[proofplan] The Ricci flow equation is only weakly parabolic because diffeomorphism invariance creates a degeneracy in its principal symbol. We remove this degeneracy by fixing a background metric and adding the DeTurck Lie derivative term, obtaining the Ricci-DeTurck equation, a strictly parabolic quasilinear system for the metric components. Standard short-time existence and uniqueness for strictly parabolic quasilinear systems gives a unique Ricci-DeTurck solution. Solving the ODE generated by the negative DeTurck vector field and pulling the Ricci-DeTurck solution back by this family of diffeomorphisms converts it into a genuine Ricci flow. Uniqueness follows by applying the same gauge construction to any Ricci flow and using uniqueness of the Ricci-DeTurck system. [/proofplan]

[step:Fix a background metric and define the DeTurck vector field] Let $\bar g := g_0$ be the fixed background metric on $M$. For any smooth Riemannian metric $h \in \Gamma(\operatorname{Sym}^2 T^*M)$, define the DeTurck vector field \begin{align*} W(h) \in \Gamma(TM) \end{align*} as follows. In a coordinate chart $(U,\varphi)$ with local coordinates $(x_1,\dots,x_n)$, write $h^{ij}$ for the components of the inverse metric $h^{-1}$, write $\Gamma(h)^k_{ij}$ for the Christoffel symbols of $h$, and write $\Gamma(\bar g)^k_{ij}$ for the Christoffel symbols of $\bar g$. Then \begin{align*} W(h)^k = h^{ij}\bigl(\Gamma(h)^k_{ij}-\Gamma(\bar g)^k_{ij}\bigr). \end{align*} The difference $\Gamma(h)^k_{ij}-\Gamma(\bar g)^k_{ij}$ is the coordinate expression of the tensor $\nabla^h-\nabla^{\bar g} \in \Gamma(T^*M \otimes T^*M \otimes TM)$, so tracing its two covariant indices with $h^{-1}$ defines a globally well-defined smooth vector field $W(h)$. The Ricci-DeTurck equation with background metric $\bar g$ is \begin{align*} \frac{\partial h}{\partial t}(t) = -2\operatorname{Ric}(h(t))+\mathcal{L}_{W(h(t))}h(t), \qquad h(0)=g_0, \end{align*} where \begin{align*} h: [0,T) &\to \Gamma(\operatorname{Sym}^2 T^*M) \end{align*} is the unknown one-parameter family of metrics, and $\mathcal{L}_{W(h(t))}h(t)$ denotes the Lie derivative of $h(t)$ along $W(h(t))$. [/step]

[step:Recognize the Ricci-DeTurck equation as a strictly parabolic quasilinear system]In a coordinate chart $(U,\varphi)$, a direct coordinate computation of the Ricci tensor, the Christoffel trace defining $W(h)$, and the Lie derivative $\mathcal{L}_{W(h)}h$ gives the Ricci-DeTurck coordinate cancellation identity \begin{align*} \frac{\partial h_{ij}}{\partial t}=h^{pq}\frac{\partial^2 h_{ij}}{\partial x_p \partial x_q}+Q_{ij}\bigl(x,h,\partial h\bigr). \end{align*} Here $Q_{ij}$ denotes a smooth expression depending on the coordinate point $x$, the metric components $h_{ab}$, the inverse metric components $h^{ab}$, the first coordinate derivatives $\partial_{x_c}h_{ab}$, and the fixed background metric $\bar g$ through its Christoffel symbols and their first derivatives. Thus the displayed second-order term is the full principal part of the system. The principal symbol of this system at a covector $\xi=\sum_{p=1}^n \xi_p\,dx_p \in T_x^*M$ is multiplication by \begin{align*} h^{pq}(x)\xi_p\xi_q \end{align*} on the [vector space](/page/Vector%20Space) $\operatorname{Sym}^2 T_x^*M$. Since $h$ is a Riemannian metric, $h^{pq}(x)\xi_p\xi_q>0$ for every non-zero covector $\xi \in T_x^*M$. Thus the Ricci-DeTurck equation is a strictly parabolic quasilinear system as long as $h(t)$ remains positive definite.[/step]

[guided]The obstruction in Ricci flow is that the equation \begin{align*} \frac{\partial g}{\partial t}=-2\operatorname{Ric}(g) \end{align*} is invariant under time-dependent diffeomorphisms, so its principal symbol has directions corresponding to infinitesimal coordinate changes. The DeTurck term is chosen to cancel exactly those degenerate directions. We now verify the cancellation with the sign convention used in this proof. Let $(U,\varphi)$ be a coordinate chart with coordinates $(x_1,\dots,x_n)$. The second-order part of the Ricci tensor of $h$ is \begin{align*} \operatorname{Ric}(h)_{ij}=\frac{1}{2}h^{pq}\left(\frac{\partial^2 h_{iq}}{\partial x_p\partial x_j}+\frac{\partial^2 h_{jq}}{\partial x_p\partial x_i}-\frac{\partial^2 h_{ij}}{\partial x_p\partial x_q}-\frac{\partial^2 h_{pq}}{\partial x_i\partial x_j}\right)+\text{terms involving at most first derivatives of }h. \end{align*} For the DeTurck vector field defined by $W(h)^k=h^{pq}(\Gamma(h)^k_{pq}-\Gamma(\bar g)^k_{pq})$, lowering the index with $h$ gives the second-order contribution \begin{align*} (\mathcal{L}_{W(h)}h)_{ij}=h^{pq}\left(\frac{\partial^2 h_{jq}}{\partial x_i\partial x_p}+\frac{\partial^2 h_{iq}}{\partial x_j\partial x_p}-\frac{\partial^2 h_{pq}}{\partial x_i\partial x_j}\right)+\text{terms involving at most first derivatives of }h. \end{align*} Combining these two displayed principal parts in $-2\operatorname{Ric}(h)+\mathcal{L}_{W(h)}h$, the mixed second derivatives cancel and the remaining second-order term is \begin{align*} h^{pq}\frac{\partial^2 h_{ij}}{\partial x_p\partial x_q}. \end{align*} All remaining terms contain at most first derivatives of $h$ and are collected into a smooth lower-order expression $Q_{ij}(x,h,\partial h)$. To check strict parabolicity, fix $x \in U$ and a non-zero covector $\xi=\sum_{p=1}^n \xi_p\,dx_p \in T_x^*M$. The principal symbol acts on each symmetric $2$-tensor component by multiplication with \begin{align*} h^{pq}(x)\xi_p\xi_q. \end{align*} Because $h(x)$ is positive definite, its inverse $h^{-1}(x)$ is positive definite on $T_x^*M$, so this quantity is strictly positive. Therefore the principal symbol is positive definite on every tensor component, which is exactly strict parabolicity for this quasilinear system.[/guided]

[step:Solve the Ricci-DeTurck equation by parabolic theory] We apply the external short-time existence and uniqueness theorem for strictly parabolic quasilinear systems on closed manifolds. Its hypotheses are satisfied here: $M$ is closed, hence compact without boundary; the coefficients of the Ricci-DeTurck system are smooth functions of $x$, $h$, $h^{-1}$, and $\partial h$ on the open cone of positive definite symmetric $2$-tensors; the initial datum $g_0$ is smooth; and the principal symbol is uniformly positive on a sufficiently small $C^0$-neighbourhood of $g_0$ because $M$ is compact and $g_0$ is positive definite. Therefore there exists $T>0$ and a unique smooth solution $h: [0,T) \to \Gamma(\operatorname{Sym}^2 T^*M)$ of the Ricci-DeTurck initial value problem. Explicitly, this solution satisfies \begin{align*} \frac{\partial h}{\partial t}(t)=-2\operatorname{Ric}(h(t))+\mathcal{L}_{W(h(t))}h(t). \end{align*} It also satisfies $h(0)=g_0$. Since $h(0)=g_0$ is positive definite and $h$ depends continuously on $t$ in $C^0$, after replacing $T$ by a smaller positive number if necessary, $h(t)$ is a Riemannian metric for every $t \in [0,T)$. [/step]

[step:Integrate the negative DeTurck vector field] Define $X: M \times [0,T) \to TM$ by $X(p,t)=-W(h(t))(p)$, so $X(\cdot,t) \in \Gamma(TM)$ is a smooth time-dependent vector field. Since $M$ is closed and $X$ is smooth, the external existence and uniqueness theorem for smooth time-dependent vector fields on compact manifolds gives a smooth flow $\phi_t: M \to M$ satisfying \begin{align*} \frac{\partial \phi_t}{\partial t}(p)=X(\phi_t(p),t). \end{align*} It also satisfies $\phi_0=\operatorname{id}_M$. The same ODE theorem applied backward from time $t$ to time $0$ gives a smooth inverse map for $\phi_t$ on a possibly smaller interval. Equivalently, uniqueness prevents two flow lines from crossing, and the backward flow supplies the inverse. After possibly shrinking $T>0$, each $\phi_t$ is therefore a smooth diffeomorphism of $M$. [/step]

[step:Pull back the Ricci-DeTurck solution to obtain a Ricci flow] Define $g: [0,T) \to \Gamma(\operatorname{Sym}^2 T^*M)$ by $g(t)=\phi_t^*h(t)$. Then $g(0)=\phi_0^*h(0)=g_0$. We compute its time derivative. The pullback derivative formula for a time-dependent tensor field $A(t)$ and a family of diffeomorphisms $\phi_t$ generated by the time-dependent vector field $X(t)$ gives \begin{align*} \frac{\partial}{\partial t}\phi_t^*A(t)=\phi_t^*\left(\frac{\partial A}{\partial t}(t)+\mathcal{L}_{X(t)}A(t)\right). \end{align*} Applying this with $A(t)=h(t)$ and $X(t)=-W(h(t))$, the DeTurck Lie derivative term cancels and yields \begin{align*} \frac{\partial g}{\partial t}(t)=-2\phi_t^*\operatorname{Ric}(h(t)). \end{align*} The Ricci tensor is natural under diffeomorphism pullback, so \begin{align*} \phi_t^*\operatorname{Ric}(h(t))=\operatorname{Ric}(\phi_t^*h(t))=\operatorname{Ric}(g(t)). \end{align*} Therefore \begin{align*} \frac{\partial g}{\partial t}(t)=-2\operatorname{Ric}(g(t)). \end{align*} Thus $g(t)$ is a smooth Ricci flow on $[0,T)$ with initial metric $g_0$. [/step]

[step:Put any Ricci flow into harmonic map gauge] Let \begin{align*} \tilde g: [0,\tilde T) &\to \Gamma(\operatorname{Sym}^2 T^*M) \end{align*} be any smooth Ricci flow with $\tilde g(0)=g_0$. We use the fixed target metric $\bar g=g_0$ and solve the harmonic map heat flow \begin{align*} F: M \times [0,\tau) &\to M \end{align*} with initial map $F_0=\operatorname{id}_M$, domain metric $\tilde g(t)$, and target metric $\bar g$. In local coordinates, this means \begin{align*} \frac{\partial F^k}{\partial t} = \tilde g^{ij}\left( \frac{\partial^2 F^k}{\partial x_i\partial x_j} -\Gamma(\tilde g)^m_{ij}\frac{\partial F^k}{\partial x_m} +\Gamma(\bar g)^k_{ab}(F)\frac{\partial F^a}{\partial x_i}\frac{\partial F^b}{\partial x_j} \right), \end{align*} where $F^k$ are the coordinate components of $F$, $\tilde g^{ij}$ are the components of $\tilde g(t)^{-1}$, $\Gamma(\tilde g)^m_{ij}$ are the Christoffel symbols of $\tilde g(t)$ on the domain, and $\Gamma(\bar g)^k_{ab}$ are the Christoffel symbols of $\bar g$ on the target. This is a strictly parabolic quasilinear system for $F$ because the principal part is $\tilde g^{ij}\partial_{x_i}\partial_{x_j}F^k$ and $\tilde g(t)$ is a Riemannian metric. By short-time existence and uniqueness for strictly parabolic quasilinear systems on closed manifolds, there exists $\tau>0$ and a unique smooth solution $F$. Since $F_0=\operatorname{id}_M$ and $F$ depends smoothly on $t$ in $C^1$, after decreasing $\tau>0$ if necessary, each map \begin{align*} F_t: M &\to M \end{align*} is a diffeomorphism. Define \begin{align*} \tilde h: [0,\tau) &\to \Gamma(\operatorname{Sym}^2 T^*M) \end{align*} by \begin{align*} \tilde h(t)=(F_t^{-1})^*\tilde g(t). \end{align*} We verify the DeTurck gauge identity with the present sign convention. Since $\tilde h(t)=(F_t^{-1})^*\tilde g(t)$, equivalently $\tilde g(t)=F_t^*\tilde h(t)$. The coordinate formula for the harmonic map heat flow, rewritten in the moving coordinates defined by $F_t$, is exactly \begin{align*} \frac{\partial F_t}{\partial t}=-W(\tilde h(t))\circ F_t. \end{align*} The minus sign comes from the convention $W(h)^k=h^{ij}(\Gamma(h)^k_{ij}-\Gamma(\bar g)^k_{ij})$: the harmonic map tension field of $F_t:(M,F_t^*\tilde h(t))\to(M,\bar g)$ has coordinate expression $-W(\tilde h(t))\circ F_t$. Applying the pullback derivative formula to $\tilde g(t)=F_t^*\tilde h(t)$ and using that $\tilde g$ is a Ricci flow gives \begin{align*} \frac{\partial \tilde h}{\partial t}(t)=-2\operatorname{Ric}(\tilde h(t))+\mathcal{L}_{W(\tilde h(t))}\tilde h(t). \end{align*} Also \begin{align*} \tilde h(0)=(F_0^{-1})^*\tilde g(0)=g_0. \end{align*} Thus every Ricci flow with initial metric $g_0$ determines, after applying harmonic map gauge, a solution of the Ricci-DeTurck equation with the same initial metric, and its original gauge map $F_t$ is the DeTurck diffeomorphism generated by $-W(\tilde h(t))$. [/step]

[step:Use uniqueness in harmonic map gauge to prove uniqueness of Ricci flow] Let $g_1: [0,\varepsilon_1) \to \Gamma(\operatorname{Sym}^2 T^*M)$ and $g_2: [0,\varepsilon_2) \to \Gamma(\operatorname{Sym}^2 T^*M)$ be smooth Ricci flows with common initial metric $g_0$. Choose $\delta>0$ smaller than $\varepsilon_1$ and $\varepsilon_2$ and small enough that the harmonic map gauge construction in the previous step applies to both flows on $[0,\delta)$. This gives diffeomorphism families $F_{1,t}$ and $F_{2,t}$ with $F_{1,0}=F_{2,0}=\operatorname{id}_M$, and Ricci-DeTurck solutions $h_1$ and $h_2$ defined by \begin{align*} h_i(t)=(F_{i,t}^{-1})^*g_i(t) \quad \text{for } i\in\{1,2\}. \end{align*} Both satisfy $h_i(0)=g_0$. By uniqueness for the strictly parabolic Ricci-DeTurck equation, \begin{align*} h_1(t)=h_2(t) \end{align*} for every $t \in [0,\delta)$. Denote this common solution by $h$. The sign verification in the previous step gives, for each $i\in\{1,2\}$, \begin{align*} \frac{\partial F_{i,t}}{\partial t}=-W(h(t))\circ F_{i,t}. \end{align*} The right-hand side is the same smooth time-dependent vector field for $i=1$ and $i=2$, and both initial maps are $\operatorname{id}_M$. Uniqueness for smooth time-dependent vector-field flows on the compact manifold $M$ therefore gives $F_{1,t}=F_{2,t}$ for every $t\in[0,\delta)$. Since $h_i(t)=(F_{i,t}^{-1})^*g_i(t)$, equivalently $g_i(t)=F_{i,t}^*h_i(t)$. Using $F_{1,t}=F_{2,t}$ and $h_1(t)=h_2(t)$, we obtain \begin{align*} g_1(t)=g_2(t) \end{align*} for every $t \in [0,\delta)$. Repeating the same argument starting at any later time at which both flows are defined extends equality throughout their common interval of existence. Therefore the Ricci flow starting from $g_0$ exists for short time and is unique among smooth Ricci flows on the closed manifold $M$ with the same initial metric. This completes the proof. [/step]