Proof. By (2.11), it is clear that critical points on $\mathcal {E}$ must lie on $\{H^2=1\}$. The list is complete by considering the vanishing of the $Y$-entry and $Z$-entry.

Proof. From the linearized solution, it is clear that each $\zeta _{s_2}$ with $s_2<0$ is initially in

(3.8)\begin{equation} \mathcal{W}:=\mathcal{E}\cap \{Z-Y\geq 0\}\cap \{X_1-X_2\geq 0\}. \end{equation}

As $\mathcal {W}$ includes all points $(\sqrt { ({n-1})/{12m}}\cosh (\lambda ),0,0,\sqrt { ({n-1})/{12nm}}\sinh (\lambda ))$ with $\lambda \geq 0$, the set is non-compact. Furthermore, since

(3.9)\begin{equation} \left\langle\nabla \left(\frac{Y}{Z}\right),V\right\rangle=-2\frac{Y}{Z}(X_1-X_2)\leq 0 \end{equation}

and

(3.10)\begin{equation} \langle\nabla(X_1-X_2),V\rangle=(X_1-X_2)H\left(G+\frac{1}{n}(1-H^2)-1\right)+2(Z-Y)((2m+3)Z-Y), \end{equation}

the set $\mathcal {W}$ is invariant. Note that the second term in (3.10) is non-negative in $\mathcal {W}\cap \{X_1-X_2=0\}$.

By (3.9), it is clear that the function $ {Y}/{Z}$ monotonically decreases from $1$ along each $\zeta _{s_2}$ with $s_2<0$. If the function $ {Y}/{Z}$ converges to some positive number, the function $X_1-X_2$ would converge to zero. From (2.9), we know that both $Y$ and $Z$ converge to some positive numbers with $ {Y}/{Z}<1$. Hence, the right-hand side of (3.10) is eventually positive as $X_1-X_2$ converges to zero, a contradiction. Hence, the function $ {Y}/{Z}$ converges to zero. Therefore, if a $\zeta _{s_2}$ with $s_2<0$ converges to a critical point in $\mathcal {E}$, it must be $q_2^-$ ($q_2^-$ or $q_3^-$ if $m=1$).

Proof. As $X_1-X_2=Y-Z$ and $X_2=-Z$ on $\mathcal {B}_{QK}$, we can eliminate $X_1$ and $X_2$ in (2.9). Hence,

(3.14)\begin{equation} \frac{12m}{n}(Y-Z)^2+6Y^2+4m(4m+8)YZ-12mZ^2=1-\frac{1}{n} \end{equation}

holds on $\mathcal {B}_{QK}$. Therefore,

(3.15)\begin{align} & \langle\nabla(X_2+Z),V\rangle|_{\mathcal{B}_{QK}} \nonumber\\ &\quad =(X_2+Z)H\left(G+\frac{1}{n}(1-H^2)\right)-\frac{1}{n}(1-H^2)-X_2H+(4m+8)YZ-6Z^2+Z(X_1-2X_2) \nonumber\\ &\quad =-\frac{1}{n}(1-H^2)-X_2H+(4m+8)YZ-6Z^2+Z(X_1-2X_2) \nonumber\\ &\quad =\frac{1}{n-1}\left(\frac{12m}{n}(Z-Y)^2+6Y^2+4m(4m+8)YZ-12mZ^2-\frac{n-1}{n}\right) \nonumber\\ &\qquad \text {eliminating} X_1 {\rm and} X_2 {\rm by the definition of} {\mathcal{B}_{QK}} \nonumber\\ &\quad =0 \quad \text{by (3.14)}. \end{align}

On the other hand, we have

(3.16)\begin{align} & \langle\nabla(X_1-X_2+Z-Y),V\rangle|_{\mathcal{B}_{QK}} \nonumber\\ &\quad =(X_1-X_2+Z-Y)\left(H\left(G+\frac{1}{n}(1-H^2)-1\right)+4Z-2Y\right)+(n-1)(X_2+Z)(Z-Y) \nonumber\\ &\quad =0. \end{align}

Therefore, the set $\mathcal {B}_{QK}$ is indeed invariant.

Since $X_1=Y-2Z$ and $X_2=-Z$ on $\mathcal {B}_{QK}$, one can realize $\mathcal {B}_{QK}$ as a hyperbola (3.14). Note that $p_0^+$ and $p_1^-$ are the only critical points in $\mathcal {B}_{QK}$ and they are in the same connected component in (3.14). Therefore, there is an integral curve that joins $p_0^+$ and $p_1^-$ and lies on $\mathcal {B}_{QK}$. Hence, the integral curve must be some $\gamma _{s_1}$. Let $v(\eta )$ be the normalized velocity of the linearized solution that uniquely corresponds to $\gamma _0$. It is clear that $\lim _{\eta \to -\infty }v(\eta )= {v_1}/{\|v_1\|}$ is tangent to $\mathcal {B}_{QK}$ at $p_0^+$. Hence, we know that $\gamma _0$ lies on $\mathcal {B}_{QK}$. By the $\mathbb {Z}_2$-symmetry on the sign of $(X_1,X_2)$, we know that $\zeta _{ {1}/({2m+6})}$ lies on the invariant set

\[ \bar{\mathcal{B}}_{QK}:=\mathcal{C}_{\Lambda\geq 0}\cap \{X_2-X_1+Z-Y=0\}\cap\{X_2-Z=0\} \]

and joins $p_1^+$ and $p_0^-$.

Proof. Since $A\geq 0$ in $\mathcal {S}$, a point in $\mathcal {S}$ with vanishing $X_2$-coordinate must have $ZX_1=0$. If we further assume $Z=0$ at that point, then from $P\geq 0$ we have $X_1(1-H^2)=X_1(1-9X_1^2)\leq 0$. Hence, we obtain $X_1=\frac {1}{3}$ and that point must be $p_0^+$. We obtain $\Gamma$ if $X_1=0$ is further assumed.

It is obvious that $\Gamma =\mathcal {E}\cap \{X_1=X_2=0\}$ for $m=1$ and the set is non-compact. We prove that $\Gamma$ is bounded for $m\geq 2$. From $B\geq 0$ and (2.9), we have

(4.2)\begin{align} & 6Y^2+4m(4m+8)YZ-12mZ^2=\frac{n-1}{n}\geq (n-1)\frac{2n^2(2m+3)(m-1)}{m(2m+1)(8m+3)}YZ \nonumber\\ &\quad \Leftrightarrow 6Y^2+\left(4m(4m+8)-(n-1)\frac{2n^2(2m+3)(m-1)}{m(2m+1)(8m+3)}\right)YZ-12mZ^2\geq 0. \end{align}

For $m\geq 2$, the inequality above implies $mY-Z\geq 0$. Then from (2.9) we have $ ({n-1})/{n}\geq 6Y^2+(4m+32)Z^2$. The first claim is clear.

Suppose there is a point in $\mathcal {S}$ with vanishing $Y$-coordinate. From $A\geq 0$ we know that $- ({3}/{m})Z(X_1+ ({2m}/{3})X_2)=0$ at that point. If $Z\neq 0$, then $X_1=X_2=Y=0$ at that point, which is impossible from (2.9). If $X_1+ ({2m}/{3})X_2\neq 0$, then $Y=Z=0$ at that point. Then from (2.9), we have $X_1-X_2\neq 0$. From $P\geq 0$ we know that $- ({1}/{n})(1-H^2)-2X_2(X_1+ ({2m}/{3})X_2)\geq 0$ at that point. The point has to be $(\frac {1}{3},0,0,0)$, which does not lie on $\mathcal {E}$. The above discussion proves the second claim.

Since $P\geq 0$ on a point with vanishing $Z$-coordinate in $\mathcal {S}$, we have

\[ (X_2-X_1)\frac{1}{n}(1-H^2)-2X_2Y^2-2X_2\left(X_1+\frac{2m}{3}X_2\right)(X_1-X_2)\geq 0. \]

By the definition of $\mathcal {S}$, each term in the above inequality is non-positive. Since $Y>0$ from the second claim, the variable $X_2$ must vanish and the third claim is clear.

Finally, from $A\geq 0$ and $X_1-X_2\geq 0$ in $\mathcal {S}$, we know that $X_2(Y- (({2m+3})/{m})Z)\geq 0$ in $\mathcal {S}$. If $X_2\neq 0$, then $mY\geq (2m+3)Z$ and the boundedness of all variables is obtained from (2.9). If $X_2=0$, then the boundedness comes from the first claim. Hence, $\mathcal {S}$ is a compact set.

Proof. Consider $\mathcal {S}$ with $m=1$. If $X_2=0$ is imposed, we obtain either the point $p_0^+$ or $\Gamma$ from Proposition 4.1(a). Hence, we assume $X_2>0$ in the following. From $A\geq 0$ it is clear that $X_2(Y-5Z)\geq YX_2-3Z(X_1+\frac {2}{3}X_2)\geq 0$ and hence $Y-5Z\geq 0$.

One can easily verify that

(4.3)\begin{equation} P=(X_1-X_2)\big(6YZ-2X_2\big(X_1+\tfrac{2}{3}X_2\big)\big)-\tfrac{1}{7}(X_1-X_2)(1-H^2)-2(Y-Z)A. \end{equation}

If $X_1=X_2=X>0$, the first two terms in (4.3) vanish while the last term is non-positive. Then we must have $A=0$, from which we deduce $Y-5Z= 0$. By (3.18) we obtain the line segment $\Phi \cap \{X_1,X_2> 0\}$.

For $X_1\neq X_2$, we rewrite (4.3) as follows:

(4.4)\begin{equation} P=(X_1-X_2)\big(6YZ-2X_2\big(X_1+\tfrac{2}{3}X_2\big)- \tfrac{10}{147}(1-H^2)\big)-\tfrac{11}{147}(X_1-X_2)(1-H^2)-2(Y-Z)A. \end{equation}

The last two terms in (4.4) are non-positive. Hence, from $P\geq 0$ we have

(4.5)\begin{align} 0&\leq 6YZ-2X_2\big(X_1+\tfrac{2}{3}X_2\big)-\tfrac{10}{147}(1-H^2) \nonumber\\ &= 6YZ-2X_2\big(X_1+\tfrac{2}{3}X_2\big)-\tfrac{10}{147} \big(\tfrac{7}{6}\big(6Y^2+48YZ-12Z^2+\tfrac{12}{7}(X_1-X_2)^2\big)-H^2\big) \nonumber\\ &\quad \text{by (2.9)} \nonumber\\ &= \tfrac{1}{21}(5X_1+4X_2+5Y+2Z)(X_1-X_2-Y+5Z) \nonumber\\ &\quad +\tfrac{1}{21}(5X_1+4X_2-5Y-2Z)(X_1-X_2+Y-5Z). \end{align}

Suppose $X_1-X_2-Y+5Z\leq 0$; then the first term in the last line of (4.5) is non-positive. As the summation above is non-negative, we know that the second term in the last line of (4.5) must be non-negative. As $X_1-X_2+Y-5Z\geq 0$, we must have $\frac {1}{5}(5X_1+4X_2-2Z)\geq Y$. From $A\geq 0$ we have

\[ \frac{X_2}{5}(5X_1+4X_2-2Z)\geq 3Z\left(X_1+\frac{2}{3}X_2\right) \Leftrightarrow (5X_1+4X_2)(X_2-3Z)\geq 0 \Leftrightarrow X_2-3Z\geq 0. \]

Then we claim that the first term in (4.3) is non-positive since

(4.6)\begin{align} 6YZ-2X_2\big(X_1+\tfrac{2}{3}X_2\big)&\leq \tfrac{6}{5}(5X_1+4X_2-2Z)Z-2X_2\big(X_1+\tfrac{2}{3}X_2\big) \nonumber\\ &\leq \tfrac{2}{5}\big(5X_1+4X_2-\tfrac{2}{3}X_2\big)X_2-2X_2\big(X_1+\tfrac{2}{3}X_2\big) \nonumber\\ &=0. \end{align}

But $P\geq 0$. Hence, assumptions $X_1\neq X_2$ and $X_1-X_2-Y+5Z\leq 0$ lead to the vanishing of $A$ and $P$.

Suppose $X_1-X_2-Y+5Z\geq 0$. Then $A\geq 0$ implies

\[ (X_1-X_2+5Z)X_2\geq 3Z\big(X_1+\tfrac{2}{3}X_2\big) \Leftrightarrow X_2\geq 3Z. \]

Then we claim that the first term in (4.3) is also non-positive since

(4.7)\begin{align} 6YZ-2X_2\big(X_1+\tfrac{2}{3}X_2\big)&\leq 6(X_1-X_2+5Z)Z-2X_2\big(X_1+\tfrac{2}{3}X_2\big) \nonumber\\ &\leq 2\big(X_1-X_2+\tfrac{5}{3}X_2\big)X_2-2X_2\big(X_1+\tfrac{2}{3}X_2\big) \nonumber\\ &=0. \end{align}

Hence, the assumption $X_1\neq X_2$ and $X_1-X_2-Y+5Z\geq 0$ also leads to the vanishing of $A$ and $P$.

Therefore, points in $\mathcal {S}$ with $X_1\neq X_2$ must have vanished $A$ and $P$, which leads to the equalities

\[ H=1,\quad 3Z=X_2,\quad 3Y=3X_1+2X_2. \]

Note that the last two equations above are equivalent to the defining equations in (3.4) for $m=1$. We obtain $\gamma _\infty$ and critical points $p_0^+$ and $p_2^+$.

Proof. We consider $\mathcal {S}\cap \{P=0\}$ as a union of slices

\[ \mathcal{S}\cap\{P= 0\}=\bigcup_{\kappa\in[0,1]}\mathcal{L}_\kappa,\quad \mathcal{L}_\kappa:=\mathcal{S}\cap\{P= 0\}\cap\{X_2-\kappa X_1=0\}. \]

Note that each $\mathcal {L}_\kappa$ contains $\Gamma$. As $X_1\geq X_2\geq 0$ in $\mathcal {S}$, the function $Q$ vanishes once $X_1$ does. We assume $X_1>0$ in the following discussion.

For $\mathcal {L}_0$, we have

\[ Q|_{\mathcal{L}_0}=-2(4m+8)X_1YZ+2X_1\frac{1-H^2}{n}. \]

From $A\geq 0$ we have $Z=0$. Then $Q|_{\mathcal {L}_0}=2X_1 (({1-H^2})/{n})$, and $P=0$ becomes $-X_1 (({1-H^2})/{n})=0$. Hence, $\mathcal {L}_0=\Gamma \cup \{p_0^+\}$ and $Q|_{\mathcal {L}_0}= 0$.

If $\kappa =1$, let $X:=X_1=X_2>0$. Then $A\geq 0$ and $P= 0$ respectively become

\[ X\left(Y-\frac{2m+3}{m}Z\right)\geq 0,\quad X(Y-(2m+3)Z)(Y-Z)= 0. \]

For $X>0$ we must have $Y= (2m+3)Z$. Then from (2.9) we find that $Y=(2m+3)Z= (2m+3)z_0$. Hence, $\mathcal {L}_1$ is a union of $\Gamma$ and a part of the invariant set (3.18). More specifically, from $B\geq 0$ we have

(4.8)\begin{equation} \frac{1}{n}-nX^2-\frac{2n^2(2m+3)(m-1)}{m(2m+1)(8m+3)}YZ= \frac{1}{n}-nX^2-\frac{2n^2(2m+3)^2(m-1)}{m(2m+1)(8m+3)}z_0^2\geq 0, \end{equation}

and it follows that

(4.9)\begin{equation} \mathcal{L}_1=\Gamma \cup \left(\Phi \cap \left\{0< X\leq \frac{1}{n}\sqrt{\frac{36m^3+90m^2+117m+54}{m(8m+3)(4m^2+14m+9)}}\leq \frac{1}{n}\right\}\right) \end{equation}

for $m\geq 2$. For $m=1$, we have

(4.10)\begin{equation} \mathcal{L}_1=\Gamma \cup \left(\Phi \cap \left\{0< X\leq \frac{1}{n}\right\}\right)\cup \{p_2^+\}. \end{equation}

And $Q|_{\mathcal {L}_1}$ becomes

(4.11) \begin{align} & Q|_{\mathcal{L}_1} \nonumber\\ &\quad =X\left(\!-4Y^2-(4m+8)(4m+2)YZ+(n+1)\left(\frac{1}{n}-nX^2\right)+4\left(1+\frac{2m}{3}\right) \left(n+\frac{2m}{3}\right)X^2\right) \nonumber\\ &\quad = X\left(\!-4(2m+3)^2z_0^2-(4m+8)(4m+2)(2m+3)z_0^2+\frac{n+1}{n}-\frac{4}{9}m(8m+3)X^2\right) \nonumber\\ &\quad = \frac{4}{9}m(8m+3)X\left(\frac{1}{n^2}\frac{36m^3+90m^2+117m+54}{m(8m+3)(4m^2+14m+9)}-X^2\right) \nonumber\\ &\quad \geq 0. \end{align}

Hence, for $m\geq 1$, the function $Q|_{\mathcal {L}_1}\geq 0$ and it only vanishes on $\Gamma$ and boundary points of $\mathcal {L}_1\cap \Phi$. Note that for $m=1$, one of the boundary points of $\mathcal {L}_1\cap \Phi$ is $p_2^+$.

For each $\mathcal {L}_\kappa$ with $\kappa \in (0,1)$, replace $X_2$ with $X_1\kappa$. Then (2.9) and $P= 0$ respectively become

(4.12)\begin{equation} \begin{aligned} & \frac{n-1}{n}-\frac{12m}{n}(1-\kappa)^2X_1^2=6Y^2+4m(4m+8)YZ-12mZ^2,\\ & \frac{\kappa-1}{n}-\frac{\kappa-1}{3n}(32\kappa^2m^2-12\kappa^2m+ 48m\kappa-18\kappa+27)X_1^2\\ & \quad = 2\kappa Y^2-(4m+8)YZ+(4m\kappa+6)Z^2. \end{aligned} \end{equation}

With the equations above, we can write the constant $1$ as a homogeneous polynomial $One(Y,Z)$ of degree $2$. Multiplying the constant term in $B$ by $One(Y, Z)$, the function $B$ restricted to each $\mathcal {L}_\kappa \cap \{P=0\}$ then becomes a homogeneous polynomial in $Y$ and $Z$. Factor out $X_1$ in $Q$ and multiply the constant term in $ {Q}/{X_1}$ by $One(Y,Z)$. We see that $Q$ restricted on each $\mathcal {L}_\kappa \cap \{P=0\}$ is a homogeneous polynomial in $X_1$, $Y$ and $Z$. In summary, we have

(4.13)\begin{equation} \begin{aligned} A|_{\mathcal{L}_\kappa\cap \{P=0\}} & =X_1\left(Y\kappa-\frac{3}{m}Z\left(1+\frac{2m}{3}\kappa\right)\right),\\ B|_{\mathcal{L}_\kappa\cap \{P=0\}} & =b_2Z^2+b_1YZ+b_0Y^2,\\ Q|_{\mathcal{L}_\kappa\cap \{P=0\}} & =X_1(q_2 Z^2+q_1YZ+q_0 Y^2). \end{aligned} \end{equation}

The coefficients $b_i$ and $q_i$ in (4.13) are rational functions in $m$ and $\kappa$. Explicit formulas for each coefficient are presented in the Appendix for the sake of simplicity. As $Y$ is positive in $\mathcal {S}$ from Proposition 4.1, we factor out $Y$ and consider $A|_{\mathcal {L}_\kappa \cap \{P=0\}}=X_1Y\tilde {A}_{m,\kappa }( {Z}/{Y})$, $B|_{\mathcal {L}_\kappa \cap \{P=0\}}=Y^2\tilde {B}_{m,\kappa }( {Z}/{Y})$ and $Q|_{\mathcal {L}_\kappa \cap \{P=0\}}=X_1Y^2\tilde {Q}_{m,\kappa }( {Z}/{Y})$, where

(4.14)\begin{equation} \tilde{A}_{m,\kappa}(x)=\kappa-\left(\frac{3}{m}+2\kappa\right)x,\quad \tilde{B}_{m,\kappa}(x)=b_2 x^2+b_1 x+b_0,\quad \tilde{Q}_{m,\kappa}(x)=q_2 x^2+q_1 x+q_0. \end{equation}

We have

(4.15)\begin{equation} q_2=-\frac{4(2\kappa m+3)(32\kappa^3m^3+\kappa^2m^2(96-68\kappa)+ \kappa m(90-84\kappa)+27(1-\kappa))}{3(1-\kappa)(2\kappa^2m(8m-5)+6\kappa(4m-1)+9)}\leq 0 \end{equation}

for any $(m,\kappa )\in [1,\infty )\times (0,1)$. Therefore, the restricted function $Q|_{\mathcal {L}_\kappa \cap \{P=0\}}$ is non-negative if it is so on the boundary of $\mathcal {L}_\kappa \cap \{P=0\}$. The upper bound and the lower bound of $ {Z}/{Y}$ on each slice are respectively provided by $A|_{\mathcal {L}_\kappa \cap \{P=0\}}\geq 0$ and $B|_{\mathcal {L}_\kappa \cap \{P=0\}}\geq 0$. Specifically, from $A|_{\mathcal {L}_\kappa \cap \{P=0\}}\geq 0$ we have $ {Z}/{Y}\leq {m\kappa }/({3+2m\kappa })$. As shown in (A.1), we have $b_2<0$. By Proposition A.1 in the Appendix, the smaller real root $\sigma (m,\kappa )$ of $\tilde {B}_{m,\kappa }$ is in the interval $(0, {m\kappa }/({3+2m\kappa }))$. Hence, from $B|_{\mathcal {L}_\kappa \cap \{P=0\}}\geq 0$ we have $ {Z}/{Y}\geq \sigma (m,\kappa )$. By the arbitrariness of $\kappa$, it is clear that the minimizing point of $Q$ on ${\mathcal {S}\cap \{P=0\}}$ lies on $\mathcal {S}\cap \{P=0\}\cap \{A=0\}$ or $\mathcal {S}\cap \{P=0\}\cap \{B=0\}$.

We have

(4.16)\begin{align} Q|_{\mathcal{L}_\kappa\cap \{A=0\}}&=X_1Y^2\tilde{Q}_{m,\kappa} \left(\frac{m\kappa}{3+2m\kappa}\right) \nonumber\\ &=X_1Y^2\frac{4(m+3)(m-1)(4\kappa^3m^2(8m-1)+4\kappa^2 m(8m-3)+18 \kappa m+9(1-\kappa))\kappa^2}{3(2\kappa m+3)(1-\kappa)(2\kappa^2m(8m-5)+6\kappa(4m-1)+9)} \nonumber\\ &\geq 0. \end{align}

Therefore, for $m\geq 2$, the function $Q$ is positive on $\mathcal {L}_\kappa \cap \{A=0\}$ for any $\kappa \in (0,1)$. For $m=1$, the function $Q$ vanishes at $\mathcal {S}\cap \{P=0\}\cap \{A=0\}$.

Proving the non-negativity of $Q|_{\mathcal {L}_\kappa \cap \{B=0\}\cap \{P=0\}}$ is a bit more computationally involved. From Proposition 4.2, we know that for $m=1$, the polynomials $A$, $B$ and $P$ identically vanish at $\gamma _\infty$. Therefore, an explicit formula for the root $\sigma (1,\kappa )$ can be obtained from $A=0$. We have

\[ \sigma(1,\kappa)=\frac{Z}{Y}=\frac{X_2}{3\left(X_1+\frac{2}{3}X_2\right)}=\frac{\kappa}{2\kappa+3}. \]

Define the function $F(m,\kappa ):=\tilde {Q}_{m,\kappa }(\sigma (m,\kappa ))$. To show that $Q|_{\mathcal {L}_\kappa \cap \{P=0\}\cap \{B=0\}}\geq 0$, it suffices to show that $F(m,\kappa )\geq 0$ for any $(m,\kappa )\in [1,\infty )\times (0,1)$. Note that the vanishing of $F(m,\kappa )$ means $\sigma (m,\kappa )$ being also a root of $\tilde {Q}_{m,\kappa }$. From the computation (4.16) we have

\[ F(1,\kappa):=\tilde{Q}_{1,\kappa}(\sigma(1,\kappa))=\tilde{Q}_{1,\kappa} \left(\frac{\kappa}{3+2\kappa}\right)=0 \]

for any $\kappa \in (0,1)$. Furthermore, by implicit derivative, we have

\[ \left(\frac{\partial \sigma}{\partial m}\right)(1,\kappa)= \left(-\frac{\frac{\partial b_2}{\partial m}\sigma^2+ \frac{\partial b_1}{\partial m}\sigma+\frac{\partial b_0}{\partial m}}{2b_2\sigma+b_1}\right) (1,\kappa)=-\frac{(92\kappa^4+1202\kappa^3+1458\kappa^2-367\kappa-537)\kappa}{66(3+4\kappa) (\kappa+1)(3+2\kappa)^2}, \]

and it follows that

\begin{align*} \left(\frac{\partial F}{\partial m}\right)(1,\kappa) &= \left(\frac{\partial q_2}{\partial m}\sigma^2+ \frac{\partial q_1}{\partial m}\sigma+\frac{\partial q_0}{\partial m}+2q_2\sigma \frac{\partial \sigma}{\partial m}+q_1\frac{\partial \sigma}{\partial m}\right)(1,\kappa)\\ &=\frac{4(\kappa-1)(184\kappa^2-244\kappa-339)\kappa^2}{99(3+4\kappa)(\kappa+1)(2\kappa+3)}>0. \end{align*}

Hence, $F\geq 0$ on a neighborhood around $\{(1,\kappa )\mid \kappa \in (0,1)\}\subset [1,\infty )\times (0,1)$. In other words, for an $m$ that is slightly larger than $1$, the root $\sigma (m,\kappa )$ of $\tilde {B}_{m,\kappa }$ is strictly between the two real roots of $\tilde {Q}_{m,\kappa }$. Hence, proving $F\geq 0$ on $[1,\infty )\times (0,1)$ is equivalent to showing that $\sigma (m,\kappa )$ stays between the two real roots of $\tilde {Q}_{m,\kappa }$ for varying $(m,\kappa )$. This idea leads us to consider the resultant $r(\tilde {Q}_{m,\kappa },\tilde {B}_{m,\kappa })$ for the two polynomials. We have

(4.17)\begin{align} r(\tilde{Q}_{m,\kappa},\tilde{B}_{m,\kappa})&= -\frac{64\kappa^2(m-1)(2\kappa m+3)}{(8m+3)^2(2m+1)^2m^2(1-\kappa)(2\kappa^2m(8m-5)+ 6\kappa(4m-1)+9)^2}\tilde{r}, \nonumber\\ \tilde{r}&=262144 \kappa ^4 m^{10}+(516096 \kappa ^4+679936 \kappa ^3) m^9 \nonumber\\ &\quad +(373760\kappa ^4+1233920 \kappa ^3+675840 \kappa ^2) m^8 \nonumber\\ &\quad +(-275904 \kappa ^4+1151040 \kappa ^3+1143744 \kappa ^2+308160 \kappa ) m^7 \nonumber\\ &\quad +(-926496 \kappa ^4+248832 \kappa ^3+1432512 \kappa ^2+507456 \kappa +54432) m^6 \nonumber\\ &\quad +(-800496 \kappa ^4-1256256 \kappa ^3+1472688 \kappa ^2+857520 \kappa +92016) m^5 \nonumber\\ &\quad +(-281880 \kappa ^4-1525392 \kappa ^3+266328 \kappa ^2+1353024 \kappa +199584) m^4 \nonumber\\ &\quad +(-33048 \kappa ^4-644436 \kappa ^3-672624 \kappa ^2+957420 \kappa +375192) m^3 \nonumber\\ &\quad +(-90396 \kappa ^3-433026 \kappa ^2 +186624 \kappa +333882) m^2 \nonumber\\ &\quad +(-74358 \kappa ^2-59049 \kappa +133407) m+19683(1-\kappa). \end{align}

As verified in Proposition A.2 in the Appendix, the inequality $r(\tilde {Q}_{m,\kappa },\tilde {B}_{m,\kappa }) \leq 0$ is valid for any $(m,\kappa )\in [1,\infty )\times (0,1)$. Furthermore, the function $r(\tilde {Q}_{m,\kappa },\tilde {B}_{m,\kappa })$ vanishes if and only if $m=1$. In particular, both $\tilde {Q}_{1,\kappa }$ and $\tilde {B}_{1,\kappa }$ have $\sigma (1,\kappa )= {\kappa }/({2\kappa +3})$ as their roots. For $m>1$, the polynomials $\tilde {Q}_{m,\kappa }$ and $\tilde {B}_{m,\kappa }$ do not share any common root. Hence, $F\geq 0$ on $[1,\infty )\times (0,1)$ and the function $F$ vanishes if and only if $m=1$. Therefore, for $m\geq 2$, the inequality $Q|_{\mathcal {L}_\kappa \cap \{P=0\}\cap \{B=0\}}\geq 0$ is valid and the equality occurs only at $p_0^+$ and a point $\Phi \cap \{B=0\}$. The proof is complete.

Proof. The boundary $\partial \mathcal {S}$ is a union of the following five sets:

\[ \mathcal{S}\cap\{X_1-X_2=0\},\quad \mathcal{S}\cap\{X_2=0\},\quad \mathcal{S}\cap \{A=0\},\quad \mathcal{S}\cap \{B=0\},\quad \mathcal{S}\cap \{P=0\}. \]

By Proposition 4.1, if a $\gamma _{s_1}$ escapes $\mathcal {S}$ through some point with vanished $X_2$-coordinate, the point must lie in $\mathcal {S}\cap \{A=0\}\cap \{P=0\}$. Hence, we aim to show that $V$ points inward when restricted to the last three parts of $\partial \mathcal {S}$.

A straightforward computation shows that

(4.18) \begin{align} & \langle\nabla A,V\rangle|_{A=0} \nonumber\\ &\quad =A\left(2H\left(G+\frac{1}{n}(1-H^2)\right)-2X_1-(4m+2)X_2\right) \nonumber\\ &\qquad +\frac{A}{X_1+(2m/3)X_2}\left(R_1+\frac{2m}{3}R_2-\left(1+\frac{2m}{3}\right) \frac{1}{n}(1-H^2)\right) +\frac{Y}{X_1+(2m/3)X_2}P \nonumber\\ &\quad =\frac{Y}{X_1+(2m/3)X_2}P\quad \text{since $A=0$} \nonumber\\ &\quad \geq 0. \end{align}

It is confirmed that $V|_{A=0}$ points to the interior of $\mathcal {S}$.

Since

(4.19)\begin{align} & \langle\nabla B,V\rangle|_{B=0} \nonumber\\ &\quad =-2\frac{H}{n}(H^2-1)\left(G+\frac{1}{n}(1-H^2)\right) \nonumber\\ &\qquad -\frac{2n^2(2m+3)(m-1)}{m(16m^2+14m+3)}YZ \left(2H\left(G+\frac{1}{n}(1-H^2)\right)-2X_2\right) \nonumber\\ &\qquad \text{by (2.11)} \nonumber\\ &\quad =2BH\left(G+\frac{1}{n}(1-H^2)\right)+\frac{4n^2(2m+3)(m-1)}{m(16m^2+14m+3)}YZ X_2 \nonumber\\ &\quad =\frac{4n^2(2m+3)(m-1)}{m(16m^2+14m+3)}YZ X_2\quad \text{since $B=0$} \nonumber\\ &\quad \geq 0, \end{align}

it is clear that $V|_{B=0}$ points to the interior of $\mathcal {S}$.

Since

(4.20)\begin{align} & \langle\nabla P,V\rangle|_{P=0} \nonumber\\ &\quad =P\left(H\left(3G+\frac{3}{n} (1-H^2)-1\right)+\frac{4m}{3}X_2\right)+(X_1-X_2)Q \nonumber\\ &\quad =(X_1-X_2)Q \quad \text{since $P=0$} \nonumber\\ &\quad \geq 0 \quad \text{by Proposition (4.4)}, \end{align}

we learn that $V|_{P=0}$ also points to the interior of $\mathcal {S}$.

Finally, we need to exclude the possibility of non-transversal passing of a $\gamma _{s_1}$ through some point with $X_1\neq X_2$. By (4.19) and Proposition 4.1, such a point does not exist on $\mathcal {S}\cap \{B=0\}$. Suppose there were such a point on $\mathcal {S}\cap \{P=0\}$; then $P=Q=0$ at that point by (4.20). Then by Proposition 4.4, we know that such a point is either $p_0^+$ or a point on $\Phi$, which is impossible. Suppose the non-transversal passing point exists on $\mathcal {S}\cap \{A=0\}$; then by (4.18) and Proposition 4.1, we must have $A=P=0$ at that point, which is also impossible.

Proof. We consider $\check {\mathcal {S}}$ as a union of slices

\[ \check{\mathcal{S}}=\bigcup_{\kappa\in (0,1)}\check{\mathcal{L}}_\kappa,\quad \check{\mathcal{L}}_\kappa:=\check{\mathcal{S}}\cap \{X_2-\kappa X_1=0\}. \]

For each $\check {\mathcal {L}}_\kappa$ with $\kappa \in (0,1)$, replace $X_2$ with $\kappa X_1$. Then from (2.9) and $P< 0$ we have

(4.21)\begin{align} 0&> \left(\frac{1-\kappa}{3n}(32\kappa^2m^2-12\kappa^2m+48m\kappa-18\kappa+27)- \frac{12m}{n(n-1)}(1-\kappa)^3\right)X_1^2 \nonumber\\ &\quad -\left(2\kappa+\frac{6(1-\kappa)}{n-1}\right)Y^2+(4m+8) \left(1-4m\frac{1-\kappa}{n-1}\right)YZ+\left(\frac{1-\kappa}{n-1}12m-(4m\kappa+6)\right)Z^2 \nonumber\\ &= \frac{(1-\kappa)(16\kappa^2m^2-10\kappa^2 m+24\kappa m-6\kappa+9)}{6m+3}X_1^2 \nonumber\\ &\quad -\left(2\kappa+\frac{6(1-\kappa)}{n-1}\right)Y^2+(4m+8)\left(1-4m\frac{1-\kappa}{n-1}\right)YZ+ \left(\frac{1-\kappa}{n-1}12m-(4m\kappa+6)\right)Z^2. \end{align}

The coefficient for $X_1^2$ in (4.21) is obviously positive for any $(m,\kappa )\in [2,\infty )\times (0,1)$. On the other hand, use (2.9) to replace the constant term in $Q$ with homogeneous polynomials in $X_1$, $Y$ and $Z$. The polynomial $Q$ restricted on $\check {\mathcal {L}}_\kappa$ becomes

(4.22)\begin{align} Q|_{\check{\mathcal{L}}_\kappa} &= \frac{-64\kappa^3 m^3+(40\kappa-96)\kappa^2 m^2+ (36\kappa^2+60\kappa-108)\kappa m+54(\kappa-1)}{18m+9}X_1^3 \nonumber\\ &\quad +X_1\left(\left(2\kappa+\frac{6}{2m+1}\right)Y^2- \frac{8m+16}{2m+1}YZ-\left(12m\kappa + \frac{12 m}{2m+1}\right)Z^2\right). \end{align}

It is obvious that the coefficient for $X_1^3$ in (4.22) is negative for any $(m,\kappa )\in [2,\infty )\times (0,1)$. Note that if (4.21) reaches equality, then one can write $X_1^2$ as a homogeneous polynomial in $Y$ and $Z$. Moreover, substituting the $X_1^2$ by the homogeneous polynomial in (4.22) gives the formula for $X_1Y^2\tilde {Q}( {Z}/{Y})$ as in (4.14). Therefore, from (4.21) we obtain

\[ Q|_{\check{\mathcal{L}}_\kappa} > X_1Y^2\tilde{Q}\left(\frac{Z}{Y}\right). \]

Note that $\tilde {Q}( {Z}/{Y})$ above is defined on $\check {\mathcal {S}}$ instead of $\mathcal {S}\cap \{P=0\}$. On the other hand, the inequalities $A> 0$ and $\check {A}< 0$ become

\[ \frac{3m\kappa}{3(m+2)+2m(m+2)\kappa}<\frac{Z}{Y}<\frac{m\kappa}{3+2m\kappa}. \]

As shown in (4.15), it is clear that the coefficient $q_2$ is negative. It suffices to show that $\tilde {Q}$ is positive at $ {m\kappa }/({3+2m\kappa })$ and $ {3m\kappa }/({3(m+2)+2m(m+2)\kappa })$ to prove that the polynomial is positive on the open interval in between. From (4.16) it is clear that $\tilde {Q}( {m\kappa }/({3+2m\kappa }))>0$. For any $(m,\kappa )\in [2,\infty )\times (0,1)$, a straightforward computation shows that

(4.23)

The proof is complete.

Proof. With the linearized solution (3.1), we have

(4.24) \begin{align} X_1(\gamma_{s_1}) & =\frac{1}{3}-(8m^2+18m+18+(4m+8)ms_1) e^{({2}/{3})\eta}+O(e^{({2}/{3}+\epsilon)\eta}),\nonumber\\ X_2(\gamma_{s_1}) & =(-9+3s_1(m+2))e^{({2}/{3})\eta}+O(e^{({2}/{3}+\epsilon)\eta}),\\ A(\gamma_{s_1}) & = \frac{m+3}{m}((m-1)s_1-3)e^{({2}/{3})\eta}+ O(e^{({2}/{3}+\epsilon)\eta}),\nonumber\\ B(\gamma_{s_1}) & = \frac{2n^2(2m+3)(m-1)}{m(2m+1)(8m+3)} \left(\frac{9(5m+3)(4m^2+4m+3)}{n^2(2m+3)(m-1)}-s_1\right) e^{({2}/{3})\eta}+O(e^{({2}/{3}+\epsilon)\eta}),\nonumber\\ P(\gamma_{s_1}) & = O(e^{({2}/{3}+\epsilon)\eta}),\nonumber\\ Q(\gamma_{s_1}) & = O(e^{({2}/{3}+\epsilon)\eta}), \end{align}

near $p_0^+$. Hence, functions $X_2$, $A$ and $B$ are positive along $\gamma _{s_1}$ near $p_0^+$ if $s_1\in ( {3}/({m-1}), {9(5m+3)(4m^2+4m+3)}/{n^2(2m+3)(m-1)})$. Furthermore, the last two equalities in (4.24) show that $\nabla P$ and $\nabla Q$ are perpendicular to the linearized solution (3.1) at $p_0$. Hence, the integral curve $\gamma _{s_1}$ is tangent to $\{P=0\}$ and $\{Q=0\}$ for any $s_1$. It takes a little bit more work to show that the function $P$ is initially positive along $\gamma _{s_1}$. Recall in (4.20), we have

(4.25)\begin{equation} P'=P\left(H\left(3G+\frac{3}{n}(1-H^2)-1\right)+\frac{4m}{3}X_2\right)+(X_1-X_2)Q. \end{equation}

Note that

\begin{align*} & \left(H\left(3G+\frac{3}{n}(1-H^2)-1\right)+\frac{4m}{3}X_2\right)(\gamma_{s_1})\\ &\quad =-(48m^2+84m+20m(m+2)s_1)e^{({2}/{3})\eta}+O(e^{({2}/{3}+\epsilon)\eta}) \end{align*}

near $p_0^+$. If $P$ were negative initially along $\gamma _{s_1}$ near $p_0^+$, the first term in (4.25) would be positive. Moreover, from the linearized solution (3.1), we have

\[ \check{A}(\gamma_{s_1})=-\frac{1}{m}(6m+6-(m+2)(m-1)s_1)e^{({2}/{3})\eta}+ O(e^{({2}/{3}+\epsilon)\eta}). \]

Since $ {9(5m+3)(4m^2+4m+3)}/{n^2(2m+3)(m-1)}< {6(m+1)}/{(m+2)(m-1)}$ for $m\geq 2$, we know that $\check {A}(\gamma _{s_1})$ is initially negative along $\gamma _{s_1}$. Based on the assumption that $P$ is initially negative along $\gamma _{s_1}$, we know that $\gamma _{s_1}$ is initially in $\check {\mathcal {S}}$. By Proposition 4.6, we know that the second term in (4.25) is also positive and so is $P'$, which is a contradiction. Therefore, the function $P$ must be positive initially along $\gamma _{s_1}$ near $p_0^+$.

Proof. An integral curve $\gamma _{s_1}$ with $s_1>-3$ (or $\zeta _{s_2}$ with $s_2>0$) is initially in the interior of the compact set $\mathcal {E}\cap \{H\geq 0\}\cap \{Y-Z\geq 0\}$. Suppose the integral curve does not have any turning point or any $W$-intersection point. Then it must be defined on $\mathbb {R}$. Furthermore, such an integral curve is in $\mathcal {E}\cap \{H<1\}$ initially. From (2.11), along the integral curve we eventually have $H^2<1-\epsilon$ for some $\epsilon >0$. Hence,

\[ H'=(H^2-1)\left(\frac{1}{n}+\frac{12m}{n}(X_1-X_2)^2\right)\leq \frac{H^2-1}{n}<-\frac{\epsilon}{n} \]

eventually, meaning the function $H$ must at vanish some point. We reach a contradiction. Each integral curve without a turning point must have a $W$-intersection point.

Proof. From the defining equations of $\mathcal {B}_{QK}$, we have $X_1=Y-2Z$ and $X_2=-Z$ along $\gamma _{0}$. Then $H^2< 1$ becomes $(3Y-(n+3)Z)^2< 1$. From (3.14) we have

(4.26)\begin{align} & \frac{12m}{n}(Y-Z)^2+6Y^2+4m(4m+8)YZ-12mZ^2> \frac{n-1}{n}(3Y-(4m+6)Z)^2 \nonumber\\ &\quad \Leftrightarrow n(n+9)(n-1)(Y-Z)Z> 0. \end{align}

Hence, $X_1-X_2=Y-Z> 0$ along $\gamma _0$. Similarly, we have $X_2-X_1=Y-Z> 0$ along $\zeta _{ {1}/({2m+6})}$. Note that $X_1-X_2$ vanishes at the critical point $p_1^-$ and stays positive along $\gamma _0$.

Proof. The compactness is derived from $Y-(2m+3)Z\geq 0$ and (2.9).

Since

(5.1)\begin{align} & \langle\nabla(Y-(2m+3)Z),V\rangle_{\mathcal{B}_{RF}\cap \{Y-(2m+3)Z=0\}} \nonumber\\ &\quad =(Y-(2m+3)Z)\left(H\left(G+\frac{1}{n}(1-H^2)\right)-X_1-(4m+6)Z\right) \nonumber\\ &\qquad +(4m+6)Z(X_2-X_1+Y-(2m+3)Z) \nonumber\\ &\quad =(4m+6)Z(X_2-X_1+Y-(2m+3)Z)\quad \text{since $Y-(2m+3)Z=0$} \nonumber\\ &\quad \geq 0, \end{align}

the vector field $V$ points inward on the boundary $\mathcal {B}_{{\rm RF}}\cap \{Y-(2m+3)Z=0\}$. As for $\mathcal {B}_{{\rm RF}}\cap \{X_2-X_1+Y-(2m+3)Z=0\}$, from (2.9) we have

(5.2)\begin{align} & \frac{12m}{n}(Y-(2m+3)Z)^2+6Y^2+4m(4m+8)YZ-12mZ^2=\frac{n-1}{n} \nonumber\\ &\quad \Leftrightarrow 18(2m+1)Y^2+8m(8m^2+16m+3)YZ+24m(2m^2+4m+3)Z^2=4m+2. \end{align}

Then we have

(5.3) \begin{align} & \langle\nabla(X_2-X_1+Y-(2m+3)Z),V\rangle_{\mathcal{B}_{{\rm RF}}\cap \{X_2-X_1+Y-(2m+3)Z)=0\}} \nonumber\\ &\quad =(X_2-X_1+Y-(2m+3)Z))\left(H\left(G+\frac{1}{n}(1-H^2)-1\right)+ \frac{4m}{n}Y+\frac{8m^2+24m+18}{n}Z\right) \nonumber\\ &\qquad +\frac{1}{n}(Y-(2m+3)Z)((4m+2)H-(12m+6)Y-(8m^2+16m+12)Z) \nonumber\\ &\quad =\frac{1}{n}(Y-(2m+3)Z)(4m+2-(12m+6)Y-(8m^2+16m+12)Z) \nonumber\\ &\text{since $H=1$ and $X_2-X_1+Y-(2m+3)Z=0$}. \end{align}

With $Y,Z\geq 0$, showing $4m+2-(12m+6)Y-(8m^2+16m+12)Z\geq 0$ is equivalent to showing $(4m+2)^2\geq ((12m+6)Y+(8m^2+16m+12)Z)^2$. Note that $(4m+2)^2$ is simply the left-hand side of (5.2) multiplied by $4m+2$. Hence, one can obtain the non-negativity by verifying

(5.4)\begin{align} & (4m+2)(18(2m+1)Y^2+8m(8m^2+16m+3)YZ+24m(2m^2+4m+3)Z^2) \nonumber\\ &\quad \geq ((12m+6)Y+(8m^2+16m+12)Z)^2 \nonumber\\ &\quad \Leftrightarrow 16 n(m-1)((4m^2+8m+3) YZ + (2m^2+ 4m+3)Z^2)\geq 0. \end{align}

Note that the equality is obtained for $m=1$. Hence, (5.3) is non-negative and identically vanishes if $m=1$. Hence, $\mathcal {B}_{{\rm RF}}$ is invariant.

Proof. The statement is clear by (3.3) and (3.13). Note that for $m=1$, the function $X_2-X_1+Y-5Z$ is identically zero on $\gamma _\infty$.

Proof. In the new coordinate, the critical point $p_0^+$ is

\[\left(1,\frac{1}{3}\sqrt{\frac{4m+5}{2m+3}},\arctan\left(\sqrt{\frac{2m+3}{2m+2}}\right),\frac{1}{3}\right).\]

As $\gamma _\infty$ joins $p_0^+$ and $p_2^+$ in the $(X_1,X_2,Y,Z)$-coordinate, it is clear that the integral curve joins $p_0^+$ and one of $A_i^+$ or $B_i^+$ in the new coordinate. From Propositions 5.1 and 5.4 we know that $r\cos (\theta )\geq 0$ and $\sqrt { ({2m+3})/({2m+2})}r\cos (\theta )-r\sin (\theta )\geq 0$ along $\gamma _\infty$. In particular, the second inequality reaches equality so that $\theta =\arctan (\sqrt { ({2m+3})/({2m+2})})$ along $\gamma _\infty$ for $m=1$. Note that $0< a_0^+<\arctan (\sqrt { ({2m+3})/({2m+2})})\leq b_0^+$ and the last inequality reaches equality only for $m=1$. The $\theta$-coordinate for $p_0^+$ is positive and $\{\theta \geq 0\}$ is clearly invariant. Hence, for $m\geq 2$, the integral curve $\gamma _\infty$ converges to $A_0^+$; for $m=1$, the integral curve $\gamma _\infty$ converges to $B_0^+$ as $\theta =b_0^+$ along $\gamma _\infty$.

In the $(H,r,\theta,Y)$-coordinate, the critical point $p_1^+$ is

\[\left(1,\sqrt{\frac{2m+2}{2m+3}}\frac{2m+2}{n},\pi,\frac{1}{n}\right).\]

It is established in [Reference ChiChi21] that $\zeta _\infty$ is an integral curve that joins $p_1^+$ and $p_2^+$ in the $(X_1,X_2,Y,Z)$-coordinate. Therefore, in the $(H,r,\theta,Y)$-coordinate, the integral curve $\zeta _\infty$ joins $p_1^+$ and one of $A_i^+$ or $B_i^+$. As $\zeta _\infty$ is in $\tilde {\mathcal {B}}_{{\rm RF}}$ initially, by Remark 5.3 we know that $\cos (\theta )=\sqrt { ({2m+2})/({2m+3})}(Y-(2m+3)Z)\leq 0$ and $\sqrt { ({2m+3})/({2m+2})}r\cos (\theta )-r\sin (\theta )\leq 0$ along $\zeta _\infty$. Hence, we know that $ {\pi }/{2}\leq \theta \leq \arctan (\sqrt { ({2m+3})/({2m+2})})+\pi \leq b_1^+$ along $\zeta _\infty$. Therefore, $\zeta _\infty$ converges to $A_1^+$ for $m\geq 2$. We claim that the integral curve $\zeta _\infty$ also converges to $A_1^+$ for $m=1$. Recall Remarks 4.3 and 5.2. The integral curve $\chi$ also converges to $p_2^+$ and along $\chi$ we have $X_2-X_1\geq 0$ and $X_2-X_1+Y-5Z=0$. Hence, $\chi$ converges to $B_1^+$ in the $(H,r,\theta,Y)$-coordinate. As the linearization at $B_1^+$ has only one stable eigenvector, we know that $\zeta _\infty$ converges to $A_1^+$.

Proof. We think of (5.6) on $\mathcal {E}$ as a dynamical system in the three-dimensional $Hr\theta$-space with $Y$ as a function in $(H,r,\theta )$. As mentioned in Remark 5.5, each $A_i^+$ is a saddle whose linearization has two stable eigenvectors and one unstable eigenvector. Furthermore, the two stable eigenvectors are parallel to $\mathcal {E}\cap \{H=1\}$, meaning that each $A_i^+$ is a sink in the Ricci-flat subsystem and $\Pi$ is the only integral curve that emanates from $A_i^+$ in $\mathcal {E}\cap \{H<1\}$. By the Hartman–Grobman theorem, there is a local homeomorphism $(\phi, \mathcal {U})$ defined around $A_0^+$ through which the system (5.6) is topologically equivalent to the linear dynamical system

(5.10)\begin{equation} \begin{bmatrix} x\\ y\\ z \end{bmatrix}'=\begin{bmatrix} -1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1\\ \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}. \end{equation}

In particular, we have

\[ \phi(A_0^+)=(0,0,0),\quad \phi(\mathcal{U}\cap\{H=1\})\subset \{(x,y,0)\mid x,y\in \mathbb{R}\},\quad \phi(\mathcal{U}\cap\Pi)\subset \{(0,0,z)\mid z\in \mathbb{R}\}. \]

Let $D_\epsilon$ be an open disk on the $xy$-plane with radius $\epsilon _1$. Let $U_1$ be the cylinder $D_{\epsilon _1}\times (-\epsilon _2,\epsilon _2]$. Note that integral curves in $U_1\cap \{z>0\}$ can only escape through the face $D_{\epsilon _1}\times \{\epsilon _2\}$. Choose small enough $\epsilon _1$ and $\epsilon _2$ so that $\mathcal {U}_1:=\phi ^{-1}(U_1)$ is contained in $\mathcal {U}$. Then $p:=\phi ^{-1}(0,0,\epsilon _2)$ is a point on $\mathcal {U}\cap \Pi$.

Let $\mathcal {U}_0$ be an open neighborhood around the point $(0,0,\theta _*)$ in the $Hr\theta$-space. By the continuous dependence, there exists an open set $\mathcal {U}_2\ni p$ in $\mathcal {U}$ such that any point in $\mathcal {U}_2$ lies on an integral curve that enters $\mathcal {U}_0$. It is clear that $p\in \mathcal {U}_1\cap \mathcal {U}_2$. Modify $U_1$ by shrinking $\epsilon _1$ so that $D_{\epsilon _1}\times \{\epsilon _2\}$ is contained in $\phi (\mathcal {U}_2)$ while leaving $\epsilon _2$ unchanged. Then correspondingly with the modified $\mathcal {U}_1$, an integral curve in $\mathcal {U}_1\cap \{H<1\}$ must enter $\mathcal {U}_2$. Since $\gamma _\infty$ converges to $A_0^+$, there exists a point $q\in \gamma _\infty \cap \mathcal {U}_1$. By the continuous dependence, there exists a large enough $N$ such that $s_1>N$ implies $\gamma _{s_1}$ must enter $\mathcal {U}_1\cap \{H<1\}$ and hence $\mathcal {U}_0$.

Proof. Consider $\mathbb {HP}^{m+1}\sharp \overline {\mathbb {HP}}^{m+1}$ for $m\geq 2$. Let $\tilde {s}_1=\max \{ {3}/({m-1}),s_\star \}$, where $\gamma _{s_\star }$ is the heterocline in the proof of Theorem 1.1 that joins $p_0^\pm$. The proof of Theorem 1.1 shows that we have $\sharp C(\gamma _{s_1})\geq 1$ for $s_1\in (\tilde {s}_1, {9(5m+3)(4m^2+4m+3)}/{n^2(2m+3)(m-1)})$. If $\theta _*< \pi$, then $\lim _{s_1\to \infty } \sharp C(\gamma _{s_1})=0$ by Lemma 5.7. By Theorem 4.11 there exists a heterocline $\gamma _{s_{\star \star }}$ for some $s_{\star \star }\in \big (\tilde {s}_1,\infty \big )$ and $s_{\star \star }\neq s_\star$.

Consider $\mathbb {S}^{4m+4}$ for $m\geq 1$. If $\theta _*>\pi$, then $\Pi +(0,0,\pi,(2m+3)z_0)$ is an integral curve that emanates from $A_1^+$ and passes $(0,0,\theta _*+\pi,(2m+3)z_0)$ and $\theta _*+\pi >2\pi$. By Proposition 4.10, any $\zeta _{s_2}$ with $s_2>0$ has either a turning point or a $W$-intersection point. We learn from the linearized solution (3.6) that along $\zeta _{s_2}$ with $s_2> 0$ the functions $Y-Z$ and $X_2-X_1$ are positive initially. Since $( {Y}/{Z})'=2 ({Y}/{Z})(X_2-X_1)$ by (3.9), the function $Y-Z=Z( {Y}/{Z}-1)$ can only have a zero after $X_2-X_1$ changes sign. In particular, the function $X_2-X_1$ must vanish first before any $W$-intersection point can occur.

Replace $(s_1,\gamma _{s_1})$ by $(s_2,\zeta _{s_2})$ in Definition 4.8 and Theorem 4.11. For $\theta _*>\pi$ we have $\lim _{s_2\to \infty }\sharp C(\zeta _{s_2})\geq 1$ by Corollary 5.8. On the other hand, from Proposition 4.12 we know that $X_2-X_1> 0$ along $\zeta _{ {1}/({2m+6})}$ and hence $\sharp C(\zeta _{ {1}/({2m+6})})=0$. By Theorem 4.11 there exists some $s_\bullet \in ( {1}/({2m+6}),\infty )$ such that $\zeta _{s_\bullet }$ is a heterocline.

Assume $\theta _*>\pi$ so that such an $s_\bullet$ exists. We claim that the Einstein metric $\hat {g}_{\mathbb {S}^{4m+4}}$ represented by $\zeta _{s_\bullet }$ is not the standard sphere metric. If $\hat {g}_{\mathbb {S}^{4m+4}}$ were the standard sphere metric, it would have constant sectional curvature. In particular, we must have $ ({\ddot {f_2}}/{f_2}) ({f_1}/{\ddot {f_1}})=1$. From (3.7) we must have $s_2=0$. Hence, $\hat {g}_{\mathbb {S}^{4m+4}}$ is not the standard sphere metric. The proof is complete.