[proofplan] We first absorb the bounded zeroth-order term into an exponential weight, reducing the inequality to one with a non-positive zeroth-order coefficient. Since multiplication by a positive factor preserves the zero set, it is enough to prove the vanishing conclusion for the weighted function. The local strong parabolic maximum principle, applied in time-dependent coordinate charts, propagates a zero at positive time backward through a parabolic neighbourhood. Compactness and connectedness then let this local propagation reach every point of $M$ at every time not exceeding $t_0$. [/proofplan] [step:Absorb the bounded zeroth-order term by an exponential weight] Since $a$ is bounded on $M \times [0,T]$, define \begin{align*} A := \sup_{(x,t) \in M \times [0,T]} |a(x,t)| < \infty. \end{align*} Define the smooth map $v: M \times [0,T] \to \mathbb{R}$ by sending each $(x,t)$ to $e^{-At}u(x,t)$. Then $v$ is nonnegative on $M \times [0,T]$, and $v(x_0,t_0)=0$. Since $e^{-At}$ is independent of $x$, the [Laplace-Beltrami operator](/page/Laplace-Beltrami%20Operator) satisfies \begin{align*} \Delta_{g(t)}v(x,t)=e^{-At}\Delta_{g(t)}u(x,t) \end{align*} for every $(x,t) \in M \times [0,T]$. Also, the product rule gives \begin{align*} \partial_t v(x,t)=e^{-At}\partial_t u(x,t)-Ae^{-At}u(x,t). \end{align*} Using the assumed differential inequality for $u$, we obtain \begin{align*} \partial_t v(x,t)\geq e^{-At}\Delta_{g(t)}u(x,t)+e^{-At}a(x,t)u(x,t)-Ae^{-At}u(x,t). \end{align*} Substituting the definition of $v$ and the spatial Laplacian identity gives \begin{align*} \partial_t v(x,t)\geq \Delta_{g(t)}v(x,t)+(a(x,t)-A)v(x,t). \end{align*} Because $a(x,t)-A \leq 0$, the function $v$ satisfies \begin{align*} \partial_t v(x,t)-\Delta_{g(t)}v(x,t)-(a(x,t)-A)v(x,t)\geq 0 \end{align*} with non-positive zeroth-order coefficient $a-A$. [guided] The zeroth-order term is harmless, but it is convenient to put it into the standard form required by the local strong parabolic maximum principle. Since $a$ is bounded on the compact spacetime cylinder $M \times [0,T]$, the number \begin{align*} A := \sup_{(x,t) \in M \times [0,T]} |a(x,t)| \end{align*} is finite. We multiply $u$ by a positive time-dependent factor and define the smooth map $v: M \times [0,T] \to \mathbb{R}$ by sending each $(x,t)$ to $e^{-At}u(x,t)$. The factor $e^{-At}$ is strictly positive, so $v \geq 0$ exactly when $u \geq 0$, and $v(x_0,t_0)=0$ exactly when $u(x_0,t_0)=0$. Now compute the differential inequality for $v$. Because the factor $e^{-At}$ does not depend on the spatial variable, the [Laplace-Beltrami operator](/page/Laplace-Beltrami%20Operator) acts only on $u$: \begin{align*} \Delta_{g(t)}v(x,t)=e^{-At}\Delta_{g(t)}u(x,t). \end{align*} For the time derivative, the product rule gives \begin{align*} \partial_t v(x,t) &= e^{-At}\partial_t u(x,t)-Ae^{-At}u(x,t). \end{align*} Using the assumed inequality for $u$ gives \begin{align*} \partial_t v(x,t)\geq e^{-At}\Delta_{g(t)}u(x,t)+e^{-At}a(x,t)u(x,t)-Ae^{-At}u(x,t). \end{align*} The identity for the spatial Laplacian and the definition of $v$ then give \begin{align*} \partial_t v(x,t)\geq \Delta_{g(t)}v(x,t)+(a(x,t)-A)v(x,t). \end{align*} The choice of $A$ ensures $a(x,t)-A \leq 0$ everywhere. This is the useful sign: at a zero of the nonnegative function $v$, the lower-order term vanishes, and away from the zero it has the sign compatible with the strong maximum principle. [/guided] [/step] [step:Apply the local strong parabolic maximum principle in coordinate cylinders] Fix a coordinate chart $(U,\varphi)$ on $M$, where $\varphi: U \to \varphi(U) \subset \mathbb{R}^n$ is a smooth coordinate map. Define the coordinate representative $\tilde v: \varphi(U) \times [0,T] \to \mathbb{R}$ by $\tilde v(z,t)=v(\varphi^{-1}(z),t)$. In these coordinates the operator $\Delta_{g(t)}$ has the uniformly parabolic local form \begin{align*} (\Delta_{g(t)}v)(\varphi^{-1}(z),t)=\sum_{i,j=1}^{n} \alpha_{ij}(z,t)\partial_{z_i}\partial_{z_j}\tilde v(z,t)+\sum_{i=1}^{n} \beta_i(z,t)\partial_{z_i}\tilde v(z,t), \end{align*} where $\alpha_{ij}:\varphi(U)\times[0,T]\to\mathbb{R}$ are the inverse-metric coefficient functions written with subscript indices, and $\beta_i:\varphi(U)\times[0,T]\to\mathbb{R}$ are smooth functions determined by $g(t)$ and its first spatial derivatives in the chart. On every compact coordinate subcylinder $K \times [s_1,s_2] \subset \varphi(U) \times (0,T]$, compactness and smoothness imply that the matrix $(\alpha_{ij}(z,t))_{i,j=1}^n$ is uniformly positive definite and that all coefficients are bounded. Thus the coordinate expression is a [uniformly parabolic operator](/page/Uniformly%20Parabolic%20Operator) on each such subcylinder. We use the [local strong parabolic maximum principle](/page/Strong%20Parabolic%20Maximum%20Principle) for uniformly parabolic operators in the following form. Let $W \subset \mathbb{R}^n$ be a connected open ball with compact closure in a coordinate domain, let $\sigma<\tau$, and let $q:W\times[\sigma,\tau]\to\mathbb{R}$ be a smooth nonnegative supersolution \begin{align*} \partial_t q-Lq-cq\geq 0 \end{align*} on $W\times(\sigma,\tau]$, where $L$ is a uniformly parabolic second-order operator with bounded smooth coefficients and $c\leq0$. If $q(z_*,\tau)=0$ for some $z_*\in W$, then $q(z,t)=0$ for every $(z,t)\in W\times[\sigma,\tau]$. The hypotheses apply to $\tilde v$ on every connected coordinate ball whose closure is compactly contained in $\varphi(U)$ because $\tilde v\geq0$, the coefficient $a-A$ is nonpositive, and the preceding paragraph verifies uniform parabolicity and boundedness of the local coefficients on its compact spacetime closure. Choose a coordinate ball $B\subset U$ with $x_0\in B$ and compact closure in $U$, and choose $\varepsilon_0\in(0,t_0]$ such that $[t_0-\varepsilon_0,t_0]\subset[0,T]$. Applying the local principle to $\varphi(B)\times[t_0-\varepsilon_0,t_0]$ gives \begin{align*} v(x,t)=0 \end{align*} for every $(x,t)\in B\times[t_0-\varepsilon_0,t_0]$. [guided] We first reduce the geometric operator to the standard local parabolic form. Fix a chart $(U,\varphi)$, where $\varphi:U\to\varphi(U)\subset\mathbb{R}^n$ is smooth. Define the coordinate representative $\tilde v: \varphi(U) \times [0,T] \to \mathbb{R}$ by $\tilde v(z,t)=v(\varphi^{-1}(z),t)$. In these coordinates, \begin{align*} (\Delta_{g(t)}v)(\varphi^{-1}(z),t)=\sum_{i,j=1}^{n} \alpha_{ij}(z,t)\partial_{z_i}\partial_{z_j}\tilde v(z,t)+\sum_{i=1}^{n} \beta_i(z,t)\partial_{z_i}\tilde v(z,t). \end{align*} Here $\alpha_{ij}$ are the inverse-metric coefficients written with subscript indices, and $\beta_i$ are the first-order coefficient functions coming from the Christoffel-symbol terms in the Laplace-Beltrami operator. Since $g(t)$ is a smooth Riemannian metric and the coordinate cylinder is compact, the matrix $(\alpha_{ij})$ has eigenvalues bounded above and below by positive constants, and the coefficient functions are bounded. This is exactly the [uniform parabolicity](/page/Uniformly%20Parabolic%20Operator) condition needed by the local theorem. The local result we use is the [local strong parabolic maximum principle](/page/Strong%20Parabolic%20Maximum%20Principle): a smooth nonnegative supersolution of $\partial_t q-Lq-cq\geq0$ on a connected coordinate ball with compact spacetime closure, with $L$ uniformly parabolic and with bounded smooth coefficients and $c\leq0$, that vanishes at an interior spatial point at the top time must vanish on the whole backward cylinder. In the present situation, $q=\tilde v$, the operator $L$ is the coordinate expression of the [Laplace-Beltrami operator](/page/Laplace-Beltrami%20Operator) $\Delta_{g(t)}$, and $c=a-A\leq0$. The zero $v(x_0,t_0)=0$ becomes the zero $\tilde v(\varphi(x_0),t_0)=0$. Choosing a coordinate ball $B\subset U$ with $x_0\in B$ and compact closure in $U$, and choosing $\varepsilon_0\in(0,t_0]$ so that $[t_0-\varepsilon_0,t_0]\subset[0,T]$, the theorem gives \begin{align*} v(x,t)=0 \end{align*} for every $(x,t)\in B\times[t_0-\varepsilon_0,t_0]$. This gives a genuine neighbourhood in space and a genuine interval backward in time, which is the local propagation mechanism used next. [/guided] [/step] [step:Propagate vanishing first across space and then down to time zero] For $s\in[0,t_0]$, define the zero set at time $s$ by \begin{align*} Z_s:=\{x\in M:v(x,s)=0\}. \end{align*} Continuity of $v(\cdot,s):M\to\mathbb{R}$ implies that $Z_s$ is closed. The previous step gives the following local implication: if $x\in Z_s$ and $s>0$, then there are an open neighbourhood $U_x\subset M$ of $x$ and a number $\varepsilon_x\in(0,s]$ such that \begin{align*} v(y,t)=0 \end{align*} for every $(y,t)\in U_x\times[s-\varepsilon_x,s]$. In particular, $U_x\subset Z_s$, so $Z_s$ is open whenever it is nonempty and $s>0$. Since $x_0\in Z_{t_0}$ and $M$ is connected, the set $Z_{t_0}$ is nonempty, open, and closed in $M$. Hence $Z_{t_0}=M$, and therefore \begin{align*} v(x,t_0)=0 \end{align*} for every $x\in M$. Define \begin{align*} \tau:=\inf\{s\in[0,t_0]:v(x,t)=0\text{ for every }(x,t)\in M\times[s,t_0]\}. \end{align*} The set in this definition is nonempty because $t_0$ belongs to it. By continuity of $v$ on the compact set $M\times[0,t_0]$, the defining property is closed under decreasing limits, so $v(x,t)=0$ for every $(x,t)\in M\times[\tau,t_0]$. If $\tau>0$, apply the local implication at time $\tau$ to every point $x\in M=Z_\tau$. The resulting neighbourhoods $U_x$ cover the compact manifold $M$, so choose finitely many points $x_1,\dots,x_N\in M$ such that $M=\bigcup_{k=1}^N U_{x_k}$. Let \begin{align*} \varepsilon:=\min_{1\leq k\leq N}\varepsilon_{x_k}>0. \end{align*} Then $v(y,t)=0$ for every $(y,t)\in M\times[\tau-\varepsilon,\tau]$, after replacing $\tau-\varepsilon$ by $0$ if necessary. Together with the already known vanishing on $M\times[\tau,t_0]$, this gives vanishing on a strictly larger interval $M\times[\max\{0,\tau-\varepsilon\},t_0]$, contradicting the definition of $\tau$ unless $\tau=0$. Therefore \begin{align*} v(y,t)=0 \end{align*} for every $(y,t)\in M\times[0,t_0]$. [guided] The local maximum principle gives more than a single zero: if $x\in M$ is a zero at a positive time $s$, then the zero spreads to a whole spatial neighbourhood and to a short interval of earlier times. Formally, for each $s\in[0,t_0]$ define \begin{align*} Z_s:=\{x\in M:v(x,s)=0\}. \end{align*} Because $v(\cdot,s)$ is continuous, $Z_s$ is closed. If $x\in Z_s$ and $s>0$, the coordinate-cylinder result supplies an [open set](/page/Open%20Set) $U_x\subset M$ and a number $\varepsilon_x\in(0,s]$ such that \begin{align*} v(y,t)=0 \end{align*} for every $(y,t)\in U_x\times[s-\varepsilon_x,s]$. Taking $t=s$ shows $U_x\subset Z_s$, so every point of $Z_s$ is an interior point. Thus $Z_s$ is open whenever $s>0$ and $Z_s$ is nonempty. At the time $t_0$, the set $Z_{t_0}$ contains $x_0$, so it is nonempty. It is also open and closed in the connected manifold $M$. Connectedness forces the only nonempty clopen subset to be the whole space, hence \begin{align*} Z_{t_0}=M. \end{align*} Equivalently, \begin{align*} v(x,t_0)=0 \end{align*} for every $x\in M$. Now we must push this vanishing all the way down to time zero. Define the earliest time from which full-slice vanishing is known by \begin{align*} \tau:=\inf\{s\in[0,t_0]:v(x,t)=0\text{ for every }(x,t)\in M\times[s,t_0]\}. \end{align*} The set is nonempty because $s=t_0$ belongs to it. Since $v$ is continuous on $M\times[0,t_0]$, if a decreasing sequence of such times tends to $\tau$, then the vanishing passes to the limit. Hence $v=0$ on $M\times[\tau,t_0]$. Suppose for contradiction that $\tau>0$. Since $v=0$ on $M\times\{\tau\}$, every point $x\in M$ lies in $Z_\tau$. Applying the local implication at each such $x$ gives open sets $U_x$ and backward time lengths $\varepsilon_x>0$. Compactness of $M$ is what prevents these time lengths from shrinking to zero globally: choose a finite subcover $U_{x_1},\dots,U_{x_N}$ of $M$ and set \begin{align*} \varepsilon:=\min_{1\leq k\leq N}\varepsilon_{x_k}>0. \end{align*} Then every point of $M$ lies in one of the finitely many $U_{x_k}$, so \begin{align*} v(y,t)=0 \end{align*} for every $(y,t)\in M\times[\tau-\varepsilon,\tau]$, with the lower endpoint replaced by $0$ if $\tau-\varepsilon<0$. Combining this with the vanishing on $M\times[\tau,t_0]$ gives full vanishing starting before $\tau$, contradicting the definition of $\tau$. Therefore $\tau=0$, and \begin{align*} v(y,t)=0 \end{align*} for every $(y,t)\in M\times[0,t_0]$. [/guided] [/step] [step:Undo the exponential weight] For every $(x,t) \in M \times [0,t_0]$, we have proved \begin{align*} v(x,t)=0. \end{align*} By definition of $v$, \begin{align*} u(x,t)=e^{At}v(x,t). \end{align*} Since $e^{At}>0$ for every $t \in [0,t_0]$, it follows that \begin{align*} u(x,t)=0 \end{align*} for every $(x,t) \in M \times [0,t_0]$. This is the desired conclusion. [/step]