\begin{thm}\label{XiaFock}
		We have
		\begin{equation*}
			C^{*} (\mathcal{WL})=\mathcal{T}^{(1)}.
		\end{equation*}
	\end{thm}

For some \(\alpha>0\), \(p>0\) and \(dV\) the standard volume measure on \(\mathbb{C}^n\), let \(L^{p}_{\alpha}(\mathbb{C}^n,dV)\) be the Lebesgue space of measurable functions \(f\) on \(\mathbb{C}^n\) such that
	\begin{equation}\label{Fnorm1}
		\left\lVert f \right\rVert^{p}_{\alpha} := \left(\frac{p\alpha}{2\pi}\right)^{n} \int_{\mathbb{C}^n} |f(z)e^{-\frac{\alpha}{2}|z|^2}|^{p} dV(z) < \infty~.
	\end{equation}
	Similarly, for \(\alpha>0\) and \(p=\infty\), we denoted by \(L^{\infty}_{\alpha}\) the space of Lebesgue measurable function \(f\) on \(\mathbb{C}^n\) such that
	\begin{equation*}
		\left\lVert f \right\rVert_{\infty,\alpha} := \text{esssup}\{|f(z)|e^{-\frac{\alpha|z|^2}{2}} : ~z \in \mathbb{C}^n\}< \infty ~.
	\end{equation*}
	The classical Fock space \(F^{p}_{\alpha}\) is the space of entire functions on \(\mathbb{C}^n\) which belong to \( L^{p}_{\alpha}(\mathbb{C}^n, dV)\). Similarly, the Fock space \(F^{\infty}_{\alpha}\) is the space of entire functions on \(\mathbb{C}^n\) which belong to \(L^{\infty}_{\alpha}\).

	Let \(d\mu\) be the Gaussian measure on \(\mathbb{C}^n\), \(n\ge 1\). In terms of the standard volume measure \(dV\) on \(\mathbb{C}^n\), it is given by
	\begin{equation*}
		d\mu(z) = \pi^{-n} e^{-|z|^2} dV(z)~.
	\end{equation*}

	 The Fock space \(H^{2}(\mathbb{C}^n,d\mu)\) is defined to be the subspace of the (Hilbert-) Lebesgue space \(L^{2}(\mathbb{C}^n,d\mu)\) consisting of entire functions. Notice that \(H^{2}(\mathbb{C}^n,d\mu) = F^{2}_{1} \). The symbol \(K_z\) denotes the reproducing kernel and the symbol \(k_z\) denotes the normalized reproducing kernel for \(H^{2}(\mathbb{C}^n,d\mu)\). That is,
	\begin{equation*}
		K_z(\zeta) = e^{\langle \zeta, z \rangle}, \quad k_{z}(\zeta) = e^{\langle \zeta, z\rangle}e^{-\frac{|z|^2}{2}}~, \quad z,\zeta \in \mathbb{C}^n.
	\end{equation*}

	In \cite{xia2015localization}, J. Xia showed in the case of the Bergman space on the unit ball of \(\mathbb{C}^n\) that the norm closure of \(\{T_{f}: f \in L^{\infty}(B,dv) \}\) coincides with the \(C^{*}\)-algebra of weakly localized operators. Also, he stated in \cite[Section 4 ]{xia2015localization}  that  the analogue of \cite[Theorem 1.5]{xia2015localization} on the Fock space \(H^{2}(\mathbb{C}^n, d\mu)\) was true. In this paper, we define the notion of weakly localized operators, state Xia's theorem for the Fock space \(H^{2}(\mathbb{C}^n, d\mu)\) and provide details of its proof. Further, we present a consequence of this theorem on the compactness of operators on \(H^{2}(\mathbb{C}^n,d\mu)\).  We begin with the following definitions and we state the main theorem, the proof of which will retain our attention in the following sections.

	\begin{defn}
		For \(f \in L^{\infty}(\mathbb{C}^n,dV) \), the \textbf{Toeplitz operator} \(T_{f}\) is defined by the formula
		\begin{equation*}
			T_{f}h = P(fh)~,\quad h \in H^{2}(\mathbb{C}^n,d\mu)~,
		\end{equation*}
		where \(P \colon L^{2}(\mathbb{C}^n,d\mu) \rightarrow H^{2}(\mathbb{C}^n,d\mu)\) is the orthogonal projection.
	\end{defn}

	The \textbf{standard lattice} in \(\mathbb{C}^n\) is denoted by
	\begin{equation*}
		\mathbb{Z}^{2n} = \{(m_{1}+il_{1}, \dots, m_{n}+il_{n}): m_{1}, l_{1}, \dots, m_{n},l_{n} \in \mathbb{Z}\}~.
	\end{equation*}
	We fix an orthonormal set \(\{e_{u}: u \in \mathbb{Z}^{2n}\}\) in \(H^{2}(\mathbb{C}^n,d\mu)\). We let \(S\) denote the \textbf{fundamental unit cube in }\(\mathbb{C}^{n}\). That is,
	\begin{equation*}
		S = \{(x_{1}+iy_{1}, \dots, x_{n}+iy_{n}) : x_{1},y_{1}, \dots , x_{n},y_{n} \in [0,1)\}.
	\end{equation*}
	With \(\mathbb{Z}^{2n}\) and \(S\), we have
	\begin{equation*}
		\cup_{u \in \mathbb{Z}^{2n}} \{S+u\} = \mathbb{C}^{n} = \cup_{u \in \mathbb{Z}^{2n}} \{u-S\} ~,
	\end{equation*}
	which is a tiling of the space, meaning that there is no overlap between \(S+u\) and \(S+v\) for \(u \neq v\) in \(\mathbb{Z}^{2n}\) (resp. between \(u-S\) and \(v-S\) for \(u \neq v \in \mathbb{Z}^{2n}\) ).

	\begin{defn}
		Let \(\mathcal{T}^{(1)}\) denote the norm closure of \(\{T_{f}: f\in L^{\infty}(\mathbb{C}^n,dV)\}\) in \(\mathcal{B}(H^2(\mathbb{C}^n,d\mu))\) with respect to the operator norm. That is
		\begin{equation*}
			\mathcal{T}^{(1)} = \{B : \lim_{k \to \infty } \norm{B-T_{b_{k}}}=0 , b_{k} \in L^{\infty}(\mathbb{C}^{n},dV)\}.
		\end{equation*}
	\end{defn}

	\begin{defn}
		We denote by \(\mathcal{S}\) the linear span of the normalized reproducing kernels \(k_z\). A linear operator \(B : \mathcal{S} \rightarrow H^{2}(\mathbb{C}^n,d\mu) \) is said to be \textbf{admissible} on \( H^{2}(\mathbb{C}^n,d\mu) \) if there exists a linear operator \( B^{*} : \mathcal{S} \rightarrow H^{2}(\mathbb{C}^n,d\mu) \) such that the duality relation
		\begin{equation}\label{R1}
			\langle Bk_{z}, k_{w} \rangle = \langle k_{z}, B^{*}k_w \rangle
		\end{equation}
		holds for all \(z,w \in \mathbb{C}^n\).
	\end{defn}
	The inner product here is with respect to \(d\mu\).

	We define below sufficiently localized operators following J. Xia and D. Zheng (XZ) in \cite{xia2013localization}.
	\begin{defn}
		A bounded linear operator \(B\) on \(H^{2}(\mathbb{C}^n,d\mu)\) is said to be XZ-\textbf{sufficiently localized} if there exist constants \(2n<\beta<\infty\) and \(0<C<\infty\) such that
		\begin{equation}\label{SL1}
			|\langle Bk_{z},k_{w}\rangle| \le \frac{C}{(1+|z-w|)^{\beta}}
		\end{equation}
		for all \(z,w\in \mathbb{C}^n\).
	\end{defn}

		\begin{defn}
		An admissible operator \(B\) on \(H^{2}(\mathbb{C}^n,d\mu)\) is said to be \textbf{weakly localized} if it satisfies the following four conditions
		\begin{equation*}
			\sup_{z\in \mathbb{C}^n}\int_{\mathbb{C}^n} |\langle Bk_{z}, k_{w} \rangle| dV(w) <\infty, \quad \sup_{z\in \mathbb{C}^n}\int_{\mathbb{C}^n} |\langle B^{*} k_{z}, k_{w} \rangle| dV(w) <\infty
		\end{equation*}

		and
		\begin{equation*}
			\lim_{r\to\infty}\sup_{z\in \mathbb{C}^n}\int_{|z-w|\ge r} |\langle Bk_{z}, k_{w} \rangle| dV(w) =0, \quad \lim_{r\to\infty}\sup_{z\in \mathbb{C}^n}\int_{|z-w|\ge r} |\langle B^{*} k_{z}, k_{w} \rangle| dV(w) =0~.
		\end{equation*}
	\end{defn}

	\begin{exa}\label{Ex1}
		If \(f\) is a \textbf{bounded measurable function} on \(\mathbb{C}^n\), then there is a positive constant \(0<C=C(f)<\infty\) such that
		\begin{equation*}
			|\langle T_{f}k_{z}, k_{w}\rangle| \le C e^{-(1/8) |z-w|^2}
		\end{equation*}
		for all \(z,w \in \mathbb{C}^n\). That is, \(T_{f} \in \text{XZ-}\mathcal{SL}\). Also, \(T_{f} \in \mathcal{L}_{p}\) for \(2<p<\frac{8}{3}\).
	\end{exa}

	\begin{proof}
		For each \(z, w, \xi \in \mathbb{C}^n\), we have
		\begin{equation*}
			|k_{z}(\xi)k_{w}(\xi)|e^{-|\xi|^2} = |e^{\langle \xi , z \rangle - \frac{|z|^2}{2} } e^{\langle \xi, w \rangle - \frac{|w|^2}{2}} | e^{-|\xi|^2} = e^{-\frac{1}{2}\left(|z-\xi|^2 + |w-\xi|^2\right)} ~.
		\end{equation*}
		By the triangular inequality we have
		\begin{equation*}
			|z-w|^2 \le |z-\xi + \xi -w|^2 \le (|z-\xi| + |w-\xi|)^2 \le 2 (|z-\xi|^2 + |w-\xi|^2)~.
		\end{equation*}
		It follows that
		\begin{eqnarray}\label{kzw}
			|k_{z}(\xi)k_{w}(\xi)|e^{-|\xi|^2} = e^{-\frac{1}{4}\left(|z-\xi|^2 + |w-\xi|^2\right)}e^{-\frac{1}{4}\left(|z-\xi|^2 + |w-\xi|^2\right)} \le e^{-\frac{1}{4}\left(|z-\xi|^2 + |w-\xi|^2\right)} e^{-\frac{1}{8}|z-w|^2}~.
		\end{eqnarray}
		Let \(f\) be a bounded measurable function on \(\mathbb{C}^n\), then for all \(g,h\in H^{2}(\mathbb{C}^n,d\mu)\), it holds
		\begin{equation}\label{Toepfgh}
			\langle T_{f}g, h \rangle = \int_{\mathbb{C}^{n}} f(\xi)g(\xi)\overline{h(\xi)} d\mu(\xi)~.
		\end{equation}
		In fact,
		\begin{eqnarray*}
			\langle T_{f}g, h \rangle &=&  \int_{\mathbb{C}^{n}} T_{f}g(w)\overline{h(w)} d\mu(w) \\
			&=& \int_{\mathbb{C}^{n}}\int_{\mathbb{C}^{n}} K(w,\xi) f(\xi)g(\xi) d\mu(\xi) \overline{h(w)} d\mu(w) \\
			&=& \int_{\mathbb{C}^{n}} f(\xi)g(\xi) \int_{\mathbb{C}^{n}} K_{\xi}(w) \overline{h(w)} d\mu(w) d\mu(\xi) \\
			&=& \int_{\mathbb{C}^{n}} f(\xi)g(\xi) \langle K_{\xi} , h \rangle d\mu(\xi) = \int_{\mathbb{C}^{n}} f(\xi)g(\xi) \overline{\langle  h, K_{\xi}  \rangle} d\mu(\xi) \\
			&=& \int_{\mathbb{C}^{n}} f(\xi)g(\xi)\overline{h(\xi)} d\mu(\xi)~.
		\end{eqnarray*}
		Therefore,
		\begin{eqnarray*}
			|\langle T_{f}k_{z}, k_{w} \rangle| &=& \left|\int_{\mathbb{C}^n}
			f(\xi)k_{z}(\xi) \overline{k_{w}(\xi)} d\mu(\xi)\right|  \\
			&\le & \frac{\left\lVert f \right\rVert_{\infty}}{\pi^n} \int_{\mathbb{C}^n} |k_{z}(\xi) k_{w}(\xi)| e^{-|\xi|^2} dV(\xi) \\
			& \le & \frac{\left\lVert f \right\rVert_{\infty}}{\pi^n}  e^{-\frac{1}{8}|z-w|^2} \int_{\mathbb{C}^n} e^{-\frac{1}{4}\left(|z-\xi|^2 + |w-\xi|^2\right)}  dV(\xi) \\
			& \le & \frac{\left\lVert f \right\rVert_{\infty}}{\pi^n}  e^{-\frac{1}{8}|z-w|^2} \left(\int_{\mathbb{C}^n} e^{-\frac{1}{2}|z-\xi|^2} dV(\xi) \right)^{\frac{1}{2}} \left(\int_{\mathbb{C}^n} e^{-\frac{1}{2}|w-\xi|^2}dV(\xi) \right)^{\frac{1}{2}} \\
			&=& \frac{\left\lVert f \right\rVert_{\infty}}{\pi^n}  e^{-\frac{1}{8}|z-w|^2} \int_{\mathbb{C}^n} e^{-\frac{1}{2}|\zeta|^2} dV(\zeta) \\
			&=& (\sqrt{2})^n \left\lVert f \right\rVert_{\infty} e^{-\frac{1}{8}|z-w|^2}~.
		\end{eqnarray*}
		This completes the proof.
	\end{proof}

	 \begin{pro}
	 	Any XZ-sufficiently localized operator is weakly localized.
	 \end{pro}
	\begin{proof}
	Let \(B\) be a XZ-sufficiently localized operator, then  for all \(z\in \mathbb{C}^n\), we have
	\begin{eqnarray*}
		\int_{\mathbb{C}^n} |\langle Bk_z, k_w \rangle| dV(w) &\le& \frac{C}{\pi^n} \int_{\mathbb{C}^n} \frac{1}{(1+|\zeta|)^{\beta}} dV(\zeta) \\
		&=& \frac{Cc_n}{\pi^n} \int_{0}^{\infty} \frac{r^{2n-1}}{(1+r)^{\beta}} dr \\
		&=& C_n \left[ \int_{0}^{1} \frac{r^{2n-1}}{(1+r)^{\beta}} dr  + \int_{1}^{\infty } \frac{r^{2n-1}}{(1+r)^{\beta}} dr\right] \\
		&\le & C_n \left[ \int_{0}^{1} \frac{1}{(1+r)^{\beta}} dr +  \int_{1}^{\infty } \frac{r^{2n-1}}{r^{\beta}} dr\right] \\
		&=& C_n \left\{\frac{1}{\beta-1}\left(1-\frac{1}{2^{\beta}-1}\right) + \frac{1}{\beta-2n}\right\} \\
		&:=& C(n,\beta)~,
	\end{eqnarray*}
	which does not depend on \(z\). Therefore,
	\begin{equation*}
		\sup_{z\in \mathbb{C}^n}\int_{\mathbb{C}^n} |\langle Bk_z, k_w \rangle| dV(w) \le C(n,\beta)  <\infty~.
	\end{equation*}
	And in the same way, we also have
	\begin{equation*}
		\sup_{z\in \mathbb{C}^n}\int_{\mathbb{C}^n} |\langle B^*k_z, k_w \rangle| dV(w) \le C(n,\beta) <\infty~.
	\end{equation*}
	We now show that \(B\) satisfies the third condition of weakly localized operator.

	Let \(z\in \mathbb{C}^n\), using the change of variables \(\zeta=z-w\) and the spherical coordinates, it holds
	\begin{eqnarray*}
		\int_{|z-w|\ge r} |\langle Bk_z, k_w \rangle dV(w) &\le& C \int_{|z-w|\ge r} \frac{1}{(1+|z-w|)^{\beta}} dV(w) \\
		&=& C \int_{|\zeta|\ge r} \frac{1}{(1+|\zeta|)^{\beta}} dV(\zeta) \\
		&=& C c_n \int_{\rho\ge r} \frac{\rho^{2n-1}}{(1+\rho)^{\beta}} d \rho \\
		&\le& C c_n \int_{\rho\ge r} \frac{\rho^{2n-1}}{\rho^{\beta}} d \rho =  C c_n \int_{\rho\ge r} \frac{1}{\rho^{\beta-2n+1}} d \rho \\
		&=& \frac{Cc_n}{\beta-2n}\frac{1}{r^{\beta-2n}}~.
	\end{eqnarray*}
	It follows that,
	\begin{equation*}
		\lim_{r\to\infty} \sup_{z\in \mathbb{C}^n} \int_{|z-w|\ge r} |\langle Bk_z, k_w \rangle dV(w) \le \frac{Cc_n}{\beta-2n} \lim_{r\to\infty} \frac{1}{r^{\beta-2n}} =0~.
	\end{equation*}
	Using the duality relation (\ref{R1}) of an admissible operator and similarly like with \(B\), we also have:
	\begin{equation*}
		\lim_{r\to\infty} \sup_{z\in \mathbb{C}^n} \int_{|z-w|\ge r} |\langle B^*k_z, k_w \rangle dV(w) \le \frac{Cc_n}{\beta-2n} \lim_{r\to\infty} \frac{1}{r^{\beta-2n}} =0~.
	\end{equation*}
\end{proof}

	We denote by \(\mathcal{WL}\) the collection of weakly localized operators on \(H^{2}(\mathbb{C}^n,d\mu)\). We recall the following definition from \cite{zhu2007}.

		\begin{defn}
		A \textbf{Banach algebra} is a complex algebra together with a complete norm satisfying the condition \(\left\lVert xy \right\rVert \le \left\lVert x \right\rVert\left\lVert y \right\rVert\).
		A \(C^{*}\)\textbf{-algebra} is a Banach algebra \(\mathcal{A}\) with an involution \(x \mapsto x^{*}\) on it satisfying the following conditions:
		\begin{enumerate}[label=$\blacktriangleright$]
			\item \(x^{**}=x\) for all vectors \(x \in \mathcal{A} \).
			\item \((ax + by)^{*} = \bar{a}x^{*} + \bar{b}y^{*} \) for all vectors \(x,y \in \mathcal{A}\) and \(a,b \in \mathbb{C}\).
			\item \((xy)^{*}=y^{*}x^{*}\) for all vectors \(x,y \in \mathcal{A}\).
			\item \(\norm{xx^{*}} = \left\lVert x \right\rVert^{2}\) for all vector \(x \in \mathcal{A}\).
		\end{enumerate}
	\end{defn}

	\begin{defn}
		A Banach algebra \(\mathcal{A}\) is called a \textbf{\(\star\)-algebra} if for every \(A\in \mathcal{A}\), we have \(A^{*} \in \mathcal{A} \).
	\end{defn}

	\begin{defn}
		We denote by \(C^{*}(\mathcal{WL})\) the \(C^{*}\)-algebra generated by weakly localized operators on \(H^{2}(\mathbb{C}^n, d\mu)\). Also, \(C^{*}(\mathcal{WL})\) is actually the norm closure of \(\mathcal{WL}\) since \(\mathcal{WL} \) is a \(\star\)-algebra.
	\end{defn}

	We will prove the following main result.

For some \(\alpha>0\), \(p>0\) and \(dV\) the standard volume measure on \(\mathbb{C}^n\), let \(L^{p}_{\alpha}(\mathbb{C}^n,dV)\) be the Lebesgue space of measurable functions \(f\) on \(\mathbb{C}^n\) such that
 \begin{equation}\label{Fnorm1}
 \left\lVert f \right\rVert^{p}_{\alpha} := \left(\frac{p\alpha}{2\pi}\right)^{n} \int_{\mathbb{C}^n} |f(z)e^{-\frac{\alpha}{2}|z|^2}|^{p} dV(z) < \infty~.
 \end{equation}
 Similarly, for \(\alpha>0\) and \(p=\infty\), we denoted by \(L^{\infty}_{\alpha}\) the space of Lebesgue measurable function \(f\) on \(\mathbb{C}^n\) such that
 \begin{equation*}
 \left\lVert f \right\rVert_{\infty,\alpha} := \text{esssup}\{|f(z)|e^{-\frac{\alpha|z|^2}{2}} : ~z \in \mathbb{C}^n\}< \infty ~.
 \end{equation*}
 The classical Fock space \(F^{p}_{\alpha}\) is the space of entire functions on \(\mathbb{C}^n\) which belong to \( L^{p}_{\alpha}(\mathbb{C}^n, dV)\). Similarly, the Fock space \(F^{\infty}_{\alpha}\) is the space of entire functions on \(\mathbb{C}^n\) which belong to \(L^{\infty}_{\alpha}\).

\begin{defn}
 We denote by \(\mathcal{S}\) the linear span of the normalized reproducing kernels \(k_z\). A linear operator \(B : \mathcal{S} \rightarrow H^{2}(\mathbb{C}^n,d\mu) \) is said to be \textbf{admissible} on \( H^{2}(\mathbb{C}^n,d\mu) \) if there exists a linear operator \( B^{*} : \mathcal{S} \rightarrow H^{2}(\mathbb{C}^n,d\mu) \) such that the duality relation
 \begin{equation}\label{R1}
 \langle Bk_{z}, k_{w} \rangle = \langle k_{z}, B^{*}k_w \rangle
 \end{equation}
 holds for all \(z,w \in \mathbb{C}^n\).
 \end{defn}
 The inner product here is with respect to \(d\mu\).

\begin{exa}\label{Ex1}
 If \(f\) is a \textbf{bounded measurable function} on \(\mathbb{C}^n\), then there is a positive constant \(0<C=C(f)<\infty\) such that
 \begin{equation*}
 |\langle T_{f}k_{z}, k_{w}\rangle| \le C e^{-(1/8) |z-w|^2}
 \end{equation*}
 for all \(z,w \in \mathbb{C}^n\). That is, \(T_{f} \in \text{XZ-}\mathcal{SL}\). Also, \(T_{f} \in \mathcal{L}_{p}\) for \(2<p<\frac{8}{3}\).
 \end{exa}

\begin{proof}
 For each \(z, w, \xi \in \mathbb{C}^n\), we have
 \begin{equation*}
 |k_{z}(\xi)k_{w}(\xi)|e^{-|\xi|^2} = |e^{\langle \xi , z \rangle - \frac{|z|^2}{2} } e^{\langle \xi, w \rangle - \frac{|w|^2}{2}} | e^{-|\xi|^2} = e^{-\frac{1}{2}\left(|z-\xi|^2 + |w-\xi|^2\right)} ~.
 \end{equation*}
 By the triangular inequality we have
 \begin{equation*}
 |z-w|^2 \le |z-\xi + \xi -w|^2 \le (|z-\xi| + |w-\xi|)^2 \le 2 (|z-\xi|^2 + |w-\xi|^2)~.
 \end{equation*}
 It follows that
 \begin{eqnarray}\label{kzw}
 |k_{z}(\xi)k_{w}(\xi)|e^{-|\xi|^2} = e^{-\frac{1}{4}\left(|z-\xi|^2 + |w-\xi|^2\right)}e^{-\frac{1}{4}\left(|z-\xi|^2 + |w-\xi|^2\right)} \le e^{-\frac{1}{4}\left(|z-\xi|^2 + |w-\xi|^2\right)} e^{-\frac{1}{8}|z-w|^2}~.
 \end{eqnarray}
 Let \(f\) be a bounded measurable function on \(\mathbb{C}^n\), then for all \(g,h\in H^{2}(\mathbb{C}^n,d\mu)\), it holds
 \begin{equation}\label{Toepfgh}
 \langle T_{f}g, h \rangle = \int_{\mathbb{C}^{n}} f(\xi)g(\xi)\overline{h(\xi)} d\mu(\xi)~.
 \end{equation}
 In fact,
 \begin{eqnarray*}
 \langle T_{f}g, h \rangle &=& \int_{\mathbb{C}^{n}} T_{f}g(w)\overline{h(w)} d\mu(w) \\
 &=& \int_{\mathbb{C}^{n}}\int_{\mathbb{C}^{n}} K(w,\xi) f(\xi)g(\xi) d\mu(\xi) \overline{h(w)} d\mu(w) \\
 &=& \int_{\mathbb{C}^{n}} f(\xi)g(\xi) \int_{\mathbb{C}^{n}} K_{\xi}(w) \overline{h(w)} d\mu(w) d\mu(\xi) \\
 &=& \int_{\mathbb{C}^{n}} f(\xi)g(\xi) \langle K_{\xi} , h \rangle d\mu(\xi) = \int_{\mathbb{C}^{n}} f(\xi)g(\xi) \overline{\langle h, K_{\xi} \rangle} d\mu(\xi) \\
 &=& \int_{\mathbb{C}^{n}} f(\xi)g(\xi)\overline{h(\xi)} d\mu(\xi)~.
 \end{eqnarray*}
 Therefore,
 \begin{eqnarray*}
 |\langle T_{f}k_{z}, k_{w} \rangle| &=& \left|\int_{\mathbb{C}^n}
 f(\xi)k_{z}(\xi) \overline{k_{w}(\xi)} d\mu(\xi)\right| \\
 &\le & \frac{\left\lVert f \right\rVert_{\infty}}{\pi^n} \int_{\mathbb{C}^n} |k_{z}(\xi) k_{w}(\xi)| e^{-|\xi|^2} dV(\xi) \\
 & \le & \frac{\left\lVert f \right\rVert_{\infty}}{\pi^n} e^{-\frac{1}{8}|z-w|^2} \int_{\mathbb{C}^n} e^{-\frac{1}{4}\left(|z-\xi|^2 + |w-\xi|^2\right)} dV(\xi) \\
 & \le & \frac{\left\lVert f \right\rVert_{\infty}}{\pi^n} e^{-\frac{1}{8}|z-w|^2} \left(\int_{\mathbb{C}^n} e^{-\frac{1}{2}|z-\xi|^2} dV(\xi) \right)^{\frac{1}{2}} \left(\int_{\mathbb{C}^n} e^{-\frac{1}{2}|w-\xi|^2}dV(\xi) \right)^{\frac{1}{2}} \\
 &=& \frac{\left\lVert f \right\rVert_{\infty}}{\pi^n} e^{-\frac{1}{8}|z-w|^2} \int_{\mathbb{C}^n} e^{-\frac{1}{2}|\zeta|^2} dV(\zeta) \\
 &=& (\sqrt{2})^n \left\lVert f \right\rVert_{\infty} e^{-\frac{1}{8}|z-w|^2}~.
 \end{eqnarray*}
 This completes the proof.
 \end{proof}

\begin{defn}
 We denote by \(C^{*}(\mathcal{WL})\) the \(C^{*}\)-algebra generated by weakly localized operators on \(H^{2}(\mathbb{C}^n, d\mu)\). Also, \(C^{*}(\mathcal{WL})\) is actually the norm closure of \(\mathcal{WL}\) since \(\mathcal{WL} \) is a \(\star\)-algebra.
 \end{defn}

We will prove the following main result.

\begin{defn}
 We denote by \(C^{*}(\mathcal{WL})\) the \(C^{*}\)-algebra generated by weakly localized operators on \(H^{2}(\C^n, d\mu)\). Also, \(C^{*}(\mathcal{WL})\) is actually the norm closure of \(\mathcal{WL}\) since \(\mathcal{WL} \) is a \(\star\)-algebra.
 \end{defn}

The organization of this paper is as follows. In Section 2, we will give propositions in the case of the Fock space \(H^{2}(\C^n, d\mu)\) which are the analogue of those given by Xia in \cite{xia2015localization} in the Bergman space case of the unit ball. Later, using these propositions in Section 3, we establish the proof of Theorem \ref{XiaFock} and present a consequence.

\begin{defn}\label{normstar}
 For an entire function \(h\) in \(\C^{n}\), we write
 \begin{equation*}
 \norm{h}_{*} = \left(\int_{\C^{n}} |h(\zeta)|^{2} e^{-\frac{1}{2}|\zeta|^2} dV(\zeta)\right)^{\frac{1}{2}}~.
 \end{equation*}
 We denote by \(\mathcal{H}_{*}\) the collection of entire functions \(h\) on \(\C^n\) satisfying \(\norm{h}_{*}<\infty\).
 \end{defn}
 \begin{rem}
 The norm \(\norm{\cdot}_{*}\) is equivalent to the norm on the Fock space \(F^{2}_{1/2}\) given in relation (\ref{Fnorm1}) which is an Hilbert space. More precisely, \(\norm{\cdot}_{*} = (2\pi)^{n/2} \norm{\cdot}_{F^{2}_{1/2}}\). This ensures the continuity of \(\norm{\cdot}_{*}\).
 \end{rem}
 In what follows, we will use the operator \(U_{z}\) defined by
 \begin{equation*}
 U_{z}f(w) = f(z-w) k_{z}(w)~, \quad f \in H^{2}(\C^n, d\mu)~.
 \end{equation*}

\begin{pro}\label{proYz}
 Suppose \(\{c_{u}: u \in \Z^{2n}\}\) are complex numbers satisfying the condition \mbox{\(\sup_{u\in \Z^{2n}}|c_{u}| < \infty\)}. Then for each \(z\in \C^{n}\), the operator
 \begin{equation}\label{Yz}
 Y_{z} = \sum_{u \in \Z^{2n}} c_{u} k_{u-z} \otimes k_{u-z}
 \end{equation}
 is bounded on \(H^{2}(\C^n, d\mu)\). Moreover, the map \(z \mapsto Y_{z}\) from \(\C^{n}\) to \(\mathcal{B}(H^{2}(\C^{n},d\mu))\) is continuous with respect to the operator norm.
 \end{pro}
 \begin{proof}
 Let \(\{c_{u} : u \in \Z^{2n}\}\) be complex numbers such that \mbox{\(\sup_{u\in \Z^{2n}}|c_{u}| < \infty\)}. For \(z\in \C^{n}\), we define
 \begin{equation*}
 A_{z} = \sum_{u \in \Z^{2n}} c_{u} (U_{u}k_{z}) \otimes e_{u} \quad \text{ and } \quad B_{z} = \sum_{u \in \Z^{2n}}(U_{u}k_{z})\otimes e_{u}~.
 \end{equation*}
 Using the relation (\ref{Uu2}), we have \(Y_{z} = A_{z}B^{*}_{z}\). Set \(h_{u} = c_{u}k_{z}\), then using Definition \ref{normstar}, we have
 \begin{equation*}
 \norm{h_{u}}_{*} = |c_{u}|\norm{k_{z}}_{*} = (2\pi)^{n/2} e^{\frac{|z|^2}{2}} |c_{u}|~.
 \end{equation*}
 Therefore, \(\sup_{u\in \Z^{2n}} \norm{h_{u}}_{*} < \infty\). Applying Lemma \ref{LemUh} to \(h_{u} = c_{u}k_{z}\), we see that \(\norm{A_{z}} \le C \sup_{u\in \Z^{2n}}|c_{u}|\). Thus each \(A_{z}\) is bounded. Since \(B_z\) is just a special case of \(A_{z}\) (with \(c_{u}=1\) for all \(u \in \Z^{2n}\)), then it is also bounded and hence \(Y_{z} = A_{z}B^{*}_{z}\) is bounded.

\begin{lem}\label{Lim=0}
 Let \(B \in \mathcal{WL} \), then for every \(z,w \in \C^n \), we have
 \begin{equation*}
 \lim_{R\to \infty} \sup_{u\in \Z^{2n}} \sum_{\underset{|u-v|>R}{v\in \Z^{2n} }} |\langle B k_{u-z}, k_{v-w} \rangle| =0 \quad \text{ and } \quad \lim_{R\to \infty} \sup_{u\in \Z^{2n}} \sum_{\underset{|u-v|>R}{v\in \Z^{2n} }} |\langle k_{u-z}, Bk_{v-w}\rangle| = 0~.
 \end{equation*}
 \end{lem}
 \begin{proof}
 By \cite[Lemma 2.32]{zhu2012}, for any entire function \(f\) on \(\C^n\), we have
 \begin{equation*}
 \left|f(z)e^{-\frac{\alpha}{2}|z|^2}\right|^p \le C \int_{B(z,\delta)} |f(w)e^{-\frac{\alpha}{2}|w|^2}|^p dV(w) \quad \text{ for } z \in \C^n.
 \end{equation*}
 Hence for \(\alpha=p=1\) and \(\delta\) small such that the balls \(\{B(v-w,\delta): v \in \Z^{2n} \}\) are mutually disjoint, we have
 \begin{equation*}
 |\langle B k_{u-z}, k_{v-w} \rangle| = |Bk_{u-z} (v-w) |e^{-\frac{|v-w|^2}{2}} \le C \int_{B(v-w,\delta)} |Bk_{u-z}(\zeta) | e^{-\frac{|\zeta|^2}{2}} dV(\zeta).
 \end{equation*}
 Indeed, for \(\delta < \frac{1}{2} \), the balls \(\{B(v-w,\delta) : v \in \Z^{2n} \}\) are mutually disjoint. Otherwise, there would exist \(v,v' \in \Z^{2n} \) such that \(v \neq v' \), and a point \(\xi\) such that \(\xi \in B(v-w,\delta) \cap B(v'-w,\delta) \). In other words:
 \begin{equation*}
 |v-w-\xi| < \delta \quad \text{ and } \quad |v'-w - \xi| < \delta~.
 \end{equation*}
 This implies that
 \begin{equation*}
 |v-v'|=|(v-w-\xi)-(v'-w-\xi)|\le |v-w-\xi| + |v'-w-\xi| < \delta + \delta= 2 \delta < 1~.
 \end{equation*}
 That is \(|v-v'|<1\). This contradicts the well-known fact that \(|v-v'|\ge 1\). This result actually implies that there exists \(N\in \N\) such that each \(\zeta\in \C^n \) belongs to at most \(N\) balls in \(\{B(v-w,\delta): v\in \Z^{2n}\}\). That is \(\sum_{v \in \Z^{2n}} \chi_{B(v-w,\delta)}(\zeta) \le N \) for each \(\zeta \in \C^n \).

\begin{proof}[\textbf{Proof of Proposition \ref{EwBEzD} }]
 From (\ref{EwBEz}), we have
 \begin{equation*}
 E_{w} BE_{z} = \frac{1}{\pi^{2n}} \sum_{u,v \in \Z^{2n}} \langle Bk_{u-z}, k_{v-w} \rangle k_{v-w} \otimes k_{u-z}~.
 \end{equation*}
 Thus for any \(R>0\), we can write \(E_{w}BE_{z} = V_{R} + W_{R}\), where
 \begin{equation*}
 V_{R} = \frac{1}{\pi^{2n}} \sum_{\underset{|u-v|\le R}{u,v \in \Z^{2n}}} \langle Bk_{u-z}, k_{v-w} \rangle k_{v-w} \otimes k_{u-z} \quad \text{ and }
 \end{equation*}
 \begin{equation*}
 \quad W_{R} = \frac{1}{\pi^{2n}} \sum_{\underset{|u-v|>R}{u,v \in \Z^{2n}}} \langle Bk_{u-z}, k_{v-w} \rangle k_{v-w} \otimes k_{u-z}~.
 \end{equation*}
 To complete the proof, it suffices to prove that:
 \begin{enumerate}[label=(\alph*)]
 \item \label{itm:first1} \(\lim_{R\to \infty} \norm{W_{R}}=0\).
 \item \label{itm:second1} \(V_{R} \in \) span(\(\mathcal{D}_{0}\)) for every \(R>0\).
 \end{enumerate}

\begin{pro}\label{DoT1}
 We have \(\mathcal{D}_{0} \subset \mathcal{T}^{(1)}\).
 \end{pro}
 To establish the proof of this proposition, we will need the next three propositions.
 \begin{pro}\label{YzT1}
 Suppose that \(\{c_{u}: u \in \Z^{2n}\}\) is a bounded set of complex coefficients. Then for each \(z \in \C^{n}\), the operator \(Y_{z}\) defined in (\ref{Yz}) belongs to \(\mathcal{T}^{(1)}\).
 \end{pro}
 \begin{proof}
 \begin{enumerate}[label=(\alph*)]
 \item Let us first show that \(Y_{0} \in \mathcal{T}^{(1)}\). We have \(|u-v|\ge 1\) for all \(u \neq v \in \Z^{2n}\). Hence \(B(u,\frac{1}{2}) \cap B(v,\frac{1}{2}) = \emptyset \) for \(u\neq v\). For each \(0 < \varepsilon < \frac{1}{2}\), define the operator
 \begin{equation*}
 A_{\varepsilon} = \frac{1}{|B(0,\varepsilon)|} \int_{B(0,\varepsilon)} Y_{z} dV(z) .
 \end{equation*}
 From Proposition \ref{proYz}, we have the norm continuity of the map \(z\mapsto Y_{z}\) and it implies that
 \begin{equation*}
 \lim_{\varepsilon \to 0} \norm{Y_{0}-A_{\varepsilon}} = 0.
 \end{equation*}
 This comes from the fact that
 \begin{eqnarray*}
 \norm{Y_{0}-A_{\varepsilon}} &=& \norm{\frac{1}{|B(0,\varepsilon)|} \int_{B(0,\varepsilon)} (Y_{0}-Y_{z}) ~dV(z)} \le \frac{1}{|B(0,\varepsilon)|} \int_{B(0,\varepsilon)} \norm{Y_{0}-Y_{z}} ~dV(z)
 \end{eqnarray*}
 and \(\lim_{z\to 0}\norm{Y_{z}-Y_{0}}=0\) .

The organization of this paper is as follows. In Section 2, we will give propositions in the case of the Fock space \(H^{2}(\mathbb{C}^n, d\mu)\) which are the analogue of those given by Xia in \cite{xia2015localization} in the Bergman space case of the unit ball. Later, using these propositions in Section 3, we establish the proof of Theorem \ref{XiaFock} and present a consequence.

To establish the proof of Theorem~\ref{XiaFock}, the paper first "give[s] propositions in the case of the Fock space \(H^{2}(\mathbb{C}^n, d\mu)\)" in Section 2, as analogues of results of Xia \cite{xia2015localization} for the Bergman space of the unit ball. Then, "using these propositions in Section 3," it "establish[es] the proof of Theorem~\ref{XiaFock}."

To establish the proof of the main theorem, the paper first "give[s] propositions in the case of the Fock space \(H^{2}(\mathbb{C}^n, d\mu)\)" in Section 2, as analogues of results of Xia \cite{xia2015localization} for the Bergman space of the unit ball. Then, "using these propositions in Section 3," it "establish[es] the proof of the main theorem."

Question: Let $H^{2}(\mathbb{C}^{n},d\mu)$ be the classical Fock space, where $d\mu(z)=\pi^{-n}e^{-|z|^{2}}\,dV(z)$. Let $\mathcal{WL}$ denote the class of weakly localized operators on $H^{2}(\mathbb{C}^{n},d\mu)$, and let $C^{*}(\mathcal{WL})$ be the $C^{*}$-algebra generated by $\mathcal{WL}$. Let $\mathcal{T}^{(1)}$ denote the Toeplitz algebra on this Fock space, i.e. the norm-closed algebra generated by Toeplitz operators $T_{f}$ with bounded measurable symbols $f$. Which statement holds about these two operator algebras?

They are equal: $C^{*}(\mathcal{WL})=\mathcal{T}^{(1)}$.

Choices:
B. One only has a strict inclusion $C^{*}(\mathcal{WL})\subsetneq \mathcal{T}^{(1)}$.
C. There is a containment $C^{*}(\mathcal{WL})\subseteq \mathcal{T}^{(1)}$.
D. They coincide only after taking compact-operator quotients, namely $C^{*}(\mathcal{WL})/\mathcal{K}=\mathcal{T}^{(1)}/\mathcal{K}$, but not necessarily before quotienting.
E. They have the same strong-operator closure, but $C^{*}(\mathcal{WL})$ and $\mathcal{T}^{(1)}$ need not be equal as norm-closed algebras.

The stem defines the two algebras and asks for their relationship, but it does not state or strongly hint that equality is the correct conclusion.

This is close to a direct theorem-recall item about the relation between the two algebras. However, it is not purely tautological because the options distinguish equality from weaker containment, strict inclusion, and quotient-level coincidence.

Some reasoning is required to select the strongest valid statement rather than the merely true weaker containment. Still, success depends mostly on recalling the known result rather than generating a substantial argument.

The distractors are mathematically plausible and target realistic confusions: settling for one-sided inclusion, mistaking equality for quotient equality, or confusing norm-closed and strong-operator-closed statements.

A solid theorem-recall MCQ with strong distractors and little answer leakage, though it tests recognition of a known result more than deep generative reasoning.


\begin{theorem}\label{thm-Serrin}
Let \((M^n, g)\) be a complete, connected, and noncompact  Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature. Assume that \(f\in C(\mathbb R_+;\mathbb R_+)\) satisfies
\begin{equation}\label{cond-Serrin}
\int_0^1[f(v)]^{-\frac{1}{2_*(q)-1}}\mathrm dv<+\infty.
\end{equation}
Then any positive supersolution to \eqref{Meq-gradient} must be constant.
\end{theorem}

\label{sec-introduction}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature.
Consider a class of semilinear equations with a gradient term
\begin{equation}\label{Meq-gradient}
-\Delta u = f(u)|\nabla u|^q\quad\text{in }M^n,
\end{equation}
where \(\Delta\) is the Laplace-Beltrami operator, \(\nabla\) is the gradient operator, and \(f:\mathbb R_+\to\mathbb R_+\) is continuous.
The equation \eqref{Meq-gradient} has a strong physical background. Such as a qualitative mathematical model studying the groundwater flow in a water-absorbing fissured porous rock in one spatial dimension (see e.g., \cite{BBCP2000} ), the porous media equation (see e.g., \cite{P2018}).

When \(f=0\), equation \eqref{Meq-gradient} becomes \(\Delta u=0\). Cheng-Yau \cite{CY1975} proved that any positive or bounded solution from above or below to \eqref{Meq-gradient} is a constant on Riemannian manifolds with nonnegative Ricci curvature.

For \(q=0\), the equation \eqref{Meq-gradient} becomes
\begin{equation}\label{Meq-q-0}
-\Delta u =f(u)\quad\text{in }M^n.
\end{equation}
For more general nonlinearity \(f(x,u)\), with additional assumptions imposed on \(f\), Gidas-Spruck \cite[Theorem 6.1]{GS1981} proved that solutions of equation \eqref{Meq-q-0} must be constant. In the Euclidean case, Serrin-Zou \cite{SZ2002} studied the \(m\)-Laplacian case. By focusing on \(m=2\), they proposed the subcritical condition on \(f\), that is
\[f(u)\geqslant0,\quad (p-1)f(u)-uf'(u)\geqslant0\]
for \(u>0, 1<p<\frac{2n}{n-2}\). Under the subcritical condition on \(f\), they obtained a Liouville theorem of equation \eqref{Meq-q-0} in \(\mathbb R^n\), that is if one of following condition holds:

(i) If \(n=2, 3\) and \(f\) is subcritical,

(ii) If \(n\geqslant 4\), \(f(u)\geqslant Cu\) for sufficiently large \(u\), and \(f\) is subcritical,

\noindent then equation \eqref{Meq-q-0} admits only constant solutions. For more Liouville theorems involving equation \eqref{Meq-q-0}, see \cite{L2025, L2025-1, LZ2003, MW2024, Z2025} and the references therein.

A typical case for equation \eqref{Meq-q-0} is
\begin{equation}\label{Meq-q-0-ty}
-\Delta u =u^p\quad\text{in }M^n.
\end{equation}
This equation is the famous Lane-Emden equation, which is related to the Yamabe problem. Gidas-Spruck \cite{GS1981} used differential identities to derive a Liouville theorem, namely, there is no positive solution when \(1<p<\frac{n+2}{n-2}\) on a complete Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci tensor. Grigor’yan-Sun \cite{GS2014} proved that equation \eqref{Meq-q-0-ty} admits no positive solution under condition \(\operatorname{vol}(B_R(x_0))\leqslant CR^{\frac{2p}{p-1}}\ln^{\frac{1}{p-1}}R\), corresponding with the Serrin subcritical case \(p<\frac{n}{n-2}\) in dimension \(n\), when \(p>1\) on a connected geodesically complete Riemannian manifold.

In the critical case  \(p=\frac{n+2}{n-2}\), the equation \eqref{Meq-q-0-ty} becomes
\[-\Delta u =u^{\frac{n+2}{n-2}}\quad\text{in }M^n, \]
which is the Euler-Lagrange equation of the Sobolev inequality. The critical case in \(\mathbb R^n\) has been successfully resolved, as can be seen in \cite{ CGS1989, CL1991,GNN1981}, all positive solutions must be of the form
\begin{equation}\label{solution-q-0}
u(x)=\Big(\frac{\sqrt{n(n-2)}\lambda}{1+(\lambda|x-x_0|)^2}\Big)^{\frac{n-2}{2}}.
\end{equation}
On generally Riemannian manifolds, the classification result turns to the rigidity result: the manifold must be \(\mathbb R^n\) if there exists a nonconstant solution. However, deriving the rigidity result is more challenging. Fogagnolo-Malchiodi-Mazzieri \cite{FMM2023} got the rigidity result under the assumption \(u=O(r^{-\frac{n-2}{2}})\) at infinity. It is worth mentioning that Catino-Monticelli \cite{CM2022} obtained the rigidity result and classified it with \(n=3\), while \(n\geqslant4\) either under the finite energy condition or under the decay assumption of the solution at infinity. Ciraolo-Farina-Polvara \cite{CFP2024} extended these results to \(n=4,5\) without any finite energy condition by the \(P\)-function method. Catino-Monticelli-Roncoroni \cite{CMR2023}, Ou \cite{O2025}, and Sun-Wang \cite{SW2025} extended the classification results on manifolds to \(m\)-Laplacian case.

Now, we return to the case with the gradient term. A crucial prototype of semilinear equation \eqref{Meq-gradient} is the case \(f(u)=u^p\), i.e.,
\begin{equation}\label{eq-u^p}
-\Delta u = u^p|\nabla u|^q\quad\text{in }M^n.
\end{equation}
There are three important curves of \((p,q)\) for this equation, namely the homogeneous curve \(A_0(p,q)=0\), the first critical curve \(A_1(p,q)=0\), and the second critical curve \(A_2(p,q)=0\), with
\[A_0(p,q)=p+q-1,\quad A_1(p,q)=(n-2)p+(n-1)q-n,\]
\[A_2(p,q)=(n-2)p+(n-1)q-\Big(n+\frac{2-q}{1-q}\Big),\quad0\leqslant q<1.\]
As shown in Figure \ref{fig-1}, we mark these three curves with \(A_0\), \(A_1\), and \(A_2\) respectively. We introduce
\[2_*(q)=\frac{2-q}{n-2}(n-1),\quad 2^*(q)=\frac{2-q}{n-2}(n+\frac{q}{1-q}),\]
thus the first subcritical range \(A_1(p,q)<0\) is equivalent with \(p<2_*(q)-1\), and the second subcritical range \(A_2(p,q)<0\) is equivalent with \(p<2^*(q)-1\) when \(0\leqslant q<1\).

In the special case \(p=0\), equation \eqref{eq-u^p} becomes \(-\Delta u=|\nabla u|^q\), called as the Hamilton-Jacobi equation. For this equation, Lions \cite{L1985} obtained a Liouville theorem for \(q>1\) in \(\mathbb R^n\). Bidaut-V\'{e}ron, Huidobro, and V\'{e}ron \cite{BV-GH-V2014} established the gradient estimate to \(m\)-Laplacian case and obtained some Liouville theorems on complete noncompact manifolds, which satisfy a lower bound estimate on the Ricci curvature and sectional curvature.

In the Euclidean case, there has already been a great deal of work on equation \eqref{eq-u^p}. In the first subcritical case \(p<2_*(q)-1\), any supersolution to \eqref{eq-u^p} must be constant, see \cite{BP-GM-Q2016, BV-GH-V2019, CHZ2022, CM1997, F2009, MP2001}. Those past proofs are usually composed of three cases split by the homogeneous curve \(A_0(p,q)=0\), which is avoided in our proof of Theorem \ref{thm-Serrin}. Besides, for \(q\geqslant2\) and \(p\geqslant0\), Filippucci-Pucci-Souple \cite{FPS2020} obtained that any positive bounded solution to equation \eqref{eq-u^p} is constant. Later, Bidaut-V\'{e}ron \cite{BV2021} extended this result to the \(m\)-Laplace equation for \(q\geqslant m\) and established a Liouville theorem with \(p\geqslant 0\).

Now we concentrate on the second subcritical range \(p<2^*(q)-1\) with \(0\leqslant q<1\) for equation \eqref{eq-u^p} in \(\mathbb R^n\). For \(0<q<2\) and \(p>0\), Bidaut-V\'{e}ron, Huidobro, and V\'{e}ron \cite{BV-GH-V2019} conjectured that any solution to \eqref{eq-u^p} must be constant in the second critical range. They gave a positive answer to the left of the curve \(G(p,q)=0\) in Figure \ref{fig-1}, where
\[G(p,q)=\big((n-1)^2q+n-2\big)p^2+c(q)p-nq^2,\]
\[c(q)=n(n-1)q^2-(n^2+n-1)q-n-2.\]
Besides, they also showed that the second critical curve is optimal by proving the existence of a nonconstant solution in the second supercritical range \(A_2(p,q)>0\). Using Bernstein's technique, Chang-Hu-Zhang \cite{CHZ2022} derived a Liouville theorem in \(m\)-Laplacian case, and the left of the curve \(C\) in Figure \ref{fig-1} shows their range when \(m=2\). An exciting result was obtained by Ma-Wu \cite{MW2023}. With the help of integral identities, they completely solved the conjecture in \cite{BV-GH-V2019} when \(0<q<\frac{1}{n-1}\), and extended the range \(G(p,q)<0\) in \cite{BV-GH-V2019} to \(H(p,q)<0\) when \(q\geqslant\frac{1}{n-1}\), where
\[H(p,q)=p^2+\Big(\frac{n-1}{n-2}q-\frac{n^2-3}{(n-2)^2}\Big)p+\frac{1-(n-1)q}{(n-2)^2}.\]

\begin{figure}[htbp]\small
\centering
\includegraphics[width=0.8\textwidth]{images/diagram.jpg}
\caption{some parameter ranges of equation \eqref{eq-u^p} in \(\mathbb R^n\) when \(n=5\)}
\label{fig-1}
\end{figure}

Return to the Riemannian manifold case. Sun-Xiao-Xu \cite{SXX2022} obtained some Liouville theorems of equation \eqref{eq-u^p} under the growth of the geodesical ball and some range \((p,q)\in\mathbb R^2\) on a complete, noncompact Riemannian manifold. He-Hu-Wang \cite{HHW2023} obtained that any \(C^1\) smooth positive solution of equation \eqref{eq-u^p} is a constant when \(p+q<\frac{n+3}{n-1}\) on complete noncompact Riemannian manifold with nonnegative Ricci curvature. More generally, both \cite{HHW2023} and \cite{SXX2022} studied \(m\)-Laplacian case.

Thus, research on Liouville theorems for equation \eqref{eq-u^p} on manifolds is relatively scarce, let alone equation \eqref{Meq-gradient} with general semilinear term \(f(u)\), which motivates us to study equation \eqref{Meq-gradient} on manifolds. First, we consider supersolutions to \eqref{Meq-gradient}.

\begin{definition}
We say that \(u\) is a supersolution to \eqref{Meq-gradient} if \(u\in H_{\mathrm{loc}}^1(M^n)\cap C(M^n)\) satisfies
\begin{align}\label{weak-supsol}
\int_{M^n} \nabla u\cdot\nabla\varphi\geqslant\int_{M^n}f(u)|\nabla u|^q\varphi,\quad\varphi\in C_c^\infty(M^n),~\varphi\geqslant 0.
\end{align}
\end{definition}

\begin{remark}
By an approximation argument, the test function \(\varphi\) may be chosen from \(H^1(M^n)\) with compact support.
\end{remark}

Under an integrable condition on \(f\), we obtain the following Liouville theorem by Serrin's technique.

\label{sec-introduction}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature.
Consider a class of semilinear equations with a gradient term
\begin{equation}\label{Meq-gradient}
-\Delta u = f(u)|\nabla u|^q\quad\text{in }M^n,
\end{equation}
where \(\Delta\) is the Laplace-Beltrami operator, \(\nabla\) is the gradient operator, and \(f:\mathbb R_+\to\mathbb R_+\) is continuous.
The equation \eqref{Meq-gradient} has a strong physical background. Such as a qualitative mathematical model studying the groundwater flow in a water-absorbing fissured porous rock in one spatial dimension (see e.g., \cite{BBCP2000} ), the porous media equation (see e.g., \cite{P2018}).

A typical case for equation \eqref{Meq-q-0} is
\begin{equation}\label{Meq-q-0-ty}
-\Delta u =u^p\quad\text{in }M^n.
\end{equation}
This equation is the famous Lane-Emden equation, which is related to the Yamabe problem. Gidas-Spruck \cite{GS1981} used differential identities to derive a Liouville theorem, namely, there is no positive solution when \(1<p<\frac{n+2}{n-2}\) on a complete Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci tensor. Grigor’yan-Sun \cite{GS2014} proved that equation \eqref{Meq-q-0-ty} admits no positive solution under condition \(\operatorname{vol}(B_R(x_0))\leqslant CR^{\frac{2p}{p-1}}\ln^{\frac{1}{p-1}}R\), corresponding with the Serrin subcritical case \(p<\frac{n}{n-2}\) in dimension \(n\), when \(p>1\) on a connected geodesically complete Riemannian manifold.

In the critical case \(p=\frac{n+2}{n-2}\), the equation \eqref{Meq-q-0-ty} becomes
\[-\Delta u =u^{\frac{n+2}{n-2}}\quad\text{in }M^n, \]
which is the Euler-Lagrange equation of the Sobolev inequality. The critical case in \(\mathbb R^n\) has been successfully resolved, as can be seen in \cite{ CGS1989, CL1991,GNN1981}, all positive solutions must be of the form
\begin{equation}\label{solution-q-0}
u(x)=\Big(\frac{\sqrt{n(n-2)}\lambda}{1+(\lambda|x-x_0|)^2}\Big)^{\frac{n-2}{2}}.
\end{equation}
On generally Riemannian manifolds, the classification result turns to the rigidity result: the manifold must be \(\mathbb R^n\) if there exists a nonconstant solution. However, deriving the rigidity result is more challenging. Fogagnolo-Malchiodi-Mazzieri \cite{FMM2023} got the rigidity result under the assumption \(u=O(r^{-\frac{n-2}{2}})\) at infinity. It is worth mentioning that Catino-Monticelli \cite{CM2022} obtained the rigidity result and classified it with \(n=3\), while \(n\geqslant4\) either under the finite energy condition or under the decay assumption of the solution at infinity. Ciraolo-Farina-Polvara \cite{CFP2024} extended these results to \(n=4,5\) without any finite energy condition by the \(P\)-function method. Catino-Monticelli-Roncoroni \cite{CMR2023}, Ou \cite{O2025}, and Sun-Wang \cite{SW2025} extended the classification results on manifolds to \(m\)-Laplacian case.

Return to the Riemannian manifold case. Sun-Xiao-Xu \cite{SXX2022} obtained some Liouville theorems of equation \eqref{eq-u^p} under the growth of the geodesical ball and some range \((p,q)\in\mathbb R^2\) on a complete, noncompact Riemannian manifold. He-Hu-Wang \cite{HHW2023} obtained that any \(C^1\) smooth positive solution of equation \eqref{eq-u^p} is a constant when \(p+q<\frac{n+3}{n-1}\) on complete noncompact Riemannian manifold with nonnegative Ricci curvature. More generally, both \cite{HHW2023} and \cite{SXX2022} studied \(m\)-Laplacian case.

\begin{remark}
By an approximation argument, the test function \(\varphi\) may be chosen from \(H^1(M^n)\) with compact support.
\end{remark}

Under an integrable condition on \(f\), we obtain the following Liouville theorem by Serrin's technique.

\begin{equation}\label{Meq-q-0}
-\Delta u =f(u)\quad\text{in }M^n.
\end{equation}

\begin{equation}\label{eq-u^p}
-\Delta u = u^p|\nabla u|^q\quad\text{in }M^n.
\end{equation}

In both cases, \(\mathbf E=0\) holds in \(Z^c\) by letting \(R\to\infty\). For \(x\in\mathring Z\), \(\nabla u(x)=0\) yields \(\mathbf E=0\). By the continuity of \(\mathbf E\), we immediately conclude that \(\mathbf E=0\) on \(M\). Together with Lemma \ref{critical-thm-pf}, the proof is complete.
\qedsections/intro.tex0000664000000000000000000004363015111220774013270 0ustar rootroot\section{Introduction}\label{sec-introduction}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature.
Consider a class of semilinear equations with a gradient term
\begin{equation}\label{Meq-gradient}
-\Delta u = f(u)|\nabla u|^q\quad\text{in }M^n,
\end{equation}
where \(\Delta\) is the Laplace-Beltrami operator, \(\nabla\) is the gradient operator, and \(f:\mathbb R_+\to\mathbb R_+\) is continuous.
The equation \eqref{Meq-gradient} has a strong physical background. Such as a qualitative mathematical model studying the groundwater flow in a water-absorbing fissured porous rock in one spatial dimension (see e.g., \cite{BBCP2000} ), the porous media equation (see e.g., \cite{P2018}).

Under an integrable condition on \(f\), we obtain the following Liouville theorem by Serrin's technique.

When \(\liminf_{v\to0^+}f(v)>0\), the condition \eqref{cond-Serrin} trivially holds.

\begin{corollary}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature. Assume that \(f\in C(\mathbb R_+;\mathbb R_+)\) satisfies
\[\liminf_{v\to0^+}f(v)>0.\]
Then any positive supersolution to \eqref{Meq-gradient} must be constant.
\end{corollary}

\begin{corollary}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature. Then any positive supersolution to \eqref{eq-u^p} must be constant if \(p<2_*(q)-1\),
\end{corollary}

\begin{theorem}\label{thm-subcritical}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature, and \(0<q<\frac{1}{n-1}\). Assume that \(f\in C^1(\mathbb R_+;\mathbb R_+)\) satisfies the following conditions:
\begin{equation}\label{cond-sub-1}
f(u)\geqslant u^{\beta_0},~u>N,\quad\text{for some }\beta_0\in(1-q,2^*(q)-1)\text{ and }N>0,
\end{equation}
\begin{equation}
uf'(u)\leqslant\alpha_0 f(u),\quad\text{for some }\alpha_0\in[\beta_0,2^*(q)-1).
\end{equation}
Then any positive \(C^3\) solution to \eqref{Meq-gradient} must be constant if \(p<2^*(q)-1\).
\end{theorem}

\begin{corollary}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature, and \(0<q<\frac{1}{n-1}\). Then any positive \(C^3\) solution to \eqref{eq-u^p} must be constant.
\end{corollary}

\begin{theorem}\label{thm-critical}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\in\{3,4,5\}\) with nonnegative Ricci curvature, \(0<q<\frac{1}{n-1}\) and \(f(u)=u^{2^*(q)-1}\) in \eqref{Meq-gradient}. Let \(u\in C^3(M)\) be a positive solution to \eqref{Meq-gradient}, then either \(u\) is constant, or \((M^n, g)\) is isometric to \(\mathbb R^n\) and \(u\) is the form given by \eqref{solution}.
\end{theorem}

\begin{equation}\label{cond-Serrin}
\int_0^1[f(v)]^{-\frac{1}{2_*(q)-1}}\mathrm dv<+\infty.
\end{equation}

\begin{equation}\label{eq-u^p}
-\Delta u = u^p|\nabla u|^q\quad\text{in }M^n.
\end{equation}

\begin{equation}\label{solution}
U_{\lambda,x_0}(x):=\Big(\frac{(n+\frac{q}{1-q})^{\frac{1}{2-q}}(n-2)^{\frac{1-q}{2-q}}\lambda}{1+(\lambda|x-x_0|)^{\frac{2-q}{1-q}}}\Big)^{\frac{(n-2)(1-q)}{2-q}},\quad\lambda>0,~x_0\in\mathbb R^n.
\end{equation}

When \(\liminf_{v\to0^+}f(v)>0\), the condition \eqref{cond-Serrin} trivially holds.

\begin{corollary}
Let \((M^n, g)\) be a complete, connected, and noncompact  Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature. Assume that \(f\in C(\mathbb R_+;\mathbb R_+)\) satisfies
\[\liminf_{v\to0^+}f(v)>0.\]
Then any positive supersolution to \eqref{Meq-gradient} must be constant.
\end{corollary}

Theorem \ref{thm-Serrin} recovers the first subcritical range \(p<2_*(q)-1\) by taking \(f(u)=u^p\).

\begin{corollary}
Let \((M^n, g)\) be a complete, connected, and noncompact  Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature. Then any positive supersolution to \eqref{eq-u^p} must be constant if \(p<2_*(q)-1\),
\end{corollary}

For solutions to \eqref{Meq-gradient}, we concentrate on the cases \(0<q<\frac{1}{n-1}\). We establish the following Liouville theorem. Our proof also works for the case \(q=0\), which recovers the \(m=2\) case in Serrin-Zou \cite{SZ2002} in \(M=\mathbb R^n\).

\begin{theorem}\label{thm-subcritical}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature, and \(0<q<\frac{1}{n-1}\). Assume that \(f\in C^1(\mathbb R_+;\mathbb R_+)\) satisfies the following conditions:
\begin{equation}\label{cond-sub-1}
f(u)\geqslant u^{\beta_0},~u>N,\quad\text{for some }\beta_0\in(1-q,2^*(q)-1)\text{ and }N>0,
\end{equation}
\begin{equation}
uf'(u)\leqslant\alpha_0 f(u),\quad\text{for some }\alpha_0\in[\beta_0,2^*(q)-1).
\end{equation}
Then any positive \(C^3\) solution to \eqref{Meq-gradient} must be constant if \(p<2^*(q)-1\).
\end{theorem}

Theorem \ref{thm-subcritical} recovers the second subcritical range \(p<2^*(q)-1\) by taking \(f(u)=u^p\).

\begin{corollary}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature, and \(0<q<\frac{1}{n-1}\). Then any positive \(C^3\) solution to \eqref{eq-u^p} must be constant.
\end{corollary}

By taking \(M=\mathbb R^n\), this corollary recovers the case \(0<q<\frac{1}{n-1}\) in \cite[Theorem 1.3]{MW2023}, where Ma and Wu solved the conjectured raised in \cite{BV-GH-V2019}.

Lastly, we consider the second critical case \(f(u) = u^{2^*(q)-1}\). Bidaut-V\'{e}ron, Huidobro, and V\'{e}ron \cite[Theorem D]{BV-GH-V2019} found an explicit 1-parameter family of positive radial solutions in \(\mathbb R^n\)
\[u_c(r)=c(Kc^{\frac{(2-q)^2}{(n-2)(1-q)}}+r^{\frac{2-q}{1-q}})^{-\frac{(n-2)(1-q)}{2-q}},\quad c>0,~K=K(n,q)>0.\]
By inserting them into equation \eqref{Meq-gradient} with \(f(u) = u^{2^*(q)-1}\), we rewrite the solutions as
\begin{equation}\label{solution}
U_{\lambda,x_0}(x):=\Big(\frac{(n+\frac{q}{1-q})^{\frac{1}{2-q}}(n-2)^{\frac{1-q}{2-q}}\lambda}{1+(\lambda|x-x_0|)^{\frac{2-q}{1-q}}}\Big)^{\frac{(n-2)(1-q)}{2-q}},\quad\lambda>0,~x_0\in\mathbb R^n.
\end{equation}
For \(q=0\), the solution \eqref{solution} is equivalent to \eqref{solution-q-0}. The rigidity of manifolds has been rigorously established in low-dimensional settings for \(q=0\), but for \(0<q<1\), the classification of equation \eqref{Meq-gradient} in the second critical case \(f(u) = u^{2^*(q)-1}\) on \(\mathbb R^n\) is open, let alone the rigidity of manifolds.

With the help of the invariant tensor \(\operatorname{\mathbf E}\) and its corresponding pseudo-invariant function \(l\) (i.e., \(\nabla l\) is only composed of terms involving the invariant tensor \(\operatorname{\mathbf E}\)), we obtain the rigidity for the ambient manifold and classify positive solutions of the second critical equation, \eqref{Meq-gradient} with \(f(u)=u^{2^*(q)-1}\), without any assumption in a low-dimensional setting. To our knowledge, this is the first rigidity result in the second critical case of equations with gradient terms.

\begin{theorem}\label{thm-critical}
Let \((M^n, g)\) be a complete, connected, and noncompact Riemannian manifold of dimension \(n\in\{3,4,5\}\) with nonnegative Ricci curvature, \(0<q<\frac{1}{n-1}\) and \(f(u)=u^{2^*(q)-1}\) in \eqref{Meq-gradient}. Let \(u\in C^3(M)\) be a positive solution to \eqref{Meq-gradient}, then either \(u\) is constant, or \((M^n, g)\) is isometric to \(\mathbb R^n\) and \(u\) is the form given by \eqref{solution}.
\end{theorem}

Differential identities are crucial in studying properties of an elliptic equation. However, establishing differential identities usually relies on some geometric background, such as the Bochner formula. Thus, finding a proper differential identity may be challenging for a sophisticated equation. In \cite{MW2024}, Ma-Wu (the third author of this paper) developed the invariant tensor technique, a powerful method of finding vector fields and differential identities without any geometric instructions, to study semilinear equations \(-\Delta u=f(u)\) on \(M^n\), which gave a new proof of Bidaut-V\'{e}ron and V\'{e}ron's rigidity result in \cite{BV-V1991}. Using the new technique, Ma-Ou-Wu \cite{MOW2025} reconstructed Jerison-Lee's identity to answer the question raised by Jerison-Lee \cite{JL1988}. Besides, Ma-Ou-Wu \cite{MOW2023} verified the subcritical rigidity conjecture raised by Wang \cite{W2022} via this new technique with sophisticated computations. Recently, Guo-Zhang \cite{GZ2025} utilised the invariant tensor technique to give a new proof of semilinear equations again. The invariant tensor technique is also widely applied to classify quasilinear cases or involving gradient terms, see \cite{LWY2025, Z2024}.

Although we concentrate only on the case \(0<q<\frac{1}{n-1}\), we make the first attempt to successfully extend the invariant tensor technique to equations with gradient terms. The pseudo-invariant function \(l\), which plays a crucial role in the classification process, is found by differential invariance. However, it becomes difficult to find \(l\) through variable transformations as previously done. The authors have extended the invariant tensor technique to the equation \eqref{Meq-gradient} to \(m\)-Laplace in \cite{WZ}, where a wider range of \(q\) is studied with more sophisticated computations.

Throughout this paper, we assume constant \(R>1\), \(\varepsilon>0\) small enough, and \(C>0\) are constants independent of \(R\). Let \(B_R\) be the geodesic ball centered at some fixed point with radius \(R\). Unless otherwise specified, we employ the summation convention for repeated indices from \(1\) to \(n\). In proofs involving manifolds, we choose a local frame, \(\varphi_i\) denotes the covariant derivative of the function \(\varphi\). Let \(\eta\) be a smooth cutoff function supported in \(B_{2R}\) satisfying \(\eta\equiv 1\) in \(B_R\), \(0\leqslant\eta\leqslant 1\), and \(|\nabla\eta|\leqslant CR^{-1}\).

This paper is structured as follows. In Section 2, we prove Theorem \ref{thm-Serrin} by using Serrin's technique with a delicate test function \(b(u)\eta^n\) and a three-term Young's inequality. In Section 3, using the invariant tensor technique, we establish a key differential identity and necessary propositions to prove Theorem \ref{thm-subcritical}. With the help of this differential identity, the proof of Theorem \ref{thm-critical}, classification of solutions in the critical case, is finished in Section 4.sections/Serrin.tex0000664000000000000000000000411015111220774013365 0ustar  rootroot

To prove Theorem~\ref{thm-Serrin}, the paper uses “Serrin's technique with a delicate test function $b(u)\eta^n$ and a three-term Young's inequality” (carried out in Section 2).

To prove the main theorem, the paper uses “Serrin's technique with a delicate test function $b(u)\eta^n$ and a three-term Young's inequality” (carried out later).

\label{thm-Serrin}
Let \((M^n, g)\) be a complete, connected, and noncompact  Riemannian manifold of dimension \(n\geqslant3\) with nonnegative Ricci curvature. Assume that \(f\in C(\mathbb R_+;\mathbb R_+)\) satisfies
\begin{equation}\label{cond-Serrin}
\int_0^1[f(v)]^{-\frac{1}{2_*(q)-1}}\mathrm dv<+\infty.
\end{equation}
Then any positive supersolution to \begin{equation}\label{Meq-gradient}
-\Delta u = f(u)|\nabla u|^q\quad\text{in }M^n,
\end{equation}
must be constant.,

Question:
Let \((M^n,g)\) be a complete, connected, noncompact Riemannian manifold of dimension \(n\ge 3\) with nonnegative Ricci curvature. Let \(f\in C(\mathbb R_+;\mathbb R_+)\) satisfy
\[
\int_0^1 [f(v)]^{-\frac{1}{2_*(q)-1}}\,dv<\infty,
\]
and consider the equation with gradient term
\[
-\Delta u=f(u)|\nabla u|^q\quad \text{in }M^n,
\]
where \(\Delta\) is the Laplace--Beltrami operator and \(\nabla u\) is the gradient of \(u\). If \(u\) is a positive supersolution, meaning \(u>0\) and \(-\Delta u\ge f(u)|\nabla u|^q\) on \(M^n\), which conclusion holds?

Every positive supersolution \(u\) is constant on \(M^n\).

Choices:
B. Every positive supersolution \(u\) is constant on \(M^n\) provided additionally that \(0<q<\frac{1}{n-1}\); without this restriction on \(q\), the conclusion need not hold.
C. Every bounded positive supersolution \(u\) is constant on \(M^n\).
D. There exists a constant \(C=C(M,n,q,f)>0\) such that every positive supersolution \(u\) satisfies \(u\equiv C\) on \(M^n\).
E. Every positive \(C^1\) solution \(u\) of \(-\Delta u=f(u)|\nabla u|^q\) on \(M^n\) is constant, but the same conclusion is not asserted for positive supersolutions satisfying only \(-\Delta u\ge f(u)|\nabla u|^q\).

no subcritical range restriction on q in the Serrin argument

dropped the full class of positive supersolutions to the smaller class of bounded positive supersolutions

constancy depends on the individual supersolution, not a universal constant determined only by the data

Serrin test-function method is stated for supersolutions, not merely for exact C^1 solutions

The stem does not explicitly reveal the answer. It states the hypotheses and asks for the resulting conclusion, without directly embedding the exact claim of choice A.

This is close to a theorem-recall item: the correct option is essentially the Liouville-type conclusion under the stated hypotheses. However, the alternatives introduce subtle variants in quantifiers, regularity, and parameter restrictions, so it is not a pure verbatim restatement.

Some reasoning is needed to distinguish the exact theorem statement from weaker true or plausibly false variants, but the item mainly tests recognition/recall rather than substantial derivation or synthesis.

The distractors are mathematically plausible and target common errors: adding unnecessary restrictions on q, weakening the conclusion to bounded supersolutions, confusing constancy with a universal constant, and restricting from supersolutions to exact solutions.

A solid MCQ with strong distractors and little answer leakage, but it remains fairly close to direct theorem recall rather than a deeply generative reasoning task.


The fourth stable homotopy group of spheres is trivial, meaning that $\pi_{n+4}(S^n)=0$ for all $n>5$. In contrast to this, it is a theorem that $S^m$ has no trivial higher homotopy groups when $m\in\{2,3,4,5\}$, as we will briefly explain.

Curtis proved that $\pi_n(S^4)\neq 0$ for all $n\geq 4$ \cite{Cur69}. Curtis also proved that $\pi_n(S^2)\neq 0$ for all $n\not\equiv 1\mod 8$. These same results were obtained (via different methods) by Mimura, Mori, and Oda \cite{MMO75}. The proof that $\pi_n(S^5)\neq 0$ for all $n\geq 5$ was given by Mori \cite{Mor75} and Mahowald \cite{Mah75,Mah82}.

Since $\pi_n(S^2)\cong\pi_n(S^3)$ for all $n\geq 3$, the remaining case was $\pi_n(S^3)$ with $n\equiv 1\mod 8$. This last case was proved by Gray \cite{Gra84}, and later by Ivanov, Mikhailov, and Wu \cite{IMW16} using different methods. In \emph{op.~cit.}, the authors note that \cite[Theorem 5.2(ii)]{Tod66} implies that
\[\mathbb{Z}/p\subseteq\pi_{2(p-1)k+1}(S^3)\]
whenever $\gcd(p,k)=1$ \cite[p.~342, Equation (B)]{IMW16}. It follows that if every positive integer $n$ admits an odd prime $p$ and an integer $k$ such that $\gcd(p,k)=1$ and $4n=(p-1)k$, then $\pi_n(S^3)\neq 0$ for all $n\equiv 1\mod 8$. This leads one to the following definition.

\begin{defn}
    Let $n$ be an integer. A \emph{Toda prime} of $n$ is an odd prime $p$ such that $p-1\mid 4n$ and $\gcd(p,\frac{4n}{p-1})=1$. Denote the set of Toda primes of $n$ by $T(n)$ (see \cref{table:toda sets}), and let $t(n):=|T(n)|$ (see \cref{fig:tn}).
\end{defn}

\begin{table}[t]
\centering
\caption{Sets of Toda primes}\label{table:toda sets}
\begin{tabular}{|r|l|}
\hline
$n$ & $T(n)$\\
\hline
1 & $3,5$\\
2 & $3,5$\\
3 & $5,7,13$\\
4 & $3,5,17$\\
5 & $3,11$\\
6 & $5,7,13$\\
7 & $3,5,29$\\
8 & $3,5,17$\\
9 & $5,7,13,19,37$\\
10 & $3,11,41$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|r|l|}
\hline
$n$ & $T(n)$\\
\hline
11 & $3,5,23$\\
12 & $5,7,13,17$\\
13 & $3,5,53$\\
14 & $3,5,29$\\
15 & $7,11,13,31,61$\\
16 & $3,5,17$\\
17 & $3,5$\\
18 & $5,7,13,19,37,73$\\
19 & $3,5$\\
20 & $3,11,17,41$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|r|l|}
\hline
$n$ & $T(n)$\\
\hline
21 & $5,13,29,43$\\
22 & $3,5,23,89$\\
23 & $3,5,47$\\
24 & $5,7,13,17,97$\\
25 & $3,11,101$\\
26 & $3,5,53$\\
27 & $5,7,13,19,37,109$\\
28 & $3,5,17,29,113$\\
29 & $3,5,59$\\
30 & $7,11,13,31,41,61$\\
\hline
\end{tabular}
\end{table}

If every positive integer has a Toda prime, then one can greatly simplify the proof that $\pi_n(S^3)\neq 0$ for $n\geq 3$. This was asked on MathOverflow \cite{MO} (and attributed to Roman Mikhailov) several years ago.

\begin{ques}[Mikhailov]\label{ques:at least one}
    Does every positive integer have a Toda prime?
\end{ques}

In fact, it appears that every positive integer has at least two Toda primes.

\begin{conj}\label{conj:at least two}
    If $n$ is a positive integer, then $t(n)\geq 2$.
\end{conj}

We tried to answer \cref{ques:at least one} in the affirmative, but our approach hits a snag. To turn our failed attempt into a theorem, we adopt the time-tested tradition of stating our snag as a conjecture (\cref{conj:main}). We will give some heuristic evidence for this conjecture in \cref{sec:heuristic}.

\begin{conj}\label{conj:main}
    Let $n$ be an odd, square-free multiple of 3. Assume that there exists $p\in\{5,7,13\}$ such that $p\mid n$, and that $r\nmid n$ for $r\in\{5,7,13\}-\{p\}$. Finally, assume there exists $q\in T(3p)-\{5,7,13\}$ such that $q\nmid n$. Then $t(n)\geq 4$.
\end{conj}

Since $\pi_n(S^2)\cong\pi_n(S^3)$ for all $n\geq 3$, the remaining case was $\pi_n(S^3)$ with $n\equiv 1\mod 8$. This last case was proved by Gray \cite{Gra84}, and later by Ivanov, Mikhailov, and Wu \cite{IMW16} using different methods. In \emph{op.~cit.}, the authors note that \cite[Theorem 5.2(ii)]{Tod66} implies that
\[\mathbb{Z}/p\subseteq\pi_{2(p-1)k+1}(S^3)\]
whenever $\gcd(p,k)=1$ \cite[p.~342, Equation (B)]{IMW16}. It follows that if every positive integer $n$ admits an odd prime $p$ and an integer $k$ such that $\gcd(p,k)=1$ and $4n=(p-1)k$, then $\pi_n(S^3)\neq 0$ for all $n\equiv 1\mod 8$. This leads one to the following definition.

\begin{defn}
 Let $n$ be an integer. A \emph{Toda prime} of $n$ is an odd prime $p$ such that $p-1\mid 4n$ and $\gcd(p,\frac{4n}{p-1})=1$. Denote the set of Toda primes of $n$ by $T(n)$ (see \cref{table:toda sets}), and let $t(n):=|T(n)|$ (see \cref{fig:tn}).
\end{defn}

If every positive integer has a Toda prime, then one can greatly simplify the proof that $\pi_n(S^3)\neq 0$ for $n\geq 3$. This was asked on MathOverflow \cite{MO} (and attributed to Roman Mikhailov) several years ago.

\begin{conj}\label{conj:at least two}
 If $n$ is a positive integer, then $t(n)\geq 2$.
\end{conj}

We tried to answer \cref{ques:at least one} in the affirmative, but our approach hits a snag. To turn our failed attempt into a theorem, we adopt the time-tested tradition of stating our snag as a conjecture (\cref{conj:main}). We will give some heuristic evidence for this conjecture in \cref{sec:heuristic}.

\begin{conj}\label{conj:main}
 Let $n$ be an odd, square-free multiple of 3. Assume that there exists $p\in\{5,7,13\}$ such that $p\mid n$, and that $r\nmid n$ for $r\in\{5,7,13\}-\{p\}$. Finally, assume there exists $q\in T(3p)-\{5,7,13\}$ such that $q\nmid n$. Then $t(n)\geq 4$.
\end{conj}

\begin{conj}\label{conj:main}
    Let $n$ be an odd, square-free multiple of 3. Assume that there exists $p\in\{5,7,13\}$ such that $p\mid n$, and that $r\nmid n$ for $r\in\{5,7,13\}-\{p\}$. Finally, assume there exists $q\in T(3p)-\{5,7,13\}$ such that $q\nmid n$. Then $t(n)\geq 4$.
\end{conj}

\begin{figure}[p]
    \includegraphics[width=.75\linewidth]{tn.png}
    \caption{$t(n)$ for $n\leq 100000$}\label{fig:tn}
\end{figure}

\begin{ques}[Mikhailov]\label{ques:at least one}
    Does every positive integer have a Toda prime?
\end{ques}

\begin{table}[t]
\centering
\caption{Sets of Toda primes}\label{table:toda sets}
\begin{tabular}{|r|l|}
\hline
$n$ & $T(n)$\\
\hline
1 & $3,5$\\
2 & $3,5$\\
3 & $5,7,13$\\
4 & $3,5,17$\\
5 & $3,11$\\
6 & $5,7,13$\\
7 & $3,5,29$\\
8 & $3,5,17$\\
9 & $5,7,13,19,37$\\
10 & $3,11,41$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|r|l|}
\hline
$n$ & $T(n)$\\
\hline
11 & $3,5,23$\\
12 & $5,7,13,17$\\
13 & $3,5,53$\\
14 & $3,5,29$\\
15 & $7,11,13,31,61$\\
16 & $3,5,17$\\
17 & $3,5$\\
18 & $5,7,13,19,37,73$\\
19 & $3,5$\\
20 & $3,11,17,41$\\
\hline
\end{tabular}
\quad
\begin{tabular}{|r|l|}
\hline
$n$ & $T(n)$\\
\hline
21 & $5,13,29,43$\\
22 & $3,5,23,89$\\
23 & $3,5,47$\\
24 & $5,7,13,17,97$\\
25 & $3,11,101$\\
26 & $3,5,53$\\
27 & $5,7,13,19,37,109$\\
28 & $3,5,17,29,113$\\
29 & $3,5,59$\\
30 & $7,11,13,31,41,61$\\
\hline
\end{tabular}
\end{table}

\begin{conj}\label{conj:main}
 Let $n$ be an odd, square-free multiple of 3. Assume that there exists $p\in\{5,7,13\}$ such that $p\mid n$, and that $r\nmid n$ for $r\in\{5,7,13\}-\{p\}$. Finally, assume there exists $q\in T(3p)-\{5,7,13\}$ such that $q\nmid n$. Then $t(n)\geq 4$.
\end{conj}

In \cref{sec:lemmas}, we state and prove a few simple lemmas. We prove \cref{thm:main} in \cref{sec:proof} by inducting on the number of odd prime factors of $n$. Essentially all of the real work happens in \cref{lem:t(n)=3}. We conclude with \cref{sec:observations}, where we pose a couple questions that arose while working on this project.

\begin{lem}\label{lem:not 5 or not 3 and 5}
 Let $p\in\{3,5\}$. If $p\nmid n$, then $p\in T(n)$. In particular:
 \begin{itemize}
 \item If $5\nmid n$, then $t(n)\geq 1$.
 \item If $3,5\nmid n$, then $t(n)\geq 2$.
 \end{itemize}
\end{lem}
\begin{proof}
 Let $p\in\{3,5\}$. Then $p-1\mid 4n$, and $\gcd(4n,p)=1$ by assumption. Thus $\gcd(\frac{4n}{p-1},p)=1$, so $p\in T(n)$.
\end{proof}

\begin{lem}\label{lem:t(n)=3}
 Assume \cref{conj:main}. Let $n$ be an odd, square-free multiple of 3. Then $t(n)\geq 3$. Moreover, $t(n)=3$ if and only if $T(n)=\{5,7,13\}$.
\end{lem}
\begin{proof}
 We will induct on $\omega(n)$. Our base cases will consist of $\omega(n)\leq 4$. Note that if $p\nmid n$ for each $p\in\{5,7,13\}$, then $T(n)\supseteq\{5,7,13\}$. In particular, we may restrict our attention to multiples of these three primes. Moreover, $t(ap)\geq t(a)$ for any prime $p\not\in T(a)$ by \cref{cor:multiply by prime}, so we may assume that every prime factor $p\mid n$ is a Toda prime of some divisor of $n$.
 \begin{itemize}
 \item The case of $\omega(n)=1$ is just the calculation $T(3)=\{5,7,13\}$.
 \item For $\omega(n)=2$, we just need to compute $t(15)=t(39)=5$ and $t(21)=4$.
 \item For $\omega(n)=3$, we first compute $t(3\cdot 5\cdot 7)=9$ and $t(3\cdot 5\cdot 13)=t(3\cdot 7\cdot 13)=8$. It remains to compute, for each $p\in\{5,7,13\}$, the Toda primes of $3pq$ for each $q\in T(3p)$. Using the code provided in \cref{sec:code}, we find that $t(3pq)\geq 4$ for all such $p,q$.
 \item For $\omega(n)=4$, we first compute $t(3\cdot 5\cdot 7\cdot 13)=16$. For the remaining computations in this case, we use the code in \cref{sec:code}.
 \begin{itemize}
 \item If $\{p,q\}\subseteq\{5,7,13\}$ and $r\in T(3pq)$, then $t(3pqr)\geq 9$.
 \item If $p\in\{5,7,13\}$ and $\{q,r\}\subseteq T(3p)$, then $t(3pqr)\geq 7$.
 \item If $p\in\{5,7,13\}$, $q\in T(3p)$, and $r\in T(3pq)$, then $t(3pqr)\geq 5$.
 \end{itemize}
 \end{itemize}

\begin{cor}\label{prop:divisible by 3}
 Assume \cref{conj:main}. If $3\mid n$, then $t(n)\geq 3$.
\end{cor}
\begin{proof}
 By \cref{cor:multiply by divisor}, we may assume that $n$ is square-free. By \cref{cor:odd}, we may further assume that $n$ is odd. The result now follows from \cref{lem:t(n)=3}.
\end{proof}

Firstly, if there exists $q'\in T(3p)-\{5,7,13,q\}$ such that $q'\nmid n$, then $\{5,7,13,q,q'\}-\{p\}\subseteq T(n)$, and we are done. In fact, the Toda primes of $n$ are precisely those primes among $\{2d+1:d\mid 2n\}$ that are not factors of $n$. Thus if
\begin{equation}\label{eq:heuristic}
\{2d+1\text{ prime}:d\mid 2n\}-(\Omega(n)\cup\{5,7,13,q\})
\end{equation}
is non-empty, then $t(n)\geq 4$. Our heuristic for \cref{conj:main} is that the set $\{2d+1:d\mid 2n\}$ consists of $2^{\omega(n)+1}$ elements, while $\Omega(n)\cup\{5,7,13,q\}$ consists of $\omega(n)+3$ elements.

\begin{prop}\label{prop:t(p)}
Assume $p\geq 7$ is a prime. Let $\vphi$ denote the totient function. If $\vphi(x)=4p$ for some integer $x$, then $T(p)=\{3,5,2p+1\}$ or $\{3,5,4p+1\}$. Otherwise, $T(p)=\{3,5\}$.
\end{prop}
\begin{proof}
 One can directly check that $3,5\in T(p)$ for all primes greater than 5. Now by Euler's product formula, we have $\vphi(x)=p_1^{e_1-1}(p_1-1)\cdots p_m^{e_m-1}(p_m-1)$, where $x=\prod_{i=1}^m p_i^{e_i}$ is the prime factorization of $x$. It follows that there exists $x$ such that $\vphi(x)=4p$ if and only if one of the following cases holds:
 \begin{enumerate}[(i)]
 \item $x=2^2\cdot q$, where $q$ is an odd prime such that $q-1=2p$. In this case, $q$ is a Toda prime of $p$ with $\frac{4p}{q-1}=2$.
 \item $x=2^r\cdot 3\cdot q$, where $r\in\{0,1\}$ and $q$ is an odd prime such that $q-1=2p$. In this case, $q$ is a Toda prime of $p$ with $\frac{4p}{q-1}=2$.
 \item $x=2^r\cdot q$, where $r\in\{0,1\}$ and $q$ is an odd prime such that $q-1=4p$. In this case, $q$ is a Toda prime of $p$ with $\frac{4p}{q-1}=1$.
 \item $x=2^r\cdot 5^2$, where $r\in\{0,1\}$ (in which case $p=5$). This case is not relevant for this lemma, as we have assumed $p\geq 7$.
 \end{enumerate}
 It remains to show that no other primes can be the Toda prime of $p$. To this end, let $q>5$ be a Toda prime of $p$. Then $q-1\mid 4p$, so we either have $q-1=4p$ or $q-1=2p$ (as $q-1$ is even and $p$ is odd). The existence of such a $q$ gives us a solution to $\vphi(x)=4p$ as outlined in cases (i), (ii), and (iii).

\begin{lem}\label{lem:strategy for denoms}
 Let $d$ be a Bernoulli denominator with $F(d)=4a$ for some integer $a$. If $\{2pi+1:i\mid 2a\}$ contains a prime number for each $p\in T(a)$, then \cref{conj:general bernoulli} \eqref{conj:toda for bernoulli} holds for this Bernoulli denominator.
\end{lem}
\begin{proof}
 We know that $p-1\mid 4a$ with $\gcd(\frac{4a}{p-1},p)=1$ for all $p\in T(a)$. Thus $p-1\mid 4am$, and we have $\gcd(\frac{4am}{p-1},p)=1$ if and only if $p\nmid m$. It therefore suffices to show that if $p\mid m$ for some $p\in T(a)$, then $D_{4am}>D_{4a}$.

\begin{conj}\label{conj:main}
    Let $n$ be an odd, square-free multiple of 3. Assume that there exists $p\in\{5,7,13\}$ such that $p\mid n$, and that $r\nmid n$ for $r\in\{5,7,13\}-\{p\}$. Finally, assume there exists $q\in T(3p)-\{5,7,13\}$ such that $q\nmid n$. Then $t(n)\geq 4$.
\end{conj}

\begin{lem}\label{lem:t(n)=3}
    Assume \cref{conj:main}. Let $n$ be an odd, square-free multiple of 3. Then $t(n)\geq 3$. Moreover, $t(n)=3$ if and only if $T(n)=\{5,7,13\}$.
\end{lem}

\begin{thm}\label{thm:main}
    Assume \cref{conj:main}. If $n$ is a positive integer, then $t(n)\geq 1$. If $5\nmid n$, then $t(n)\geq 2$. If $3\mid n$, then $t(n)\geq 3$.
\end{thm}

In \cref{sec:lemmas}, we state and prove a few simple lemmas. We prove \cref{thm:main} in \cref{sec:proof} by inducting on the number of odd prime factors of $n$. Essentially all of the real work happens in \cref{lem:t(n)=3}. We conclude with \cref{sec:observations}, where we pose a couple questions that arose while working on this project.

\subsection*{Acknowledgements}
We thank Nick Andersen, Pace Nielsen, and Kyle Pratt for helpful conversations. The author was partially supported by the NSF (DMS-2502365) and the Simons Foundation.

\label{thm:main}
Assume
\begin{conj}\label{conj:main}
    Let $n$ be an odd, square-free multiple of 3. Assume that there exists $p\in\{5,7,13\}$ such that $p\mid n$, and that $r\nmid n$ for $r\in\{5,7,13\}-\{p\}$. Finally, assume there exists $q\in T(3p)-\{5,7,13\}$ such that $q\nmid n$. Then $t(n)\geq 4$.
\end{conj}
If $n$ is a positive integer, then $t(n)\geq 1$. If $5\nmid n$, then $t(n)\geq 2$. If $3\mid n$, then $t(n)\geq 3$.,

Question: For a positive integer m, let T(m) be the set of odd primes p such that p-1 divides 4m and \(\gcd\!\left(p,\frac{4m}{p-1}\right)=1\), and let \(t(m)=|T(m)|\). Assume the following statement is true: whenever m is an odd, square-free multiple of 3, there exists some \(p\in\{5,7,13\}\) with \(p\mid m\), no prime \(r\in\{5,7,13\}\setminus\{p\}\) divides m, and there exists \(q\in T(3p)\setminus\{5,7,13\}\) such that \(q\nmid m\), then \(t(m)\ge 4\). Under this assumption, which statement about a positive integer n holds?

For every positive integer n, \(t(n)\ge 1\). In addition, if \(5\nmid n\), then \(t(n)\ge 2\); and if \(3\mid n\), then \(t(n)\ge 3\).

Choices:
B. For every positive integer n, \(t(n)\ge 1\). In addition, if \(5\nmid n\), then \(t(n)\ge 2\); and if \(3\nmid n\), then \(t(n)\ge 3\).
C. For every positive integer n, \(t(n)\ge 1\). In addition, if \(5\nmid n\), then \(t(n)\ge 2\).
D. For every positive integer n, \(t(n)\ge 2\). In addition, if \(5\nmid n\), then \(t(n)\ge 3\); and if \(3\mid n\), then \(t(n)\ge 4\).
E. For every positive integer n, \(t(n)\ge 1\). In addition, if \(3\nmid n\), then \(t(n)\ge 2\); and if \(3\mid n\), then \(t(n)\ge 3\).

divisibility condition on 3 in the final clause

dropped the strongest conclusion \(3\mid n\Rightarrow t(n)\ge 3\)

replaced the specific hypothesis \(5\nmid n\) by the unrelated \(3\nmid n\)

The stem gives a technical assumed implication about special integers m, but it does not explicitly state the broader conclusion in choice A. There is no direct wording overlap that reveals the correct option.

The question is not a restatement of the assumption. It asks the test-taker to infer a global conclusion about t(n) from a conditional statement on a restricted class of m, so the correct answer is not tautologically built into the stem.

Selecting the correct option requires distinguishing between a strongest valid conclusion, a weaker true-looking alternative, and several subtle overgeneralizations or condition swaps. This creates real pressure to reason about how the assumption propagates to statements about arbitrary n.

The distractors are plausible and target common failure modes: reversing a divisibility condition, choosing a weaker-but-safe statement, and overstrengthening all bounds. They are distinct and mathematically aligned with likely errors.

A strong MCQ on these criteria: it avoids answer leakage, is clearly non-tautological, and uses high-quality distractors that force genuine comparative reasoning, though the stem is quite technically dense.