{3}),\mathbb{Z}_{2})\rightarrow
H_{3}((S^{3}\times S^{3}\times S^{3})_{-}\times
D^{2},\mathbb{Z}_{2})\rightarrow 0\end{gathered}$

So, we obtain

$\begin{gathered}H_{3}(\mathcal{G}(S^{3}\times S^{3}\times
S^{3}),\mathbb{Z}_{2})=\mathbb{Z}^{3}_{2},\quad H_{4}(\mathcal{G}(S^{3}\times
S^{3}\times S^{3}),\mathbb{Z}_{2})=\mathbb{Z}_{2}^{3},\\\
H_{6}(\mathcal{G}(S^{3}\times S^{3}\times
S^{3}),\mathbb{Z}_{2})=\mathbb{Z}^{3}_{2},\quad H_{7}(\mathcal{G}(S^{3}\times
S^{3}\times S^{3}),\mathbb{Z}_{2})=\mathbb{Z}_{2}^{3},\\\
H_{10}(\mathcal{G}(S^{3}\times S^{3}\times
S^{3}),\mathbb{Z}_{2})=\mathbb{Z}_{2},\\\ H_{k}(\mathcal{G}(S^{3}\times
S^{3}\times S^{3}),\mathbb{Z}_{2})=0\quad if\;\;k\neq
0,3,4,6,7,10\end{gathered}$

Repeating this procedure one more time for the pair $(\mathcal{G}(S^{3}\times
S^{3}\times S^{3})_{-}\times D^{2},\;\;\mathcal{G}(\mathcal{G}(S^{3}\times
S^{3}\times S^{3})))$ we prove the lemma. ∎

Let us study the face ring of $P$ and use Theorem 2.16. We denote by
$v_{1},...,v_{6}$ the variables of $[R(P),d]$ coming from the cube $I^{3}$ and
by $w_{1},w_{2}$ the variables coming from the new facets. Denote the
corresponding exterior variables of $[R(P),d]$ by
$y_{1},\ldots,y_{6},t_{1},t_{2}$ such that $dy_{i}=v_{i}$, $dt_{j}=w_{j}$ (see
Figure 6).

Figure 6: truncted cube

We see that

$\begin{gathered}\widetilde{a}_{1}=y_{1}v_{4},\;\;\widetilde{a}_{2}=y_{2}v_{5},\;\;\widetilde{a}_{3}=y_{3}v_{6},\;\;\widetilde{s}=t_{2}w_{1}\end{gathered}$

represent nontrivial elements in
$H^{3}(\widetilde{\mathcal{Z}}_{P},\mathbb{Z})$. Since $v_{4}\cap v_{5}\cap
v_{6}=\emptyset$, we have

$\begin{gathered}\widetilde{a}_{1}\widetilde{a}_{2}\widetilde{a}_{3}=y_{1}y_{2}y_{3}v_{4}v_{5}v_{6}=0,\\\
\widetilde{a}_{1}\widetilde{a}_{2},\;\widetilde{a}_{1}\widetilde{a}_{2},\;\widetilde{a}_{2}\widetilde{a}_{3}\neq
0.\end{gathered}$

Therefore, $\widetilde{a}_{1},\widetilde{a}_{2},\widetilde{a}_{3}$ represent
classes of $H^{*}(\mathcal{G}(\mathcal{G}(S^{3}\times S^{3}\times
S^{3})),\mathbb{Z})$ (see formula $(\ref{cubecomplexangle}))$. Then, elements

$\begin{gathered}\widetilde{b}_{1}=y_{1}t_{2}v_{4},\;\widetilde{b}_{2}=y_{2}t_{2}v_{5},\;\widetilde{b}_{3}=y_{3}t_{2}v_{6},\\\
\widetilde{c}_{1}=y_{4}t_{1}v_{1},\;\widetilde{c}_{2}=y_{5}t_{1}v_{2},\;\widetilde{c}_{3}=y_{6}t_{1}v_{3}\end{gathered}$

belong to $H^{4}(\widetilde{\mathcal{Z}}_{P},\mathbb{Z})$. Moreover,

$\widetilde{a}_{i}\widetilde{a}_{j}\widetilde{b}_{m},\;\widetilde{a}_{i}\widetilde{a}_{j}\widetilde{c}_{m}\neq
0\;\;\;if\;and\;only\;if\;\;i\neq j,i\neq m,j\neq m,$

Hence, classes $\widetilde{a}_{i},\widetilde{b}_{j}$ belong to cohomology ring
$H^{*}(\mathcal{G}(\mathcal{G}(S^{3}\times S^{3}\times S^{3})),\mathbb{Z})$.

Since $\widetilde{s}=t_{2}w_{1}$ and $t_{1}w_{2}$ represent the same classes
(because $d(t_{1}t_{2})=t_{2}w_{1}-t_{2}w_{1}$), we see that

$\widetilde{s}\widetilde{b}_{i}=\widetilde{s}\widetilde{c}_{i}=0\;\;\;\forall
i.\\\ $

Let us consider

$\widetilde{g}_{1}=y_{1}y_{2}y_{3}w_{1},\;\;\widetilde{g}_{2}=y_{4}y_{5}y_{6}w_{2}.$

Since $v_{1}\cap w_{1}=v_{2}\cap w_{1}=v_{3}\cap w_{1}=v_{4}\cap
w_{2}=v_{5}\cap w_{2}=v_{6}\cap w_{2}=\emptyset$, we have

$\widetilde{a}_{i}\widetilde{g}_{1}=\widetilde{b}_{i}\widetilde{g}_{1}=\widetilde{b}_{i}\widetilde{g}_{2}=\widetilde{b}_{i}\widetilde{g}_{2}=0\;\;\;\forall
i.$

Let us study the face ring of $Q=P^{(k,\ldots,2k,2k)}$. Theorem 2.21 says that
$H^{*}(\widetilde{\mathcal{Z}}_{P},\mathbb{Z})$ and
$H^{*}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z})$ are isomorphic as ungraded
rings. We denote the variables of $[R(Q),d]$ by
$v_{1,1},\ldots,v_{1,k},\ldots,v_{6,1},\ldots,v_{6,k}$,
$v_{7,1},\ldots,v_{7,2k},v_{8,1},\ldots,v_{8,2k}$. From Lemma 2.22 and from
the compuations above we get that

$\begin{gathered}a_{1}=y_{1,1}v_{1,2}\ldots v_{4,1}\ldots
v_{4,k},\;\;a_{2}=y_{2,1}v_{2,2}\ldots v_{5,1}\ldots v_{5,k},\\\
a_{3}=y_{3,1}v_{3,2}\ldots v_{6,1}\ldots
v_{6,k},\;\;b_{1}=y_{1,1}t_{2,1}v_{1,2}\ldots
v_{1,k}w_{2,2}\ldots,w_{2,2k}v_{4,1}\ldots v_{4,k},\\\
b_{2}=y_{2,1}t_{2,1}v_{2,2}\ldots v_{2,k}w_{2,2}\ldots,w_{2,2k}v_{5,1}\ldots
v_{5,k},\\\ b_{3}=y_{3,1}t_{2,1}v_{3,2}\ldots
v_{3,k}w_{2,2}\ldots,w_{2,2k}v_{6,1}\ldots v_{6,k},\\\
c_{1}=y_{4,1}t_{1,1}v_{4,2}\ldots v_{4,k}w_{1,2}\ldots,w_{1,2k}v_{1,1}\ldots
v_{1,k},\\\ c_{2}=y_{5,1}t_{1,1}v_{5,2}\ldots
v_{5,k}w_{1,2}\ldots,w_{1,2k}v_{2,1}\ldots v_{2,k},\\\
c_{3}=y_{6,1}t_{1,1}v_{6,2}\ldots v_{6,k}w_{1,2}\ldots,w_{1,2k}v_{3,1}\ldots
v_{3,k},\\\ g_{1}=y_{1,1}y_{2,1}y_{3,1}v_{1,2}\ldots
v_{1,k}v_{2,2}\ldots,v_{2,k}v_{3,2}\ldots,v_{3,k}w_{1,1},\ldots,w_{1,2k},\\\
g_{2}=y_{4,1}y_{5,1}y_{6,1}v_{4,2}\ldots
v_{4,k}v_{5,2}\ldots,v_{5,k}v_{6,2}\ldots,v_{6,k}w_{2,1},\ldots,w_{2,2k},\\\
s=t_{2,1}w_{2,2}\ldots w_{2,2k}w_{1,1}\ldots w_{1,2k}.\end{gathered}$

represent nontrivial elements. Moreover,

$\begin{gathered}a_{1},a_{2},a_{3}\;\in
H^{4k-1}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}),\\\
b_{1},b_{2},b_{3},c_{1},c_{2},c_{3}\;\in
H^{8k-2}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}),\\\ g_{1},g_{2}\;\in
H^{10k-3}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}),\\\
a_{1}a_{2},\;a_{1}a_{3},\;a_{2}a_{3}\neq 0,\;\;\;a_{1}a_{2}a_{3}=0,\\\
a_{i}a_{j}b_{m},\;a_{i}a_{j}c_{m}\neq 0\;\;\;if\;and\;only\;if\;\;i\neq
j,i\neq m,j\neq m,\\\
a_{i}g_{1}=b_{i}g_{1}=b_{i}g_{2}=b_{i}g_{2}=0,\;sa_{i}=sb_{i}=0\;\;\;\forall
i.\end{gathered}$ (5.23)

######  Lemma 5.19.

Assume that $r\leqslant 12k-3$, and $r\neq 0,4k-1,6k-1,8k-2,8k-1,10k-3,12k-3$.
If $k\geqslant 2$, then we have:
$H^{r}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z})=0$;
$H^{4k-1}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z})$ is generated by
$a_{1},a_{2},a_{3}$;
$H^{8k-1}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z})$ is generated by $s$;
$H^{10k-3}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z})$ is generated by
$g_{1},g_{2}$.

###### Proof.

We need to consider all possible combinations of the variables of $[R(Q),d]$.
For example, $deg(t_{2}w_{2,2}\ldots w_{2,2k}w_{1,1}\ldots
w_{1,2k}v_{5,1}\ldots v_{5,k})=10k-1$ but

$\begin{gathered}d(y_{5,1}t_{2,1}w_{2,2}\ldots w_{2,2k}w_{1,1}\ldots
w_{1,2k}\ldots v_{5,k})=\\\ t_{2}w_{2,2}\ldots w_{2,2k}w_{1,1}\ldots
w_{1,2k}v_{5,1}\ldots v_{5,k}\end{gathered}$

because $w_{1,1}\cap\ldots w_{1,2k}\cap w_{2,1}\cap\ldots w_{2,2k}=\emptyset$.
This means that the considered element is exact and represent trivial element
in the cohomology group. Arguing in the same way we can prove the lemma. ∎

Since $\widetilde{\mathcal{Z}}_{Q}$ does not have torsion we get that

$H^{*}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}_{2})=H^{*}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z})\otimes\mathbb{Z}_{2}.$

Let us denote elements in $H^{*}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}_{2})$
corresponding to $a_{i},b_{i},c_{i},g_{i},s$ by the same symbols. Denote

$\mathcal{Z}_{Q}=\widetilde{\mathcal{Z}}_{Q}/\mathbb{Z}_{2}.$

We have the Cartan-Leray spectral sequence

$E_{2}^{p,q}=H^{p}(\mathbb{R}P^{\infty},H^{q}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}_{2}))\Rightarrow
H^{*}(\mathcal{Z}_{Q},\mathbb{Z}_{2}),$

Denote the differential of $E_{k}$ by $\partial_{k}$. Recall that

$m(Q)=min\\{\ell\in\mathbb{N}\;|\;v_{i_{1}}\cap\ldots\cap
v_{i_{\ell}}=\emptyset\\}.$

We see that $m(P)=2$. It follows from Theorem 2.22 that

$m(Q)=m(P^{(k,\ldots,k,2k,2k)})=2k.$

Let $\beta$ be the generator of
$H^{*}(\mathbb{R}P^{\infty},\mathbb{Z}_{2})=E_{2}^{*,0}$. Let us show that
$\partial_{4k}(E_{4k}^{0,4k-1})=0$, First, note that
$E_{2}^{0,4k-1}=E_{4k}^{0,4k-1}$. Then, from formulas
$(\ref{widenarrowrelations})$ we have (we work over $\mathbb{Z}_{2}$)

$\begin{gathered}a_{1}a_{2}a_{3}=0\Rightarrow\\\
0=\partial_{4k}(a_{1}a_{2}a_{3})=\partial_{4k}(a_{1})a_{2}a_{3}+\partial_{4k}(a_{2})a_{1}a_{3}+\partial_{4k}(a_{1})a_{2}a_{3},\\\
\partial_{4k}(a_{1}),\partial_{4k}(a_{1}),\partial_{4k}(a_{1})\in\\{0,\beta^{4k}\\}.\end{gathered}$
(5.24)

If $\beta^{4k}a_{1}a_{2},\beta^{4k}a_{1}a_{3},\beta^{4k}a_{2}a_{3}$ are
nontrivial in $E_{4k}^{4k,8k-2}$, then $\partial_{4k}(a_{i})=0$ for all $i$.

The second differential $\partial_{2}:E_{2}^{0,8k-1}\rightarrow
E_{2}^{2,8k-2}$ may be nontrivial. By Lemma 5.19,
$H^{8k-1}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}_{2})$ is generated by $s$.
Assume that

$\partial_{2}(s)=\lambda_{1}\beta^{2}a_{1}a_{2}+\lambda_{2}\beta^{2}a_{1}a_{3}+\lambda_{3}\beta^{2}a_{2}a_{3}+\ldots,\quad\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{Z}_{2}.$

Since $b_{1}s=0$ and $b_{1}$ represents the cohomology class of
$\mathcal{G}(\mathcal{G}(S^{3}\times S^{3}\times S^{3}))$, we have

$\begin{gathered}\partial_{2}(b_{1})\in E_{2}^{2,8k-3}=0\Rightarrow\\\
0=\partial_{2}(b_{1}s)=b_{1}\partial_{2}(s)=\lambda_{3}\beta^{2}b_{1}a_{2}a_{3}\Rightarrow\lambda_{3}=0.\end{gathered}$

In the same way, using $b_{2},b_{3}$ we can prove that
$\lambda_{1}=\lambda_{2}=0$. So, $\partial_{2}$ does not kill
$\beta^{4k}a_{i}a_{j}$. Since $\partial_{2}(a_{i})=0$, we get that
$\partial_{2}(a_{i}a_{j})=0$ for any $i,j$. This means that elements
$a_{i}a_{j}$ are not trivial in $E_{3}$.

It follows from Lemma 5.19 that $E_{3}^{4k,8k-2}=E_{2k}^{4k,8k-2}$. Since
$H^{10k-3}(\widetilde{\mathcal{Z}}_{Q},\mathbb{Z}_{2})=E_{2k}^{0,10k-3}$ is
generated by $g_{1},g_{2}$ and $b_{i}g_{j}=0$ for all $i,j$, we have (recall
that $b_{i}\in H^{*}(\mathcal{G}(\mathcal{G}(S^{3}\times S^{3}\times
S^{3})),\mathbb{Z}_{2})$)

$\begin{gathered}\partial_{2k}(g_{1})=\lambda_{1}\beta^{2k}a_{1}a_{2}+\lambda_{2}\beta^{2k}a_{1}a_{3}+\lambda_{3}\beta^{2k}a_{2}a_{3}+\ldots,\quad\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{Z}_{2},\\\
0=\partial_{2}(b_{1}g_{1})=b_{1}\partial_{2}(g_{1})=\lambda_{3}\beta^{2k}b_{1}a_{2}a_{3}\Rightarrow\lambda_{3}=0.\end{gathered}$

Analogously, $\lambda_{1}=\lambda_{2}=0$. Since $\partial_{2k}(a_{i})\in
E_{2k}^{2k,2k}=0$, we get that $\partial_{2k}(a_{i}a_{j})=0$. This yields that
$\partial_{2k}$ does not kill $\beta^{4k}a_{i}a_{j}$.

Finally, we see that $a_{1}a_{2},a_{1}a_{3},a_{2}a_{3}$ are nontrivial in
$E_{4k}^{0,8k-1}$. So, we get from formulas $(\ref{vanishprovethe})$ that

$\partial_{4k}(E_{4k}^{0,4k-1})=0.$

It follows from Theorem 1.3 that $QH(L,\mathbb{Z}_{2}[T,T^{-1}])=0$.

Let $G$ be a ring in which $2$ is invertible. Lemma 1.3 says that
$QH(L,G[T,T^{-1}])=0$.

###  5.8 Proof of Theorem 1.8

Let $P$ be a $6-$gon defined by inequalities

$P=\left\\{\begin{array}[]{l}x_{1}+1\geqslant 0\;\;\;x_{2}+1\geqslant 0\\\
-x_{1}+1\geqslant 0\;\;\;-x_{2}+1\geqslant 0\\\ x_{1}+x_{2}+1\geqslant
0\;\;\;-x_{1}-x_{2}+1\geqslant 0\end{array}\right.$

The real moment-angle manifold associated to $P$ is given by

$\widetilde{\mathcal{R}}_{P}=\left\\{\begin{array}[]{l}u_{1}^{2}+u_{3}^{2}=2\\\
u_{2}^{2}+u_{4}^{2}=2\\\ u_{1}^{2}+u_{2}^{2}-u_{5}^{2}=1\\\
u_{1}^{2}+u_{2}^{2}+u_{6}^{2}=3\end{array}\right.$ (5.25)

The manifold $\widetilde{\mathcal{R}}_{P}=S_{17}$, where $S_{17}$ is an
orientable surface of genus $17$ (see [11] Proposition $4.1.8$). The sum of
normal vectors to the facets is equal to zero. Moreover, $P$ is Fano and
Delzant. From Section 4.2 and Lemma 4.6 we get that there is monotone embedded
Lagrangian $L\subset\mathbb{C}P^{6}$ associated to $P$. By our construction
$L$ fibers over $T^{3}$

$\begin{gathered}L\xrightarrow{S_{17}/\mathbb{Z}_{2}}T^{3},\\\
(u_{1},\ldots,u_{6})\sim(-u_{1},\ldots,-u_{6}),\quad(u_{1},\ldots,u_{6})\in
S_{17}.\end{gathered}$

From the exact sequence of the fibration we have

$1\rightarrow\pi_{1}(S_{17}/\mathbb{Z}_{2})\rightarrow\pi_{1}(L)\rightarrow\pi_{1}(T^{3})\rightarrow
1.$

###  5.9 Proof of Theorem 1.10

We argue as in the previous theorems. Let us consider a submanifold
$\widetilde{\mathcal{R}}$ of $\mathbb{R}^{n}$ defined by a system

$\begin{gathered}\widetilde{\mathcal{R}}=\left\\{\begin{array}[]{l}\sum\limits_{r=1}^{k}u_{r}^{2}+\sum\limits_{r=k+1}^{p_{1}}u_{r}^{2}=p_{1}\\\
\sum\limits_{r=1}^{k}u_{r}^{2}+\sum\limits_{r=p_{1}+1}^{p_{2}}u_{r}^{2}=p_{2}-p_{1}+k\\\
-\sum\limits_{r=1}^{k}u_{r}^{2}+\sum\limits_{r=p_{2}+1}^{n}u_{r}^{2}=n-p_{2}-k\\\
\end{array}\right.\\\ k<p_{1}-1,\quad p_{2}-p_{1}+k>p_{1},\quad
n-p_{2}-k>p_{1},\\\ p_{1},\;p_{2}-p_{1},\;n-p_{2}>2\end{gathered}$ (5.26)

Since $p_{2}-p_{1}+k>p_{1}$ and $n-p_{2}-k>p_{1}$, we have a diffeomorphism

$\begin{gathered}f:\widetilde{\mathcal{R}}\rightarrow S^{p_{1}-1}\times
S^{p_{2}-p_{1}-1}\times S^{n-p_{2}-1},\\\
f(u_{1},\ldots,u_{n})=\bigg{(}u_{1},\ldots,u_{p_{1}},\frac{u_{p_{1}+1}}{\sqrt{p_{2}-p_{1}+k-\sum\limits_{r=1}^{k}u_{r}^{2}}},\ldots,\frac{u_{p_{2}}}{\sqrt{p_{2}-p_{1}+k-\sum\limits_{r=1}^{k}u_{r}^{2}}},\\\
\frac{u_{p_{2}+1}}{\sqrt{n-p_{2}-k+\sum\limits_{r=1}^{k}u_{r}^{2}}},\ldots,\frac{u_{n}}{\sqrt{n-p_{2}-k+\sum\limits_{r=1}^{k}u_{r}^{2}}}\bigg{)}.\end{gathered}$

Denote by $\widetilde{\Lambda}$ the lattice generated by columns of
$(\ref{threesphereeq})$. Direct computation shows that
$\widetilde{\Lambda}_{u}=\widetilde{\Lambda}$ and conditions of Lemma 4.4 are
satisfied (for $C=1$). This means that the polytope $P$ associated to system
$(\ref{threesphereeq})$ is Delzant and Fano. Theorem 4.12 says that there
exists an embedded monotone Lagrangian $L\subset\mathbb{C}P^{n-1}$ associated
to the polytope $P$.

If $k,p_{1},p_{2},n$ are even, then from Lemma 5.6 we get that

$L=(S^{p_{1}-1}\times S^{p_{2}-p_{1}-1}\times
S^{n-p_{2}-1})/\mathbb{Z}_{2}\times T^{2}.$

It follows from Lemma 5.5 that the minimal Maslov number

$N_{L}=gcd(p_{1},p_{2}-p_{1}+k,n-p_{2}-k).$

##  6 From symplectic geometry to algebraic topology

In this section we explain how to use symplectic geometry and Lagrangian
quantum cohomology to study the singular cohomology of manifolds. We plan to
prove a general theorem in a further paper. In this Section we consider an
example and show how the new method works.

Let $\mathcal{R}\subset\mathbb{R}P^{n-1}$ be a smooth manifold defined by a
system of quadrics

$\begin{gathered}\mathcal{R}=\left\\{\begin{array}[]{l}\gamma_{1,r}u_{1}^{2}+...+\gamma_{n,r}u_{n}^{2}=0,\quad
r=1,...,n-m-1.\end{array}\right.\\\
[u_{1}:\ldots:u_{n}]\in\mathbb{R}P^{n-1}.\end{gathered}$

We would like to know $H^{*}(\mathcal{R},\mathbb{Z}_{2})$. In general, the
topology of $\mathcal{R}$ can be highly complicted (see [25] for examples).
The Agrachev-Lerario spectral sequence( see [1]) can be used to study the
Betti numbers of $\mathcal{R}$. Unfortunately, the Agrachev-Lerario spectral
sequence often requires extremely heavy computations.

The manifold $\mathcal{R}$ can be considered as a real toric space. There are
explicit formulas for $H^{*}(\mathcal{R},\mathbb{Q})$ (see [16]), but formula
for $H^{*}(\mathcal{R},\mathbb{Z}_{2})$ is not known.

Let us study a manifold $\widetilde{\mathcal{R}}\subset\mathbb{R}^{n}$ given
by

$\begin{gathered}\widetilde{\mathcal{R}}=\left\\{\begin{array}[]{l}u_{1}^{2}+\ldots+u_{n}^{2}=1\\\
\gamma_{1,r}u_{1}^{2}+...+\gamma_{n,r}u_{n}^{2}=0,\quad
r=1,...,n-m-1.\end{array}\right.\\\
(u_{1},\ldots,u_{n})\in\mathbb{R}^{n}.\end{gathered}$

Easy to see that

$\mathcal{R}=\widetilde{\mathcal{R}}/\mathbb{Z}_{2},\quad(u_{1},\ldots,u_{n})\sim(-u_{1},\ldots,-u_{n}).$

Since the Eilenberg-MacLane space $K(\mathbb{Z}_{2},1)=\mathbb{R}P^{\infty}$,
we have the Cartan-Leray spectral sequence (see [41, p. 206])

$E_{2}^{p,q}=H^{p}(\mathbb{R}P^{\infty},H^{q}(\widetilde{\mathcal{R}},\mathbb{Z}_{2}))\Rightarrow
H^{*}(\mathcal{R},\mathbb{Z}_{2}).$

In general, it is not easy to find $E_{\infty}$. Moreover, $E_{\infty}$ does
not give enough information to find the singular cohomology ring of
$\mathcal{R}$. So, we need some luck or extra information to get
$H^{*}(\mathcal{R},\mathbb{Z}_{2})$ from $E_{\infty}$.

From Section 2.1 we know that there is a polytope $P$ associated to
$\widetilde{\mathcal{R}}$. From Section 4.2 we know that there is a Lagrangian
$L\subset\mathbb{C}P^{n-1}$ associated to the polytope $P$. If $P$ is Delzant
and Fano, then the Lagrangian is embedded and monotone. Therefore,
$QH^{*}(L,\mathbb{Z}_{2}[T,T^{-1}])$ is defined. We have the spectral sequence
of Oh, where the first page is defined by $H^{*}(L,\mathbb{Z}_{2})$. Since the
spectral sequence of Oh is multiplicative and converges to
$QH^{*}(L,\mathbb{Z}_{2}[T,T^{-1}])$, we get some restrictions on
$\mathcal{R}$ (because $L$ is constructed using $\mathcal{R}$). It follows
from Lemma 1.2 that for a rich set of polytopes we have

$L=\mathcal{R}\times T^{n-m-1}.$

If $N_{L}>2$, then we can “ignore” the torus in the spectral sequence of Oh
and work with $H^{*}(\mathcal{R},\mathbb{Z}_{2})$. Theorem 1.3 says that
$QH^{*}(L,\mathbb{Z}_{2}[T,T^{-1}])=0$ for some set of polytopes. Hence, the
spectral sequence of Oh converges to zero and this gives us some extra
information for studying $E_{\infty}$ and $H^{*}(\mathcal{R},\mathbb{Z}_{2})$.

Remark. We consider Delzant and Fano polytopes, but as we can see from proofs
of the previous theorems many polytopes can be made Fano by applying
simplicial multiwedge operation. So, we can make the polytope Fano, study
cohomology ring of its real toric space, and use Theorem 2.21 to study the
cohomology ring of the original real toric space.

First, let us consider a simple artificial example. Assume that
$L=\mathcal{R}\times S^{1}$, where $dim(M)=6$, $N_{L}=4$, and
$QH(L,\mathbb{Z}_{2}[T,T^{-1}])=0$. Assume that $E_{\infty}$ page of the
Cartan-Leray spectral sequence has the form shown in the figure 7 Then,
$y^{2}$ equals either $yx^{3}$, or $0$. We see that $H^{*}(L,\mathbb{Z}_{2})$
is generated by $x,y,c$, where $c\in H^{1}(S^{1},\mathbb{Z}_{2})$. The
spectral sequence of Oh is defined in the end of Section 3. Let $\delta_{1}$
be the differential on the first page of Oh’s spectral sequence. Since
$N_{L}=4$, we get that the degree of $\delta_{1}$ equals $3$ and
$\delta_{1}(x)=\delta_{1}(c)=0$. If $y^{2}=x^{3}y$, then

$0=2\delta_{1}(y)=\delta_{1}(y^{2})=\delta_{1}(x^{3}y)=x^{3}\delta_{1}(y)\;\;\Rightarrow\;\;\delta_{1}(y)=0.$

This yields that $QH(L,\mathbb{Z}_{2}[T,T^{-1}])$ is isomorphic to
$H^{*}(L,\mathbb{Z}_{2})$, but we know that
$QH(L,\mathbb{Z}_{2}[T,T^{-1}])=0$. Hence, $y^{2}=0$.

Figure 7:  $E_{\infty}$ page

Let us consider a real example. In Section 1.5 we constructed the Lagrangian
(see Lemma 5.16)

$\begin{gathered}L=\mathcal{R}_{P}\times
T^{3},\quad\mathcal{R}_{P}=\widetilde{\mathcal{R}}_{P}/\mathbb{Z}_{2},\\\
H^{*}(\widetilde{\mathcal{R}}_{P},\mathbb{Z}_{2})\cong
H^{*}(\\#9(S^{4k-1}\times S^{8k-3})\\#8(S^{6k-2}\times
S^{6k-2}),\;\mathbb{Z}_{2}),\\\
QH^{*}(L,\mathbb{Z}_{2}[T,T^{-1}])=0.\end{gathered}$

In the end of Section 1.5 we show that $E_{\infty}$ page of the Cartan-Leray
spectral sequence has one of the following forms (see the figures below).

Figure 8:  $E_{\infty}$ page  Figure 9:  $E_{\infty}$ page Figure 10:
$E_{\infty}$ page

It is not clear how to determine $E_{\infty}$ page using standard methods of
algebraic topology. Let us show that the spectral sequence of Oh helps to find
$E_{\infty}$ page and provides some information about
$H^{*}(\mathcal{R}_{P},\mathbb{Z}_{2})$.

######  Lemma 6.1.

Assume that $k\geqslant 3$. Then, Figure 8 is the $E_{\infty}$ page of the
Cartan-Leray spectral sequence.

Remark. The goal of this Lemma is to demonstrate how the spectral sequence of
Oh can be applied to algebraic topology. The rigorous proof of the lemma above
is long. We give general ideas and skip all details.

_Sketch proof._ Denote the nontrivial element of $E_{\infty}^{1,0}$ by
$\alpha$. Let us recall that

$L=\mathcal{R}_{P}\times T^{3}.$

Let $c_{1},c_{2},c_{3}$ be generators of $H^{1}(T^{3},\mathbb{Z}_{2})$.

Let us consider Figure 10. Denote generators of $E_{\infty}^{0,4k-1}$ by
$x_{1},\ldots,x_{9}$, generators of $E_{\infty}^{0,6k-2}$ by
$y_{1},\ldots,y_{8}$.

$\begin{gathered}E_{\infty}^{m,6k-2}=\mathbb{Z}_{2}\langle\alpha^{m}y_{1},\ldots,\alpha^{m}y_{8}\rangle,\quad
m\leqslant 2k-1,\\\
E_{\infty}^{m,4k-1}=\mathbb{Z}_{2}\langle\alpha^{m}x_{1},\ldots,\alpha^{m}x_{9}\rangle,\quad
m\leqslant 2k-1,\\\
E_{\infty}^{r,4k-1}=\mathbb{Z}_{2}\langle\alpha^{r}x_{1}\rangle,\quad
2k\leqslant r\leqslant 4k-2,\\\
E_{\infty}^{\ell,0}=\mathbb{Z}_{2}\langle\alpha^{\ell}\rangle,\quad
0\leqslant\ell\leqslant 12k-4.\end{gathered}$

Let us denote elements of $H^{*}(\mathcal{R},\mathbb{Z}_{2})$ by the same
symbols.

Denote the differentials of the spectral sequence of Oh by $\delta_{k}$. Since
the minimal Maslov number $N_{L}=2k$, we get that the differential
$\delta_{1}$ has degree $2k-1$.

We see that $\alpha^{6k-2}x_{1}$ equals either $\alpha^{10k-3}$, or $0$. Since
$k>1$ we have $\delta_{1}(\alpha)=\delta_{1}(\alpha^{\ell})=0$. Therefore
(here we should say more to make the expression below rigorous)

$0=\delta_{1}(\alpha^{6k-2}x_{1})=\alpha^{6k-2}\delta_{1}(x_{1}).$

where
$\delta_{1}(x_{1})\in\mathbb{Z}_{2}\langle\alpha^{2k-2},\alpha^{2k-3}c_{i},\alpha^{2k-4}c_{i}c_{j},\alpha^{2k-5}c_{1}c_{2}c_{3}\rangle$,
$i,j=1,\ldots,3$. We see that

$\alpha^{6k-2}\delta_{1}(x_{1})=0\Leftrightarrow\delta_{1}(x_{1})=0.$

We get that $\delta_{1}(x_{1})=0$. Analogously, let $y\in
H^{6k-2}(L,\mathbb{Z}_{2})$ be an element such that $\delta_{1}(y)=x_{1}$.
Then, $\alpha^{4k-2}y$ equals either $\alpha^{10k-4}$, or $0$. We get

$0=\delta_{1}(\alpha^{4k-2}y)=\alpha^{4k-2}\delta_{1}(y)=\alpha^{4k-2}x_{1}\neq
0.$

This contradiction proves that there is no $y$ such that
$\delta_{1}(y)=x_{1}$. Hence, $x_{1}$ survives and is nontrivial element in
the second page of Oh’s spectral sequence.

Arguing as before, we can show that $x_{1}$ survives on all other pages, but
we know that $QH^{*}(L,\mathbb{Z}_{2}[T,T^{-1}])=0$ and the spectral sequence
of Oh converges to zero. Therefore, Figure 10 is not $E_{\infty}$ page of
$\mathcal{R}$.

In the same way we can show that Figure 9 is not $E_{\infty}$ page of
$\mathcal{R}$.

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Department of Mathematics and Statistics, University of Montreal,
C.P. 6128 Succ. Centre-Ville Montreal, QC, H3C 3J7, Canada.
Department of Mathematics and Mechanics, Novosibirsk State University,
Novosibirsk, 630090, Russia.
E-mail address<EMAIL_ADDRESS>


# Path integral approach to theories of diffusion-influenced reactions

Thorsten Prüstel  Laboratory of Systems Biology
National Institute of Allergy and Infectious Diseases
National Institutes of Health  Martin Meier-Schellersheim  Laboratory of
Systems Biology
National Institute of Allergy and Infectious Diseases
National Institutes of Health

###### Abstract

The path decomposition expansion represents the propagator of the irreversible
reaction as a convolution of the first-passage, last-passage and rebinding
time probability densities. Using path integral technique, we give an
elementary, yet rigorous, proof of the path decomposition expansion of the
Green’s functions describing the non-reactive case and the irreversible
reaction of an isolated pair of molecules. To this end, we exploit the
connection between boundary value problems and interaction potential problems
with $\delta$\- and $\delta^{\prime}$-function perturbation. In particular, we
employ a known exact summation of a perturbation series to derive exact
relations between the Green’s functions of the perturbed and unperturbed
problem. Along the way, we are able to derive a number of additional exact
identities that relate the propagators describing the free-space, the non-
reactive as well as the completely and partially reactive case.

11footnotetext: Email<EMAIL_ADDRESS><EMAIL_ADDRESS>
##  1 Introduction

The transition probability density function (PDF) $p(x,t|x_{0})$ is the
fundamental dynamical quantity that describes the diffusion-influenced
reaction of an isolated pair of molecules. Traditionally, $p(x,t|x_{0})$
appears as Green’s function (GF) of the diffusion equation supplemented by
suitable boundary conditions (BC) imposed at the encounter distance $x=a$ that
implement the actual reaction [34, 8, 12, 13, 28, 2, 20, 23]. However,
imposing BC are not the only way to implement reactions. Alternatively, sink
terms may be added to the diffusion equation. It has been discussed that,
under certain conditions, sink terms are equivalent to a description in terms
of BC, but they provide a more general and more flexible approach than the
traditional one [35, 9, 10, 22, 19, 24, 25, 1]. Also, they offer another
adavantage: The diffusion equation with sink terms assumes a form that is
quite similar to the Schroedinger equation. In fact, it is well-known that
both equations are related by a Wick-transformation, that is by $t\rightarrow-
it$ [29, 5, 30]. In this way, methods that have been used in a quantum-
mechanical context, can be borrowed to study the theory of diffusion-
influenced reactions. A case in point is the path decomposition expansion
(PDX), which was originally employed to study quantum-mechanical problems [4,
18, 33, 30, 36, 37]. Recently, it has been shown that in the context of
diffusion-influenced reactions the GF of the irreversible reaction can be
factorized in the Laplace domain as a product of the first-, last-passage and
rebinding time probability densities [26]. In this manuscript, we present a
straightforward, yet rigorous, proof of the PDX of the propagators that play
an important role in the theory of diffusion-influenced reactions. The proof
is based on another technique imported from quantum mechanics. In a series of
papers [14, 15, 16], Grosche discussed the implementation of Dirichlet and
Neumann BC in terms of interaction potentials that are $\delta$\- and
$\delta^{\prime}$-function perturbed. By using an exact summation of the
expansion of the corresponding path integral, an exact convolution relation
between the perturbed and unperturbed GF can be established. Because Dirichlet
and Neumann BC correspond to absorbing (completely reactive) and reflecting
(non-reactive) BC in the theory of diffusion-influenced reactions, the Wick-
rotated (Euclidean) version of Grosche’s results can be immediately applied to
the case of an isolated pair. By considering different choices for the
unperturbed potential, we can derive several important relations between the
GF corresponding to various BC, which are far from obvious from a differential
equation point of view [3]. With these relations at our disposal, the actual
proof of the PDX becomes rather accessible. We point out that some of the
relations that we will derive have been obtained before by using the general
form of the GF of the Smoluchowski equation [3]. Alternative proofs of the PDX
have been given in Refs. [4, 18, 33, 17]. The application of quantum field
theoretical methods to diffusion-limited reactions and path integral
techniques to stochastic differential equation are reviewed in Ref. [21] and
[6], respectively. The manuscript is structured as follows. First, we will
briefly describe the traditional BC approach and introduce the alternative
formulation given in terms of sink potentials. Then, we will briefly recall
the PDX and summarize Grosche’s results. Equipped with those, we will derive
various convolution relations, from which the PDX results. In the following,
we will consider an isolated pair in one dimension (1D) for convenience. We
will study the 2D and 3D cases in a forthcoming manuscript.

##  2 Theory

###  2.1 Boundary vs potential problems

The conditional PDF $p(x,t|x_{0})$, which yields the probability to find the
molecule at position $x$ at time $t$, given that it was initially localized at
$x_{0}$, is the GF of the diffusion equation

$\frac{\partial p(x,t|x_{0})}{\partial
t}=D\frac{\partial^{2}p(x,t|x_{0})}{\partial x^{2}},$ (1)

subject to the initial condition (IC)

$p(x,t=0|x_{0})=\delta(x-x_{0}).$ (2)

The traditional, Smoluchowski-Collins-Kimball (SCK) approach [34, 8, 13, 28]
to incorporate the actual binding reaction, amounts to impose a BC at the
encounter distance $x=a$. The Collins-Kimball, or radiation BC reads [8, 13,
28, 2, 20, 23]

$D\frac{\partial p(x,t|x_{0})}{\partial x}|_{x=a}=\kappa_{a}p(x=a,t|x_{0}).$
(3)

The absorbing BC is recovered for $\kappa_{a}\rightarrow\infty$, while the
non-reactive, reflective BC is obtained for $\kappa_{a}=0$. We will denote the
GF corresponding to free-space, absorbing, reflecting and radiation BC as
$p_{\text{free}}(x,t|x_{0})$, $p_{\text{abs}}(x,t|x_{0})$,
$p_{\text{ref}}(x,t|x_{0})$ and $p_{\text{rad}}(x,t|x_{0})$, respectively.

Instead of imposing BC at the encounter distance, the actual binding reaction
may alternatively be taken into account by introducing sink terms in the
diffusion equation. For instance, under certain conditions, the radiation BC
may be recovered from the diffusion equation with an additional
$\delta$-function interaction potential [35]

$\frac{\partial p(x,t|x_{0})}{\partial
x}=D\frac{\partial^{2}p(x,t|x_{0})}{\partial x^{2}}-\gamma\delta(x-a).$ (4)

Note that Eq. (4) does not necessarily come with any BC imposed at the
encounter distance. Also, we emphasize that the solution of Eq. (4) is not
identical to the solution of the radiation BC problem [32], unless one makes
additional assumptions [35]. One obvious difference is that the GF solution to
Eq. (4) is also defined for $x<a$ and $x_{0}<a$, in contrast to
$p_{\text{rad}}(x,t|x_{0})$. We will use superscripts $p_{\delta}^{++}$ to
refer to the solution of Eq. (4) for $x,x_{0}>a$. Note that even
$p_{\delta}^{++}$ is not equal to $p_{\text{rad}}$, although both are defined
on the same domain $x,x_{0}>a$ [32].

###  2.2 Path decomposition expansion

The PDX [4, 18, 30] allows to break space into different regions and to study
the individual contributions from the propagators restricted to one region to
the complete propagator across all regions. While the decomposition of space
may be chosen arbitrarily, the existence of a non-vanishing potential may
suggest a natural decomposition. This is in particular true for a $\delta$\-
and Heaviside step-function potential. These potentials are of special
interest in the theory of diffusion-influenced reactions, because they
correspond to the SCK and Doi model, respectively [9, 10, 19, 25, 1]. It has
been shown by explicit calculation that the PDX indeed is satisfied by the
associated propagators (in 1D and 2D) [26]. Also, it was pointed out that in
the context of theories of diffusion-influenced reactions, the PDX corresponds
to the convolution of three prominent time PDF: the first- and last-passage as
well as the rebinding time PDF [26]. Furthermore, the PDX has been employed in
a stochastic simulation algorithm that does not require explicit analytical
representations of the exact GF to propagate an isolated pair correctly [27].
Here, we shall prove the following form of the PDX

$\tilde{p}(x,s|x_{0})=\tilde{p}_{\text{abs}}(x,s|x_{0})+D^{2}\frac{\partial\tilde{p}_{\text{abs}}(x,s|\xi)}{\partial\xi}\bigg{|}_{\xi=a}\tilde{p}(a,s|a)\frac{\partial\tilde{p}_{\text{abs}}(\xi,s|x_{0})}{\partial\xi}\bigg{|}_{\xi=a},$
(5)

where $\tilde{p}(x,s|x_{0})$ refers either to
$\tilde{p}_{\delta}^{++}(x,s|x_{0})$, $\tilde{p}_{\text{ref}}(x,s|x_{0})$ or
$\tilde{p}_{\text{rad}}(x,s|x_{0})$. The proof we shall present here is based
on an exactly summable perturbative expansion of a path integral that gives
the propagator associated with $\delta$\- and $\delta^{\prime}$-function
perturbed interaction potentials. In the following, we will always consider
$x,x_{0}\geq a$. Also, without loss of generality, we may assume that $a=0$.

###  2.3 Path integral representation of $\delta$-function perturbed
potentials

Following Grosche [14, 15, 16], we consider a potential $W(x)$ comprised of an
arbitrary potential $V(x)$ that is perturbed by a $\delta$-function term:

$W(x)\equiv V(x)+\mathcal{V}(x)\equiv V(x)-\gamma\delta(x-a).$ (6)

The corresponding GF enjoys the following path integral representation

$p_{W}(x,t|x_{0})=\int\mathcal{D}x(\tau)e^{-\int^{t}_{0}[\frac{1}{4D}\dot{x}^{2}(\tau)+W(x)]d\tau},$
(7)

while the GF for the unperturbed problem is given by

$p_{V}(x,t|x_{0})=\int\mathcal{D}x(\tau)e^{-\int^{t}_{0}[\frac{1}{4D}\dot{x}^{2}(\tau)+V(x)]d\tau}.$
(8)

The path integral can be expanded in a perturbation series as follows

$\displaystyle p_{W}(x,t|x_{0})$ $\displaystyle=$ $\displaystyle
p_{V}(x,t|x_{0})+\sum^{\infty}_{n=1}\gamma^{n}\int^{t}_{0}d\tau^{(n)}\int^{\tau^{(n)}}_{0}d\tau^{(n-1)}\ldots\int^{\tau^{(2)}}_{0}d\tau^{(1)}$
(9) $\displaystyle\times$ $\displaystyle p_{V}(a,\tau^{(1)}|x_{0})\ldots
p_{V}(a,\tau^{(n)}-\tau^{(n-1)}|a)p_{V}(x,t-\tau^{(n)}|a).$

The form of this expansion suggests to use the Laplace transformed GF
$\tilde{p}(x,s|x_{0})$, which is defined by

$\tilde{p}(x,s|x_{0})=\int^{\infty}_{0}e^{-st}p(x,t|x_{0})dt.$ (10)

Then, by virtue of the convolution theorem of the Laplace tranform, one
obtains in the Laplace domain

$\displaystyle\tilde{p}_{W}(x,s|x_{0})$ $\displaystyle=$
$\displaystyle\tilde{p}_{V}(x,s|x_{0})+\sum^{\infty}_{n=1}\gamma^{n}\tilde{p}_{V}(x,s|a)\tilde{p}_{V}(a,s|x_{0})[\tilde{p}_{V}(a,s|a)]^{n-1}$
(11) $\displaystyle=$
$\displaystyle\tilde{p}_{V}(x,s|x_{0})+\gamma\frac{\tilde{p}_{V}(x,s|a)\tilde{p}_{V}(a,s|x_{0})}{1-\gamma\tilde{p}_{V}(a,s|a)}.$

As we will see, this convolution relation is sufficient to prove the PDX
relation by suitable choice of $V(x)$ and $W(x)$ and by taking certain limits.
For instance, the limit of an infinitely repulsive $\delta$-perturbation
$\gamma\rightarrow-\infty$ implements a Dirichlet, that is, absorbing BC [14,
15].

###  2.4 Proof of the PDX

As a prerequisite, we shall derive convolution relations that connect
$\tilde{p}_{\text{free}}$ with $\tilde{p}_{\text{abs}}$ and
$\tilde{p}_{\delta}^{++}$. We point out that $p_{\text{free}}(x,t|x_{0})$,
$p_{\text{abs}}(x,t|x_{0})$, and $p_{\delta}^{++}(x,t|x_{0})$ correspond to
the potentials $V(x)=0,\gamma=0$, $V(x)=0,\gamma\rightarrow-\infty$ and
$V(x)=0,-\infty<\gamma<0$, respectively (see Eq. (6)).

#### Relation between $\tilde{p}_{\delta}^{++}$ and $\tilde{p}_{\text{free}}$

We consider first the free-particle case, that is $V(x)=0$, perturbed by
$\mathcal{V}(x)=-\gamma\delta(x-a),-\infty<\gamma<0$. Then, by virtue of Eq.
(11) it immediately follows that

$\displaystyle\tilde{p}_{\delta}^{++}(x,s|x_{0})=\tilde{p}_{\text{free}}(x,s|x_{0})+\gamma\frac{\tilde{p}_{\text{free}}(x,s|a)\tilde{p}_{\text{free}}(a,s|x_{0})}{1-\gamma\tilde{p}_{\text{free}}(a,s|a)}.$
(12)

As a consequence of Eq. (12), we arrive at

$\frac{\tilde{p}_{\delta}^{++}(a,s|x_{0})}{\tilde{p}_{\delta}^{++}(a,s|a)}=\frac{\tilde{p}_{\text{free}}(a,s|x_{0})}{\tilde{p}_{\text{free}}(a,s|a)}.$
(13)

#### Relation between $\tilde{p}_{\text{abs}}$ and $\tilde{p}_{\text{free}}$

If we, in addition, consider the limiting case of an infinitely repulsive
$\delta$-function perturbation, that is $\gamma\rightarrow-\infty$, we obtain

$\tilde{p}_{\text{abs}}(x,s|x_{0})=\tilde{p}_{\text{free}}(x,s|x_{0})-\frac{\tilde{p}_{\text{free}}(x,s|a)\tilde{p}_{\text{free}}(a,s|x_{0})}{\tilde{p}_{\text{free}}(a,s|a)}.$
(14)

Clearly, Eq. (14) leads to

$\frac{\partial\tilde{p}_{\text{abs}}(\xi,s|x_{0})}{\partial\xi}\bigg{|}_{\xi=a}=\frac{\partial\tilde{p}_{\text{free}}(\xi,s|x_{0})}{\partial\xi}\bigg{|}_{\xi=a}-\frac{\partial\tilde{p}_{\text{free}}(\xi,s|a)}{\partial\xi}\bigg{|}_{\xi\rightarrow
a}\frac{\tilde{p}_{\text{free}}(a,s|x_{0})}{\tilde{p}_{\text{free}}(a,s|a)}.$
(15)

Combining this equation with an identity that results from properties of the
free-space GF only (Eq. (45)), we obtain the relations

$\displaystyle
D\frac{\partial\tilde{p}_{\text{abs}}(\xi,s|x_{0})}{\partial\xi}\bigg{|}_{\xi=a}$
$\displaystyle=$
$\displaystyle\frac{\tilde{p}_{\text{free}}(a,s|x_{0})}{\tilde{p}_{\text{free}}(a,s|a)}=\frac{D\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0})|_{\xi=a}}{1+D\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|a)|_{\xi\rightarrow
a}}$ (16) $\displaystyle=$
$\displaystyle\frac{\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0})|_{\xi=a}}{\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0}\rightarrow
a)|_{\xi=a}},$

where we introduced the notation $\partial_{\xi}\equiv\partial/\partial\xi$.
Eq. (16) gives rise to

$D\frac{\partial^{2}\tilde{p}_{\text{abs}}(\xi,s|\phi)}{\partial\xi\partial\phi}\bigg{|}_{\xi=\phi=a}=\frac{\partial_{\xi}\partial_{\phi}\tilde{p}_{\text{free}}(\xi,s|\phi)|_{\xi=\phi=a}}{\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0}\rightarrow
a)|_{\xi=a}}=\frac{\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|a)|_{\xi\rightarrow
a}}{\tilde{p}_{\text{free}}(a,s|a)}.$ (17)

Finally, combining Eqs. (13) and (16), we are led to

$D\frac{\partial\tilde{p}_{\text{abs}}(\xi,s|x_{0})}{\partial\xi}\bigg{|}_{\xi=a}=\frac{\tilde{p}_{\delta}^{++}(a,s|x_{0})}{\tilde{p}_{\delta}^{++}(a,s|a)}.$
(18)

#### Relation between $\tilde{p}_{\text{ref}}$ and $\tilde{p}_{\text{free}}$

We follow Ref. [16] and treat reflective BC by considering potentials that are
perturbed by the derivative of the $\delta$-function

$W(x)=V(x)-\gamma\delta^{\prime}(x-a).$ (19)

Analogously to the absorbing BC case, we first consider the free particle as
unperturbed problem, that is $V(x)=0$, and take the limit of an infinitely
repulsive perturbation $\gamma\rightarrow-\infty$. Then, one has [16]

$\tilde{p}_{\text{ref}}(x,s|x_{0})=\tilde{p}_{\text{free}}(x,s|x_{0})-\frac{\partial_{\xi}p_{\text{free}}(x,s|\xi)|_{\xi=a}\partial_{\xi}p_{\text{free}}(\xi,s|x_{0})|_{\xi=a}}{\partial_{\xi}\partial_{\phi}p_{\text{free}}(\xi,s|\phi)|_{\xi=\phi=a}}.$
(20)

Employing Eqs. (48) and (45), we conclude from Eq. (20) that

$\tilde{p}_{\text{ref}}(a,s|x_{0})=-\frac{1}{D}\frac{\tilde{p}_{\text{free}}(a,s|x_{0})}{\partial_{\xi}\tilde{p}_{\text{free}}(a,s|\xi)|_{\xi\rightarrow
a}},$ (21)

and hence, using Eq. (16), we arrive at

$\frac{\tilde{p}_{\text{ref}}(a,s|x_{0})}{\tilde{p}_{\text{ref}}(a,s|a)}=\frac{\tilde{p}_{\text{free}}(a,s|x_{0})}{\tilde{p}_{\text{free}}(a,s|a)}=D\frac{\partial\tilde{p}_{\text{abs}}(\xi,s|x_{0})}{\partial\xi}\bigg{|}_{\xi=a}=\frac{\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0})|_{\xi=a}}{\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0}\rightarrow
a)|_{\xi=a}}.$ (22)

In addition, Eq. (21) tells us that

$\partial_{\xi}\tilde{p}_{\text{ref}}(a,s|\xi)|_{\xi\rightarrow
a}=-\frac{1}{D},$ (23)

Another consequence of Eq. (20) is the relation

$\tilde{p}_{\text{ref}}(a,s|a)=-\frac{1}{D^{2}}\frac{1}{\partial_{\xi}\partial_{\phi}\tilde{p}_{\text{abs}}(\xi,s|\phi)|_{\xi=\phi=a}},$
(24)

where we have used Eqs. (21) and (17). We note that Eqs. (23), (24) and the
relation

$\frac{\tilde{p}_{\text{ref}}(a,s|x_{0})}{\tilde{p}_{\text{ref}}(a,s|a)}=D\frac{\partial\tilde{p}_{\text{abs}}(\xi,s|x_{0})}{\partial\xi}\bigg{|}_{\xi=a}$
(25)

(see Eq. (22)) have also been obtained by other means [3].

#### Relation between $\tilde{p}_{\delta}^{++}$ and $\tilde{p}_{\text{abs}}$

Now, we are sufficiently prepared to turn to the actual proof of the PDX
version that relates $\tilde{p}_{\text{abs}}$ with $\tilde{p}_{\delta}^{++}$.
To this end, we make the following construction. Starting point is the
potential $W(x)=V(x)-\gamma_{\text{abs}}\delta(x-a_{\text{abs}})$, where
$V(x)=-\gamma\delta(x-a)$, that is, we have a $\delta$-function potential
perturbed by a second $\delta$-function potential localized at a different
position. We assume that $a_{\text{abs}}<a$. Then, the propagators
$\tilde{p}_{W}(x,s|x_{0})$ and $\tilde{p}_{V}(x,s|x_{0})$ satisfy

$\displaystyle\tilde{p}_{W}(x,s|x_{0})=\tilde{p}_{V}(x,s|x_{0})+\gamma_{\text{abs}}\frac{\tilde{p}_{V}(x,s|a_{\text{abs}})\tilde{p}_{V}(a_{\text{abs}},s|x_{0})}{1-\gamma_{\text{abs}}\tilde{p}_{V}(a_{\text{abs}},s|a_{\text{abs}})}.$
(26)

Note that we have for the unperturbed GF
$\tilde{p}_{V}(x,s|x_{0})=\tilde{p}_{\delta}^{++}(x,s|x_{0})$, while
$\tilde{p}_{W}(x,s|x_{0})$ is not equal to any of the GF we introduced before.
However, as a next step, we consider the limit $a_{\text{abs}}\rightarrow a$
and $\gamma_{\text{abs}}\rightarrow-\infty$. In this limit, the potential
$W(x)$ assumes the form corresponding to a purely absorbing BC, that is
$W(x)=-\lim_{\gamma\rightarrow-\infty}\gamma\delta(x-a)$. Therefore, Eq. (26)
becomes

$\tilde{p}_{\text{abs}}(x,s|x_{0})=\tilde{p}_{\delta}^{++}(x,s|x_{0})-\frac{\tilde{p}_{\delta}^{++}(x,s|a)\tilde{p}_{\delta}^{++}(a,s|x_{0})}{\tilde{p}_{\delta}^{++}(a,s|a)}.$
(27)

Now, the PDX formula for $\tilde{p}_{\delta}^{++}$ immediately results from
the just established convolution relation by virtue of Eq. (18):

$\tilde{p}_{\delta}^{++}(x,s|x_{0})=\tilde{p}_{\text{abs}}(x,s|x_{0})+D^{2}\frac{\partial\tilde{p}_{\text{abs}}(x,s|\xi)}{\partial\xi}\bigg{|}_{\xi=a}\tilde{p}_{\delta}^{++}(a,s|a)\frac{\partial\tilde{p}_{\text{abs}}(\xi,s|x_{0})}{\partial\xi}\bigg{|}_{\xi=a}.$
(28)

#### Relation between $\tilde{p}_{\text{ref}}$ and $\tilde{p}_{\text{abs}}$

To analyze this case we consider a potential of the form
$W(x)=V(x)-\gamma_{\text{ref}}\delta^{\prime}(x-a)$, where the unperturbed
potential is given by a $\delta$-function, that is,
$V=-\gamma_{\text{abs}}\delta(x-a)$. We take the limit of an infinitely
repulsive $\delta$\- and $\delta^{\prime}$\- potential. More precisely, we
consider

$\lim_{\gamma_{\text{abs}},\gamma_{\text{ref}}\rightarrow-\infty}W(x)$ (29)

and obtain

$\tilde{p}_{\text{ref}}(x,s|x_{0})=\tilde{p}_{\text{abs}}(x,s|x_{0})-\frac{\partial_{\xi}\tilde{p}_{\text{abs}}(x,s|\xi)|_{\xi=a}\partial_{\phi}\tilde{p}_{\text{abs}}(\phi,s|x_{0})|_{\phi=a}}{\partial_{\xi}\partial_{\phi}\tilde{p}_{\text{abs}}(\xi,s|\phi)|_{\xi=\phi=a}}.$
(30)

Then, by making use of Eq. (24), we obtain the PDX for
$\tilde{p}_{\text{ref}}(x,s|x_{0})$

$\tilde{p}_{\text{ref}}(x,s|x_{0})=\tilde{p}_{\text{abs}}(x,s|x_{0})+D^{2}\partial_{\xi}\tilde{p}_{\text{abs}}(x,s|\xi)|_{\xi=a}\tilde{p}_{\text{ref}}(a,s|a)\partial_{\phi}\tilde{p}_{\text{abs}}(\phi,s|x_{0})|_{\phi=a}.$
(31)

We note that the considered potential (Eq. (29)) also gives rise to the
relation

$\tilde{p}_{\text{abs}}(x,s|x_{0})=\tilde{p}_{\text{ref}}(x,s|x_{0})-\frac{\tilde{p}_{\text{ref}}(x,s|a)\tilde{p}_{\text{ref}}(a,s|x_{0})}{\tilde{p}_{\text{ref}}(a,s|a)}.$
(32)

#### Relation between $\tilde{p}_{\text{rad}}$ and $\tilde{p}_{\text{abs}}$

It has been shwon that the GF $\tilde{p}_{\text{rad}}$ corresponding to a
radiation BC enjoys a path integral representation, given one considers a
$\delta$-function potential in the presence of a Neumann BC [7, 11]. Put
differently, this means that $\tilde{p}_{\text{rad}}$ is obtained from the
potential $W(x)=V(x)+\mathcal{V}(x)$, where the unperturbed potential and the
perturbation are given by

$\displaystyle V(x)$ $\displaystyle=$
$\displaystyle-\lim_{\gamma\rightarrow-\infty}\delta^{\prime}(x-a),$ (33)
$\displaystyle\mathcal{V}(x)$ $\displaystyle=$
$\displaystyle\kappa_{a}\delta(x-a),$ (34)

respectively. Using once again the general convolution formula Eq. (11), we
arrive at

$\tilde{p}_{\text{rad}}(x,s|x_{0})=\tilde{p}_{\text{ref}}(x,s|x_{0})-\kappa_{a}\frac{\tilde{p}_{\text{ref}}(x,s|a)\tilde{p}_{\text{ref}}(a,s|x_{0})}{1+\kappa_{a}\tilde{p}_{\text{ref}}(a,s|a)}.$
(35)

Eq. (35) has been derived before [31, 3] in a different way. From Eq. (35) we
obtain

$\tilde{p}_{\text{rad}}(a,s|x_{0})=\frac{\tilde{p}_{\text{ref}}(a,s|x_{0})}{1+\kappa_{a}\tilde{p}_{\text{ref}}(a,s|a)}.$
(36)

Using the established identity Eq. (36) as well as Eq. (22), we obtain from
Eq. (35)

$\displaystyle\tilde{p}_{\text{rad}}(x,s|x_{0})=\tilde{p}_{\text{ref}}(x,s|x_{0})$
(37) $\displaystyle-$ $\displaystyle
D^{2}\kappa_{a}\partial_{\xi}\tilde{p}_{\text{abs}}(x,s|\xi)|_{\xi=a}\partial_{\phi}\tilde{p}_{\text{abs}}(\phi,s|x_{0})|_{\phi=a}\tilde{p}_{\text{rad}}(a,s|a)\tilde{p}_{\text{ref}}(a,s|a).$

By virtue of Eqs. (31), (37) and again (36), we finally arrive at the PDX for
$\tilde{p}_{\text{rad}}(x,s|x_{0})$

$\tilde{p}_{\text{rad}}(x,s|x_{0})=\tilde{p}_{\text{abs}}(x,s|x_{0})+D^{2}\partial_{\xi}\tilde{p}_{\text{abs}}(x,s|\xi)|_{\xi=a}\tilde{p}_{\text{rad}}(a,s|a)\partial_{\phi}\tilde{p}_{\text{abs}}(\phi,s|x_{0})|_{\phi=a}.$
(38)

We finally note that, due to

$1-\frac{D^{2}}{\kappa_{a}}\partial_{\xi}\partial_{\phi}\tilde{p}_{\text{abs}}(\xi,s|\phi)|_{\xi=\phi=a}=\frac{1}{\kappa_{a}\tilde{p}_{\text{rad}}(a,s|a)},$
(39)

see Eqs. (24) and (36), the PDX for $\tilde{p}_{\text{rad}}$ gives rise to a
representation of $\tilde{p}_{\text{rad}}$ in terms of
$\tilde{p}_{\text{abs}}$ only

$\tilde{p}_{\text{rad}}(x,s|x_{0})=\tilde{p}_{\text{abs}}(x,s|x_{0})+\frac{D^{2}}{\kappa_{a}}\frac{\partial_{\xi}\tilde{p}_{\text{abs}}(x,s|\xi)|_{\xi=a}\partial_{\phi}\tilde{p}_{\text{abs}}(\phi,s|x_{0})|_{\phi=a}}{1-\frac{D^{2}}{\kappa_{a}}\partial_{\xi}\partial_{\phi}\tilde{p}_{\text{abs}}(\xi,s|\phi)|_{\xi=\phi=a}}.$
(40)

This shows that, dependent on which term of the potential $W(x)$ is chosen as
perturbation, we obtain either Eq. (35) or (40), see also Ref. [11].

##  3 Appendix

Let us define

$\displaystyle u(x,s)$ $\displaystyle\equiv$
$\displaystyle\tilde{p}_{\text{free}}(x,s|x_{0}),$ (41) $\displaystyle v(x,s)$
$\displaystyle\equiv$ $\displaystyle\tilde{p}_{\text{free}}(x,s|a).$ (42)

Then, $u(x,s)$and $v(x,s)$ satisfy the free-space diffusion equation,

$\displaystyle su(x,s)-\delta(x-x_{0})$ $\displaystyle=$ $\displaystyle
D\frac{\partial^{2}u(x,s)}{\partial x^{2}},$ (43) $\displaystyle
sv(x,s)-\delta(x-a)$ $\displaystyle=$ $\displaystyle
D\frac{\partial^{2}v(x,s)}{\partial x^{2}}.$ (44)

Next, assuming $x_{0}>a$, multiplying Eq. (43) with $v(x,s)$ and Eq. (44) with
$u(x,s)$, as well as substracting the resulting equations and integrating the
outcome from $a$ to $\infty$ leads to

$\tilde{p}_{\text{free}}(a,s|x_{0})=D\bigg{[}\tilde{p}_{\text{free}}(a,s|a)\frac{\partial\tilde{p}_{\text{free}}(\xi,s|x_{0})}{\partial\xi}\bigg{|}_{\xi=a}-\,\tilde{p}_{\text{free}}(a,s|x_{0})\frac{\partial\tilde{p}_{\text{free}}(\xi,s|a)}{\partial\xi}\bigg{|}_{\xi\rightarrow
a}\bigg{]},$ (45)

where we have used the definitions Eqs. (41) and (42), the BC at infinity,
$\lim_{x\rightarrow\infty}\tilde{p}_{\text{free}}(x,s|x_{0})\rightarrow 0$,
and $\tilde{p}_{\text{free}}(x,s|x_{0})=\tilde{p}_{\text{free}}(x_{0},s|x)$.
Taking the limit $x_{0}\rightarrow a$, it follows that

$\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0}\rightarrow
a)|_{\xi=a}-\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|a)|_{\xi\rightarrow
a}=D^{-1}.$ (46)

Furthermore, dividing Eq. (45) by $\tilde{p}_{\text{free}}(a,s|a)$ and using
Eq. (46), we obtain the chain of identities

$\frac{\tilde{p}_{\text{free}}(a,s|x_{0})}{\tilde{p}_{\text{free}}(a,s|a)}=\frac{D\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0})|_{\xi=a}}{1+D\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|a)|_{\xi\rightarrow
a}}=\frac{\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0})|_{\xi=a}}{\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0}\rightarrow
a)|_{\xi=a}}.$ (47)

Finally, as a consequence of Eq. (47), we arrive at

$\frac{\partial_{\xi}\partial_{\phi}\tilde{p}_{\text{free}}(\xi,s|\phi)|_{\xi=\phi=a}}{\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|x_{0}\rightarrow
a)|_{\xi=a}}=\frac{\partial_{\xi}\tilde{p}_{\text{free}}(\xi,s|a)|_{\xi\rightarrow
a}}{\tilde{p}_{\text{free}}(a,s|a)}.$ (48)

## Acknowledgments

This research was supported by the Intramural Research Program of the NIH,
National Institute of Allergy and Infectious Diseases.

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