0.0pt\hbox{$\textstyle{X_{-}}$}}}}}}}\ignorespaces}}}}\ignorespaces$ is a
discrepant resolution induced from a Type II toric wall-crossing, then
crossing the first wall $\tilde{W}_{1}$ induces a birational transformation
between $Y_{+}$ (resp. $Y_{+,X_{\infty},r}$) and a toric stack $\tilde{Y}_{-}$
(resp. $\tilde{Y}_{-,X_{\infty},r}$). This birational transformation is a
flop.

Crossing the wall $\tilde{W}_{2}$ gives a birational transformation between
$\tilde{Y}_{-}$ (resp. $\tilde{Y}_{-,X_{\infty},r}$) and $Y_{-}$ (resp.
$Y_{-,X_{\infty},r}$). This birational transformation is a discrepant
resolution.

###### Proof.

The first wall-crossing is similar to the wall-crossing in Proposition 7.4. We
have $m+1\in\tilde{M}_{-}$, then $\\#\\{\tilde{M}_{+}\\}\geq 2$ and
$\\#\\{\tilde{M}_{-}\\}\geq 2$. Since it is crepant, we conclude that it is a
flop.

For the second wall-crossing, $m+1\in\tilde{M}_{0}$ and $m+2\in\tilde{M}_{+}$.
then $\\#\\{\tilde{M}_{+}\\}=\\#\\{M_{+}\\}+1\geq 3$ and
$\\#\\{\tilde{M}_{-}\\}=\\#\\{M_{-}\\}=1$. Therefore, it is a discrepant
resolution. ∎

######  Remark 7.16.

The geometric construction in this section also gives another explanation of
the relation between $\overline{T}_{-}$ and $T_{-}$ in Section 7.1. Recall
that $\overline{T}_{-}$ is a partial compactification of $T_{-}$. The first
wall-crossing here is related to the wall-crossing in Section 7.1 that gives a
crepant transformation between $T_{+}$ and $\overline{T}_{-}$. The second
wall-crossing here is related to the partial compactification as follows. Note
that $\overline{T}_{-}=\tilde{Y}_{-}\setminus\bar{\tilde{D}}_{m+2}$ and
$T_{-}=Y_{-}\setminus\bar{\tilde{D}}_{m+2}=Y_{-}\setminus X_{\infty}$.
Moreover, the loci of indeterminacy are
$\bar{\tilde{D}}_{i}\subset\tilde{Y}_{-}$ where $i\in\tilde{M}_{-}$ and
$\bigcap_{i\in\tilde{M}_{+}}\bar{D}_{i}\subset X_{\infty}\subset Y_{-}$. By
removing $X_{\infty}$, one removes the loci of indeterminacy in $Y_{-}$.
Therefore, the complements are just related by partial compactification and
the partial compactification is coming from the exceptional divisor in
$\tilde{Y}_{-}$. The complement $\overline{T}_{-}\setminus T_{-}$ is
codimension one.

The extended $I$-function of relative Gromov–Witten theory is given in
[FTY]*Section 4.3. It is taken as a limit of the extended $I$-function of root
stacks which is obtained from the hypersurface construction as described in
Section 7.2. The extended $I$-function for a smooth pair $(X,D)$ is defined as
follows. We take the extended data

$S:=\\{a_{1},\ldots,a_{m}\\}.$

The $S$-extended $I$-function for $(X,D)$ is

$I_{(X,D)}^{S}(Q,x,t,z)=I_{\operatorname{pos}}+I_{\operatorname{neg}},$

where

$\displaystyle I_{\operatorname{pos}}:=$ $\displaystyle
ze^{t/z}\sum_{\begin{subarray}{c}d\in\mathbb{K},(k_{1},\ldots,k_{m})\in(\mathbb{Z}_{\geq
0})^{m}\\\ \sum_{i=1}^{m}k_{i}a_{i}<D\cdot
d\end{subarray}}y^{d}\left(\prod_{i=0}^{m}\frac{\prod_{a\leq 0,\langle
a\rangle=\langle D_{i}\cdot d\rangle}(\bar{D}_{i}+az)}{\prod_{a\leq D_{i}\cdot
d,\langle a\rangle=\langle D_{i}\cdot
d\rangle}(\bar{D}_{i}+az)}\right)\frac{\prod_{i=1}^{m}x_{i}^{k_{i}}}{z^{\sum_{i=1}^{m}k_{i}}\prod_{i=1}^{m}(k_{i}!)}$
$\displaystyle\frac{\prod_{0<a\leq D\cdot d}(D+az)}{D+(D\cdot
d-\sum_{i=1}^{m}k_{i}a_{i})z}\textbf{1}_{[-d]}[{\mathbf{1}}]_{-D\cdot
d+\sum_{i=1}^{m}k_{i}a_{i}},$

and

$\displaystyle I_{\operatorname{neg}}:=$ $\displaystyle
ze^{t/z}\sum_{\begin{subarray}{c}d\in\mathbb{K},(k_{1},\ldots,k_{m})\in(\mathbb{Z}_{\geq
0})^{m}\\\ \sum_{i=1}^{m}k_{i}a_{i}\geq D\cdot
d\end{subarray}}y^{d}\left(\prod_{i=0}^{m}\frac{\prod_{a\leq 0,\langle
a\rangle=\langle D_{i}\cdot d\rangle}(\bar{D}_{i}+az)}{\prod_{a\leq D_{i}\cdot
d,\langle a\rangle=\langle D_{i}\cdot
d\rangle}(\bar{D}_{i}+az)}\right)\frac{\prod_{i=1}^{m}x_{i}^{k_{i}}}{z^{\sum_{i=1}^{m}k_{i}}\prod_{i=1}^{m}(k_{i}!)}$
$\displaystyle\left(\prod_{0<a\leq D\cdot
d}(D+az)\right)\textbf{1}_{[-d]}[{\mathbf{1}}]_{-D\cdot
d+\sum_{i=1}^{m}k_{i}a_{i}}.$

One can simplify the expression by considering the $H$-function. The
$H$-function is

(31) $\displaystyle H_{(X,D)}(y)=e^{\frac{t}{2\pi
i}}\sum_{\begin{subarray}{c}d\in\mathbb{K}\\\
(k_{1},\ldots,k_{m})\in(\mathbb{Z}_{\geq
0})^{m}\end{subarray}}y^{d}\left(\frac{\Gamma(1+\frac{D}{2\pi i}+D\cdot
d)}{\prod_{i=1}^{m}\Gamma(1+\frac{\bar{D}_{i}}{2\pi i}+D_{i}\cdot
d)}\right)\textbf{1}_{[d]}[\textbf{1}]_{D\cdot
d-\sum_{i=1}^{m}k_{i}a_{i}}\frac{\prod_{i=1}^{m}x_{i}^{k_{i}}}{\prod_{i=1}^{m}(k_{i}!)}.$

We can state the relation between the quantum cohomology of $(X_{+},D_{+})$
and $(X_{-},D_{-})$ in terms of $H$-functions.

######  Theorem 7.17.

The (extended) $H$-function of $(X_{+},D_{+})$ can be analytically continued
to the (extended) $H$-function of $(X_{-},D_{-})$ under the specialization
$y^{\prime}=0$.

###### Proof.

The proof consists of the following steps.

  * Step 1

We consider the $I$-functions of $X_{D,r}$. By the hypersurface construction,
We just need to study the $I$-function of $Y_{X_{\infty},r}$ and apply the
orbifold quantum Lefschetz principle of [Tseng]. By the orbifold quantum Serre
duality of [Tseng], it is equivalent to considering the $I$-function of the
total space of the line bundle $p^{*}\mathcal{O}_{Y}(-1)$. This is a toric
stack and we will denote it by $L_{r}$. We use $\tilde{D}_{m+2}/r$ to denote
the toric divisor with the $r$-th root construction.

  * Step 2

Recall that $Y_{+}$ and $Y_{-}$ are related by two wall-crossings: the first
wall-crossing is a flop and the second wall-crossing is a discrepant
resolution. We consider how the toric stack $L_{+,r}$ changes under these two
wall-crossings. The first wall-crossing relates the $H$-function of $L_{+,r}$
and the $H$-function of $\tilde{L}_{-,r}$ via analytic continuation in
[CIJ]*Section 6.2, where $\tilde{L}_{-,r}$ is the toric stack obtained from
crossing the first wall. Note that the toric divisor $\tilde{D}_{m+2}/r$ is on
the wall. The analytic continuation in [CIJ]*Section 6.2 essentially does not
affect the factor involving $\tilde{D}_{m+2}/r$. Hence the analytic
continuation is compatible with the limit $r\rightarrow\infty$.

  * Step 3

The second wall-crossing is a flip (it “becomes crepant” when
$r\rightarrow\infty$). The toric stack obtained by crossing the second wall is
denoted by $\overline{L}_{-,r}$. The $H$-function of $\tilde{L}_{-,r}$ is
analytic. Note that $\tilde{D}_{m+2}/r$ is the $r$-th root of
${\tilde{D}}_{m+2}$ and it is on the positive side of the wall (i.e.
$m+2\in\tilde{M}_{+}$). There is a factor of
$\frac{1}{\Gamma(1+\bar{\tilde{D}}_{m+2}/r+(\tilde{D}_{m+2}\cdot d)/r)}$ in
the $H$-function and it becomes $1/\Gamma(1)=1$ when $r\rightarrow\infty$.
Follow the analytic continuation computation of [CIJ]*Section 6.2 carefully,
the factor $\frac{1}{\Gamma(1+\bar{\tilde{D}}_{m+2}/r+(\tilde{D}_{m+2}\cdot
d)/r)}$ will again be $1$ as $r\rightarrow\infty$ after analytic continuation.
In other words, this analytic continuation is again compatible with the limit
$r\rightarrow\infty$ and one obtains the analytic continuation between the
$H$-function of $\tilde{L}_{-,\infty}$ and the $H$-function of
$\overline{L}_{-,\infty}$.

  * Step 4

Proposition 7.8 can be applied here and implies that $\tilde{L}_{-,r}$ is a
partial compactification of $L_{-,r}$. Then Theorem 7.10 implies that their
$I$-functions are related by specialization of $y^{\prime}=0$. This
specialization is also compatible with the limit $r\rightarrow\infty$. This
concludes the proof.

∎

######  Remark 7.18.

The case of toric complete intersections works similarly, by combining the
discussion in this section and the discussion in Section 6.1. So we will not
repeat it here.

##  8\. Connection to transitions

Let $\phi:\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
9.7125pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{X_{+}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces{}\ignorespaces\ignorespaces{\hbox{\lx@xy@drawline@}}\ignorespaces{\hbox{\kern
33.7125pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}\ignorespaces\ignorespaces{\hbox{\lx@xy@drawline@}}\ignorespaces{\hbox{\lx@xy@drawline@}}{\hbox{\kern
33.7125pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{X_{-}}$}}}}}}}\ignorespaces}}}}\ignorespaces$ be a
toric blow-up given by a Type II toric wall-crossing. Let $D_{+}$ and $D_{-}$
be smooth and given by

$D_{+}=\sum_{i\in(M_{+}\cup\\{j\\})}\bar{D}_{i}\subset X_{+}\text{ and
}D_{-}=\sum_{i\in M_{+}}\bar{D}_{i}\subset X_{-}.$

When $D_{+}$ and $D_{-}$ are nef, Theorem 7.11 states that the narrow relative
quantum D-modules of $(X_{+},D_{+})$ and $(X_{-},D_{-})$ are related by
analytic continuation and specialization. On the other hand, the divisors
$D_{+}\subset X_{+}$ and $D_{-}\subset X_{-}$ are related by transitions
[MS]*Proposition 4.5. [MS]*Theorem 1.1 states that the ambient quantum
D-modules of $D_{+}$ and $D_{-}$ are related by analytic continuation and
specialization. Indeed, Theorem 7.11 (in the case of toric blow-ups) follows
from [MS]*Theorem 1.1 by the local-relative correspondence [vGGR], [TY20b],
[BNTY] and the quantum Serre duality.

There is a rank reduction phenomenon caused by the specialization. We provide
a partial answer to it. It is not surprising that the rank reduction has to do
with the curves that are contracted by the map $\phi$. In this section, we
will discuss the rank reduction phenomenon for transitions, the rank reduction
phenomenon for the log-crepant transformation of smooth pairs works the same.

In this section, we use slightly different notation to emphasize that we are
considering blow-ups. We write

$X:=X_{-},\quad\tilde{X}:=X_{+},\quad,D:=D_{-},\quad\tilde{D}:=D_{+}.$

We can simply consider the fan sequence (19) for the blow-up:

(32)
$0\longrightarrow\mathbb{L}\oplus\mathbb{Z}\longrightarrow\mathbb{Z}^{m+1}\stackrel{{\scriptstyle\tilde{\beta}}}{{\longrightarrow}}N\longrightarrow
0,$

such that the map $\tilde{\beta}:\mathbb{Z}^{m+1}\rightarrow N$ is given by
$\\{\beta_{1},\ldots,\beta_{m},\beta_{m+1}\\}$ and the map
$\mathbb{L}\oplus\mathbb{Z}\rightarrow\mathbb{Z}^{m+1}$ is given by
$(\tilde{D}_{1},\ldots,\tilde{D}_{m},\tilde{D}_{m+1})$ where

$\tilde{D}_{i}=\left\\{\begin{array}[]{ll}D_{i}\oplus 0&\text{if
$i\in\\{1,\ldots,m\\}$ and $D_{i}\cdot e\leq 0$}\\\ D_{i}\oplus(-D_{i}\cdot
e)&\text{if $i\in\\{1,\ldots,m\\}$ and $D_{i}\cdot e>0$}\\\ 0\oplus 1&\text{If
$i=m+1$}.\end{array}\right.$

The blown-up variety $\tilde{X}$ is the toric Deligne–Mumford stack given by
the stability condition $\tilde{\omega}:=(\omega_{0},-\varepsilon)$, where
$\omega_{0}$ is in the relative interior of
$W\cap\overline{C_{+}}=W\cap\overline{C_{-}}$ and $\varepsilon$ is a positive
and sufficiently small real number.

Let

$Z=\cap_{i\in M_{+}}\\{z_{i}=0\\}\subset X$

be the center of the blow-up. Then the exceptional divisor $E\subset\tilde{X}$
is a projectivization of the normal bundle of $Z$ in $X$:

$\pi:E:=\mathbb{P}(N_{Z/X})\rightarrow Z.$

Then the fiber curve classes are contracted by $\pi$. Let $w_{i}=D_{i}\cdot e$
for $i\in M_{+}$, then the fiber of $\pi$ is $\mathbb{P}[w_{1},\ldots,w_{k}]$,
where $k=\operatorname{codim}Z=\\#\\{M_{+}\\}$. We would like to know what is
the subvariety that is contracted under the transition.

  * •

If $\dim Z=0$, we can assume $Z$ is a point without loss of generality. Then
$E=\mathbb{P}[w_{1},\ldots,w_{n}]$. The intersection of the hypersurface
$D_{+}$ and $E$ is of dimension $n-2$. The transition
$\textstyle{\tilde{D}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{D}$
contracts a hypersurface $S$ in the weighted projective space
$\mathbb{P}[w_{1},\ldots,w_{n}]$ of degree $(-1+\sum_{i=1}^{n}w_{i})$, where
$n=\dim X_{\pm}=\\#\\{M_{+}\\}$.

  * •

If $\dim Z>0$, let $k=\operatorname{codim}Z=\\#\\{M_{+}\\}$. Then $\dim(Z\cap
D)=n-1-k$ and $\dim(E\cap\tilde{D})=n-2$, the transition
$\textstyle{\tilde{D}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{D}$
contracts a $(k-1)$-dimensional space which is the fibers
$\mathbb{P}[w_{1},\ldots,w_{k}]$ over $Z$.

######  Example 8.1.

Let

$X=\mathbb{P}^{4}\text{ and,
}\tilde{X}=\operatorname{Bl}_{\operatorname{pt}}\mathbb{P}^{4}.$

Let $Q_{5}\subset\mathbb{P}^{4}$ be a smooth quintic threefold and
$\tilde{Q}_{5}\subset\tilde{X}$ be a smooth Calabi–Yau threefold defined by a
regular section of the anticanonical bundle $-K_{\tilde{X}}$. The map
$\textstyle{\tilde{Q}_{5}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Q_{5}}$
contracts a cubic surface $S$ in $\mathbb{P}^{3}$. Then $Q_{5}$ and
$\tilde{Q}_{5}$ are related by cubic extremal transition which was studied by
[Mi].

######  Example 8.2.

Let

$X=\mathbb{P}^{4}\text{ and,
}\tilde{X}=\operatorname{Bl}_{\mathbb{P}^{2}}\mathbb{P}^{4}.$

Let $Q_{5}\subset\mathbb{P}^{4}$ be a smooth quintic threefold and
$\tilde{Q}_{5}\subset\tilde{X}$ be a smooth Calabi–Yau threefold defined by a
regular section of the anticanonical bundle $-K_{\tilde{X}}$. Then $Q_{5}$ and
$\tilde{Q}_{5}$ are related by a conifold transition. The map
$\textstyle{\tilde{Q}_{5}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Q_{5}}$
contracts
$\mathcal{O}_{\mathbb{P}^{1}}(-1)\oplus\mathcal{O}_{\mathbb{P}^{1}}(-1)$-curves.

We would like to consider the local Gromov–Witten theory of the total space of
the normal bundle of $S$ or $\mathbb{P}[w_{1},\ldots,w_{k}]$ in $\tilde{D}$.
On can apply quantum Serre duality to the first case. Then, in both cases, we
can consider the local Gromov–Witten theory of
$\mathcal{O}_{\mathbb{P}}(-1)\oplus\mathcal{O}_{\mathbb{P}}(1-\sum_{i=1}^{k}w_{i})$,
where $\mathbb{P}:=\mathbb{P}[w_{1},\ldots,w_{k}]$ and
$k=\operatorname{codim}Z=\\#\\{M_{+}\\}$.

Recall that the main result of [MS] (see also [Mi] for a cubic extremal
transition) can be stated in terms of $I$-functions as follows. There is an
explicit degree-preserving linear transformation
$L:H^{*}(\tilde{D})\rightarrow H^{*}(D)$ such that $I_{D}(y)$ is recovered by

$I_{D}(y)=\lim_{y_{\mathrm{r}+1}\rightarrow 0}L\circ\bar{I}_{\tilde{D}}(y),$

where $\bar{I}_{\tilde{D}}(y)$ is obtained from $I_{\tilde{D}}(\tilde{y})$ via
analytic continuation.

######  Theorem 8.3.

Let $\iota$ be the inclusion map $\iota:\mathbb{P}\hookrightarrow\tilde{X}$.
Then
$\iota^{*}I_{\tilde{D}}(\tilde{y})|_{\tilde{y}_{i}=0,i\in\\{1,\ldots,\mathrm{r}\\}}$
coincides with
$I_{\mathcal{O}_{\mathbb{P}}(-1)\oplus\mathcal{O}_{\mathbb{P}}(1-\sum_{i=1}^{k}w_{i})}(\tilde{y}_{\mathrm{r}+1})$.
Under this identification, we have

$\lim_{y_{\mathrm{r}+1}\rightarrow
0}\bar{I}_{\mathcal{O}_{\mathbb{P}}(-1)\oplus\mathcal{O}_{\mathbb{P}}(1-\sum_{i=1}^{k}w_{i})}(y)=0,$

where
$\bar{I}_{\mathcal{O}_{\mathbb{P}}(-1)\oplus\mathcal{O}_{\mathbb{P}}(1-\sum_{i=1}^{k}w_{i})}(y)$
is obtained from
$I_{\mathcal{O}_{\mathbb{P}}(-1)\oplus\mathcal{O}_{\mathbb{P}}(1-\sum_{i=1}^{k}w_{i})}(\tilde{y}_{\mathrm{r}+1})$
after the analytic continuation in [MS].

###### Proof.

Recall that the local $I$-function for $\mathcal{O}_{\tilde{X}}(-\tilde{D})$
is

$I_{\mathcal{O}_{\tilde{X}}(-\tilde{D})}(\tilde{y},z)=ze^{\tilde{t}/z}\sum_{d\in\tilde{\mathbb{K}}}\tilde{y}^{d}\left(\prod_{i=0}^{m+1}\frac{\prod_{a\leq
0,\langle a\rangle=\langle\tilde{D}_{i}\cdot
d\rangle}(\bar{\tilde{D}}_{i}+az)}{\prod_{a\leq\tilde{D}_{i}\cdot d,\langle
a\rangle=\langle\tilde{D}_{i}\cdot
d\rangle}(\bar{\tilde{D}}_{i}+az)}\right)\frac{\prod_{a\leq 0,\langle
a\rangle=\langle-\tilde{D}\cdot
d\rangle}(-\bar{\tilde{D}}+az)}{\prod_{a\leq-\tilde{D}\cdot d,\langle
a\rangle=\langle-\tilde{D}\cdot
d\rangle}(-\bar{\tilde{D}}_{i}+az)}\textbf{1}_{[-d]},$

where $\tilde{\mathbb{K}}=\mathbb{K}\oplus\mathbb{R}(0,-1)$,
$\tilde{t}=\sum_{a=1}^{\mathrm{r}+1}\bar{p}_{a}\log\tilde{y}_{a}$,
$\tilde{y}^{d}=\tilde{y}_{1}^{p_{1}\cdot
d}\cdots\tilde{y}_{\mathrm{r}+1}^{p_{\mathrm{r}+1}\cdot d}$ and, $[-d]$ is the
equivalence class of $-d$ in $\mathbb{K}/\mathbb{L}$.

We restrict to the $I$-function to $\tilde{y}_{a}=0$ for $1\leq
a\leq\mathrm{r}$ and consider the pullback of it under the inclusion map
$\iota:\mathbb{P}\hookrightarrow\tilde{X}$. Then we have the $I$-function for
$\mathcal{O}_{\mathbb{P}}(-1)\oplus\mathcal{O}_{\mathbb{P}}(1-\sum_{i=1}^{k}w_{i})$:

$\displaystyle
I_{\mathcal{O}_{\mathbb{P}}(-1)\oplus\mathcal{O}_{\mathbb{P}}(1-\sum_{i=1}^{k}w_{i})}(\tilde{y}_{\mathrm{r}+1},z)=ze^{P/z}\sum_{\begin{subarray}{c}n\in\mathbb{Q}_{\geq
0}\\\ \exists
j:nw_{j}\in\mathbb{Z}\end{subarray}}\tilde{y}_{\mathrm{r}+1}^{n}\left(\prod_{i\in
M_{+}}\frac{1}{\prod_{a\leq w_{i}n,\langle a\rangle=\langle
w_{i}n\rangle}(w_{i}P+az)}\right)$
$\displaystyle\cdot\frac{1}{\prod_{a\leq-n,\langle
a\rangle=\langle-n\rangle}(-P+az)}\frac{1}{\prod_{a\leq
1-\sum_{i=1}^{k}w_{i}n,\langle a\rangle=\langle
1-\sum_{i=1}^{k}w_{i}n\rangle}((1-\sum_{i=1}^{k}w_{i})P+az)}\textbf{1}_{[-n]},$

where $P\in H^{2}(\mathbb{P})$ is the hyperplane class.

Moreover, after analytic continuation,
$I_{\mathcal{O}_{\mathbb{P}}(-1)\oplus\mathcal{O}_{\mathbb{P}}(1-\sum_{i=1}^{k}w_{i})}(\tilde{y}_{\mathrm{r}+1},z)$
becomes a function of $y_{\mathrm{r}+1}$, hence specializes to zero under the
specialization in [MS]*Theorem 1.3 where the specialization
$y_{\mathrm{r}+1}=0$ was considered.

∎

######  Remark 8.4.

The main result of [MS] is obtained via quantum Serre duality and the
specialization of a variable is related to a partial compactification of the
corresponding line bundle. The geometric explanation in remark 7.16 also
provides an indication of Theorem 8.3.

######  Remark 8.5.

In [Mi], the rank reduction phenomenon was partially explained as the FJRW
theory for an example of cubic extremal transitions. It was shown that the
regularized FJRW $I$-function satisfies a Picard–Fuchs equation which, after
analytic continuation, is satisfied by the $I$-function of the cubic extremal
transitions. Here, we give a general explanation of it directly in terms of
Gromov–Witten theory. One may wonder how the $I$-function for local/relative
Gromov–Witten theory that we considered here is related to the regularized
FJRW $I$-function. We will answer this question in the next section and
understand it as a result of the _Landau–Ginzburg/ log Calabi–Yau
correspondence_. The analytic continuation for the example in [Mi] will be
explicitly worked out.

##  9\. Connection to FJRW theory

Landau–Ginzburg/Calabi–Yau (LG/CY) correspondence arises from a variation of
the GIT quotient in Witten’s gauged linear sigma model (GLSM) [Witten]. The
LG/CY correspondence describes a relationship between sigma models based on
Calabi-Yau hypersurfaces in weighted projective spaces and the Landau-Ginzburg
model of the defining equation of the Calabi-Yau. The LG/CY correspondence was
generalized to the Fano case and general type case by [Acosta]. In the
Calabi–Yau case, the $I$-functions of Gromov–Witten theory and FJRW theory are
related by analytic continuation. However, the analytic continuation is
impossible in the Fano case because the Picard–Fuchs operator develops an
irregular singularity at the Landau–Ginzburg point and the $I$-function has
radius of convergence equal to zero at the irregular singular point. In
[Acosta], the author used asymptotic expansions to relate the $I$-functions.
This is known as the Landau–Ginzburg/Fano (LG/Fano) correspondence.

We propose a different version of the LG/Fano correspondence by using relative
Gromov–Witten invariants instead of absolute Gromov–Witten invariants of the
Fano variety. We believe it is natural when we consider mirror symmetry for
Fano varieties. As mentioned in Section 6.2, when we consider a
Landau–Ginzburg model $(X^{\vee},W)$ as a mirror to a Fano variety $X$, we
usually expect the generic fiber of the Landau–Ginzburg model is mirror the
the smooth anticanonical divisor $D$ of the Fano variety $X$. When we consider
mirror symmetry for a Fano variety $X$, it is usually natural to consider
mirror symmetry for the log Calabi–Yau pair $(X,D)$ instead. Hence, relative
Gromov–Witten invariants of $(X,D)$ appear naturally in mirror symmetry.
Instead of considering the LG/Fano correspondence, we consider the
_Landau–Ginzburg/log Calabi–Yau (LG/(log CY)) correspondence_. In terms of
$I$-functions, we claim that the $I$-function for the log Calabi–Yau pair
$(X,D)$ can be analytically continued to the regularized $I$-function for the
FJRW theory.

In Section 8, we explained that the rank reduction phenomenon is related to
the local Gromov–Witten invariants of weighted projective spaces. On the other
hand, [Mi] explains how the rank reduction phenomenon is related to the FJRW
theory in an example of cubic extremal transition. By the local-orbifold
correspondence of [vGGR]. [TY20b] and [BNTY], it is natural to expect the
corresponding relative Gromov–Witten invariants are related to the FJRW theory
of the singularity.

###  9.1. Example: cubic surface

We begin with the example in [Mi]. We claim that the regularized FJRW
$I$-function is actually obtained by the analytic continuation of the local
Gromov–Witten theory of the total space of the canonical bundle $K_{S}$, where
$S$ is a cubic surface. Recall that the $I$-function for $K_{S}$ is

(33) $\displaystyle I_{K_{S}}(q)=(3H)q^{H/z}\sum_{d\geq
0}q^{d}\frac{\prod_{k=1}^{3d}(3H+kz)(-1)^{d}\prod_{k=0}^{d-1}(H+kz)}{\prod_{k=1}^{d}(H+kz)^{4}}.$

To study analytic continuation, we write the $I$-function in terms of
$\Gamma$-functions:

$\displaystyle I_{K_{S}}(q)$
$\displaystyle=3Hq^{H/z}\frac{\Gamma(1+\frac{H}{z})^{4}\Gamma(1-\frac{H}{z})}{\Gamma(1+\frac{3H}{z})}\sum_{d\geq
0}q^{d}\frac{\Gamma(1+\frac{3H}{z}+3d)}{\Gamma(1+\frac{H}{z}+d)^{4}\Gamma(1-\frac{H}{z}-d)}$
$\displaystyle=3Hq^{H/z}\frac{\Gamma(1+\frac{H}{z})^{4}\Gamma(1-\frac{H}{z})}{\Gamma(1+\frac{3H}{z})}\sum_{d\geq
0}\operatorname{Res}_{s=d}\frac{2\pi i}{e^{2\pi
is}-1}q^{s}\frac{\Gamma(1+\frac{3H}{z}+3s)}{\Gamma(1+\frac{H}{z}+s)^{4}\Gamma(1-\frac{H}{z}-s)}.$

We consider the contour integral

(34) $\displaystyle
3Hq^{H/z}\frac{\Gamma(1+\frac{H}{z})^{4}\Gamma(1-\frac{H}{z})}{\Gamma(1+\frac{3H}{z})}\int_{C}\frac{1}{e^{2\pi
is}-1}q^{s}\frac{\Gamma(1+\frac{3H}{z}+3s)}{\Gamma(1+\frac{H}{z}+s)^{4}\Gamma(1-\frac{H}{z}-s)},$

where the contour $C$ is chosen such that the poles at $s=l$ are on the right
of $C$ and the poles at $s=-1-l$ and at

$s=-\frac{H}{z}-\frac{m}{3},m\geq 1$

are on the left of $C$, where $l$ is a non-negative integer.

Then (34) is the sum of residue at

$s=-1-l,l\geq 0\quad\text{and}\quad s=-\frac{H}{z}-\frac{m}{3},m\geq 1.$

The residue at $s=-1-l$ vanishes because they are multiples of $H^{4}=0$. Note
that

$\operatorname{Res}_{s=-\frac{H}{z}-\frac{m}{3}}\Gamma(1+\frac{3H}{z}+3s)=-\frac{1}{3}\frac{(-1)^{m}}{\Gamma(m)}.$

We obtain the following analytic continuation of $I_{K_{S}}$:

(35)
$\displaystyle\bar{I}_{K_{S}}=-H\frac{\Gamma(1+\frac{H}{z})^{4}\Gamma(1-\frac{H}{z})}{\Gamma(1+\frac{3H}{z})}\sum_{m\geq
0}(-1)^{m}\frac{(2\pi i)e^{\frac{2\pi im}{3}}}{e^{-2\pi
i\frac{H}{z}}-e^{\frac{2\pi
im}{3}}}\frac{q^{-\frac{m}{3}}}{\Gamma(m)\Gamma(1-\frac{m}{3})^{4}\Gamma(1+\frac{m}{3})}.$

Let $t^{3}=q^{-1}$, then we have

$\displaystyle\bar{I}_{K_{S}}$
$\displaystyle=-H\frac{\Gamma(1+\frac{H}{z})^{4}\Gamma(1-\frac{H}{z})}{\Gamma(1+\frac{3H}{z})}\sum_{m\geq
0}(-1)^{m}\frac{(2\pi i)e^{\frac{2\pi im}{3}}}{e^{-2\pi
i\frac{H}{z}}-e^{\frac{2\pi
im}{3}}}\frac{t^{m}}{\Gamma(m)\Gamma(1-\frac{m}{3})^{4}\Gamma(1+\frac{m}{3})}$
$\displaystyle=-H\frac{\Gamma(1+\frac{H}{z})^{4}\Gamma(1-\frac{H}{z})}{\Gamma(1+\frac{3H}{z})}\sum_{m\geq
0}(-1)^{m}\frac{(2\pi i)e^{\frac{2\pi im}{3}}}{e^{-2\pi
i\frac{H}{z}}-e^{\frac{2\pi
im}{3}}}\frac{t^{m}\Gamma(\frac{m}{3})^{4}\sin^{4}(\frac{m}{3}\pi)}{\pi^{4}\Gamma(m)\Gamma(1+\frac{m}{3})}$
$\displaystyle=-H\frac{\Gamma(1+\frac{H}{z})^{4}\Gamma(1-\frac{H}{z})}{\Gamma(1+\frac{3H}{z})}\sum_{k=1,2}\frac{(2\pi
i)e^{\frac{2\pi ik}{3}}}{e^{-2\pi i\frac{H}{z}}-e^{\frac{2\pi
ik}{3}}}\frac{(-1)^{k}}{\Gamma(\frac{k}{3})^{4}\Gamma(1-\frac{k}{3})^{4}}\sum_{l\geq
0}\frac{t^{k+3l}\Gamma(\frac{k+3l}{3})^{4}}{\Gamma(k+3l)\Gamma(1+\frac{k}{3}+l)}$
$\displaystyle=-H\frac{\Gamma(1+\frac{H}{z})^{4}\Gamma(1-\frac{H}{z})}{\Gamma(1+\frac{3H}{z})}\sum_{k=1,2}(-1)^{k}\frac{(2\pi
i)e^{\frac{2\pi ik}{3}}}{e^{-2\pi i\frac{H}{z}}-e^{\frac{2\pi
ik}{3}}}\frac{(-1)^{k}}{\Gamma(1-\frac{k}{3})^{4}}\sum_{l\geq
0}\frac{t^{k+3l}\Gamma(\frac{k}{3}+l)^{4}}{\Gamma(\frac{k}{3})^{4}\Gamma(k+3l)\Gamma(1+\frac{k}{3}+l)}.$

Let $W=x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}$, we consider the FJRW theory
of the pair $(W,G)$, where $G=\langle J_{W}\rangle$ is generated by

$J_{W}:=\left(\exp\left(\frac{2\pi i}{3}\right),\exp\left(\frac{2\pi
i}{3}\right),\exp\left(\frac{2\pi i}{3}\right),\exp\left(\frac{2\pi
i}{3}\right)\right)\in(\mathbb{C}^{\times})^{4}.$

Recall that the regularized $I$-function for FJRW theory of $(W,G)$ is given
in [Acosta] as follows

$\displaystyle I^{\operatorname{reg}}_{\operatorname{FJRW}}(t):=$
$\displaystyle
I^{\operatorname{reg}}_{\operatorname{FJRW},1}\phi_{0}+I^{\operatorname{reg}}_{\operatorname{FJRW},2}\phi_{1}$
$\displaystyle=$ $\displaystyle\sum_{l\geq
0}\frac{t^{1+3l}\Gamma(\frac{1}{3}+l)^{4}}{\Gamma(\frac{1}{3})^{4}(3l)!\Gamma(\frac{4}{3}+l)}\phi_{0}+\sum_{l\geq
0}\frac{t^{2+3l}\Gamma(\frac{2}{3}+l)^{4}}{\Gamma(\frac{2}{3})^{4}(1+3l)!\Gamma(\frac{5}{3}+l)}\phi_{1},$

where $\phi_{0}$ and $\phi_{1}$ are generators of the narrow state space of
the FJRW theory of $(W,G)$. Therefore, we have

$\displaystyle\bar{I}_{K_{S}}=-H\frac{\Gamma(1+\frac{H}{z})^{4}\Gamma(1-\frac{H}{z})}{\Gamma(1+\frac{3H}{z})}\sum_{k=1,2}(-1)^{k}\frac{(2\pi
i)e^{\frac{2\pi ik}{3}}}{e^{-2\pi i\frac{H}{z}}-e^{\frac{2\pi
ik}{3}}}\frac{(-1)^{k}}{\Gamma(1-\frac{k}{3})^{4}}I^{\operatorname{reg}}_{\operatorname{FJRW},k}.$

Hence, the local $I$-function of $K_{S}$ can be analytically continued to the
regularized $I$-function of the FJRW theory of cubic singularity. By the
local-orbifold correspondence and the relative-orbifold correspondence [ACW],
[vGGR], [TY20b], [TY18], we may identify the narrow relative quantum D-module
of $(S,-K_{S})$ with the narrow quantum D-module of the FJRW theory of
$(W,G)$. Therefore, we conclude the following.

######  Theorem 9.1.

the narrow relative quantum D-module of $(S,-K_{S})$ can be identified with
the narrow quantum D-module of $(W,G)$ via the analytic continuation.

######  Remark 9.2.

Since $(S,-K_{S})$ is log Calabi–Yau, we may consider Theorem 9.1 as a LG/(log
CY) correspondence. The meaning on the Gromov–Witten side is clear: one
considers the relative Gromov–Witten theory of log Calabi–Yau pairs. On the
other hand, the meaning on the FJRW side is not clear to us at this moment. It
may be interesting to find out if there is a direct enumerative meaning of the
regularized FJRW $I$-function in the quantum singularity theory related to the
log Calabi–Yau pair $(X,D)$.

###  9.2. Fano hypersurfaces in weighted projective spaces

In general, we can consider a degree $d$ Fano hypersurface $X$ in a weighted
projective space $\mathbb{P}[w_{1},\ldots,w_{N}]$ along with the smooth
anticanonical divisor $D$ of $X$ and study the relation between relative
Gromov–Witten theory of $(X,D)$ and the FJRW theory. The set-up is the
following.

Given a Fermat polynomial

$W=x_{1}^{d/w_{1}}+\cdots+x_{n}^{d/w_{N}},$

where $\operatorname{gcd}(w_{1},\ldots,w_{N})=1$ and $W$ has a unique
singularity at the origin. We also assume that $w_{i}$ divides $d$ for all
$i\in\\{1,\ldots,N\\}$. The polynomial $W$ is quasi-homogeneous of degree $d$:

$W(\lambda^{w_{1}}x_{1},\ldots,\lambda^{w_{N}}x_{N})=\lambda^{d}W(x_{1},\ldots,x_{N}),$

for all $\lambda\in\mathbb{C}$. Let $q_{i}:=w_{i}/d$ for
$i\in\\{1,\ldots,N\\}$ and $G=\langle J_{W}\rangle$, where

$J_{W}:=\left(\exp\left(2\pi iq_{1}\right),\ldots,\exp\left(2\pi
iq_{N}\right)\right)\in(\mathbb{C}^{\times})^{N}.$

Define the set

$\operatorname{Nar}:=\\{k\in\\{0,\ldots,d-1\\}|(k+1)w_{i}\not\in
d\mathbb{Z},\text{ for all }i=1,\ldots,N\\}.$

The regularized FJRW $I$-function is given in [Acosta]:

$I^{\operatorname{reg}}_{\operatorname{FJRW}}(\tau)=\sum_{k\in\operatorname{Nar}}\sum_{l=0}^{\infty}\frac{\tau^{d^{\prime}(l+\frac{k+1}{d})}}{(dl+k)!}\frac{(-1)^{dl+k+1}}{\Gamma(1+d^{\prime}\frac{dl+k+1}{d})}\prod_{i=1}^{N}\frac{(-1)^{\lfloor
q_{i}(dl+k)\rfloor}\Gamma(q_{i}(dl+k+1))}{\Gamma(q_{i}+\langle
q_{i}k\rangle)}\phi_{k},$

where $d^{\prime}:=-d+\sum_{i=1}^{N}w_{i}$ and
$\\{\phi_{k}\\}_{k\in\operatorname{Nar}}$ is a basis of the narrow state space
for the FJRW theory of $(W,G)$.

On the Gromov–Witten side, we consider the relative Gromov–Witten theory of a
degree $d$ Fano hypersurface $X$ with its smooth anticanonical divisor. In
here, we just need genus zero relative Gromov–Witten invariants with maximal
contact orders. We further assume that $w_{i}$ divides
$d^{\prime}:=-d+\sum_{i=1}^{N}w_{i}$ for all $i\in\\{1,\ldots,N\\}$.

######  Theorem 9.3.

The narrow relative quantum D-module of $(X,-K_{X})$ can be identified with
the narrow quantum D-module of $(W,G)$ via analytic continuation.

###### Proof.

By the local-orbifold correspondence [vGGR], [TY20b] and [BNTY] and the
relative-orbifold correspondence[ACW] and [TY18], we can consider the genus
zero local Gromov–Witten theory of the total space of the canonical bundle
$K_{X}$ of $X$.

The $I$-function for $K_{X}$ is

(36) $\displaystyle
I_{K_{X}}(q)=(dH)q^{H/z}\sum_{\begin{subarray}{c}n\in\mathbb{Q}_{\geq 0}\\\
\exists
j:nw_{j}\in\mathbb{Z}\end{subarray}}q^{n}\frac{\prod_{k=1}^{nd}(dH+kz)(-1)^{nd^{\prime}}\prod_{k=1}^{nd^{\prime}}(d^{\prime}H+kz)}{\prod_{i=1}^{N}\prod_{0<k\leq
nw_{i},\langle k\rangle=\langle
nw_{i}\rangle}(w_{i}H+kz)}\textbf{1}_{\langle-n\rangle}.$

The $I$-function can be written in terms of $\Gamma$-functions

$\displaystyle I_{K_{X}}(q)$
$\displaystyle=dHq^{H/z}\frac{\prod_{i=1}^{N}\Gamma(1+\frac{w_{i}H}{z}-\langle-
w_{i}n\rangle)\Gamma(1-d^{\prime}\frac{H}{z})}{\Gamma(1+\frac{dH}{z})}$
$\displaystyle\cdot\sum_{\begin{subarray}{c}n\in\mathbb{Q}_{\geq 0}\\\ \exists
j:nw_{j}\in\mathbb{Z}\end{subarray}}q^{n}\frac{\Gamma(1+\frac{dH}{z}+nd)}{\Gamma(1-\frac{d^{\prime}H}{z}-nd^{\prime})\prod_{i=1}^{N}\Gamma(1+\frac{w_{i}H}{z}+w_{i}n)}\textbf{1}_{\langle-n\rangle}$
$\displaystyle=dHq^{H/z}\frac{\prod_{i=1}^{N}\Gamma(1+\frac{w_{i}H}{z}-\langle-
w_{i}n\rangle)\Gamma(1-d^{\prime}\frac{H}{z})}{\Gamma(1+\frac{dH}{z})}$
$\displaystyle\cdot\sum_{\begin{subarray}{c}n\in\mathbb{Q}_{\geq 0}\\\ \exists
j:nw_{j}\in\mathbb{Z}\end{subarray}}\operatorname{Res}_{s=n}\frac{2\pi
i}{e^{2\pi
is}-1}q^{s}\frac{\Gamma(1+\frac{dH}{z}+sd)}{\Gamma(1-\frac{d^{\prime}H}{z}-sd^{\prime})\prod_{i=1}^{N}\Gamma(1+\frac{w_{i}H}{z}+w_{i}s)}\textbf{1}_{\langle-n\rangle}.$

Following [CIR], we consider the $H$-functions. The regularized $H$-function
for the FJRW theory is

$\displaystyle H^{\operatorname{reg}}_{\operatorname{FJRW}}(\tau)$
$\displaystyle=\sum_{k\in\\{0,\ldots,d-1\\}}\sum_{l=0}^{\infty}\frac{\tau^{d^{\prime}(l+\frac{k+1}{d})}}{(dl+k)!}\frac{(-1)^{dl+k+1}}{\Gamma(1+d^{\prime}\frac{dl+k+1}{d})}\prod_{j=1}^{N}(-1)^{\lfloor
q_{j}(dl+k)\rfloor}\Gamma(q_{j}(dl+k+1))\phi_{k}$
$\displaystyle=\sum_{k\in\\{0,\ldots,d-1\\}}H^{\operatorname{reg}}_{\operatorname{FJRW},k}\phi_{k}.$

The $H$-function for $I_{K_{X}}$ is

$\displaystyle H_{K_{X}}(q)$ $\displaystyle=dHe^{\frac{H}{2\pi i}\log
q}\sum_{\begin{subarray}{c}n\in\mathbb{Q}_{\geq 0}\\\ \exists
j:nw_{j}\in\mathbb{Z}\end{subarray}}q^{n}\frac{\Gamma(1+\frac{dH}{2\pi
i}+nd)}{\Gamma(1-\frac{d^{\prime}H}{2\pi
i}-nd^{\prime})\prod_{i=1}^{N}\Gamma(1+\frac{w_{i}H}{2\pi
i}+w_{i}n)}\textbf{1}_{\langle-n\rangle}$ $\displaystyle=dHe^{\frac{H}{2\pi
i}\log q}\sum_{\begin{subarray}{c}n\in\mathbb{Q}_{\geq 0}\\\ \exists
j:nw_{j}\in\mathbb{Z}\end{subarray}}\operatorname{Res}_{s=n}\frac{2\pi
i}{e^{2\pi is}-1}q^{s}\frac{\Gamma(1+\frac{dH}{2\pi
i}+sd)}{\Gamma(1-\frac{d^{\prime}H}{2\pi
i}-sd^{\prime})\prod_{i=1}^{N}\Gamma(1+\frac{w_{i}H}{2\pi
i}+w_{i}s)}\textbf{1}_{\langle-n\rangle}$
$\displaystyle=\sum_{f}H_{K_{X},f}\textbf{1}_{f},$

where $f$ runs over the set

$\\{0\leq f<1|fw_{j}\in\mathbb{Z}\text{ for some }1\leq j\leq N\\},$

$\bar{f}=\langle 1-f\rangle$ and

$H_{K_{X},f}=dHe^{\frac{H}{2\pi i}\log q}\sum_{n\in\mathbb{Z}_{\geq
0}}\operatorname{Res}_{s=n}\frac{2\pi i}{e^{2\pi
i(s+\bar{f})}-1}q^{s+\bar{f}}\frac{\Gamma(1+\frac{dH}{2\pi
i}+sd+\bar{f}d)}{\Gamma(1-\frac{d^{\prime}H}{2\pi
i}-sd^{\prime}+\bar{f}d^{\prime})\prod_{i=1}^{N}\Gamma(1+\frac{w_{i}H}{2\pi
i}+w_{i}s+\bar{f}w_{i})}.$

We consider the contour integral

$\int_{C}\frac{1}{e^{2\pi
i(s+\bar{f})}-1}q^{s+\bar{f}}\frac{\Gamma(1+\frac{dH}{2\pi
i}+sd+\bar{f}d)}{\Gamma(1-\frac{d^{\prime}H}{2\pi
i}-sd^{\prime}+\bar{f}d^{\prime})\prod_{i=1}^{N}\Gamma(1+\frac{w_{i}H}{2\pi
i}+w_{i}s+\bar{f}w_{i})},$

where the contour $C$ is chosen such that the poles at $s=l$ are on the right
of $C$ and the poles at $s=-1-l$ and at

$s=-\frac{H}{2\pi i}-\frac{m}{d}-\bar{f},m\geq 1$

are on the left of $C$, where $l$ is a non-negative integer. Then the contour
integral is the sum of residue at

$s=-1-l,l\geq 0\text{ and }s=-\frac{H}{2\pi i}-\frac{m}{d}-\bar{f},m\geq 1.$

The residue at $s=-1-l$ vanishes.

We have

$\operatorname{Res}_{s=-\frac{H}{2\pi i}-\frac{m}{d}}\Gamma(1+\frac{dH}{2\pi
i}+ds)=-\frac{1}{d}\frac{(-1)^{m}}{\Gamma(m)}.$

Therefore, we obtain an analytic continuation of $H_{K_{X},f}$:

$\displaystyle dHe^{\frac{H}{2\pi i}}\sum_{m\geq 1}(-1)^{m}\frac{(2\pi
i)e^{\frac{2\pi im}{d}}}{e^{-H}-e^{\frac{2\pi
im}{d}}}\frac{q^{-\frac{m}{d}}}{\Gamma(m)\Gamma(1+\frac{d^{\prime}}{d}m)\prod_{i=1}^{N}\Gamma(1-q_{i}m)}.$

Let $\tau^{d}=q^{-1}$ and $m=k+dl+1$, where $l\in\mathbb{Z}_{\geq 0}$. Then we
have

$\displaystyle dHe^{\frac{H}{2\pi i}}\sum_{k\in\\{0,\ldots
d-1\\}}\sum_{l=0}^{\infty}(-1)^{k+dl+1}\frac{(2\pi i)e^{\frac{2\pi
i(k+1)}{d}}}{e^{-H}-e^{\frac{2\pi
i(k+1)}{d}}}\frac{q^{-\frac{k+dl+1}{d}}\prod_{i=1}^{N}\Gamma(q_{i}(k+dl+1))\sin(\pi
q_{i}(k+dl+1))}{\pi^{N}(k+dl)!\Gamma(1+\frac{d^{\prime}}{d}(k+dl+1))}$
$\displaystyle=$ $\displaystyle dHe^{\frac{H}{2\pi i}}\sum_{k\in\\{0,\ldots
d-1\\}}(-1)^{k+1}\frac{(2\pi i)e^{\frac{2\pi i(k+1)}{d}}}{e^{-H}-e^{\frac{2\pi
i(k+1)}{d}}}\frac{1}{\Gamma(1-q_{i}(k+1))\Gamma(q_{i}(k+1))}\sum_{l=0}^{\infty}\frac{q^{-\frac{k+dl+1}{d}}\prod_{i=1}^{N}\Gamma(q_{i}(k+dl+1))}{(k+dl)!\Gamma(1+\frac{d^{\prime}}{d}(k+dl+1))}$
$\displaystyle=$ $\displaystyle dHe^{\frac{H}{2\pi i}}\sum_{k\in\\{0,\ldots
d-1\\}}(-1)^{k+1}\frac{(2\pi i)e^{\frac{2\pi i(k+1)}{d}}}{e^{-H}-e^{\frac{2\pi
i(k+1)}{d}}}\frac{1}{\Gamma(1-q_{i}(k+1))\Gamma(q_{i}(k+1))}H^{\operatorname{reg}}_{\operatorname{FJRW},k}.$

Therefore, we obtained an analytic continuation between the $I$-function for
$K_{X}$ and the regularized $I$-function for the FJRW theory of $(W,G)$. ∎

## References


# Certain classes of vector fields on the torus and some of their periodic
orbits

Supriyo Jana  Department of Mathematics, Indian Institute of Technology
Madras, India<EMAIL_ADDRESS>
###### Abstract.

In this paper, we classify all cubic vector fields in $\mathbb{R}^{3}$ such
that their flow makes the torus

$\mathbb{T}^{2}=\\{(x,y,z)\in\mathbb{R}^{3}:(x^{2}+y^{2}-a^{2})^{2}+z^{2}-1=0\\}~{}\mbox{with}~{}a\in(1,\infty)$

invariant. We also classify cubic Kolmogorov vector fields on
$\mathbb{T}^{2}$. We investigate invariant meridians and parallels of cubic
vector fields and present some necessary and sufficient conditions when they
will be periodic orbits or limit cycles. We study ‘Pseudo Type-$n$’ vector
fields and cubic vector fields generated by Lie bracket of two arbitrary
quadratic vector fields on $\mathbb{T}^{2}$. We discuss invariant meridians
and parallels of Pseudo Type-$n$ vector fields also.

###### Key words and phrases:

Polynomial vector fields, First integrals, Kolmogorov vector fields, Pseudo
Type-$n$ vector fields, Invariant meridians, Invariant parallels, Periodic
orbit, Limit cycle, Lie bracket

###### 2020 Mathematics Subject Classification:

34A34, 34C14, 34C40, 34C45

##  1\. Introduction

Polynomial vector fields have been intensely treated since 1900 due to the
second part of Hilbert’s 16th problem [4, 7, 8, 11]. People also studied
invariant algebraic curves of polynomial vector fields in the plane due to
Darboux Integrability Theory [3], which states that a sufficient number of
invariant algebraic curves guarantee the existence of a first integral. Later,
Llibre and Bolanos [13] extends the Darboux Integrability Theory to vector
fields on any regular algebraic hypersurface in $\mathbb{R}^{n}$. First
integral is also an important part in the qualitative theory of differential
systems, see for instance [6, 12, 18, 19] and references therein. A first
integral of the associated vector field of a differential system reduces the
dimension of the system by one, which makes the analysis of the system easier.

Planar Kolmogorov systems were introduced by A. Kolmogorov [10] to describe
the interaction between two species occupying the same ecological habitat.
Later polynomial Kolmogorov systems were generalized in higher dimensions in
the following form.

$\frac{dx_{i}}{dt}=x_{i}P_{i}(x_{1},...,x_{n})~{}\mbox{where}~{}P_{i}\in\mathbb{R}[x_{1},...,x_{n}].$

These systems may exhibit quite complex dynamical behavior as explained in the
series of papers by Hirsch (see [5]).

We consider the polynomial differential system

(1.1) $\frac{dx}{dt}=P,~{}\frac{dy}{dt}=Q,~{}\frac{dz}{dt}=R$

and its associated polynomial vector field

(1.2) $\chi=P\frac{\partial}{\partial x}+Q\frac{\partial}{\partial
y}+R\frac{\partial}{\partial z}$

where $P,Q,R\in\mathbb{R}[x,y,z]$. For simplicity, we may write
$\chi:=(P,Q,R)$. The number $n:=\max\\{\deg P,~{}\deg Q,~{}\deg R\\}$ is
called the degree of the polynomial vector field (1.2).

An invariant algebraic set for $\chi$ is a set $A\subset\mathbb{R}^{3}$ such
that $A$ is the zero set of some $f\in\mathbb{R}[x,y,z]$ and $\mathcal{X}f=Kf$
for some $K\in\mathbb{R}[x,y,z]$. The polynomial $K$ is called cofactor of
$\chi$ for $A$. Moreover, $\mathcal{X}$ is also called a vector field on $A$.
In this case, if any solution curve of (1.1) intersect $A$ then the whole
curve lies in $A$.

We consider the two-dimensional torus [15],

$\mathbb{T}^{2}=\\{(x,y,z)\in\mathbb{R}^{3}:(x^{2}+y^{2}-a^{2})^{2}+z^{2}-1=0\\}~{}\mbox{where}~{}a\in(1,\infty).$

We note that Llibre and Medrado [15] studied polynomial vector fields on
$\mathbb{T}^{2}$, in particular, they investigated the sharp upper bounds of
invariant meridians and parallels. Also, they discussed limit cycles of
quadratic vector field on $\mathbb{T}^{2}$. In this article, we study cubic
and Pseudo Type-$n$ vector fields on $\mathbb{T}^{2}$. The definition of
Pseudo Type-$n$ vector field can be seen in section 4.

This paper is organized in the following way. In Section 2, we present some
basic definitions and useful results on polynomial vector fields. We recall
the criteria to search for the invariant hypersurfaces of a vector field. We
have also presented a SageMath code to compute Lie bracket of two vector
fields in Euclidean space.

In Section 3, we give a necessary and sufficient condition when a cubic vector
field in $\mathbb{R}^{3}$ is a vector field on $\mathbb{T}^{2}$, see Theorem
3.1. As a consequence of this theorem, we also classify all cubic Kolmogorov
vector fields on $\mathbb{T}^{2}$, see Corollary 3.2. We show that every cubic
Kolmogorov vector field on $\mathbb{T}^{2}$ possesses a rational first
integral, see Proposition 3.3. If a cubic Kolmogorov vector field on
$\mathbb{T}^{2}$ has finitely many invariant meridians and parallels then we
show that the only meridian and parallel planes are the coordinate planes, see
Theorem 3.5. We provide a necessary and sufficient condition when a cubic
vector field on $\mathbb{T}^{2}$ having finitely many invariant meridians
reaches the sharp bound for number of invariant meridians, see Theorem 3.7.
Moreover, we give a necessary and sufficient condition for these invariant
meridians to be limit cycles, see Theorem 3.8. We classify all the cubic
vector fields on $\mathbb{T}^{2}$ having two invariant parallel planes. Also,
we give a necessary and sufficient condition for the corresponding invariant
parallels to be periodic orbits, see Theorem 3.10. We present a necessary and
sufficient condition for a cubic vector field on $\mathbb{T}^{2}$ to have
infinitely many invariant parallels, see Theorem 3.11.

In Section 4, we classify all Pseudo Type-$n$ vector fields on
$\mathbb{T}^{2}$ and prove that any such Pseudo Type-$n$ vector field is
completely integrable, see Theorem 4.3. We prove that an invariant meridian of
a Pseudo Type-$n$ vector field on $\mathbb{T}^{2}$ is a subset of the set of
all singular points of the vector field, see Proposition 4.4. We characterize
the Lie bracket of any two quadratic vector fields on $\mathbb{T}^{2}$, see
Theorem 4.8. We show that Lie bracket of two Pseudo Type-2 vector fields on
$\mathbb{T}^{2}$ is either zero or a Pseudo Type-3 vector field, see Corollary
4.9. We show that Lie bracket of two Pseudo Type-2 vector fields on
$\mathbb{T}^{2}$ can have maximum two invariant meridians, see Proposition
4.11. We prove that for a quadratic vector field $(P,Q,R)$ on $\mathbb{T}^{2}$
with $R\neq 0$, singular points lie on the parallels given by $z=\pm 1$, see
Theorem 4.12.

##  2\. Preliminaries

In this section, we present some basic definitions and useful results. We
recall the following from [15]. The curves obtained by intersection of a plane
containing the $z$-axis and a plane orthogonal to the $z$-axis with
$\mathbb{T}^{2}$ are called meridians and parallels respectively. More
precisely, meridians can be obtained by $\\{ax+by=0\\}\cap\mathbb{T}^{2}$
where $a,b\in\mathbb{R}$. So, meridians always appear in pairs. We call the
plane $ax+by=0$ a meridian plane. Also, parallels can be obtained by
$\\{z-k=0\\}\cap\mathbb{T}^{2}$ with $|k|\leq 1$. So, if $|k|\neq 1$ then
parallels appear in pairs. We call the plane $z-k=0$ a parallel plane.

We say that a meridian obtained by the intersection with a plane $ax+by=0$ is
invariant by the flow of $\chi$ if $ax+by=0$ is an invariant algebraic set for
$\chi$. Similarly, we say that a parallel obtained by the intersection with a
plane $z-k=0$ is invariant if $z-k=0$ is an invariant algebraic set for
$\chi$.

We will use the tool extactic polynomial to find invariant meridians and
parallels. Let $W$ be a vector subspace of $\mathbb{R}[x_{1},...,x_{n}]$
generated by the independent polynomials $v_{1},...,v_{l}$, i.e., $W=\langle
v_{1},...,v_{\ell}\rangle$. The extactic polynomial of $\chi$ associated with
$W$ is the polynomial

$\mathcal{E}_{W}(\chi)=\begin{vmatrix}v_{1}&\dots&v_{\ell}\\\
\chi(v_{1})&\dots&\chi(v_{\ell})\\\ \cdot&&\cdot\\\ \cdot&\cdots&\cdot\\\
\cdot&&\cdot\\\
\chi^{{\ell}-1}(v_{1})&\dots&\chi^{{\ell}-1}(v_{\ell})\end{vmatrix}$

where $\chi^{j}(v_{i})=\chi^{j-1}(\chi(v_{i}))$ for all $i,j$. From the
properties of determinant, it follows that the definition of the extactic
polynomial is independent of the chosen basis of $W$. The proof of the
following is similar to the proof of [14, Proposition 1].

######  Proposition 2.1.

Let $\chi$ be a polynomial vector field in $\mathbb{R}^{n}$ and $W$ be a
finite dimensional vector subspace of $\mathbb{R}[x_{1},x_{2},\cdots,x_{n}]$
with $\dim(W)>1$. If $f=0$ is an invariant algebraic hypersurface for the
vector field $\chi$ with $f\in W$, then $f$ is a factor of
$\mathcal{E}_{W}(\chi)$.

The multiplicity of an invariant algebraic hypersurface $f=0$ with $f\in W$,
is the largest positive integer $k$ such that $f^{k}$ divides the extactic
polynomial $\mathcal{E}_{W}(\chi)$ when $\mathcal{E}_{W}(\chi)\neq 0$,
otherwise the multiplicity is infinite. For more details regarding the
multiplicity, see [2, 17]. In this paper, by finitely many invariant algebraic
set, we will always consider taking into account multiplicities.

######  Definition 2.2.

Let $U$ be an open subset of $\mathbb{R}^{3}$. A non-constant analytic map
$H\colon U\to\mathbb{R}$ is called a first integral of the vector field (1.2)
on $U$ if $H$ is constant on all solution curves of the system (1.1) contained
in $U$ ; i.e. $H(x(t),y(t),z(t))=$ constant for all values of $t$ for which
the solution $(x(t),y(t),z(t))$ is defined and contained in $U$.

Note that $H$ is a first integral of the vector field (1.2) on $U$ if and only
if $\chi H=0$ on $U$. A vector field in $\mathbb{R}^{n}$ is called completely
integrable if it has $n-1$ functionally independent first integrals. We recall
the following two results.

######  Lemma 2.3.

[9, Lemma 3.1] Let $Q_{1},Q_{2},R_{1},R_{2}\in\mathbb{R}[x_{1},...,x_{n}]$
with $R_{1},R_{2}\neq 0$ be such that ${Q_{1}R_{1}+Q_{2}R_{2}}$ is the zero
polynomial and $\gcd(R_{1},R_{2})=1$. Then $Q_{1}=AR_{2},~{}Q_{2}=-AR_{1}$ for
some polynomial ${A\in\mathbb{R}[x_{1},...,x_{n}].}$

######  Lemma 2.4.

[9, Lemma 4.1] Let $Q_{1},Q_{2},Q_{3}\in\mathbb{R}[x,y,z]$ such that
$Q_{1}x+Q_{2}y+Q_{3}z$ is the zero polynomial. Then
$Q_{1}=Ay+Bz,Q_{2}=-Ax+Cz$, and $Q_{3}=-Bx-Cy$ for some polynomials
$A,B,C\in\mathbb{R}[x,y,z]$.

Suppose $X,Y$ are two vector fields on a smooth manifold $M$. Then, we know
that the Lie bracket of $X$ and $Y$ defined by $[X,Y]:=XY-YX$ is also a vector
field on $M$. More precisely, if
$X=\sum\limits_{i=1}^{n}P_{i}\frac{\partial}{\partial x_{i}}$ and
$Y=\sum\limits_{i=1}^{n}Q_{i}\frac{\partial}{\partial x_{i}}$ are two vector
fields then the Lie bracket of $X$ and $Y$ is defined as
$[X,Y]:=\sum\limits_{i=1}^{n}(X(Q_{i})-Y(P_{i}))\frac{\partial}{\partial
x_{i}}$.

Computing Lie bracket of two vector fields is quite tedious. One can use the
following SageMath111Official website of the open-source software SageMath:
https://www.sagemath.org/ code to compute Lie bracket of two vector fields in
$\mathbb{R}^{3}$. For example, we have taken two vector fields $X=(P1,P2,P3)$
and $Y=(Q1,Q2,Q3)$ to find their Lie bracket.

⬇

1from sage.manifolds.operators import *

2E.<x,y,z> = EuclideanSpace()

3def commutator(X,Y):

4com1 = X.dot(grad(E.scalar_field(Y[1])))-Y.dot(grad(E.scalar_field(X[1])))

5com2 = X.dot(grad(E.scalar_field(Y[2])))-Y.dot(grad(E.scalar_field(X[2])))

6com3 = X.dot(grad(E.scalar_field(Y[3])))-Y.dot(grad(E.scalar_field(X[3])))

7comm = E.vector_field(com1, com2, com3, name=‘com’)

8return comm

9X=E.vector_field(P1,P2,P3, name=‘X’)

10Y=E.vector_field(Q1,Q2,Q3, name=‘Y’)

11commutator(X,Y).display()’

##  3\. Cubic vector fields on $\mathbb{T}^{2}$

In this section, we classify and study cubic vector fields on
$\mathbb{T}^{2}$. We also study invariant meridians and parallels of a cubic
vector field on $\mathbb{T}^{2}$ and their periodicity. We discuss all
possible invariant meridians and parallels of a cubic Kolmogorov vector field
on $\mathbb{T}^{2}$.

######  Theorem 3.1.

Let $\chi=(P,Q,R)$ be a cubic vector field in $\mathbb{R}^{3}$. Then $\chi$ is
a vector field on $\mathbb{T}^{2}$ if and only if there exist quadratic
polynomials $f,K\in\mathbb{R}[x,y,z]$ and $\beta,\gamma\in\mathbb{R}$ such
that

(3.1) $\begin{split}P=&\frac{1}{4}Kx+fy+\beta z,\\\ Q=&\frac{1}{4}Ky-fx+\gamma
z,\text{ and }\\\
R=&\frac{1}{2}K^{\prime}(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)-2(\beta x+\gamma
y)(x^{2}+y^{2}-a^{2})\end{split}$

where $K=K^{\prime}z$ for some linear polynomial $K^{\prime}$. Moreover, $K$
is the cofactor of $\chi$.

###### Proof.

Suppose $\chi$ is a vector field on $\mathbb{T}^{2}$. Then $\chi$ satisfies
the following.

(3.2) $4(x^{2}+y^{2}-a^{2})(Px+Qy)+2Rz=K((x^{2}+y^{2}-a^{2})^{2}+z^{2}-1)$

for some quadratic polynomial $K\in\mathbb{R}[x,y,z]$. Observe that the
constant term of $K$ is zero because left side of (3.2) has no constant term.

We write $P=P^{(3)}+P^{(2)}+P^{(1)}+P^{(0)}$ where $P^{(j)}$ denotes the
degree $j$ homogeneous component of $P$. We also write $Q$, $R$ and $K$ in
this manner. Note that $K^{(0)}=0$. Now, comparing the degree 6 terms in
(3.2), we get

$4(x^{2}+y^{2})(P^{(3)}x+Q^{(3)}y)=K^{(2)}(x^{2}+y^{2})^{2}.$

Hence,

(3.3) $P^{(3)}x+Q^{(3)}y=\frac{1}{4}K^{(2)}(x^{2}+y^{2}).$

We re-write the above equation as
$(P^{(3)}-\frac{1}{4}K^{(2)}x)x+(Q^{(3)}-\frac{1}{4}K^{(2)}y)y=0$. Then, by
Lemma 2.3,

$P^{(3)}=\frac{1}{4}K^{(2)}x+Ay~{}\mbox{and}~{}Q^{(3)}=\frac{1}{4}K^{(2)}y-Ax$

for some quadratic homogeneous polynomial $A$.

Again, comparing the degree 5 terms in (3.2), we get

(3.4) $4(x^{2}+y^{2})(P^{(2)}x+Q^{(2)}y)=K^{(1)}(x^{2}+y^{2})^{2}.$

Hence, $(P^{(2)}-\frac{1}{4}K^{(1)}x)x+(Q^{(2)}-\frac{1}{4}K^{(1)}y)y=0.$
Then, by Lemma 2.3,

$P^{(2)}=\frac{1}{4}K^{(1)}x+By~{}\mbox{and}~{}Q^{(2)}=\frac{1}{4}K^{(1)}y-Bx$

for some linear homogeneous polynomial $B$.

The equation (3.2) is true for any $x,y,z\in\mathbb{R}$. In particular, for
$z=0$, the equation (3.2) becomes

(3.5)
$4(x^{2}+y^{2}-a^{2})(P(x,y,0)x+Q(x,y,0)y)=K(x,y,0)((x^{2}+y^{2}-a^{2})^{2}-1).$

Comparing the degree 4 terms in (3.5), we get

$4(x^{2}+y^{2})(P^{(1)}(x,y,0)x+Q^{(1)}(x,y,0)y)-4a^{2}(P^{(3)}(x,y,0)x+Q^{(3)}(x,y,0)y)=-2a^{2}K^{(2)}(x,y,0)(x^{2}+y^{2}).$

Now, using (3.3) in the above equation, we obtain

$P^{(1)}(x,y,0)x+Q^{(1)}(x,y,0)y=-\frac{a^{2}}{4}K^{(2)}(x,y,0).$

Also, comparing the degree 2 terms in (3.5), we get

$-4a^{2}(P^{(1)}(x,y,0)x+Q^{(1)}(x,y,0)y)=(a^{4}-1)K^{(2)}(x,y,0).$

Hence, from last two equations, we have $K^{(2)}(x,y,0)=0$. So,
$K^{(2)}=L_{1}z$ where $L_{1}$ is a linear homogenous polynomial.

Now, comparing the degree 3 terms in (3.5), we get

$4(x^{2}+y^{2})(P^{(0)}x+Q^{(0)}y)-4a^{2}(P^{(2)}(x,y,0)x+Q^{(2)}(x,y,0)y)=-2a^{2}K^{(1)}(x,y,0)(x^{2}+y^{2}).$

Hence, using (3.4) in the above equation, we have

$P^{(0)}x+Q^{(0)}y=-\frac{a^{2}}{4}K^{(1)}(x,y,0).$

Again, comparing the degree 1 terms in (3.5), we have

$-4a^{2}(P^{(0)}x+Q^{(0)}y)=(a^{4}-1)K^{(1)}(x,y,0).$

Hence, from the last two equations, we obtain $P^{(0)}x+Q^{(0)}y=0$ as well as
$K^{(1)}(x,y,0)=0$. So, $P^{(0)}=Q^{(0)}=0$ and $K^{(1)}=L_{0}z$ for some
$L_{0}\in\mathbb{R}$. Hence, the cofactor $K=K^{\prime}z$ where
$K^{\prime}:=L_{1}+L_{0}$.

Now, comparing the degree 1 terms in (3.2), we get that
$2R^{(0)}z=(a^{4}-1)L_{0}z$. This gives $R^{(0)}=\frac{a^{4}-1}{2}L_{0}$. Now,
comparing the degree 2 terms in (3.2), we get

$\begin{split}&-4a^{2}(P^{(1)}x+Q^{(1)}y)+2R^{(1)}z=(a^{4}-1)L_{1}z\\\
\implies&P^{(1)}x+Q^{(1)}y+(-\frac{1}{2a^{2}}R^{(1)}+\frac{a^{4}-1}{4a^{2}}L_{1})z=0.\end{split}$

Then, by Lemma 2.4,

$P^{(1)}=\alpha y+\beta z,~{}Q^{(1)}=-\alpha x+\gamma
z~{}\mbox{and}~{}R^{(1)}=2a^{2}(\frac{a^{4}-1}{4a^{2}}L_{1}+\beta x+\gamma y)$

for some $\alpha,\beta,\gamma\in\mathbb{R}$. Now, comparing the degree 3 terms
in (3.2), we get

$-4a^{2}(P^{(2)}x+Q^{(2)}y)+2R^{(2)}z=L_{0}z(-2a^{2}(x^{2}+y^{2})+z^{2}).$

Then, using (3.4) in the above equation, we get

$R^{(2)}=\frac{1}{2}L_{0}(-a^{2}(x^{2}+y^{2})+z^{2}).$

Next, comparing the degree 4 terms in (3.2), we get

(3.6)
$4(x^{2}+y^{2})(P^{(1)}x+Q^{(1)}y)-4a^{2}(P^{(3)}x+Q^{(3)}y)+2R^{(3)}z=L_{1}z(-2a^{2}(x^{2}+y^{2})+z^{2}).$

Substituting the values of $P^{(1)},Q^{(1)}$ in (3.6) along with (3.3), we
obtain

$R^{(3)}=\frac{1}{2}L_{1}(-a^{2}(x^{2}+y^{2})+z^{2})-2(x^{2}+y^{2})(\beta
x+\gamma y).$

Let us now construct $P,Q,R$.

$\begin{split}P&=P^{(3)}+P^{(2)}+P^{(1)}+P^{(0)}\\\
&=\frac{1}{4}K^{(2)}x+Ay+\frac{1}{4}K^{(1)}x+By+\alpha y+\beta z\\\
&=\frac{1}{4}Kx+(A+B+\alpha)y+\beta z.\end{split}$

Assuming $f:=A+B+\alpha$, we get $P=\frac{1}{4}Kx+fy+\beta z$. Again,

$\begin{split}Q&=\frac{1}{4}K^{(2)}y-Ax+\frac{1}{4}K^{(1)}y-Bx-\alpha x+\gamma
z\\\ &=\frac{1}{4}Ky-fx+\gamma z.\end{split}$

Now,

$\begin{split}R=&\frac{1}{2}L_{1}(-a^{2}(x^{2}+y^{2})+z^{2})-2(x^{2}+y^{2})(\beta
x+\gamma y)+\frac{1}{2}L_{0}(-a^{2}(x^{2}+y^{2})+z^{2})+\\\
&2a^{2}(\frac{a^{4}-1}{4a^{2}}L_{1}+\beta x+\gamma
y)+\frac{a^{4}-1}{2}L_{0}\\\
=&\frac{1}{2}(L_{1}+L_{0})(-a^{2}(x^{2}+y^{2})+z^{2})-2(\beta x+\gamma
y)(x^{2}+y^{2}-a^{2})+\frac{a^{4}-1}{2}(L_{1}+L_{0})\\\
=&\frac{1}{2}K^{\prime}(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)-2(\beta x+\gamma
y)(x^{2}+y^{2}-a^{2}).\end{split}$

Hence, if $\chi=(P,Q,R)$ is a cubic vector field on $\mathbb{T}^{2}$, we get
$P,Q,R$ in the form of (3.1).

Again, if $P,Q,R$ is given by (3.1), then they satisfy (3.2). Hence, the
converse part also follows. ∎

######  Corollary 3.2.

Let $\chi=(P,Q,R)$ be a cubic Kolmogorov vector field in $\mathbb{R}^{3}$.
Then $\chi$ is a vector field on $\mathbb{T}^{2}$ if and only if there exist
$c_{1},c_{2}\in\mathbb{R}$ such that

(3.7) $\begin{split}P=&\frac{1}{4}Kx+c_{1}xy^{2},\\\ Q=&\frac{1}{4}Ky-
c_{1}x^{2}y,\text{ and }\\\
R=&\frac{c_{2}}{2}z(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)\end{split}$

where $K=c_{2}z^{2}$ is the cofactor of $\chi$.

###### Proof.

Suppose $\chi=(P,Q,R)$ is a vector field on $\mathbb{T}^{2}$ given by (3.1).
Since $\chi$ is Kolmogorov vector field, $x,y,z$ must divide $P,Q,R$,
respectively. If $x$ divides $P$ then $x$ divides $fy+\beta z$. Also, $y$
divides $Q$ implies $y$ divides $-fx+\gamma z$. Suppose $fy+\beta z=K_{1}x$
and $-fx+\gamma z=K_{2}y$ for some $K_{1},K_{2}\in\mathbb{R}[x,y,z]$. Now,
$K_{1}x^{2}+K_{2}y^{2}=(fy+\beta z)x+(-fx+\gamma z)y=z(\beta x+\gamma y)$.
Hence, $\beta=\gamma=0$ and so $K_{1}=c_{1}y^{2}$ and $K_{2}=-c_{1}x^{2}$ for
some $c_{1}\in\mathbb{R}$. So,

$P=\frac{1}{4}Kx+c_{1}xy^{2}~{}\text{and}~{}Q=\frac{1}{4}Ky-c_{1}x^{2}y.$

Also, $R=\frac{1}{2}K^{\prime}(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)$ since
$\beta=\gamma=0$. As $z$ divides $R$, $z$ divides $K^{\prime}$, say
$K^{\prime}=c_{2}z$ for some $c_{2}\in\mathbb{R}$. So,
$R=\frac{c_{2}}{2}z(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)$.

One can check that the converse part is also true. ∎

######  Proposition 3.3.

Every cubic Kolmogorov vector field on $\mathbb{T}^{2}$ has a rational first
integral.

###### Proof.

An arbitrary cubic Kolmogorov vector field on $\mathbb{T}^{2}$ is given by
(3.7). Note that $x^{2}+y^{2}=0$ and the torus
$(x^{2}+y^{2}-a^{2})^{2}+z^{2}-1=0$ are invariant hypersurfaces with dependent
cofactor $\frac{1}{2}K$ and $K$ respectively. Hence, by the Darboux
Integrability Theory [16, Theorem 5],
$((x^{2}+y^{2}-a^{2})^{2}+z^{2}-1)(x^{2}+y^{2})^{-2}$ is a rational first
integral of any cubic Kolmogorov vector field on $\mathbb{T}^{2}$. ∎

######  Remark 3.4.

In (3.1), by vanishing the cubic terms, we can get any quadratic vector field
on $\mathbb{T}^{2}$. That is, for a quadratic vector field, $\beta=\gamma=0$
and $K,f$ are linear polynomials. Hence, by using similar arguments of the
proof of Corollary LABEL:cor:kolm-rational-fi, the function given by
$f:=((x^{2}+y^{2}-a^{2})^{2}+z^{2}-1)(x^{2}+y^{2})^{-2}$ is a rational first
integral of any quadratic vector field on $\mathbb{T}^{2}$ also.

######  Theorem 3.5.

Suppose that a cubic Kolmogorov vector field on $\mathbb{T}^{2}$ has finitely
many invariant meridians and parallels. Then $x=0$ and $y=0$ are the only
invariant meridian planes and $z=0$ is the only invariant parallel plane.

###### Proof.

Consider the cubic Kolmogorov vector field $\chi=(P,Q,R)$ on $\mathbb{T}^{2}$
given by (3.7). Then the extactic polynomial associated with $\langle
x,y\rangle$ is

$\mathcal{E}_{\langle x,y\rangle}(\chi)=\begin{vmatrix}x&y\\\
\frac{1}{4}Kx+c_{1}xy^{2}&\frac{1}{4}Ky-
c_{1}x^{2}y\end{vmatrix}=-c_{1}xy(x^{2}+y^{2}).$

The only divisors of the form $ax+by$ of the above extactic polynomial are $x$
and $y$. Notice that $x=0$ and $y=0$ produce invariant meridians. By
Proposition 2.1, these are the only invariant meridian planes of $\chi$.

Note that $z-k=0$ is an invariant parallel plane of $\chi$ if and only if
$z-k$ divides $R$. But the only divisor of $R$ of the form is $z-k$ is $z$.
Hence, $z=0$ is the only invariant parallel plane of $\chi$. ∎

######  Remark 3.6.

For a cubic Kolmogorov vector field on $\mathbb{T}^{2}$, no invariant meridian
or parallel is periodic orbit since each invariant meridian and parallel
contains some singular points (say $(0,\pm\sqrt{a^{2}\pm 1},0)$ or
$(\pm\sqrt{a^{2}\pm 1},0,0)$).

Llibre and Medrado [15] proved that if $\chi$ of degree $m>1$ on
$\mathbb{T}^{2}$ has finitely many invariant meridians then it has at most
$2(m-1)$ invariant meridians. Next, we look at the cubic vector fields on
$\mathbb{T}^{2}$ which have 2(3-1)=4 invariant meridians.

######  Theorem 3.7.

Suppose $\chi$ is a cubic vector field on $\mathbb{T}^{2}$ given by (3.1).
Then $\chi$ have four invariant meridians given by the planes
$a_{i}x+b_{i}y=0$ for $i=1,2$ if and only if
$f=c\prod\limits_{i=1}^{2}(a_{i}x+b_{i}y)$ for some
$c\in\mathbb{R}\setminus\\{0\\}$ and $\beta=\gamma=0$.

###### Proof.

The extactic polynomial of $\chi=(P,Q,R)$ given by (3.1) associated with
$W=\langle x,y\rangle$ is

$\begin{split}\mathcal{E}_{W}(\chi)=\begin{vmatrix}x&y\\\
P&Q\end{vmatrix}&=\begin{vmatrix}x&y\\\ \frac{1}{4}Kx+fy+\beta
z&\frac{1}{4}Ky-fx+\gamma z\end{vmatrix}\\\ &=-f(x^{2}+y^{2})+z(\gamma x-\beta
y).\end{split}$

Notice that $\mathcal{E}_{W}(\chi)$ is a degree 4 polynomial. By Proposition
2.1, there exists $g\in\mathbb{R}[x,y,z]$ of degree 2 such that

(3.8) $-f(x^{2}+y^{2})+z(\gamma x-\beta
y)=\mathcal{E}_{W}(\chi)=g\prod\limits_{i=1}^{2}(a_{i}x+b_{i}y).$

We denote $f^{(j)}$ and $g^{(j)}$ as the degree $j$ homogeneous part of $f$
and $g$, respectively. Hence,
$-f^{(2)}(x^{2}+y^{2})=g^{(2)}\prod\limits_{i=1}^{2}(a_{i}x+b_{i}y)$. We know
that $x^{2}+y^{2}$ has no linear factor over $\mathbb{R}$. Hence,
$\prod\limits_{i=1}^{2}(a_{i}x+b_{i}y)$ must divide $f^{(2)}$. So,
$f^{(2)}=c\prod\limits_{i=1}^{2}(a_{i}x+b_{i}y)$ for some $c\in\mathbb{R}$.
Similarly, $\prod\limits_{i=1}^{2}(a_{i}x+b_{i}y)$ divides $f^{(1)}$ which
implies that $f^{(1)}=0$. Now comparing degree 2 terms in (3.8), we obtain

$-f^{(0)}(x^{2}+y^{2})+z(\gamma x-\beta
y)=g^{(0)}\prod\limits_{i=1}^{2}(a_{i}x+b_{i}y).$

This gives $\beta=\gamma=0$ since the coefficients of $xz,yz$ at the right
side are zero. Hence,
$-f^{(0)}(x^{2}+y^{2})=g^{(0)}\prod\limits_{i=1}^{2}(a_{i}x+b_{i}y)$, which is
possible only when $f^{(0)}=g^{(0)}=0$. Hence,
$f=c\prod\limits_{i=1}^{2}(a_{i}x+b_{i}y)$. Again, $c$ must be non-zero since
the extactic polynomial $\mathcal{E}_{W}(\chi)$ is non-zero.

One can check that the converse part also holds. ∎

We recall the following from [15]. In cylindrical coordinates
$(x,y,z)=(r\cos\theta,r\sin\theta,z)$ the vector field $\chi=(P,Q,R)$ becomes

$\chi=(P(r\cos\theta,r\sin\theta,z)\cos\theta+Q(r\cos\theta,r\sin\theta,z)\sin\theta)\frac{\partial}{\partial
r}-\frac{1}{r}(P(r\cos\theta,r\sin\theta,z)\sin\theta-Q(r\cos\theta,r\sin\theta,z)\cos\theta)\frac{\partial}{\partial\theta}+R(r\cos\theta,r\sin\theta,z)\frac{\partial}{\partial
z}.$

######  Theorem 3.8.

Suppose the cubic vector field (3.1) on $\mathbb{T}^{2}$ has exactly four
invariant meridians. Then the invariant meridians are limit cycles if and only
if $K^{\prime}=0$ has no solution on those meridians. Moreover, the limit
cycles appear as stable or unstable alternately.

###### Proof.

Consider the cubic vector field $\chi$ on $\mathbb{T}^{2}$ given by (3.1).
Suppose $\chi$ has exactly four invariant meridians given by the planes
$a_{i}x+b_{i}y=0$ for $i=1,2$. Then, by using Theorem 3.7, we get
$f=c\prod\limits_{i=1}^{2}(a_{i}x+b_{i}y)$ for some
$c\in\mathbb{R}\setminus\\{0\\}$ and $\beta=\gamma=0$. In cylindrical
coordinates the vector field $\chi$ becomes $(\dot{r},\dot{\theta},\dot{z})$
where

$\begin{split}\dot{r}&=\frac{r}{4}K,\\\
\dot{\theta}&=-c\prod\limits_{i=1}^{2}(a_{i}r\cos\theta+b_{i}r\sin\theta),\\\
\dot{z}&=\frac{1}{2}K^{\prime}(-a^{2}r^{2}+z^{2}+a^{4}-1).\end{split}$

Suppose $(r_{0},\theta_{0},z_{0})\in\mathbb{T}^{2}$ is a singular point of
$\chi$. Observe that $r_{0}\neq 0$. Hence, from $\dot{r}=0$, we obtain
$K(r_{0}\cos\theta_{0},r_{0}\sin\theta_{0},z_{0})=0$. As $K=K^{\prime}z$, if
$K^{\prime}(r_{0}\cos\theta_{0},r_{0}\sin\theta_{0},z_{0})\neq 0$ then
$z_{0}=0$. In this case, $(r_{0}^{2}-a^{2})^{2}=1$ since
$(r_{0},\theta_{0},z_{0})\in\mathbb{T}^{2}$. Also, from $\dot{z}=0$, we get
$-a^{2}r_{0}^{2}+a^{4}-1=0$ which gives
$(r_{0}^{2}-a^{2})^{2}=\frac{1}{a^{4}}$. Thus, $1=\frac{1}{a^{4}}$ which
contradicts the fact that $a>1$. Hence,
$K^{\prime}(r_{0}\cos\theta_{0},r_{0}\sin\theta_{0},z_{0})$ must be zero.
Also, notice that $\dot{\theta}=0$ on the invariant meridians. So, a point on
the invariant meridians is a singular point if and only if $K^{\prime}=0$ at
that point. It is well known that an invariant closed curve is periodic orbit
if and only if the curve does not contain any singular point. Hence, the
invariant meridians are periodic orbit if and only if $K^{\prime}=0$ has no
solution on the meridians.

If any other periodic orbit on $\mathbb{T}^{2}$ exists except the invariant
meridians, it must lie in between two consecutive invariant meridians. We
observe that $\dot{\theta}=0$ only on the invariant meridians. So, there are
no periodic orbits on $\mathbb{T}^{2}$ except the invariant meridians. Hence,
the invariant meridians are limit cycles. Between two consecutive invariant
meridians $\dot{\theta}$ is either positive or negative. Also, sign of
$\dot{\theta}$ changes only when we cross one invariant meridian because
$\dot{\theta}=\rho\cos(\theta-\alpha)\cos(\theta-\beta)$ for some
$\rho,\alpha,\beta\in\mathbb{R}$. So, the invariant meridians are stable or
unstable limit cycles alternately. ∎

######  Example 3.9.

Consider the cubic vector field $\chi=(P,Q,R)$ with

$\begin{split}P&=\frac{1}{4}xz(x^{2}+y^{2}+1)+xy^{2},\\\
Q&=\frac{1}{4}yz(x^{2}+y^{2}+1)-x^{2}y,\\\
R&=\frac{1}{2}(x^{2}+y^{2}+1)(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1).\end{split}$

By Theorem 3.1, $\chi$ is a vector field on $\mathbb{T}^{2}$. Observe that
$\chi$ has exactly four invariant meridians generated by the planes $x=0$ and
$y=0$. Hence, by Theorem 3.8, the invariant meridians are stable or unstable
limit cycles alternately. Notice that in cylindrical coordinate system
$(r,\theta,z)$, the angular velocity $\dot{\theta}=-\frac{r^{2}}{2}\sin
2\theta$.

Figure 1. Left and right meridians are stable limit cycles, whereas top and
bottom meridians are unstable limit cycles.

It is proved in [15, Theorem 1(a)] that a degree $m>1$ vector field on
$\mathbb{T}^{2}$ has maximum $m-2$ parallel planes. However, Benny and Sarkar
[1, Proposition 6.2] have pointed out that there exists vector field of degree
$m$ on $S^{p}\times S^{1}$ having exactly $m-1$ invariant parallel planes and
proved that it is the sharp bound.

Hence, a cubic vector field on $\mathbb{T}^{2}$ has at most two invariant
parallel planes. We characterize cubic vector fields on $\mathbb{T}^{2}$ which
have exactly two invariant parallel planes.

######  Theorem 3.10.

If a cubic vector field $\chi=(P,Q,R)$ on $\mathbb{T}^{2}$ has exactly two
invariant parallel planes, then it will be of the following form.

(3.9) $\begin{split}P=&\frac{1}{2}xz(px+qy)+fy-\frac{a^{2}p}{2}z,\\\
Q=&\frac{1}{2}yz(px+qy)-fx-\frac{a^{2}q}{2}z,\text{ and }\\\
R=&(px+qy)(z^{2}-1)\end{split}$

for some quadratic polynomial $f\in\mathbb{R}[x,y,z]$ and $p,q\in\mathbb{R}$.

Moreover, $z-1=0$ and $z+1=0$ are the invariant parallel planes of this vector
field. Also, the parallels on $z-1=0$ are periodic orbits if and only if
$f(x,y,1)-\frac{1}{2}(py-qx)=0$ has no solution on the circle
$x^{2}+y^{2}=a^{2}$ and the parallels on $z+1=0$ are periodic orbits if and
only if $f(x,y,-1)+\frac{1}{2}(py-qx)=0$ has no solution on the circle
$x^{2}+y^{2}=a^{2}$.

###### Proof.

Suppose $\chi$ is given by (3.1). Then the extactic polynomial of $\chi$
associated with $\langle 1,z\rangle$ is

$\mathcal{E}_{\langle 1,z\rangle}(\chi)=\begin{vmatrix}1&z\\\
0&R\end{vmatrix}=\frac{1}{2}K^{\prime}(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)-2(\beta
x+\gamma y)(x^{2}+y^{2}-a^{2}).$

Suppose $z-k_{1}=0$ and $z-k_{2}=0$ are two invariant parallel planes. Then,
$\mathcal{E}_{\langle 1,z\rangle}(\chi)$ is divisible by $(z-k_{1})(z-k_{2})$.
Assume that $\frac{1}{2}K^{\prime}=px+qy+rz+s$. So,

(3.10) $(px+qy+rz+s)(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)-2(\beta x+\gamma
y)(x^{2}+y^{2}-a^{2})=(z-k_{1})(z-k_{2})L$

for some linear polynomial $L\in\mathbb{R}[x,y,z]$. We compare the
coefficients of $x^{3},y^{3}$ in (3.10) and get $-a^{2}p-2\beta=0$ and
$-a^{2}q-2\gamma=0$, respectively. This gives $\beta=-\frac{a^{2}p}{2}$ and
$\gamma=-\frac{a^{2}q}{2}$. Also, comparing coefficients of $x^{2}z,x^{2}$ in
(3.10), we get $r=s=0$. Hence, (3.10) becomes

(3.11)
$\begin{split}&(px+qy)(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)+a^{2}(px+qy)(x^{2}+y^{2}-a^{2})=(z-k_{1})(z-k_{2})L\\\
\implies&(px+qy)(z^{2}-1)=((z-k_{1})(z-k_{2})L\end{split}$

Hence, without loss of generality, $k_{1}=-1,k_{2}=1$. So,
$K=K^{\prime}z=2(px+qy)z$. Hence, we get $P,Q,R$ as mentioned in (3.10).

One can check that $z-1=0$ and $z+1=0$ are the only invariant parallel planes
of $\chi$.

Now, we prove the necessary and sufficient condition for the parallels on
$z-1=0$ to be periodic orbits. The proof for the parallels on $z+1=0$ is
similar. In cylindrical coordinates the vector field $\chi$ becomes
$(\dot{r},\dot{\theta},\dot{z})$ where

$\begin{split}\dot{r}&=\frac{r^{2}-a^{2}}{2}(p\cos\theta+q\sin\theta)z,\\\
\dot{\theta}&=-f+\frac{a^{2}}{2r^{2}}(pr\sin\theta-qr\cos\theta)z,\\\
\dot{z}&=r(p\cos\theta+q\sin\theta)(z^{2}-1).\end{split}$

Since $z-1=0$ is invariant then the parallels generated by this plane are
periodic orbits if and only if the plane does not contain any singular point.
Suppose $(r_{0},\theta_{0},z_{0})\in\mathbb{T}^{2}$ is a singular point such
that the point lies on $z-1=0$. Hence, $z_{0}=1$ and $r_{0}=a$. So, at
$(r_{0},\theta_{0},z_{0})$, $\dot{r}=\dot{z}=0$. Again, at
$(r_{0},\theta_{0},z_{0})$, $\dot{\theta}=0$ if and only if
$f(r_{0}\cos\theta_{0},r_{0}\sin\theta_{0},1)=\frac{1}{2}(pr_{0}\sin\theta_{0}-qr_{0}\cos\theta_{0})$.
Hence, the result follows. ∎

######  Theorem 3.11.

Suppose $\chi=(P,Q,R)$ is a cubic vector field on $\mathbb{T}^{2}$. Then
$\chi$ has infinitely many invariant parallels if and only if
$P=fx,~{}Q=-fy,~{}R=0$ for some quadratic polynomial $f\in\mathbb{R}[x,y,z]$.

###### Proof.

Suppose $\chi$ is given by (3.1) and it has infinitely many invariant
parallels. So, we obtain

$\mathcal{E}_{\langle
1,z\rangle}(\chi)=R=\frac{1}{2}K^{\prime}(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)-2(\beta
x+\gamma y)(x^{2}+y^{2}-a^{2})=0.$

Then,

$\frac{1}{2}K^{\prime}(z^{2}-1)=(x^{2}+y^{2}-a^{2})(\frac{a^{2}}{2}K^{\prime}+2(\beta
x+\gamma y)).$

So, $x^{2}+y^{2}-a^{2}$ divides $K^{\prime}$ and this is possible only when
$K^{\prime}=0$ since $K^{\prime}$ is a linear polynomial. Consequently, $K=0$
and $\beta=\gamma=0$. So, we get $P=fy,Q=-fx$.

One can check that the converse part is also true since $R=0$. ∎

##  4\. Pseudo Type-$n$ vector fields and Lie bracket of quadratic vector
fields

In this section, we classify and study Pseudo Type-$n$ vector fields and cubic
vector fields generated by Lie bracket of two quadratic vector fields on
$\mathbb{T}^{2}$. We study invariant meridians and parallels of these vector
fields.

######  Proposition 4.1.

Suppose $\chi=(P,Q,R)$ is a vector field on $\mathbb{T}^{2}$ such that $P,Q,R$
are homogeneous polynomials with $\deg P=\deg Q=n$ and $\deg R\leq n$. Then
$R=0$.

###### Proof.

Since $\chi$ is a vector field on $\mathbb{T}^{2}$, the following must be
satisfied.

(4.1) $4(x^{2}+y^{2}-a^{2})(Px+Qy)+2Rz=K((x^{2}+y^{2}-a^{2})^{2}+z^{2}-1)$

for some polynomial $K$ with $\deg K\leq n-1$.

Suppose that $Px+Qy\neq 0$. Then $\deg K=n-1$. Observe that degree of each
monomial on the left side of (4.1) is either $n+1,~{}n+3$ or $\deg R+1$. Also,
notice that $\deg K(a^{4}-1)=n-1$. Hence, $\deg R=n-2$ and $K$ is a
homogeneous polynomial. Note that $2Rz=K(a^{4}-1)$. So, rewriting (4.1), we
get

$4(x^{2}+y^{2}-a^{2})(Px+Qy)+K(a^{4}-1)=K((x^{2}+y^{2}-a^{2})^{2}+z^{2}-1).$

Hence,

$4(x^{2}+y^{2}-a^{2})(Px+Qy)=K((x^{2}+y^{2}-a^{2})^{2}+z^{2}-a^{4}).$

So, $x^{2}+y^{2}-a^{2}$ must divide $K(z^{2}-a^{4})$, which is not possible.
So, $Px+Qy$ must be zero. Then (4.1) implies that

$2Rz=K((x^{2}+y^{2}-a^{2})^{2}+z^{2}-1).$

The above equation is valid only when $R=0$ because $R$ is a homogeneous
polynomial. Hence, the result is proved. ∎

We recall the definition of Pseudo Type-$n$ vector fields from [9].

######  Definition 4.2.

Suppose $\chi=(P,Q,R)$ is a polynomial vector field in $\mathbb{R}^{3}$. We
say $\chi$ is Pseudo Type-$n$ vector field if $P,Q$ are homogeneous
polynomials of degree $n$ but $R=0$.

######  Theorem 4.3.

Suppose $\chi=(P,Q,0)$ is a Pseudo Type-$n$ vector field in $\mathbb{R}^{3}$.
Then $\chi$ is a vector field on $\mathbb{T}^{2}$ if and only if there exists
homogeneous $A\in\mathbb{R}[x,y,z]$ of degree $n-1$ such that $P=Ay,~{}Q=-Ax$.
Also, any Pseudo Type-$n$ vector field on $\mathbb{T}^{2}$ is completely
integrable.

###### Proof.

Proof of the first part is similar to the proof of [9, Theorem 4.11].

To prove the second part, notice that $f:=x^{2}+y^{2}$ and $g:=z$ are two
functionally independent first integrals for $\chi$. ∎

We denote $s(\chi)$ as the set of all singular points of the vector field
$\chi$. Note that for a Pseudo Type-$n$ vector field on $\mathbb{T}^{2}$, any
parallel is invariant. Next, we characterize invariant meridians of a Pseudo
Type-$n$ vector field on $\mathbb{T}^{2}$.

######  Proposition 4.4.

Any invariant meridian of a Pseudo Type-$n$ vector field $\chi$ on
$\mathbb{T}^{2}$ is a subset of $s(\chi)$.

###### Proof.

Suppose $\chi=(P,Q,0)$ is a Pseudo Type-$n$ vector field on $\mathbb{T}^{2}$.
By Theorem 4.3, $P=Ay$, $Q=-Ax$ for some degree $n-1$ homogeneous polynomial
$A$. The extactic polynomial associated with $\langle x,y\rangle$ is
$\mathcal{E}_{\langle x,y\rangle}(\chi)=\begin{vmatrix}x&y\\\
Ay&-Ax\end{vmatrix}=-A(x^{2}+y^{2})$. Hence, if $ax+by=0$ is an invariant
meridian, then by Proposition 2.1, $ax+by$ must divide $A$. So, any point on
the invariant meridian is a singular point of the vector field $\chi$. ∎

######  Remark 4.5.

No invariant meridian of a Pseudo Type-$n$ vector field on $\mathbb{T}^{2}$ is
periodic orbit.

######  Proposition 4.6.

If $f=0$ is an invariant algebraic hypersurface of two vector fields $X$ and
$Y$ in $\mathbb{R}^{n}$ then $f=0$ is an invariant algebraic hypersurface of
Lie bracket $[X,Y]$ also.

###### Proof.

The hypersurface $f=0$ is invariant under the vector fields $X$ and $Y$. So,
there exist $K_{1},K_{2}\in\mathbb{R}[x_{1},...,x_{n}]$ such that $Xf=K_{1}f$
and $Yf=K_{2}f$. Now,

$f=(XY-
YX)(f)=X(Yf)-Y(Xf)=X(K_{2}f)-Y(K_{1}f)=X(K_{2})f+K_{2}X(f)-Y(K_{1})f-K_{1}Y(f)=(X(K_{2})-Y(K_{1}))f$

Hence, $f=0$ is invariant under $[X,Y]$. ∎

######  Remark 4.7.

If two vector fields possess a common first integral $f$ then their Lie
bracket also has first integral $f$.

Observe that we can get the general form of a quadratic vector field on
$\mathbb{T}^{2}$ by vanishing the cubic terms in (3.1), i.e., for a quadratic
vector field, $\beta=\gamma=0$ and $K,f$ are linear polynomials. Hence, if
$(P,Q,R)$ is a quadratic vector field on $\mathbb{T}^{2}$ then $P,Q,R$ must
satisfy the following.

$\begin{split}P=&\frac{\alpha}{4}xz+fy,\\\ Q=&\frac{\alpha}{4}yz-fx\text{ and
}\\\ R=&\frac{\alpha}{2}(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)\end{split}$

where $\alpha\in\mathbb{R}$ and $f$ is a linear polynomial.

######  Theorem 4.8.

Lie bracket of any two quadratic vector fields on $\mathbb{T}^{2}$ is a cubic
vector field $\chi=(P,Q,0)$ with $P=Ay$ and $Q=-Ax$ for some quadratic
polynomial $A$. Moreover, $\chi$ is completely integrable.

###### Proof.

Suppose $X=(P_{0},Q_{0},R_{0})$ and $Y=(P_{1},Q_{1},R_{1})$ are two quadratic
vector fields on $\mathbb{T}^{2}$ such that

$P_{0}=\frac{\alpha_{0}}{4}xz+f_{0}y,~{}Q_{0}=\frac{\alpha_{0}}{4}yz-
f_{0}x,~{}R_{0}=\frac{\alpha_{0}}{2}(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)$

and

$P_{1}=\frac{\alpha_{1}}{4}xz+f_{1}y,~{}Q_{1}=\frac{\alpha_{1}}{4}yz-
f_{1}x,~{}R_{1}=\frac{\alpha_{1}}{2}(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)$

for some $\alpha_{0},\alpha_{1}\in\mathbb{R}$ and linear polynomials
$f_{0},f_{1}$. Then we compute the Lie bracket $[X,Y]$. If $[X,Y]=(P,Q,R)$
then one can find that $R=0$. By Proposition 4.6, $[X,Y]$ is a vector field on
$\mathbb{T}^{2}$. So, we have the following.

(4.2) $4(x^{2}+y^{2}-a^{2})(Px+Qy)=K((x^{2}+y^{2}-a^{2})^{2}+z^{2}-1)$

for some $K\in\mathbb{R}[x,y,z]$.

Hence, $x^{2}+y^{2}-a^{2}$ divides $K(z^{2}-1)$ and so divides $K$. Observe
that $[X,Y]$ is a cubic vector field and hence $\deg(K)\leq 2$. So,
$K=c(x^{2}+y^{2}-a^{2})$ for some $c\in\mathbb{R}$. Then (4.2) becomes

$4(Px+Qy)=c((x^{2}+y^{2}-a^{2})^{2}+z^{2}-1).$

Since left side has no term $z^{2}$, we obtain $c=0$. This gives $Px+Qy=0$.
Then by Lemma 2.3, $P=Ay,~{}Q=-Ax$ for some quadratic polynomial
$A\in\mathbb{R}[x,y,z]$.

The functions $f:=x^{2}+y^{2}$ and $g:=z$ are two functionally independent
first integrals of $[X,Y]$, which makes $[X,Y]$ completely integrable. ∎

######  Corollary 4.9.

Lie bracket of two Pseudo Type-2 vector fields on $\mathbb{T}^{2}$ is either
zero or a Pseudo Type-3 vector field.

###### Proof.

One can check that in Lie bracket of two Pseudo Type-2 vector field in
$\mathbb{R}^{3}$, $\dot{x}$ and $\dot{y}$ components are either zero or
homogeneous polynomials of degree 3. Hence, by Theorem 4.8, Lie bracket of two
Pseudo Type-2 vector fields on $\mathbb{T}^{2}$ is either zero or a Pseudo
Type-3 vector field. ∎

We can deduce the general form of a degree one vector field on
$\mathbb{T}^{2}$ by vanishing the cubic and quadratic terms in (3.1). We note
that any degree one vector field on $\mathbb{T}^{2}$ is a Pseudo Type-1 vector
field. Hence, Lie bracket of any two degree one vector fields on
$\mathbb{T}^{2}$ is zero. So, Lie bracket of two degree one vector fields and
Lie bracket of two degree two vector fields on $\mathbb{T}^{2}$ has the
$\dot{z}$-component always zero. But this is not true in general.

######  Example 4.10.

Consider following two vector fields in $\mathbb{R}^{3}$.

$X=(x^{2}z,xyz,2x(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1))~{}\mbox{and}~{}Y=(y^{3},-xy^{2},0).$

$X$ and $Y$ are vector fields on $\mathbb{T}^{2}$ due to Theorem 3.1 with the
parameter sets
{$K=4xz,f=\beta=\gamma=0$} and {$f=y^{2},K=\beta=\gamma=0$}, respectively.
Then the $\dot{z}$-component of $[X,Y]$ is
$-2y^{3}(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)$.

######  Proposition 4.11.

Lie bracket of two Pseudo Type-2 vector fields on $\mathbb{T}^{2}$ having
finitely many invariant meridians can have maximum two invariant meridians.

###### Proof.

Suppose $X=(Ay,-Ax,0)$ and $Y=(By,-Bx,0)$ are two Pseudo Type-2 vector fields
on $\mathbb{T}^{2}$ with $A=a_{1}x+a_{2}y+a_{3}z$ and
$B=b_{1}x+b_{2}y+b_{3}z$. Then $[X,Y]=(Cy,-Cx,0)$ where
$C=(a_{2}b_{1}-a_{1}b_{2})(x^{2}+y^{2})-(a_{3}b_{2}-a_{2}b_{3})xz+(a_{3}b_{1}-a_{1}b_{3})yz$.
Hence, the extactic polynomial $\mathcal{E}_{\langle
x,y\rangle}([X,Y])=-C(x^{2}+y^{2})$. One can check that if $C\neq 0$ then $C$
can have maximum one factor of the form $ax+by$. Hence, the result follows. ∎

######  Theorem 4.12.

Suppose $\chi=(P,Q,R)$ is a quadratic vector field on $\mathbb{T}^{2}$ such
that $R$ is not identically zero. Then any singular point (if exists) of
$\chi$ on $\mathbb{T}^{2}$ lies on either of the parallels
$\\{z-1=0\\}\cap\mathbb{T}^{2}$ and $\\{z+1=0\\}\cap\mathbb{T}^{2}$.

###### Proof.

Consider a quadratic vector field $\chi=(P,Q,R)$ on $\mathbb{T}^{2}$ with

$P=\frac{\alpha}{4}xz+fy,~{}Q=\frac{\alpha}{4}yz-
fx,~{}R=\frac{\alpha}{2}(-a^{2}(x^{2}+y^{2})+z^{2}+a^{4}-1)$

for some linear polynomial $f$ and $\alpha\in\mathbb{R}$. If $\alpha=0$ then
$R$ is identically zero. So, we assume that $\alpha\neq 0$. Let
$(x_{0},y_{0},z_{0})\in\mathbb{T}^{2}$. Then
$(x_{0}^{2}+y_{0}^{2}-a^{2})^{2}+z_{0}^{2}=1$. If $(x_{0},y_{0},z_{0})$ is a
singular point of $\chi$ then $R(x_{0},y_{0},z_{0})=0$. Hence,

$\begin{split}&-a^{2}(x_{0}^{2}+y_{0}^{2})+z_{0}^{2}+a^{4}-1=0\\\
\implies&(-a^{2}(x_{0}^{2}+y_{0}^{2}-a^{2}))^{2}=(1-z_{0}^{2})^{2}\\\
\implies&a^{4}(1-z_{0}^{2})=(1-z_{0}^{2})^{2}\end{split}$

If $1-z_{0}^{2}\neq 0$ then from the last implication, we obtain
$z_{0}^{2}=1-a^{4}$. But this is not possible since $a>1$. So, $z_{0}=\pm 1$.
Hence, the result follows. ∎

Acknowledgments. The author is grateful to Prof. Soumen Sarkar for
encouragement, many fruitful discussions and providing several insightful
comments after carefully going through the initial draft of this paper. Author
thanks Joji Benny for helpful discussions. The author is supported by the
Prime Minister’s Research Fellowship, Government of India.

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