$\displaystyle(E_{n_{1},n_{1}+1}-E_{n+n_{1}+1,n+n_{1}})^{r_{1}}(E_{n+n_{2}+1,n_{1}+1}+E_{n+n_{1}+1,n_{2}+1})^{r_{2}}(x_{n_{1}+1}^{2m_{3}}x_{n_{1}}^{m_{4}}y_{n_{2}+1}^{m_{5}})$
$\displaystyle=$
$\displaystyle(2m_{3})![\prod_{s_{1}=0}^{r_{1}-1}(m_{4}-s_{1})][\prod_{s_{2}=0}^{r_{2}-1}(m_{5}-s_{2})]x_{n_{1}}^{m_{4}-r_{1}}y_{n_{2}+1}^{m_{5}-r_{2}}\in
M.\hskip 91.04872pt(7.46)$

Furthermore, (7.42) yields

$\displaystyle(E_{n+n_{2}+1,n_{1}}+E_{n+n_{1},n_{2}+1})^{m_{5}-r_{2}}(x_{n_{1}}^{m_{4}-r_{1}}y_{n_{2}+1}^{m_{5}-r_{2}})$
$\displaystyle=$
$\displaystyle(-1)^{m_{5}-r_{2}}(m_{5}-r_{2})!x_{n_{1}}^{k}\in M.\hskip
187.78836pt(7.47)$

Thus we always have $x_{n_{1}}^{k}\in M$.

Note that ${\cal M}^{1}_{\langle-k\rangle}\ni x_{n_{1}}^{k}$ is an irreducible
${\cal G}_{1}$-module (cf. (7.27) and (7.29)) by Step 1. So

${\cal M}^{1}_{\langle-k\rangle}\subset M.$ $None$

Let $r\in\mathbb{Z}$. According to (7.32),

${\cal M}^{3}_{\langle r\rangle}{\cal M}^{4}_{\langle-k-r\rangle}\subset{\cal
M}^{1}_{\langle-k\rangle}\subset M.$ $None$

Moreover, ${\cal M}^{2}_{\langle-k-r\rangle}\supset{\cal M}^{4}_{\langle-
k-r\rangle}$ is an irreducible ${\cal G}_{2}$-module (cf. (7.28) and (7.29))
by Step 1. Thus

${\cal M}^{3}_{\langle r\rangle}{\cal M}^{2}_{\langle-k-r\rangle}=U({\cal
G}_{2})({\cal M}^{3}_{\langle r\rangle}{\cal M}^{4}_{\langle-
k-r\rangle})\subset M.$ $None$

Then

${\cal B}_{\langle-k\rangle}=\bigoplus_{r\in\mathbb{Z}}{\cal M}^{3}_{\langle
r\rangle}{\cal M}^{2}_{\langle-k-r\rangle}\subset M$ $None$

by (7.29) and (7.30). Therefore, $M={\cal B}_{\langle-k\rangle}$, that is,
${\cal B}_{\langle-k\rangle}$ is an irreducible $sp(2n,\mathbb{F})$-submodule.

Fix $0<k\in\mathbb{N}$. Then the singular vectors of $sl(n,\mathbb{F})$ in
${\cal B}_{\langle k\rangle}$ are

$\displaystyle\\{\eta^{m_{2}}(x_{n_{1}+1}^{k+m_{1}-2m_{2}}y_{n_{2}+1}^{m_{1}}),\eta^{m_{2}}(x_{n_{1}}^{m_{1}}y_{n_{2}}^{k+m_{1}-2m_{2}}),\eta^{m_{3}}(x_{n_{1}+1}^{m_{4}}y_{n_{2}}^{m_{5}})$
$\displaystyle\mid m_{i}\in\mathbb{N};2m_{2}\leq
k+m_{1};m_{4}+m_{5}+2m_{3}=k\\}\hskip 145.10922pt(7.52)$

by (7.33). Let $M$ be a nonzero $sp(2n,\mathbb{F})$-submodule of ${\cal
B}_{\langle k\rangle}$. Then $M$ contains a singular of $sl(n,\mathbb{F})$.
Suppose some $\eta^{m_{2}}(x_{n_{1}+1}^{k+m_{1}-2m_{2}}y_{n_{2}+1}^{m_{1}})\in
M$ with $2m_{2}\leq k+m_{1}$. We have $E_{n_{1}+1,n+n_{1}+1}|_{\cal
B}=x_{n_{1}+1}\partial_{y_{n_{1}+1}}$ and

$E_{n_{1}+1,n+n_{1}+1}^{m_{2}}[\eta^{m_{2}}(x_{n_{1}+1}^{k+m_{1}-2m_{2}}y_{n_{2}}^{m_{1}})]=m_{2}!x_{n_{1}+1}^{k+m_{1}}y_{n_{2}+1}^{m_{1}}\in
M$ $None$

by (7.3) and (7.5). Moreover, (7.40) gives

$(E_{n+n_{2}+1,n_{1}+1}+E_{n+n_{1}+1,n_{2}+1})^{m_{1}}(x_{n_{1}+1}^{k+m_{1}}y_{n_{2}+1}^{m_{1}})=m_{1}![\prod_{r=1}^{m_{1}}(k+r)]x_{n_{1}+1}^{k}\in
M.$ $None$

Thus

$x_{n_{1}+1}^{k}\in M.$ $None$

Assume some $\eta^{m_{2}}(x_{n_{1}}^{m_{1}}y_{n_{2}}^{k+m_{1}-2m_{2}})\in M$
with $2m_{2}\leq k+m_{1}$. Observe
$E_{n+n_{2},n_{2}}=y_{n_{2}}\partial_{x_{n_{2}}}$ by (1.16). So

$E_{n+n_{2},n_{2}}^{m_{2}}[\eta^{m_{2}}(x_{n_{1}}^{m_{1}}y_{n_{2}}^{k+m_{1}-2m_{2}})]=m_{2}!x_{n_{1}}^{m_{1}}y_{n_{2}}^{k+m_{1}}\in
M.$ $None$

Moreover, (7.4) gives that $(E_{n_{1},n+n_{2}}+E_{n_{2},n+n_{1}})|_{\cal
B}=\partial_{x_{n_{1}}}\partial_{y_{n_{2}}}+x_{n_{2}}\partial_{y_{n_{1}}}$ and

$(E_{n_{1},n+n_{2}}+E_{n_{2},n+n_{1}})^{m_{1}}(x_{n_{1}}^{m_{1}}y_{n_{2}}^{k+m_{1}})=m_{1}![\prod_{r=1}^{m_{1}}(k+r)]y_{n_{2}}^{k}\in
M.$ $None$

Furthermore, (7.5) yields that
$(E_{n_{1}+1,n+n_{2}}+E_{n_{2},n+n_{1}+1})|_{\cal
B}=x_{n_{1}+1}\partial_{y_{n_{2}}}+x_{n_{2}}\partial_{y_{n_{1}+1}}$ and

$(E_{n_{1}+1,n+n_{2}}+E_{n_{2},n+n_{1}+1})^{k}(y_{n_{2}}^{k})=k!x_{n_{1}+1}^{k}\in
M.$ $None$

Thus (7.55) holds again.

Consider $\eta^{m_{3}}(x_{n_{1}+1}^{m_{4}}y_{n_{2}}^{m_{5}})$ for some
$m_{3},m_{3},m_{4}\in\mathbb{N}$ such that $m_{4}+m_{5}+2m_{3}=k$. Note
$E_{n_{1}+1,n+n_{1}+1}=x_{n_{1}+1}\partial_{y_{n_{1}+1}}$ by (7.5). So

$E_{n_{1}+1,n+n_{1}+1}^{m_{3}}[\eta^{m_{3}}(x_{n_{1}+1}^{m_{4}}y_{n_{2}}^{m_{5}})]=m_{3}!x_{n_{1}+1}^{m_{4}+2m_{3}}y_{n_{2}}^{m_{5}}\in
M.$ $None$

According to (7.5),

$(E_{n_{1}+1,n+n_{2}}+E_{n_{2},n+n_{1}+1})^{m_{5}}(x_{n_{1}+1}^{m_{4}+2m_{3}}y_{n_{2}}^{m_{5}})=m_{5}!x_{n_{1}+1}^{k}\in
M.$ $None$

Therefore, we always have $x_{n_{1}+1}^{k}\in M$.

Observe that ${\cal M}^{2}_{\langle k\rangle}\ni x_{n_{1}+1}^{k}$ is an
irreducible ${\cal G}_{2}$-module (cf. (7.28) and (7.29)) by Step 1. So

${\cal M}^{2}_{\langle k\rangle}\subset M.$ $None$

Let $r\in\mathbb{Z}$. Denote

${\cal
M}^{5}=\mathbb{F}[x_{n_{2}+1},...,x_{n},y_{n_{2}+1},...,y_{n}],\qquad{\cal
M}^{5}_{\langle k\rangle}={\cal M}^{5}\bigcap{\cal B}_{\langle
k\rangle},\;\;k\in\mathbb{Z}.$ $None$

Then

${\cal M}^{2}_{\langle k\rangle}=\bigoplus_{r\in\mathbb{Z}}{\cal
M}^{4}_{\langle r\rangle}{\cal M}^{5}_{\langle k-r\rangle}$ $None$

(cf. (7.30)). Fix $r\in\mathbb{Z}$.

${\cal M}^{4}_{\langle r\rangle}{\cal M}^{5}_{\langle k-r\rangle}\subset{\cal
M}^{2}_{\langle k\rangle}\subset M.$ $None$

Moreover, ${\cal M}^{1}_{\langle r\rangle}\supset{\cal M}^{4}_{\langle
r\rangle}$ is an irreducible ${\cal G}_{1}$-module (cf. (7.27) and (7.29)) by
Step 1. Thus

${\cal M}^{1}_{\langle r\rangle}{\cal M}^{5}_{\langle k-r\rangle}=U({\cal
G}_{1})({\cal M}^{4}_{\langle r\rangle}{\cal M}^{5}_{\langle
k-r\rangle})\subset M.$ $None$

Furthermore,

${\cal B}_{\langle k\rangle}=\bigoplus_{r\in\mathbb{Z}}{\cal M}^{1}_{\langle
r\rangle}{\cal M}^{5}_{\langle k-r\rangle}\subset M$ $None$

by (7.27) and (7.65). Therefore, $M={\cal B}_{\langle k\rangle}$, that is,
${\cal B}_{\langle k\rangle}$ is an irreducible $sp(2n,\mathbb{F})$-submodule.

Case 2. $n_{2}=n_{1}+1$.

According to (3.104), the nonzero weight vectors in

$\displaystyle\mbox{Span}\\{\eta^{m_{2}}(x_{i}^{m_{1}}y_{j}^{m_{3}}),x_{n_{1}+1}^{m_{1}}y_{n_{1}+1}^{m_{2}},\eta^{m_{1}+m_{2}}(x_{n_{1}}^{m_{2}}y_{n_{1}+1}^{m_{3}-m_{1}}),\eta^{m_{1}+m_{2}}(y_{n_{1}+2}^{m_{2}}x_{n_{1}+1}^{m_{3}-m_{1}})$
$\displaystyle\qquad\;\;\mid
m_{r}\in\mathbb{N};(i,j)=(n_{1},n_{1}+1),(n_{1},n_{1}+2),(n_{1}+1,n_{1}+2)\\}.\hskip
71.13188pt(7.67)$

are all the singular vectors of $sl(n,\mathbb{F})$ in ${\cal B}$. Fix
$k\in\mathbb{N}$. Then the singular vectors of $sl(n,\mathbb{F})$ in ${\cal
B}_{\langle-k\rangle}$ are those in (7.34). According to the arguments in Case
1, ${\cal B}_{\langle-k\rangle}$ is an irreducible
$sp(2n,\mathbb{F})$-submodule. Let $0<k\in\mathbb{N}$. Then the singular
vectors of $sl(n,\mathbb{F})$ in ${\cal B}_{\langle k\rangle}$ are

$\displaystyle\\{\eta^{m_{2}}(x_{n_{1}+1}^{k+m_{1}-2m_{2}}y_{n_{1}+2}^{m_{1}}),\eta^{m_{2}}(x_{n_{1}}^{m_{1}}y_{n_{1}+1}^{k+m_{1}-2m_{2}}),\eta^{m_{5}+m_{6}}(x_{n_{1}}^{m_{6}}y_{n_{1}+1}^{m_{7}-m_{5}}),\eta^{m_{5}+m_{6}}(y_{n_{1}+2}^{m_{6}}x_{n_{1}+1}^{m_{7}-m_{5}}),\hskip
17.07182pt$
$\displaystyle\\!\\!\\!\\!\\!x_{n_{1}+1}^{m_{3}}y_{n_{1}+1}^{m_{4}}\mid
m_{i}\in\mathbb{N};2m_{2}\leq k+m_{1};m_{3}+m_{4}=k=m_{5}+m_{5}+m_{7}\\}\hskip
59.75095pt(7.68)$

by (7.67). Let $M$ be a nonzero $sp(2n,\mathbb{F})$-submodule of ${\cal
B}_{\langle k\rangle}$. As an $sl(n,\mathbb{F})$-module, $M$ contains a
singular of $sl(n,\mathbb{F})$. If $x_{n_{1}+1}^{m_{3}}y_{n_{1}+1}^{m_{4}}\in
M$ with $m_{3}+m_{4}=k$, then $E_{n_{1}+1,n+n_{1}+1}|_{\cal
B}=x_{n_{1}+1}\partial_{y_{n_{1}+1}}$ and

$E_{n_{1}+1,n+n_{1}+1}^{m_{4}}(x_{n_{1}+1}^{m_{3}}y_{n_{1}+1}^{m_{4}})=m_{4}!x_{n_{1}+1}^{k}\in
M\Longrightarrow x_{n_{1}+1}^{k}\in M$ $None$

by (7.5). Suppose some
$\eta^{m_{5}+m_{6}}(x_{n_{1}}^{m_{6}}y_{n_{1}+1}^{m_{7}-m_{5}})\in M$ with
$m_{5}+m_{5}+m_{7}=k$. According to (1.16),
$E_{n+n_{1}+1,n_{1}+1}=y_{n_{1}+1}\partial_{x_{n_{1}+1}}$. So

$E_{n+n_{1}+1,n_{1}+1}^{m_{5}+m_{6}}[\eta^{m_{5}+m_{6}}(x_{n_{1}}^{m_{6}}y_{n_{1}+1}^{m_{7}-m_{5}})]=(m_{5}+m_{6})!x_{n_{1}}^{m_{6}}y_{n_{1}+1}^{k+m_{6}}\in
M.$ $None$

Moreover, (7.4) yields that $(E_{n_{1},n+n_{1}+1}+E_{n_{1}+1,n+n_{1}})|_{\cal
B}=\partial_{x_{n_{1}}}\partial_{y_{n_{1}+1}}+x_{n_{1}+1}\partial_{y_{n_{1}}}$
and

$(E_{n_{1},n+n_{1}+1}+E_{n_{1}+1,n+n_{1}})^{m_{6}}(x_{n_{1}}^{m_{6}}y_{n_{1}+1}^{k+m_{6}})=m_{6}![\prod_{r=1}^{m_{6}}(k+r)]y_{n_{1}+1}^{k}\in
M.$ $None$

Assume some
$\eta^{m_{5}+m_{6}}(y_{n_{1}+2}^{m_{6}}x_{n_{1}+1}^{m_{7}-m_{5}})\in M$ with
$m_{5}+m_{5}+m_{7}=k$. By (7.3) and (7.5),

$E_{n_{1}+1,n+n_{1}+1}^{m_{5}+m_{6}}[\eta^{m_{5}+m_{6}}(y_{n_{1}+2}^{m_{6}}x_{n_{1}+1}^{m_{7}-m_{5}})]=(m_{5}+m_{6})!y_{n_{1}+2}^{m_{6}}x_{n_{1}+1}^{k+m_{6}}\in
M.$ $None$

Observe

$(E_{n+n_{1}+2,n_{1}+1}+E_{n+n_{1}+1,n_{1}+2})|_{\cal
B}=\partial_{x_{n_{1}+1}}\partial_{y_{n_{1}+2}}+y_{n_{1}+1}\partial_{x_{n_{1}+2}}$
$None$

by (1.16). Hence

$(E_{n+n_{1}+2,n_{1}+1}+E_{n+n_{1}+1,n_{1}+2})^{m_{6}}(y_{n_{1}+2}^{m_{6}}x_{n_{1}+1}^{k+m_{6}})=m_{6}![\prod_{r=1}^{m_{6}}(k+r)]x_{n_{1}+1}^{k}\in
M.$ $None$

Expressions (7.53)-(7.60), (7.69), (7.71) and (7.74) show that we always have
$x_{n_{1}+1}^{k}\in M$. Furthermore, (7.61)-(7.66) imply that ${\cal
B}_{\langle k\rangle}$ is an irreducible $sp(2n,\mathbb{F})$-module.

Case 3. $n_{1}=n_{2}$.

In this case,

$\eta=\sum_{i=1}^{n_{1}}y_{i}\partial_{x_{i}}+\sum_{s=n_{2}+1}^{n}x_{s}\partial_{y_{s}}.$
$None$

First we consider the subcase $1<n_{1}<n-1$. Expression (4.17) says that the
nonzero weight vectors in

$\mbox{Span}\\{x_{n_{1}}^{m_{1}}y_{n_{1}}^{m_{2}}\zeta_{1}^{m_{3}+1},x_{n_{1}+1}^{m_{1}}y_{n_{1}+1}^{m_{2}}\zeta_{2}^{m_{3}+1},\eta^{m_{3}}(x_{n_{1}}^{m_{1}}y_{n_{1}+1}^{m_{2}})\mid
m_{i}\in\mathbb{N}\\}$ $None$

are all the singular vectors of $sl(n,\mathbb{F})$ in ${\cal B}$, where

$\zeta_{1}=x_{n_{1}-1}y_{n_{1}}-x_{n_{1}}y_{n_{1}-1},\;\;\zeta_{2}=x_{n_{1}+1}y_{n_{1}+2}-x_{n_{1}+2}y_{n_{1}+1}.$
$None$

Fix $k\in\mathbb{N}+1$. Then the singular vectors of $sl(n,\mathbb{F})$ in
${\cal B}_{\langle-k\rangle}$ are

$\displaystyle\\{x_{n_{1}}^{k+m_{1}}y_{n_{1}}^{m_{1}}\zeta_{1}^{m_{2}+1},x_{n_{1}+1}^{m_{1}}y_{n_{1}+1}^{k+m_{1}}\zeta_{2}^{m_{2}+1},\eta^{m_{3}}(x_{n_{1}}^{m_{4}}y_{n_{1}+1}^{m_{5}})$
$\displaystyle\mid m_{i}\in\mathbb{N};m_{4}+m_{5}-2m_{3}=k\\}.\hskip
219.08612pt(7.78)$

Let $M$ be a nonzero $sp(2n,\mathbb{F})$-submodule of ${\cal
B}_{\langle-k\rangle}$. As an $sl(n,\mathbb{F})$-module, $M$ contains a
singular vector of $sl(n,\mathbb{F})$. Suppose some
$x_{n_{1}}^{k+m_{1}}y_{n_{1}}^{m_{1}}\zeta_{1}^{m_{2}+1}\in M$. Note
$E_{n_{1},n+n_{1}}|_{\cal B}=\partial_{x_{n_{1}}}\partial_{y_{n_{1}}}$ by
(7.4), and so

$\displaystyle
E_{n_{1},n+n_{1}}(x_{n_{1}}^{k+m_{1}}y_{n_{1}}^{m_{1}}\zeta_{1}^{m_{2}})$
$\displaystyle=$
$\displaystyle(k+m_{1})m_{1}x_{n_{1}}^{k+m_{1}-1}y_{n_{1}}^{m_{1}-1}\zeta_{1}^{m_{2}}-m_{2}(m_{2}-1)x_{n_{1}}^{k+m_{1}}y_{n_{1}}^{m_{1}}x_{n_{1}-1}y_{n_{1}-1}\zeta_{1}^{m_{2}-2}$
$\displaystyle+(k+m_{1})m_{2}x_{n_{1}}^{k+m_{1}-1}y_{n_{1}}^{m_{1}}x_{n_{1}-1}\zeta_{1}^{m_{2}-1}-m_{1}m_{2}x_{n_{1}}^{k+m_{1}}y_{n_{1}}^{m_{1}-1}y_{n_{1}-1}\zeta_{1}^{m_{2}-1}.\hskip
48.36958pt(7.79)$

Moreover,

$(E_{n_{1}-1,n_{1}}-E_{n+n_{1},n+n_{1}-1})|_{\cal
B}=-(x_{n_{1}}\partial_{x_{n_{1}-1}}+y_{n_{1}}\partial_{y_{n_{1}-1}})$ $None$

by (1.7), (1.8) and (1.14). Thus

$\displaystyle(E_{n_{1}-1,n_{1}}-E_{n+n_{1},n+n_{1}-1})^{2}E_{n_{1},n+n_{1}}(x_{n_{1}}^{k+m_{1}}y_{n_{1}}^{m_{1}}\zeta_{1}^{m_{2}})$
$\displaystyle=$
$\displaystyle-2m_{2}(m_{2}-1)x_{n_{1}}^{k+m_{1}+1}y_{n_{1}}^{m_{1}+1}\zeta_{1}^{m_{2}-2}\in
M.\hskip 145.10922pt(7.81)$

Hence

$x_{n_{1}}^{k+m_{1}+1}y_{n_{1}}^{m_{1}+1}\zeta_{1}^{m_{2}-2}\in
M\qquad\mbox{if}\;\;m_{2}>1.$ $None$

Furthermore,

$(E_{n_{1}-1,n_{1}}-E_{n+n_{1},n+n_{1}-1})E_{n_{1},n+n_{1}}(x_{n_{1}}^{k+m_{1}}y_{n_{1}}^{m_{1}}\zeta_{1})=-kx_{n_{1}}^{k+m_{1}}y_{n_{1}}^{m_{1}}\in
M.$ $None$

So we always have $x_{n_{1}}^{k+m}y_{n_{1}}^{m}\in M$ for some
$m\in\mathbb{N}$ by induction on $m_{2}$.

Observe

$E_{n_{1},n+n_{1}}(x_{n_{1}}^{k+m}y_{n_{1}}^{m})=\partial_{x_{n_{1}}}\partial_{y_{n_{1}}}(x_{n_{1}}^{k+m}y_{n_{1}}^{m})=m![\prod_{r=1}^{m}(k+r)]x_{n_{1}}^{k}$
$None$

by (7.4). Thus

$x_{n_{1}}^{k}\in M.$ $None$

Symmetrically, if some
$x_{n_{1}+1}^{m_{1}}y_{n_{1}+1}^{k+m_{1}}\zeta_{2}^{m_{2}+1}\in M$, we have
$y_{n_{1}+1}^{k}\in M$. But

$(E_{n+n_{1}+1,n_{1}}+E_{n+n_{1},n_{1}+1})|_{\cal
B}=-x_{n_{1}}\partial_{y_{n_{1}+1}}+y_{n_{1}}\partial_{x_{n_{1}+1}}$ $None$

by (1.16), which gives

$(E_{n+n_{1}+1,n_{1}}+E_{n+n_{1},n_{1}+1})^{k}(y_{n_{1}+1}^{k})=(-1)^{k}k!x_{n_{1}}^{k}\in
M.$ $None$

Thus (7.85) holds again.

Assume that some $\eta^{m_{3}}(x_{n_{1}}^{m_{4}}y_{n_{1}+1}^{m_{5}})\in M$
with $m_{4}+m_{5}-2m_{3}=k$. Note there exists $r_{1},r_{2}\in\mathbb{N}$ such
that $r_{1}+r_{2}=m_{3}$ and $2r_{1}\leq m_{4},\;2r_{2}\leq m_{5}$. Moreover,
$E_{n_{1},n+n_{1}}|_{\cal B}=\partial_{x_{n_{1}}}\partial_{y_{n_{1}}}$ by
(7.4) and $E_{n+n_{1}+1,n_{1}+1}|_{\cal
B}=\partial_{x_{n_{1}+1}}\partial_{y_{n_{1}+1}}$ by (1.16). Thus

$\displaystyle
E_{n_{1},n+n_{1}}^{r_{1}}E_{n+n_{1}+1,n_{1}+1}^{r_{2}}[\eta^{m_{3}}(x_{n_{1}}^{m_{4}}y_{n_{1}+1}^{m_{5}})]$
$\displaystyle=$ $\displaystyle
m_{3}![\prod_{s_{1}=0}^{2r_{1}-1}(m_{4}-s_{1})][\prod_{s_{2}=0}^{2r_{2}-1}(m_{5}-s_{2})]x_{n_{1}}^{m_{4}-2r_{1}}y_{n_{1}+1}^{m_{5}-2r_{2}}\in
M.\hskip 91.04872pt(7.88)$

Furthermore, (1.16) gives $(E_{n+n_{1}+1,n_{1}}+E_{n+n_{1},n_{1}+1})|_{\cal
B}=-x_{n_{1}}\partial_{y_{n_{1}+1}}+y_{n_{1}}\partial_{x_{n_{1}+1}}$ , and so

$\displaystyle(E_{n+n_{1}+1,n_{1}}+E_{n+n_{1},n_{1}+1})^{m_{5}-2r_{2}}(x_{n_{1}}^{m_{4}-2r_{1}}y_{n_{1}+1}^{m_{5}-2r_{2}})$
$\displaystyle=$
$\displaystyle(-1)^{m_{5}-2r_{2}}(m_{5}-2r_{2})!x_{n_{1}}^{k}\in M.\hskip
176.407pt(7.89)$

Thus we always have $x_{n_{1}}^{k}\in M$.

Now

$(E_{n_{1},n+n_{1}+1}+E_{n_{1}+1,n+n_{1}})|_{\cal
B}=-y_{n_{1}+1}\partial_{x_{n_{1}}}+x_{n_{1}+1}\partial_{y_{n_{1}}}$ $None$

by (7.4). For any $r\in\mathbb{N}+1$,

$\frac{(-1)^{r}}{\prod_{s=0}^{r-1}(k-r)}(E_{n_{1},n+n_{1}+1}+E_{n_{1}+1,n+n_{1}})^{r}(x_{n_{1}}^{k})=x_{n_{1}}^{k-r}y_{n_{1}+1}^{r}\in
M.$ $None$

If $k\geq 2$ and $r\in\overline{1,k-1}$, then

${\cal M}^{1}_{\langle-k+r\rangle}{\cal M}^{2}_{\langle-r\rangle}=U({\cal
G}_{1})U({\cal G}_{2})(x_{n_{1}}^{k-r}y_{n_{1}+1}^{r})\subset M$ $None$

because ${\cal M}^{1}_{\langle-k+r\rangle}$ is an irreducible ${\cal
G}_{1}$-module and ${\cal M}^{2}_{\langle-r\rangle}$ is an irreducible ${\cal
G}_{2}$-module by Step 1. Moreover,

${\cal M}^{1}_{\langle-k\rangle}=U({\cal G}_{1})(x_{n_{1}}^{k}),\;{\cal
M}^{2}_{\langle-k\rangle}=U({\cal G}_{2})(y_{n_{1}+1}^{k})\subset M.$ $None$

Furthermore,

${\cal M}^{1}_{\langle-k\rangle}{\cal M}^{2}_{\langle 0\rangle}=U({\cal
G}_{1})U({\cal G}_{2})(x_{n_{1}}^{k})\subset M\;\;\mbox{if}\;\;n_{1}=n-1$
$None$

and

${\cal M}^{1}_{\langle 0\rangle}{\cal M}^{2}_{\langle-k\rangle}=U({\cal
G}_{1})U({\cal G}_{2})(y_{n_{1}+1}^{k})\subset M\;\;\mbox{if}\;\;n_{1}=1.$
$None$

Note

$(E_{r,i}-E_{n+i,n+r})|_{\cal
B}=y_{i}y_{r}-x_{i}x_{r}\qquad\mbox{for}\;\;i\in\overline{1,n_{1}},\;r\in\overline{n_{1}+1,n}$
$None$

by (1.7), (1.8) and (1.14). In particular, if $k>1$ or $n_{1}=1$, we have

$(E_{n_{1}+1,n_{1}}-E_{n+n_{1},n+n_{1}+1})(x_{n_{1}}^{k})=y_{n_{1}}x_{n_{1}}^{k}y_{n_{1}+1}-x_{n_{1}}^{k+1}x_{n_{1}+1}\in
M.$ $None$

Since

$y_{n_{1}}x_{n_{1}}^{k}y_{n_{1}+1}\in{\cal M}^{1}_{\langle-k+1\rangle}{\cal
M}^{2}_{\langle-1\rangle}\subset M,$ $None$

we get

$x_{n_{1}}^{k+1}x_{n_{1}+1}\in M.$ $None$

Suppose $k=1$ and $n_{1}>1$. By (7.93),

$\zeta_{1}x_{n_{1}}=(x_{n_{1}-1}y_{n_{1}}-x_{n_{1}}y_{n_{1}-1})x_{n_{1}}\in
M.$ $None$

Observe

$(E_{n_{1}+1,n+n_{1}-1}+E_{n_{1}-1,n+n_{1}+1})|_{\cal
B}=x_{n_{1}+1}\partial_{y_{n_{1}-1}}-y_{n_{1}+1}\partial_{x_{n_{1}-1}}$ $None$

by (1.15). So

$-(E_{n_{1}+1,n+n_{1}-1}+E_{n_{1}-1,n+n_{1}+1})(\zeta_{1}x_{n_{1}})=x_{n_{1}}^{2}x_{n_{1}+1}-x_{n_{1}}y_{n_{1}}y_{n_{1}+1}\in
M.$ $None$

On the other hand, (1.16) gives

$(E_{n+i,j}+E_{n+j,i})|_{\cal
B}=-(x_{i}y_{j}+x_{j}y_{i})\qquad\mbox{for}\;\;i,j\in\overline{1,n_{1}},$
$None$

which implies

$-E_{n+n_{1},n_{1}}(y_{n_{1}+1})=x_{n_{1}}y_{n_{1}}y_{n_{1}+1}\in M.$ $None$

By (7.102), we have $x_{n_{1}}^{2}x_{n_{1}+1}\in M.$ So (7.99) always holds.

By Step 1,

${\cal M}^{1}_{\langle-k-1\rangle}{\cal M}^{2}_{\langle 1\rangle}=U({\cal
G}_{1})U({\cal G}_{2})(x_{n_{1}}^{k+1}x_{n_{1}+1})\subset M.$ $None$

Suppose

${\cal M}^{1}_{\langle-k-i\rangle}{\cal M}^{2}_{\langle i\rangle}\subset M$
$None$

for $1\leq i\leq m$. Then

$\displaystyle(E_{n_{1}+1,n_{1}}-E_{n+n_{1},n+n_{1}+1})(x_{n_{1}}^{k+m}x_{n_{1}+1}^{m})$
$\displaystyle=$ $\displaystyle
y_{n_{1}}x_{n_{1}}^{k+m}x_{n_{1}+1}^{m}y_{n_{1}+1}-x_{n_{1}}^{k+m+1}x_{n_{1}+1}^{m+1}\in
M\hskip 142.26378pt(7.107)$

by (7.96). If $m>1$, we have

$y_{n_{1}}x_{n_{1}}^{k+m}x_{n_{1}+1}^{m}y_{n_{1}+1}\in{\cal
M}^{1}_{\langle-k-(m-1)\rangle}{\cal M}^{2}_{\langle m-1\rangle}\subset M.$
$None$

Note

$(E_{r,n+s}+E_{s,n+r})|_{\cal
B}=-(x_{r}y_{s}+x_{s}y_{r})\qquad\mbox{for}\;\;r,s\in\overline{n_{1}+1,n}$
$None$

by (1.15). If $m=1$, we have

$y_{n_{1}}x_{n_{1}}^{k+1}x_{n_{1}+1}y_{n_{1}+1}=-E_{n_{1}+1,n+n_{1}+1}(y_{n_{1}}x_{n_{1}}^{k+1})\subset
E_{n_{1}+1,n+n_{1}+1}({\cal M}^{1}_{\langle-k\rangle})\subset M.$ $None$

Then (7.107), (7.108) and (7.110) give

$x_{n_{1}}^{k+m+1}x_{n_{1}+1}^{m+1}\in M.$ $None$

Furthermore,

${\cal M}^{1}_{\langle-k-m-1\rangle}{\cal M}^{2}_{\langle m+1\rangle}=U({\cal
G}_{1})U({\cal G}_{2})(x_{n_{1}}^{k+m+1}x_{n_{1}+m+1})\subset M.$ $None$

Thus (7.106) holds for any $i\in\mathbb{N}+1$. Symmetrically, we have

${\cal M}^{1}_{\langle i\rangle}{\cal M}^{2}_{\langle-k-i\rangle}\subset
M\qquad\mbox{for}\;\;i\in\mathbb{N}+1.$ $None$

Suppose $n_{1}<n-1$. Then $x_{n_{1}}^{k+1}x_{n_{1}+1}\zeta_{2}\in M$ by
(7.105). Moreover,

$(k+1)y_{n_{1}}x_{n_{1}}^{k}y_{n_{1}+1}\zeta_{2}=-(k+1)E_{n+n_{1},n_{1}}(x_{n_{1}}^{k-1}y_{n_{1}+1}\zeta_{2})\in
M$ $None$

by (7.92) and (7.103). According (1.7), (1.8) and (1.14),

$(E_{n_{1},n_{1}+1}-E_{n+n_{1}+1,n+n_{1}})|_{\cal
B}=\partial_{x_{n_{1}}}\partial_{x_{n_{1}+1}}-\partial_{y_{n_{1}}}\partial_{y_{n_{1}+1}}.$
$None$

Thus

$\displaystyle(E_{n_{1},n_{1}+1}-E_{n+n_{1}+1,n+n_{1}})[(x_{n_{1}}^{k+1}x_{n_{1}+1}-(k+1)y_{n_{1}}x_{n_{1}}^{k}y_{n_{1}+1})\zeta_{2}]$
$\displaystyle=$ $\displaystyle 3(k+1)x_{n_{1}}^{k}\zeta_{2}\in M\hskip
273.14662pt(7.116)$

by (7.77). Hence

${\cal M}^{1}_{\langle-k\rangle}{\cal M}^{2}_{\langle 0\rangle}=U({\cal
G}_{1})U({\cal G}_{2})(x_{n_{1}}^{k})+U({\cal G}_{1})U({\cal
G}_{2})(x_{n_{1}}^{k}\zeta_{2})\subset M$ $None$

by (7.26) and(7.85). Symmetrically,

${\cal M}^{1}_{\langle 0\rangle}{\cal M}^{2}_{\langle-k\rangle}\subset M.$
$None$

By (7.92)-(7.95), (7.106), (7.112), (7.113), (7.117) and (7.118),

${\cal M}^{1}_{\langle-k-r\rangle}{\cal M}^{2}_{\langle r\rangle}\subset
M\qquad\mbox{for}\;\;r\in\mathbb{Z}.$ $None$

Therefore,

${\cal B}_{\langle-k\rangle}=\bigoplus_{r\in\mathbb{Z}}{\cal M}^{1}_{\langle-
k-r\rangle}{\cal M}^{2}_{\langle r\rangle}\subset M.$ $None$

We get $M={\cal B}_{\langle-k\rangle}$, that is, ${\cal B}_{\langle-k\rangle}$
is an irreducible $sp(2n,\mathbb{F})$-module. We can similarly prove that
${\cal B}_{\langle k\rangle}$ is an irreducible $sp(2n,\mathbb{F})$-module.

Finally, we study ${\cal B}_{\langle 0\rangle}$. We first consider the generic
case $1<n_{1}<n-1$. Set

$\displaystyle\qquad{\cal B}_{\langle 0,1\rangle}$ $\displaystyle=$
$\displaystyle\mbox{Span}\\{[\prod_{1\leq r\leq s\leq
n_{1}\;\mbox{or}\;n_{1}+1\leq r\leq s\leq
n}(x_{r}y_{s}+x_{s}y_{r})^{l_{r,s}}]$
$\displaystyle\qquad\;\;\times[\prod_{p=1}^{n_{1}}\prod_{q=n_{1}+1}^{n}(x_{p}x_{q}-y_{p}y_{q})^{k_{p,q}}]\mid
l_{r,s},k_{p,q}\in\mathbb{N}\\}\hskip 85.35826pt(7.121)$

and

${\cal B}_{\langle 0,2\rangle}=\sum_{1\leq r<s\leq
n_{1}\;\mbox{or}\;n_{1}+1\leq r<s\leq n}{\cal B}_{\langle
0,1\rangle}(x_{r}y_{s}-x_{s}y_{r})+\sum_{p=1}^{n_{1}}\sum_{q=n_{1}+1}^{n}{\cal
B}_{\langle 0,1\rangle}(x_{p}x_{q}+y_{p}y_{q}).$ $None$

We want to prove that ${\cal B}_{\langle 0,1\rangle}$ and ${\cal B}_{\langle
0,2\rangle}$ forms $sp(2n,\mathbb{F})$-submodules.

Let

$\displaystyle\qquad{\cal G}^{\prime}$ $\displaystyle=$
$\displaystyle\sum_{1\leq r\leq s\leq
n_{1}}\mathbb{F}(E_{n+s,r}+E_{n+r,s})+\sum_{n_{1}+1\leq p\leq q\leq
n}\mathbb{F}(E_{p,n+q}+E_{q,n+p})$
$\displaystyle+\sum_{r=1}^{n_{1}}\sum_{p=n_{1}+1}^{n}\mathbb{F}(E_{p,r}-E_{n+r,n+p})\hskip
199.16928pt(7.123)$

and

$\displaystyle\hat{\cal G}$ $\displaystyle=$
$\displaystyle\sum_{i,j=1}^{n_{1}}\mathbb{F}(E_{i,j}-E_{n+j,n+i})+\sum_{r,s=n_{1}+1}^{n}\mathbb{F}(E_{r,s}-E_{n+s,n+r})+\sum_{1\leq
r\leq s\leq n_{1}}\mathbb{F}(E_{r,n+s}+E_{s,n+r})$
$\displaystyle+\sum_{n_{1}+1\leq p\leq q\leq
n}\mathbb{F}(E_{n+q,p}+E_{n+p,q})+\sum_{r=1}^{n_{1}}\sum_{p=n_{1}+1}^{n}[\mathbb{F}(E_{r,p}-E_{n+p,n+r})$
$\displaystyle+\mathbb{F}(E_{r,n+p}+E_{p,n+r})+\mathbb{F}(E_{n+r,p}-E_{n+p,r})].\hskip
165.02606pt(7.124)$

Then ${\cal G}^{\prime}$ and $\hat{\cal G}$ are Lie subalgebras of
$sp(2n,\mathbb{F})$ and $sp(2n,\mathbb{F})={\cal G}^{\prime}\oplus\hat{\cal
G}.$ By PBW Theorem $U(sp(2n,\mathbb{F}))=U({\cal G}^{\prime})U(\hat{\cal
G}).$

By (7.96), (7.103) and (7.109),

$U({\cal G}^{\prime})|_{\cal B}={\cal B}_{\langle 0,1\rangle}\;\;\mbox{as
multiplication operators on}\;\;{\cal B}.$ $None$

Moreover,

$(E_{r,s}-E_{n+s,n+r})|_{\cal
B}=x_{r}\partial_{x_{s}}+y_{r}\partial_{y_{s}}+\delta_{r,s},$ $None$
$(E_{n+r,s}+E_{n+s,r})|_{\cal
B}=\partial_{x_{r}}\partial{y_{s}}+\partial_{x_{s}}\partial{y_{r}},$ $None$
$(E_{n+r,i}+E_{n+i,r})|_{\cal B}=-x_{i}\partial{y_{r}}+y_{i}\partial{x_{r}},$
$None$ $(E_{i,n+r}+E_{r,n+i})|_{\cal
B}=-y_{r}\partial{x_{i}}+x_{r}\partial{y_{i}},$ $None$
$(E_{i,r}-E_{n+r,n+i})|_{\cal
B}=\partial_{x_{i}}\partial{x_{r}}-\partial_{y_{i}}\partial{y_{r}}$ $None$

for $i\in\overline{1,n_{1}}$ and $r,s\in\overline{n_{1}+1,n}$. According to
(7.4), (7.6), (7.124) and (7.126)-(7.130), $U(\hat{\cal G})(1)=\mathbb{F}.$
Thus

${\cal B}_{\langle 0,1\rangle}=U({\cal
G}^{\prime})(1)=U(sp(2n,\mathbb{F}))(1)$ $None$

forms an $sp(2n,\mathbb{F})$-submodule.

Let

$W=\sum_{1\leq r<s\leq n_{1}\;\mbox{or}\;n_{1}+1\leq r<s\leq
n}\mathbb{F}(x_{r}y_{s}-x_{s}y_{r})+\sum_{p=1}^{n_{1}}\sum_{q=n_{1}+1}^{n}\mathbb{F}(x_{p}x_{q}+y_{p}y_{q}).$
$None$

By (7.4), (7.6) and (7.126)-(7.130), we can verify that $W$ forms an
irreducible $\hat{\cal G}$-submodule. Hence

${\cal B}_{\langle 0,2\rangle}=U({\cal
G}^{\prime})(W)=U(sp(2n,\mathbb{F}))(W)$ $None$

forms an $sp(2n,\mathbb{F})$-submodule. Moreover,

${\cal B}_{\langle 0,1\rangle}\bigcap W=\\{0\\}.$ $None$

Next we want to prove that ${\cal B}_{\langle 0,1\rangle}$ and ${\cal
B}_{\langle 0,2\rangle}$ are irreducible $sp(2n,\mathbb{F})$-submodules.
According to (7.78), the singular vectors of $sl(n,\mathbb{F})$ in ${\cal
B}_{\langle 0\rangle}$ are

$\displaystyle\\{x_{n_{1}}^{m_{1}}y_{n_{1}}^{m_{1}}\zeta_{1}^{m_{2}+1},x_{n_{1}+1}^{m_{1}}y_{n_{1}+1}^{m_{1}}\zeta_{2}^{m_{2}+1},\eta^{m_{3}}(x_{n_{1}}^{m_{4}}y_{n_{1}+1}^{m_{5}})$
$\displaystyle\mid m_{i}\in\mathbb{N};m_{4}+m_{5}=2m_{3}\\}.\hskip
233.3125pt(7.135)$

Let $M$ be a nonzero submodule of ${\cal B}_{\langle 0,1\rangle}$. Then $M$
contains a singular vector of $sl(n,\mathbb{F})$. Suppose some
$x_{n_{1}}^{m_{1}}y_{n_{1}}^{m_{1}}\zeta_{1}^{m_{2}}\in M$. By (7.79)-(7.82),
we can assume $m_{2}=0,1$. If $m_{2}=0$, (7.84) yields $1\in M$. Then $M={\cal
B}_{\langle 0,1\rangle}$ by (7.131). Suppose $m_{2}=1$. We have
$E_{n_{1},n+n_{1}}|_{\cal B}=\partial_{x_{n_{1}}}\partial_{y_{n_{1}}}$ by
(7.4), and

$E_{n_{1},n+n_{1}}[x_{n_{1}}^{m_{1}}y_{n_{1}}^{m_{1}}\zeta_{1}]=m_{1}(m_{1}+1)x_{n_{1}}^{m_{1}-1}y_{n_{1}}^{m_{1}-1}\zeta_{1}$
$None$

by (7.79). By induction on $m_{1}$, we have $\zeta_{1}\in M\subset{\cal
B}_{\langle 0,1\rangle}$, which contradicts (7.134). Similarly, if some
$x_{n_{1}+1}^{m_{1}}y_{n_{1}+1}^{m_{1}}\zeta_{2}^{m_{2}+1}\in M$, we have
$M={\cal B}_{\langle 0,1\rangle}$. Assume some
$\eta^{m_{3}}(x_{n_{1}}^{m_{4}}y_{n_{1}+1}^{m_{5}})\in M$ with
$m_{4}+m_{5}=2m_{3}$. Note $m_{4}$ and $m_{5}$ are both even or odd. If
$m_{4}=2r_{1}$ and $m_{5}=2r_{2}$ are even, then (7.88) gives $1\in M$,
equivalently $M={\cal B}_{\langle 0,1\rangle}$. Suppose that $m_{4}=2r_{1}+1$
and $m_{5}=2r_{2}+1$ are odd. Expression (7.75) yields

$\eta(x_{n_{1}}y_{n_{1}+1})=x_{n_{1}}x_{n_{1}+1}+y_{n_{1}}y_{n_{1}+1}\in
M\subset{\cal B}_{\langle 0,1\rangle},$ $None$

which contradicts (7.134) again. Thus we always have $M={\cal B}_{\langle
0,1\rangle}$, that is, ${\cal B}_{\langle 0,1\rangle}$ is irreducible.
Similarly, we can prove that ${\cal B}_{\langle 0,2\rangle}$ is irreducible.

If $n_{1}=1$ and $n=2$, we let

${\cal B}_{\langle
0,1\rangle}=\mbox{Span}\\{[\prod_{i=1}^{n}(x_{i}y_{i})^{m_{i}}](x_{1}x_{2}-y_{1}y_{2})^{m_{3}}]\mid
m_{i}\in\mathbb{N}\\}$ $None$

and ${\cal B}_{\langle 0,2\rangle}={\cal B}_{\langle
0,1\rangle}(x_{1}x_{2}+y_{1}y_{2}).$ When $n_{1}=1$ and $n>2$, we set

${\cal B}_{\langle 0,1\rangle}=\mbox{Span}\\{[(x_{1}y_{1})^{l}\prod_{2\leq
r\leq s\leq
n}(x_{r}y_{s}+x_{s}y_{r})^{l_{r,s}}][\prod_{q=2}^{n}(x_{1}x_{q}-y_{1}y_{q})^{k_{q}}]\mid
l,l_{r,s},k_{q}\in\mathbb{N}\\}$ $None$

and

${\cal B}_{\langle 0,2\rangle}=\sum_{2\leq r<s\leq n}{\cal B}_{\langle
0,1\rangle}(x_{r}y_{s}-x_{s}y_{r})+\sum_{q=2}^{n}{\cal B}_{\langle
0,1\rangle}(x_{1}x_{q}+y_{1}y_{q}).$ $None$

In the case $1<n_{1}=n-1$, we put

$\displaystyle\qquad{\cal B}_{\langle 0,1\rangle}$ $\displaystyle=$
$\displaystyle\mbox{Span}\\{(x_{n}y_{n})^{l}[\prod_{1\leq r\leq s\leq
n-1}(x_{r}y_{s}+x_{s}y_{r})^{l_{r,s}}]$
$\displaystyle\qquad\;\;\times[\prod_{p=1}^{n-1}(x_{p}x_{n}-y_{p}y_{n})^{k_{p}}]\mid
l,l_{r,s},k_{p}\in\mathbb{N}\\}\hskip 119.50148pt(7.141)$

and

${\cal B}_{\langle 0,2\rangle}=\sum_{1\leq r<s\leq n_{1}}{\cal B}_{\langle
0,1\rangle}(x_{r}y_{s}-x_{s}y_{r})+\sum_{p=1}^{n_{1}}{\cal B}_{\langle
0,1\rangle}(x_{p}x_{n}+y_{p}y_{n}).$ $None$

The above corresponding partial arguments show that ${\cal B}_{\langle
0,1\rangle}$ and ${\cal B}_{\langle 0,2\rangle}$ are irreducible in the
corresponding case.

Now $1$ is a non-isotropic element in ${\cal B}_{\langle 0,1\rangle}$ and
$x_{n_{1}}x_{n_{1}+1}+y_{n_{1}}y_{n_{1}+1}$ a non-isotropic element in ${\cal
B}_{\langle 0,2\rangle}$ by (3.54). By Lemma 2.3, the symmetric bilinear form
$(\cdot|\cdot)$ restricted to them are nondegenrate. Since $(1|{\cal
B}_{\langle 0,2\rangle})=\\{0\\}$, ${\cal B}_{\langle 0,1\rangle}$ is
orthogonal to ${\cal B}_{\langle 0,2\rangle}$. Thus the symmetric bilinear
form $(\cdot|\cdot)$ restricted ${\cal B}_{\langle 0,1\rangle}+{\cal
B}_{\langle 0,2\rangle}$ is nondegenerate. Then

${\cal B}_{\langle 0\rangle}=({\cal B}_{\langle 0,1\rangle}+{\cal B}_{\langle
0,2\rangle})\oplus({\cal B}_{\langle 0,1\rangle}+{\cal B}_{\langle
0,2\rangle})^{\perp}\bigcap{\cal B}_{\langle 0\rangle}.$ $None$

If $({\cal B}_{\langle 0,1\rangle}+{\cal B}_{\langle
0,2\rangle})^{\perp}\bigcap{\cal B}_{\langle 0\rangle}\neq\\{0\\}$, then it
contains a singular vector of $sl(n,\mathbb{F})$. Our above arguments in
proving the irreducibility of ${\cal B}_{\langle 0,1\rangle}$ show that it
contains either ${\cal B}_{\langle 0,1\rangle}$ or ${\cal B}_{\langle
0,2\rangle}$, which is absurd. Therefore, ${\cal B}_{\langle 0\rangle}={\cal
B}_{\langle 0,1\rangle}\oplus{\cal B}_{\langle 0,2\rangle}$ is an orthogonal
decomposition of irreducible $sp(2n,\mathbb{F})$-submodules.

Suppose $n_{1}=n_{2}=n$. For $k\in\mathbb{N}+1$, (4.21) and (4.31) imply that

${\cal B}_{\langle
k\rangle}=\bigoplus_{m=0}^{\infty}\bigoplus_{r=\llbracket(k+1)/2\rrbracket}^{\infty}\eta^{r}({\cal
H}_{\langle k-2r-m,m\rangle})$ $None$

and

${\cal B}_{\langle-k\rangle}=\bigoplus_{m,r=0}^{\infty}\eta^{r}({\cal
H}_{\langle-k-2r-m,m\rangle})$ $None$

are $({\cal G},{\cal K})$-structures, where ${\cal H}_{\langle-
m_{1}-m_{2},m_{2}\rangle}$ is given in (4.42). Moreover,

${\cal B}_{\langle 0,1\rangle}=\bigoplus_{m,r=0}^{\infty}\eta^{r}({\cal
H}_{\langle-2r-2m,2m\rangle})$ $None$

and

${\cal B}_{\langle 0,2\rangle}=\bigoplus_{m,r=0}^{\infty}\eta^{r}({\cal
H}_{\langle-2r-2m-1,2m+1\rangle})$ $None$

are $({\cal G},{\cal K})$-structures by the arguments in (7.79)-(7.82), (7.84)
and (7.136) (cf. (7.24), (7.25)). $\qquad\Box$

References

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B. Cao, Solutions of Navier Equations and Their Representation Structure,
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[DES]

M. Davidson, T. Enright, and R. Stanke, Differential Operators and Highest
Weight Representations, Memoirs of American Mathematical Society 94, no. 455,
1991.

[FC]

F. M. Fernández and E. A. Castro, Algebraic Methods in Quantum Chemistry and
Physics, CRC Press, Inc., 1996.

[FSS]

L.Frappat, A.Sciarrino and P.sorba, Dictionary on Lie Algebras and
Superalgebras, Academic Press,2000

[G]

H. Georgi, Lie Algebras in Particle Physics, Second Edition, Perseus Books
Group, 1999.

[Ho]

R. Howe, Perspectives on invariant theory: Schur duality, multiplicity-free
actions and beyond, The Schur lectures (1992) (Tel Aviv), 1-182, Israel Math.
Conf. Proc., 8, Bar-Ilan Univ., Ramat Gan, 1995.

[Hu]

J. E. Humphreys, Introduction to Lie Algebras and Representation Theory,
Springer-Verlag New York Inc., 1972.

[K]

V. G. Kac, Infinite Dimensional Lie Algebras, Third edition, Cambridge
University Press, 1990.

[I1]

N. H. Ibragimov, Transformation groups applied to mathematical physics, Nauka,
1983.

[I2]

N. H. Ibragimov, Lie Group Analysis of Differential Equations, Volume 2, CRC
Handbook, CRC Press, 1995.

[L]

F. S. Levin, An Introduction to Quantum Theory, Cambridge University Press,
2002.

[LF]

W. Ludwig and C. Falter, Symmetries in Physics, Second Edition, Springer-
Verlag, Berlin/Heidelberg, 1996.

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X.Xu, Flag partial differential equations and representations of Lie algebras,
Acta Appl Math 102 (2008), 249–280.


# An Analogue to Infinitery Hales-Jewett Theorem

ANINDA CHAKRABORTY AND SAYAN GOSWAMI<EMAIL_ADDRESS>Government
General Degree College at Chapra, Chapra<EMAIL_ADDRESS>Department of
Mathematics, University of Kalyani, Kalyani.

###### Abstract.

In a recent work, N. Hindman, D. Strauss and L. Zamboni have shown that the
Hales-Jewett theorem can be combined with a sufficiently well behaved
homomorphisms. In this paper we will show that those combined extensions can
be made if we replace the alphabet by an increasing sequence of alphabets,
infact it holds for some Ramsey theoretic small sets. To obtained this we
achieved some interesting cofigurations.

##  1\. Preliminaaries

Let $\omega=\mathbb{N}\cup\left\\{0\right\\}$, where $\mathbb{N}$ is the set
of positive integers. Then $\omega$ is the first infinite ordinal. Let
$\mathcal{P}_{f}\left(X\right)$ be the set of all finite subsets of a set $X$.

Given a nonempty set $\mathbb{A}$ (or alphabet) we let
$w\left(\mathbb{A}\right)$ be the set of all finite words $w=a_{1}a_{2}\ldots
a_{n}$ with $n\geq 1$ and $a_{i}\in\mathbb{A}$. The quantity $n$ is called the
length of $w$ and denoted $\left|w\right|$. The set $w\left(\mathbb{A}\right)$
is naturally semigroup under the operation of concatenation of words. We will
denote the empty word by $\theta.$ For each $u\in w\left(\mathbb{A}\right)$
and $a\in\mathbb{A}$, we let $\left|u\right|_{a}$ be the number of occurrences
of $a$ in $u$, we will identify the elements of $\mathbb{A}$ with the length-
one words over $\mathbb{A}$.

Let $v$ (a variable) be a letter not belonging to $\mathbb{A}$. By a variable
word over $\mathbb{A}$ we mean a word $w$ over
$\mathbb{A}\cup\left\\{v\right\\}$ with $\left|w\right|_{v}\geq 1$. We let
$S_{1}\left(\mathbb{A}\right)$ be the set of one-variable words over
$\mathbb{A}$. If $w\in S_{1}\left(\mathbb{A}\right)$ and $a\in\mathbb{A}$,
then $w\left(a\right)\in w\left(\mathbb{A}\right)$ is the result of replacing
each occurrence of $v$ by $a$.

A finite coloring of a set $A$ is a function from $A$ to a finite set
$\left\\{1,2,\ldots,n\right\\}$. A subset $B$ of $A$ is monochromatic if the
function is constant on $B$. If $\mathbb{A}$ be any finite nonempty set and
$S$ be the free semigroup of all words over the alphabet $\mathbb{A}$, then
the Hales-Jewett Theorem states that for any finite coloring of the $S$ there
is a variable word over $\mathbb{A}$ all of whose instances are the same
color.

######  Theorem 1.1.

[HJ] Assume that $\mathbb{A}$ is finite. For each finite coloring of $S_{0}$
there exists a variable word $w$ such that
$\left\\{w\left(a\right):a\in\mathbb{A}\right\\}$ is monochromatic.

Now, we need to recall some definitions from [HSZ].

######  Definition 1.2.

Let $n\in\mathbb{N}$ and $v_{1},v_{2},\ldots,v_{n}$ be distinct variables
which are not members of $\mathbb{A}$.

(a) An $n$-variable word over $\mathbb{A}$ is a word $w$ over
$\mathbb{A}\cup\left\\{v_{1},v_{2},\ldots,v_{n}\right\\}$ such that
$\left|w\right|_{v_{i}}\geq 1$ for each $i\in\left\\{1,2,\ldots,n\right\\}$.

(b) If $w$ is an $n$-variable word over $\mathbb{A}$ and
$\vec{x}=\left(x_{1},x_{2},\ldots,x_{n}\right)$, then $w\left(\vec{x}\right)$
is the result of replacing each occurrence of $v_{i}$ in $w$ by $x_{i}$ for
each $i\in\left\\{1,2,\ldots,n\right\\}$.

(c) If $w$ is an $n$-variable word over $\mathbb{A}$ and $u=a_{1}a_{2}\ldots
a_{n}$ is a length $n$ word, then $w\left(u\right)$ is the result of replacing
each occurrence of $v_{i}$ in $w$ by $a_{i}$ for each
$i\in\left\\{1,2,\ldots,n\right\\}$.

This particular homomorphism is very useful to us.

######  Definition 1.3.

Let $n\in\mathbb{N}$, for a set of alphabet $\mathbb{A}$, let
$\vec{a}\in\mathbb{A}^{n}$. Than the homomorphism
$h_{\vec{a}}:S_{n}\left(\mathbb{A}\right)\rightarrow w\left(\mathbb{A}\right)$
is defined by $h_{\vec{a}}\left(w\right)=w\left(\vec{a}\right)$ for all $w\in
h_{\vec{a}}:S_{n}\left(\mathbb{A}\right)$.

######  Definition 1.4.

Let $S,T\text{ and }R$ be semigroups (or partial semigroups) such that $S\cup
T$ is a semigroup (or partial semigroup) and $T$ is an ideal of $S\cup T$.
Then a homomorphism $\tau:T\rightarrow R$ is said to be $S$-independent if,
for every $w\in T$ and every $u\in S$,

$\tau\left(uw\right)=\tau\left(w\right)=\tau\left(wu\right).$

For an example, let $T$ be a semigroup with identity $e$. Then for any $n\geq
1$, a homomorphism $\tau:S_{n}\left(\mathbb{A}\right)\cup
S_{0}\left(\mathbb{A}\right)\rightarrow T$ is
$S_{0}\left(\mathbb{A}\right)$-independent if
$\tau\left[S_{0}\left(\mathbb{A}\right)\right]=\left\\{e\right\\}$.

The following is a version of a multi-variable extension of the Hales-Jewett
Theorem:

######  Theorem 1.5.

Assume that $\mathbb{A}$ is finite. Let $w\left(\mathbb{A}\right)$ be finitely
colored and let $n\in\mathbb{N}$. There exists $w\in
S_{n}\left(\mathbb{A}\right)$ such that
$\left\\{w\left(\vec{x}\right):\vec{x}\in\mathbb{A}^{n}\right\\}$ is
monochromatic.

Let,

$\mathbb{A}_{1}\subseteq\mathbb{A}_{2}\subseteq\mathbb{A}_{3}\subseteq\ldots$

is an increasing sequence of alphabets. Then for each $n,i\in\mathbb{N}$,
$S_{n}\left(\mathbb{A}_{i}\right)$ is the set of all $n$-variable words over
$\mathbb{A}_{i}$.

Let, $S_{n}=\bigcup_{i=1}^{\infty}S_{n}\left(\mathbb{A}_{i}\right)$ be the set
of all $n$-variable words over
$\mathbb{A}=\bigcup_{i=1}^{\infty}\mathbb{A}_{i}$ and
$S_{0}=\bigcup_{i=1}^{\infty}w\left(\mathbb{A}_{i}\right)$ be the set of all
words of finite length over $\mathbb{A}=\bigcup_{i=1}^{\infty}\mathbb{A}_{i}$.

The following theorem is due to N.Karagiannis, which is a stronger version of
weak Carlson-Simpson theorem.

######  Theorem 1.6.

[K, Theorem 3] Let $\left(\mathbb{A}_{i}\right)_{i=1}^{\infty}$ be an
increasing sequence of finite alphabets and let
$\mathbb{A}=\bigcup_{i=1}^{\infty}\mathbb{A}_{i}$. Then for every finite
coloring of $S_{0}$ there exists a sequence
$\left(w_{n}\left(x\right)\right)_{n=0}^{\infty}$ of variable words over
$\mathbb{A}$ such that for every $n\in\mathbb{N}$ and every
$m_{1}<m_{2}<\ldots<m_{n}$, the words of the form
$w_{m_{0}}\left(a_{0}\right)w_{m_{1}}\left(a_{1}\right)\ldots
w_{m_{n}}\left(a_{n}\right)$ with $a_{i}\in\mathbb{A}_{m_{i}}$,
$i\in\left\\{i,2,\ldots,n\right\\}$ are of the same color.

In [HSZ], authors have shown various combining extension of Hales-Jewett
theorem. After that N. Karagiannis has proved that weak Carlson-Simpson
theorem holds if one replaced the set of alphabet $\mathbb{A}$ by an
increasing chain of alphabets
$\mathbb{A}_{1}\subseteq\mathbb{A}_{2}\subseteq\mathbb{A}_{3}\subseteq\ldots$.

In this paper, we used the Stone-Čech compactification of a semigroup (or
partial semigroup) $S$ to develop some results from [HSZ] in the infinite
alphabet settings.

Let $\left(S,\cdot\right)$ be any discrete semigroup and denote its Stone-Čech
compactification as $\beta S$. $\beta S$ is the set of all ultrafilters on
$S$, where the points of $S$ are identified with the principal ultrafilters.
The basis for the topology is $\left\\{\bar{A}:A\subseteq S\right\\}$, where
$\bar{A}=\left\\{p\in\beta S:A\in p\right\\}$. The operation of $S$ can be
extended to $\beta S$ making $\left(\beta S,\cdot\right)$ a compact, right
topological semigroup with $S$ contained in its topological center. That is,
for all $p\in\beta S$ the function $\rho_{p}:\beta S\rightarrow\beta S$ is
continuous, where $\rho_{p}\left(q\right)=q\cdot p$ and for all $x\in S$, the
function $\lambda_{x}:\beta S\rightarrow\beta S$ is continuous, where
$\lambda_{x}\left(q\right)=x\cdot q$. For $p,q\in\beta S$ and $A\subseteq S$,
$A\in p\cdot q$ if and only if $\left\\{x\in S:x^{-1}A\in q\right\\}\in p$,
where $x^{-1}A=\left\\{y\in S:x\cdot y\in A\right\\}$.

Since $\beta S$ is a compact Hausdorff right topological semigroup, it has a
smallest two sided ideal denoted $K\left(\beta S\right)$, which is the union
of all of the minimal right ideals of $S$, as well as the union of all of the
minimal left ideals of $S$. Every left ideal of $\beta S$ contains a minimal
left ideal and every right ideal of $\beta S$ contains a minimal right ideal.
The intersection of any minimal left ideal and any minimal right ideal is a
group and any two such groups are isomorphic. Any idempotent $p$ in $\beta S$
is said to be minimal if and only if $p\in K\left(\beta S\right)$. A subset
$A$ of $S$ is then central if and only if there is some minimal idempotent $p$
such that $A\in p$. For more details, the reader can see [HS].

We need to use some elementary structure of partial semigroup. Here we recall
some definitions of our need from [HP].

######  Definition 1.7.

A partial semigroup is defined as a pair $\left(G,\ast\right)$ where $\ast$ is
an operation defined on a subset $X$ of $G\times G$ and satisfies the
statement that for all $x,y,z$ in $G$, $\left(x\ast y\right)\ast
z=x\ast\left(y\ast z\right)$ in the sense that if either side is defined, so
is the other and they are equal.

If $\left(G,\ast\right)$ is a partial semigroup, we will denote it by $G$,
when the operation $\ast$ is clear from the context. Now, we give an example
which will be useful in our work.

######  Example 1.8.

Let us consider any sequence $\left\langle x_{n}\right\rangle_{n=1}^{\infty}$
in $\omega$ and let

$G=\text{FS}\left(\left\langle
x_{n}\right\rangle_{n=1}^{\infty}\right)=\left\\{\sum_{j\in
H}x_{j}:H\in\mathcal{P}_{f}\left(\mathbb{N}\right)\right\\}$

and

$X=\left\\{\left(\sum_{j\in H_{1}}x_{j},\sum_{j\in
H_{2}}x_{j}\right):H_{1}\cap H_{2}=\emptyset\right\\}.$

Define $\ast:X\rightarrow G$ by

$\left(\sum_{j\in H_{1}}x_{j},\sum_{j\in
H_{2}}x_{j}\right)\longrightarrow\sum_{j\in H_{1}}x_{j}+\sum_{j\in
H_{2}}x_{j}.$

It is easy to check that $G$ is a commutative partial semigroup. One can
similarly check the same for

$G=\text{FP}\left(\left\langle
x_{n}\right\rangle_{n=1}^{\infty}\right)=\left\\{\prod_{j\in
H}x_{j}:H\in\mathcal{P}_{f}\left(\mathbb{N}\right)\right\\}.$

######  Definition 1.9.

Let $G$ be a partial semigroup.

(a) For $g\in G$, $\varphi\left(g\right)=\left\\{h\in G:g\ast h\text{ is
defined}\right\\}$.

(b) For $H\in\mathcal{P}_{f}\left(G\right)$,
$\sigma\left(H\right)=\bigcap_{h\in H}\varphi\left(h\right)$.

(c) $\left(G,\ast\right)$ is adequate if and only if
$\sigma\left(H\right)\neq\emptyset$ for all
$H\in\mathcal{P}_{f}\left(G\right)$.

Now, for a semigroup $S$, let ${}^{\mathbb{N}}S$ be the set of all sequences
in $S$ and let

$\mathcal{J}_{m}=\left\\{t=\left(t_{1},t_{2},\ldots,t_{m}\right)\in\mathbb{N}^{m}:t_{1}<t_{2}<...<t_{m}\right\\}.$

######  Definition 1.10.

Let $\left(S,\cdot\right)$ be a semigroup and $A\subseteq S$.

  1. (1)

Then $A$ is a $J$-set if and only if for each
$F\in\mathcal{P}_{f}\left({}^{\mathbb{N}}S\right)$ there exist
$m\in\mathbb{N}$, $a=\left(a_{1},a_{2},\ldots,a_{m+1}\right)\in S^{m+1}$ and
$t=\left(t_{1},t_{2},\ldots,t_{m}\right)\in\mathcal{J}_{m}$ such that for each
$f\in F$,

$\left(\prod_{j=1}^{m}a_{j}\cdot f\left(t_{j}\right)\right)\cdot a_{m+1}\in
A.$

  2. (2)

$J\left(S\right)=\left\\{p\in\beta S:\left(\forall A\in p\right)\left(A\text{
is a }J\text{-set}\right)\right\\}.$

  3. (3)

A set $A\subseteq S$ is called $C$-set if it is an element of some idempotent
in $J\left(S\right)$.

Now, we need to recall the definition of $J$-set for partial semigroups.
First, let $\mathcal{F}$ be the set of all adequate sequences in $S$.

######  Definition 1.11.

Let $G$ be an adequate partial semigroup. Then a set $A\subseteq G$ is a
$J$-set if and only if for all $F\in\mathcal{P}_{f}\left(\mathcal{F}\right)$
and all $L\in\mathcal{P}_{f}\left(S\right)$, there exist $m\in\mathbb{N}$,
$a=\left(a_{1},a_{2},\ldots,a_{m+1}\right)\in S^{m+1}$ and
$t=\left(t_{1},t_{2},\ldots,t_{m}\right)\in\mathcal{J}_{m}$ such that for all
$f\in F$,

$\left(\prod_{i=1}^{m}a_{i}\ast f\left(t_{i}\right)\right)\ast a_{m+1}\in
A\cap\sigma\left(L\right).$

Then we have proved one of the main theorem from [HSZ, Theorem 17] for
infinite sequence of alphabets.

##  2\. Our Results

Throughout this section, we have developed various results from [HSZ] using
infinite sequence of alphabets. Some of the results also hold in various small
sets, like $C$-sets, $J$-sets etc. The following fact is very useful to us:

######  Fact 2.1.

Now, let us take the homomorphism $\tau:S_{0}\cup S_{1}\rightarrow\omega$ be
defined by $\tau\left(w\right)=\left|w\right|_{v}$. So,
$\tau\left(w\right)=0\text{ if and only if }w\in S_{0}$. Let $D\subseteq
S_{0}$ is a $J$-set and let $\left\langle x_{n}\right\rangle_{n=1}^{\infty}$
be a sequence in $\mathbb{N}$. Let $y_{1}=0,\,y_{n+1}=x_{n}$, for all
$n\in\mathbb{N}$. Then $A=\text{FS}\left(\left\langle
y_{n}\right\rangle_{n=1}^{\infty}\right)$ is an adequate partial semigroup.
As, $\tau$ is a homomorphism, one can easily check that
$\tau^{-1}\left[A\right]$ is also an adequate partial semigroup. Now, let $T$
is an adequate partial semigroup and $S$ be any partial subsemigroup of $T$.
Let $F$ be a nonempty finite set of partial semigroup homomorphisms from $T$
to $S$, which are $S$-preserving. Then, proceeding similar to the proof of
[CG, Lemma2.1], it can be easily be verified that for any $J$-set $D\subseteq
S$, $\bigcap_{\nu\in F}\nu^{-1}\left[D\right]$ is a $J$-set in $T$. So, let
$T=S_{0}\cup S_{1}$, $\tau^{-1}\left[A\right]$ is also a $J$-set in $T$. Also,
it is easy to check that $S_{0}$ is not a $J$-set in $S_{n}\cup S_{0}$.

The following result are a variation of Hales-Jewett theorem for piecewise
syndetic set:

######  Lemma 2.2.

Let $m,n\in\mathbb{N}$ and
$F=\left\\{h_{\vec{a}}:\vec{a}\in\mathbb{A}_{m}^{n}\right\\}$ be a finite
nonempty set of homomorphism mappings from $T=S_{n}\cup S_{0}\longrightarrow
S_{0}$ which are equal to the identity mapping on $S_{0}$. Let $D\subseteq
S_{0}$ is a piecewise syndetic set in $S_{0}$, then $\bigcap_{\nu\in
F}h_{\vec{a}}^{-1}\left[D\right]$ contains $w_{n}\left(\vec{a}\right)$, for
all $\vec{a}\in\mathbb{A}_{m}^{n}$ and $w_{n}\in
S_{n}\left(\mathbb{A}\right)$.

###### Proof.

Let $\in$$\mathbb{N}$, then from [HSZ, Corollary 6],
$\bigcap_{h_{\vec{a}}^{-1}\in F}h_{\vec{a}}^{-1}\left[D\right]$ is piecewise
syndetic in $T$. Now, $S_{0}$ is not piecewise syndetic in $T$.
So,$S_{n}\cap\bigcap_{h_{\vec{a}}\in F}h_{\vec{a}}^{-1}\left[D\right]$ is
piecewise syndetic . Now, choose $N>m$ such that $w_{n}\in
S_{n}\left(A_{N}\right)\cap\bigcap_{h_{\vec{a}}\in
F}h_{\vec{a}}^{-1}\left[D\right]$. So, $w_{n}\in\bigcap_{h_{\vec{a}}\in
F}h_{\vec{a}}^{-1}\left[D\right]$ and $w_{n}\left(\vec{a}\right)\in D$ for all
$\vec{a}\in A_{m}^{n}$. ∎

Proceeding as above for the $J$-sets, we get the following corollary:

######  Corollary 2.3.

Let $m,n\in\mathbb{N}$ and
$F=\left\\{h_{\vec{a}}:\vec{a}\in\mathbb{A}_{m}^{n}\right\\}$ be a finite
nonempty set of homomorphism mappings from $T=S_{n}\cup S_{0}\longrightarrow
S_{0}$ which are equal to the identity mapping on $S_{0}$. Let $D\subseteq
S_{0}$ is a $J$-set in $S_{0}$, then $\bigcap_{h_{\vec{a}}\in
F}h_{\vec{a}}^{-1}\left[D\right]$ contains $w_{n}\left(\vec{a}\right)$, for
all $\vec{a}\in A_{m}^{n}$ and $w_{n}\in S_{n}\left(A\right)$.

###### Proof.

From [CG, Theorem 2.3] and Lemma 2.2, the result follows. ∎

In the next theorem we have proved [K, Theorem 3] for central sets:

######  Theorem 2.4.

Let $D\subseteq S_{0}$ be central then there exists a sequence
$\left(w_{i}\left(x\right)\right)_{i=1}^{\infty}$ of variable words over $A$
such that for every $n\in\mathbb{N}$ and every

$m_{1}<m_{2}<\ldots<m_{l}$

the words of the form

$w_{m_{1}}\left(\vec{a_{1}}\right)w_{m_{2}}\left(\vec{a_{2}}\right)\ldots
w_{m_{l}}\left(\vec{a_{l}}\right)\in D$

with $a_{1}\in A_{m_{1}}^{n},a_{2}\in A_{m_{2}}^{n},\ldots,a_{l}\in
A_{m_{l}}^{n}$.

###### Proof.

Let us choose a minimal idempotent $p$ such that $D\in p$. Then $D^{*}\in p$,
where $D^{*}=\left\\{x\in D:x^{-1}D\in p\right\\}$. Now, let
$F_{1}=\left\\{h_{\vec{a_{1}}}:\,\vec{a_{1}}\in A_{1}^{n}\right\\}$. So, from
Lemma 2.2, there exists $w\in S_{n}\left(A_{1}\right)$ such that $w_{1}\in
S_{n}\cap\bigcap_{\vec{a_{1}}\in
A_{1}^{n}}h_{\vec{a_{1}}}^{-1}\left[D^{*}\right]\Rightarrow
w_{1}\left(\vec{a_{1}}\right)\in D^{*}$ for all $\vec{a_{1}}\in A_{1}^{n}$.

So, $D^{*}\cap\bigcap_{\vec{a_{1}}\in
A_{1}^{n}}w_{1}\left(\vec{a_{1}}\right)^{-1}\left[D^{*}\right]$.

Let, $p\in\mathbb{N}$ and suppose we have a sequence
$\left(w_{i}\left(x\right)\right)_{i=1}^{p}$ such that

$w_{m_{1}}\left(\vec{a_{1}}\right)\ldots w_{m_{l}}\left(\vec{a_{l}}\right)\in
D$

for all $m_{1}<m_{2}<\ldots<m_{l}\leq p$.

Now, choose

$E=D^{*}\cap\bigcap_{m_{1}<m_{2}<\ldots<m_{l}\leq
p}w_{m_{1}}\left(\vec{a_{1}}\right)\ldots
w_{m_{l}}\left(\vec{a_{l}}\right)^{-1}\left[D^{*}\right]$

. Then, choose $w_{p+1}\in S_{n}\cap\bigcap_{\overrightarrow{a_{p+1}}\in
A_{p+1}^{n}}h_{\overrightarrow{a_{p+1}}}^{-1}\left[E\right]$.

So,

$w_{m_{1}}\left(\vec{a_{1}}\right)\ldots w_{m_{l}}\left(\vec{a_{l}}\right)\in
D^{*}$

, where $m_{1}<m_{2}<\ldots<m_{l}\leq p+1$.

Hence by induction we can choose such infinite sequence
$\left(w_{i}\left(x\right)\right)_{i=1}^{\infty}$. ∎

In fact, an easy combinatorial approach can be followed to prove the following
generalized result of the above theorem, using $C^{*}\in p$, whenever $C\in
p$.

######  Corollary 2.5.

Let $D\subseteq S_{0}$ be $C$-set then there exists a sequence
$\left(w_{n}\left(x\right)\right)_{n=1}^{\infty}$ of variable words over $A$
such that for every $n\in\mathbb{N}$ and every

$m_{1}<m_{2}<\ldots<m_{l}$

the words of the form

$w_{m_{1}}\left(\vec{a_{1}}\right)w_{m_{2}}\left(\vec{a_{2}}\right)\cdots
w_{m_{l}}\left(\vec{a_{l}}\right)\in D$

with $a_{1}\in A_{m_{1}}^{n},a_{2}\in A_{m_{2}}^{n},\ldots,a_{l}\in
A_{m_{l}}^{n}$.

###### Proof.

Same as in Theorem 2.4. ∎

Note that, for a $J$-set $D\subseteq S_{0}$ and a sequence $\left\langle
x_{n}\right\rangle_{n=1}^{\infty}$ in $\mathbb{N}$, let $y_{1}=0$ and
$y_{n+1}=x_{n}$ and $A=\text{FS}\left(\left\langle
y_{n}\right\rangle_{n=1}^{\infty}\right)$ is an adequate partial semigroup.

######  Theorem 2.6.

Let $\tau:T=S_{0}\cup S_{1}\rightarrow\omega$ by
$\tau\left(w\right)=\left|w\right|_{v_{1}}$. Let, $D\subseteq S_{0}$ be a
$J$-set and $\left\langle x_{n}\right\rangle_{n=1}^{\infty}$ be a sequence in
$\mathbb{N}$.Then, there exists $\left\langle
w_{n}\right\rangle_{n=1}^{\infty}$ such that

$\left\\{w_{n}\left(a_{n}\right):a_{n}\in A_{n}\right\\}\subseteq D\ \forall
n\in\mathbb{N}$

and for each $G\in\mathcal{P}_{f}\left(\mathbb{N}\right)$,

$\sum_{n\in G}\tau\left(w_{n}\right)\in\text{FS}\left(\left\langle
x_{n}\right\rangle_{n=1}^{\infty}\right).$

###### Proof.

Let $F_{1}=\left\\{h_{a_{1}}:a_{1}\in A_{1}\right\\}$ be a finite set of
partial semigroup homomorphism from $\tau^{-1}\left(A_{1}\right)$ to $S_{0}$
where $A_{1}=\text{FS}\left(\left\langle
y_{n}\right\rangle_{n=1}^{\infty}\right)$.

Let $D\subseteq S_{0}$ is a $J$-set then one can easily verify that
$\bigcap_{a_{1}\in A_{1}}h_{a_{1}}^{-1}\left[D\right]$ is a $J$-set in
$\tau^{-1}\left(A_{1}\right)$. As, $S_{0}$ is not a $J$-set in
$\tau^{-1}\left(A_{1}\right)$, there exists $w_{1}\in
S_{1}\left(A_{m}\right)\cap\tau^{-1}\left(A_{1}\right)$ for some $m\geq 1$such
that $w_{1}\in\bigcap_{a_{1}\in A_{1}}h_{a_{1}}^{-1}\left[D\right]$.

So, $\tau\left(w_{1}\right)\in\text{FS}\left(\left\langle
x_{n}\right\rangle_{n=1}^{\infty}\right)\subseteq
A_{1}=\text{FS}\left(\left\langle y_{n}\right\rangle_{n=1}^{\infty}\right)$
and $w_{1}\left(a_{1}\right)\in D$ $\forall a_{1}\in A_{1}$. Now by induction,
suppose we have proved that for some $p\in\mathbb{N}$,
$\left\\{w_{n}\left(a_{n}\right):a_{n}\in A_{n}\right\\}\subseteq D$, for all
$n\in\left[1,p\right]$ and for $G\subseteq\left\\{1,2,\ldots,p\right\\}$,
$\sum_{n\in G}\tau\left(w_{n}\right)\in\text{FS}\left(\left\langle
x_{n}\right\rangle_{n=1}^{\infty}\right)$. Now let us take a sequence
$\left\langle z_{n}\right\rangle_{n=1}^{\infty}$ such that $z_{1}=0$ and

$A_{p+1}=\text{FS}\left(\left\langle
z_{n}\right\rangle_{n=1}^{\infty}\right)\subseteq\bigcap_{x\in\left\\{\sum_{n\in
G}\tau\left(w_{n}\right):G\subseteq\left\\{1,2,\ldots,p\right\\}\right\\}}\left(-x+\text{FS}\left(\left\langle
x_{n}\right\rangle_{n=1}^{\infty}\right)\right).$

Then, $\tau^{-1}\left[A_{p+1}\right]$ is adequate partial semigroup and as
above there exists $m\geq p+1$ and a $w_{p+1}\in
S_{1}\left(A_{m}\right)\cap\tau^{-1}\left[A_{p+1}\right]$ such that
$w\in\bigcap_{a_{p+1}\in A_{p+1}}h_{a_{p+1}}^{-1}\left[D\right]$.

So, $\tau\left(w_{p+1}\right)\in A_{p+1}$ and
$\left\\{w\left(a_{p+1}\right):a_{p+1}\in A_{p+1}\right\\}$. This proves the
result. ∎

The following corollary is one of the interesting result:

######  Corollary 2.7.

Let $\tau:T=S_{0}\cup S_{1}\rightarrow\omega$ by
$\tau\left(w\right)=\left|w\right|_{v_{1}}$. Let, $D\subseteq S_{0}$ be a
$C$-set and $\left\langle x_{n}\right\rangle_{n=1}^{\infty}$ be a sequence in
$\mathbb{N}$. Then, there exists $\left\langle
w_{n}\right\rangle_{n=1}^{\infty}$ and for any
$\left\\{n_{1},n_{2},\ldots,n_{i}\right\\}=G\in\mathcal{P}_{f}\left(\mathbb{N}\right)$,
such that

$\left\\{w_{n_{1}}\left(a_{n_{1}}\right)w_{n_{2}}\left(a_{n_{2}}\right)\ldots
w_{n_{l}}\left(a_{n_{l}}\right):a_{n_{i}}\in A_{n_{i}},\text{ where }n_{i}\in
G\right\\}\subseteq D$

and $\sum_{n_{i}\in G}\tau\left(w_{n_{i}}\right)\in\text{FS}\left(\left\langle
x_{n}\right\rangle_{n=1}^{\infty}\right).$

###### Proof.

As $D\subseteq S_{0}$ is a $C$-set, let us choose an idempotent
$p\in\delta\left(S_{0}\right)$ such that $D\in p$. Then, $D^{*}\in p$. Now,
proceeding as the above proof, assume,

$E=D^{*}\cap\bigcap_{G=\left\\{n_{1},n_{2},\ldots,n_{i}\right\\}\subseteq\left[1,p\right]}w_{n_{1}}\left(a_{n_{1}}\right)w_{n_{2}}\left(a_{n_{2}}\right)\ldots
w_{n_{l}}\left(a_{n_{l}}\right)^{-1}\left[D^{*}\right]$

and $A_{p+1}$ as similar as above.

Then, as $\bigcap_{a_{p+1}\in A_{p+1}}h_{a_{p+1}}^{-1}\left[E\right]$ is a
$J$-set, we have the result. ∎

In [CG, Theorem 2.5], authors proved [HSZ, Corollary 16] for $J$-sets. In the
following theorem, we developed [CG, Theorem 2.5] for infinite sequence of
alphabets.

######  Theorem 2.8.

Let $k,n\in\mathbb{N}$ with $k<n$ and let $T$ be the set of words over
$\left\\{v_{1},v_{2},\ldots,v_{k}\right\\}$ in which $v_{i}$ occurs for each
$i\in\left\\{1,2,\ldots,k\right\\}$. Given, $w\in S_{n}$, let
$\tau\left(w\right)$ be obtained from $w$ by deleting all occurrence of
elements of $A$ as well as occurrences of $v_{i}$, $k<i\leq n$. Let
$\left\langle y_{t}\right\rangle_{t=1}^{\infty}$ be a sequence in $T$ and let
$D\subseteq S_{0}$ is a $J$-set. Then there exists a sequence of variable
words $\left\langle w_{t}\right\rangle_{t=1}^{\infty}$ over $A$ such that
$\left\\{w_{m}\left(a_{m}\right):a_{m}\in A_{m}^{n}\right\\}\subseteq D$ for
all $m\in\mathbb{N}$ and for every
$G\in\mathcal{P}_{f}\left(\mathbb{N}\right)$,

$\prod_{i=1}^{i}\tau\left(w_{m_{i}}\right)\in\text{FP}\left(\left\langle
y_{t}\right\rangle_{t=1}^{\infty}\right),$

where $G=\left\\{m_{1}<m_{2}<\ldots<m_{l}\right\\}$.

###### Proof.

We will proceed same as in Theorem 2.6.

Let $T^{*}=T\cup\left\\{\theta\right\\}$ and $\tau^{*}:S_{n}\cup
S_{0}\rightarrow T^{*}$ be defined by

$\tau^{*}\left(w\right)=\left\\{\begin{array}[]{c}\tau\left(w\right),\text{if
}w\in S_{n}\\\ \theta,\text{ if }w\in S_{0}\end{array}\right..$

Then $\left(\tau^{*}\right)^{-1}\left[\text{FP}\left(\left\langle
y_{t}\right\rangle_{t=1}^{\infty}\right)\cup\theta\right]$ is a partial
semigroup in $S_{n}$.

Let $A_{1}=\left(\tau^{*}\right)^{-1}\left[\text{FP}\left(\left\langle
y_{t}\right\rangle_{t=1}^{\infty}\right)\cup\theta\right]$ and
$F_{1}=\left\\{h_{\vec{a_{1}}}:\vec{a_{1}}\in A_{1}^{n}\right\\}$.

Then, $\bigcap_{\vec{a_{1}}\in A_{1}^{n}}h_{\vec{a_{1}}}^{-1}\left[D\right]$
is a $J$-set in $S_{n}\cup S_{0}$ and so in $S_{n}$ as $S_{0}$ is not a
$J$-set in $S_{n}\cup S_{0}$.

So, there exists $w_{1}\in S_{n}\cap\bigcap_{\vec{a_{1}}\in
A_{1}^{n}}h_{\vec{a_{1}}}^{-1}\left[D\right]$ such that
$\tau\left(w_{1}\right)\in\text{FP}\left(\left\langle
y_{t}\right\rangle_{t=1}^{\infty}\right)$. i.e;
$w_{1}\left(\vec{a_{1}}\right)\in D$ for all $\vec{a_{1}}\in A_{1}^{n}$ and
$\tau\left(w_{1}\right)\in\text{FP}\left(\left\langle
y_{t}\right\rangle_{t=1}^{\infty}\right)$.

from here proceeding as Theorem 2.6, we get the result. ∎

Similarly, we get the following corollary.

######  Corollary 2.9.

Let $k,n\in\mathbb{N}$ with $k<n$ and let $T$ be the set of words over
$\left\\{v_{1},v_{2},\ldots,v_{k}\right\\}$ in which $v_{i}$ occurs for each
$i\in\left\\{1,2,\ldots,k\right\\}$. Given, $w\in S_{n}$, let
$\tau\left(w\right)$ be obtained from $w$ by deleting all occurrence of
elements of $A$ as well as occurrences of $v_{i}$, $k<i\leq n$. Let
$\left\langle y_{t}\right\rangle_{t=1}^{\infty}$ be a sequence in $T$ and let
$C\subseteq S_{0}$ is a $C$-set. Then there exists a sequence of variable
words $\left\langle w_{t}\right\rangle_{t=1}^{\infty}$ over $A$ such that
$\left\\{w_{m}\left(a_{m}\right):a_{m}\in A_{m}^{n}\right\\}\subseteq C$ for
all $m\in\mathbb{N}$ and for every
$G\in\mathcal{P}_{f}\left(\mathbb{N}\right)$,

$\prod_{i=1}^{l}\tau\left(w_{m_{i}}\right)\in\text{FP}\left(\left\langle
y_{t}\right\rangle_{t=1}^{\infty}\right),$

where $G=\left\\{m_{1}<m_{2}<\ldots<m_{l}\right\\}$.

###### Proof.

Proceeding same as in Theorem 2.6, we get the result. ∎

Similar to [HSZ, Lemma 17], in Theorem 2.10, the semigroup $T$ and the matrix
$M$ satisfy all the appropriate hypotheses to matrix multiplication makes
sense and distributive.

######  Theorem 2.10.

Let $\left(T,+\right)$ be a commutative semigroup with identity element $0$.
Let $k,m,n\in\mathbb{N}$, and $M$ be a $k\times m$ matrix. For
$i\in\left\\{1,2,\ldots,m\right\\}$, let $\tau_{i}$ be an $S_{0}$\-
independent homomorphism from $S_{n}$ to $T$. Define a function $\psi$ on
$S_{n}$ by

$\psi\left(w\right)=\left(\begin{array}[]{c}\tau_{1}\left(w\right)\\\
\tau_{2}\left(w\right)\\\ \vdots\\\ \tau_{m}\left(w\right)\end{array}\right),$

with the property that for any collection of IP-sets
$\left\\{C_{i}:i\in\left\\{1,2,\ldots,k\right\\}\right\\}$ in $T$, there
exists $a\in S_{n}$ such that $M\psi\left(a\right)\in\times_{i=1}^{k}C_{i}$.
Let $B_{i}=\text{FS}\left(\left\langle
x_{n}^{\left(i\right)}\right\rangle_{n=1}^{\infty}\right)$ for $1\leq i\leq k$
be $k$ IP sets in $T$ and $D\subseteq S_{0}$ is a $C$-set in $S_{0}$, then,
there exists $\left\langle w_{i}\right\rangle_{i=1}^{\infty}\subseteq S_{n}$
such that for each
$\left\\{m_{1},m_{2},\ldots,m_{l}\right\\}=G\in\mathcal{P}_{f}\left(\mathbb{N}\right)$,
we have

$w_{m_{1}}\left(\overrightarrow{a_{m_{1}}}\right)w_{m_{2}}\left(\overrightarrow{a_{m_{2}}}\right)\cdots
w_{m_{l}}\left(\overrightarrow{a_{m_{l}}}\right)\in D$

and

$M\circ\psi\left(w_{m_{1}}w_{m_{2}}\ldots
w_{m_{l}}\right)\in\times_{i=1}^{k}B_{i}$

.

###### Proof.

Let $\phi:S_{n}\cup S_{0}\rightarrow
B=\times_{i=1}^{k}\left(B_{i}\cup\left\\{0\right\\}\right)$, where
$\phi\left(w\right)=M\psi\left(w\right)$, so as in [HSZ, Lemma 17], $\phi$ is
a homomorphism. Now, let us choose $p\in\beta S_{0}$ such that $D\in p$ so
then $D^{*}\in p$.

Now, $h_{\vec{a_{1}}}:\phi^{-1}\left[B\right]\rightarrow S_{0}$,
$\vec{a_{1}}\in A_{1}^{n}$ is a collection of partial semigroup homomorphism
fixed on $S_{0}$.

Now, $\bigcap_{\vec{a_{1}}\in
A_{1}^{n}}h_{\vec{a_{1}}}^{-1}\left[D^{*}\right]\setminus S_{0}$ is a $J$-set
in $\phi^{-1}\left[B\right]$ and so there exists
$w_{1}\in\phi^{-1}\left[B\right]\setminus S_{0}$ such that
$w_{1}\left(\vec{a_{1}}\right)\in D^{*}\forall\vec{a_{1}}\in A_{1}^{n}$ and
$\phi\left(w_{1}\right)=M\psi\left(w_{1}\right)\in B$.

Now, take an IP set $B_{1}\subseteq-M\psi\left(w_{1}\right)+B$ and clearly
$\left(0,0,\ldots,0\right)\in B_{1}$. So,
$S_{0}\subseteq\phi^{-1}\left[B_{1}\right]$. Now,
$\phi^{-1}\left(B_{1}\right)$ is again a partial semigroup, and hence

$h_{\vec{a_{2}}}:\phi^{-1}\left[B_{1}\right]\rightarrow
E=D^{*}\cap\bigcap_{\vec{a_{1}}\in
A_{1}^{n}}w_{1}\left(\vec{a_{1}}\right)^{-1}D^{*},\ \vec{a_{2}}\in A_{2}^{n}$

is again a partial semigroup homomorphism fixed on $S_{0}$.

So, there exists $w_{2}\in\bigcap_{\vec{a_{2}}\in
A_{2}^{n}}h_{\vec{a_{2}}}^{-1}\left[E\right]\cap\left(\phi^{-1}\left[B_{1}\right]\setminus
S_{0}\right)$. Thus,

$h_{\vec{a_{2}}}\left(w_{2}\right)\in E=D^{*}\cap\bigcap_{\vec{a_{1}}\in
A_{1}^{n}}w_{1}\left(\vec{a_{1}}\right)^{-1}D^{*}$

for all $\vec{a_{2}}\in A_{2}^{n}$. Thus, $w_{2}\left(\vec{a_{2}}\right)\in
D^{*}$, $w_{1}\left(\vec{a_{1}}\right)=w_{2}\left(\vec{a_{2}}\right)\in D^{*}$
and $\phi\left(w_{2}\right)=M\psi\left(w_{2}\right)\in B_{1}$.

So, we have,

$\left.\begin{array}[]{c}w_{1}\left(\vec{a_{1}}\right)\in D^{*}\text{ for all
}\vec{a_{1}}\in A_{1}^{n}\\\ M\psi\left(w_{1}\right)\in
B\end{array}\right\\}----\left(\text{1}\right),$
$\left.\begin{array}[]{c}w_{2}\left(\vec{a_{2}}\right)\in D^{*}\text{ for all
}\vec{a_{2}}\in A_{1}^{n}\\\ M\psi\left(w_{2}\right)\in
B_{1}\subseteq-M\psi\left(w_{1}\right)+B\end{array}\right\\}----\left(\text{2}\right),$

$\left.\begin{array}[]{c}\begin{array}[]{c}w_{1}\left(\vec{a_{1}}\right)w_{2}\left(\vec{a_{2}}\right)\in
D^{*}\text{ for all }\vec{a_{1}}\in A_{1}^{n}\text{ and }\vec{a_{2}}\in
A_{2}^{n}\\\
M\psi\left(w_{2}\right)\in-M\psi\left(w_{1}\right)+B\end{array}\end{array}\right\\}----\left(\text{3}\right)$

and so
$M\circ\psi\left(w_{1}w_{2}\right)=M\left(\psi\left(w_{1}\right)+\psi\left(w_{2}\right)\right)\in
B$.

Similarly, by proceeding further with this iteration, we will get the result.
∎

Acknowledgment: The second author of the paper acknowledges the grant UGC-NET
SRF fellowship with id no. 421333 of CSIR-UGC NET December 2016. We would like
to thank Prof. Dibyendu De for his helpful comments on this paper.

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