[proofplan]
We compute the Ricci-DeTurck equation in a fixed coordinate chart and isolate exactly the terms containing second derivatives of the evolving metric $\hat{g}$. The second-order part of the Ricci tensor term $-2\operatorname{Ric}(\hat{g})$ contains the desired rough Laplacian term together with gauge terms involving derivatives of the traced Christoffel expression. The Lie derivative along the DeTurck vector field contributes precisely the opposite gauge terms; the remaining background contributions contain only the fixed metric $\bar{g}$ and therefore enter the lower-order expression $Q_{ij}$. The principal symbol is then multiplication by the positive quadratic form $\hat{g}^{ab}\xi_a\xi_b$, which gives strict parabolicity and, under two-sided eigenvalue bounds, uniform strict parabolicity.
[/proofplan]
[step:Fix local notation for the two metrics and their Christoffel symbols]
Work in a coordinate chart $(U,\varphi)$ with coordinates $(x_1,\dots,x_n)$. Write
\begin{align*}
\hat{g}_{ij}: U\times [0,T]&\to \mathbb{R}
\end{align*}
for the coordinate components of $\hat{g}$, and write $\hat{g}^{ij}$ for the components of the inverse matrix $(\hat{g}_{ij})^{-1}$. Similarly, write $\bar{g}_{ij}:U\to\mathbb{R}$ and $\bar{g}^{ij}$ for the background metric and its inverse.
For a metric $g$, define the Christoffel symbols of the second kind by
\begin{align*}
\Gamma(g)^k_{ij}
=
\frac{1}{2}\sum_{\ell=1}^n g^{k\ell}
\bigl(
\partial_{x_i}g_{j\ell}
+
\partial_{x_j}g_{i\ell}
-
\partial_{x_\ell}g_{ij}
\bigr),
\end{align*}
and define the Christoffel symbols of the first kind by
\begin{align*}
\Gamma(g)_{ij,k}
=
\sum_{\ell=1}^n g_{k\ell}\Gamma(g)^\ell_{ij}
=
\frac{1}{2}
\bigl(
\partial_{x_i}g_{jk}
+
\partial_{x_j}g_{ik}
-
\partial_{x_k}g_{ij}
\bigr).
\end{align*}
The DeTurck vector field $W(\hat{g},\bar{g})$ is the smooth time-dependent vector field on $U$ along $U\times[0,T]$, viewed as a map $W(\hat{g},\bar{g}):U\times[0,T]\to TU$. Its components are
\begin{align*}
W^k
=
\sum_{a,b=1}^n \hat{g}^{ab}
\bigl(
\Gamma(\hat{g})^k_{ab}
-
\Gamma(\bar{g})^k_{ab}
\bigr).
\end{align*}
Its covariant components with respect to $\hat{g}$ are
\begin{align*}
W_i
=
\sum_{k=1}^n \hat{g}_{ik}W^k
=
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,i}
-
\sum_{a,b,k,\ell=1}^n
\hat{g}_{ik}\hat{g}^{ab}\bar{g}^{k\ell}\Gamma(\bar{g})_{ab,\ell}.
\end{align*}
The first summand contains first derivatives of $\hat{g}$; the second contains no derivatives of $\hat{g}$ and at most first derivatives of $\bar{g}$.
[/step]
[step:Separate the second-order part of the Ricci tensor]
The Ricci tensor of $\hat{g}$ has coordinate expression
\begin{align*}
\operatorname{Ric}(\hat{g})_{ij}
=
\sum_{a=1}^n \partial_{x_a}\Gamma(\hat{g})^a_{ij}
-
\partial_{x_j}\Gamma(\hat{g})^a_{ia}
+
\sum_{a,b=1}^n
\Gamma(\hat{g})^a_{ij}\Gamma(\hat{g})^b_{ab}
-
\Gamma(\hat{g})^a_{ib}\Gamma(\hat{g})^b_{ja}.
\end{align*}
When the derivatives fall on the inverse coefficients $\hat{g}^{ab}$, the resulting terms contain only first derivatives of $\hat{g}$ because
\begin{align*}
\partial_{x_c}\hat{g}^{ab}
=
-\sum_{p,q=1}^n \hat{g}^{ap}\hat{g}^{bq}\partial_{x_c}\hat{g}_{pq}.
\end{align*}
Thus the only second derivatives of $\hat{g}$ arise by differentiating the first derivatives inside the Christoffel symbols. The genuine second-order part of $2\sum_a\partial_{x_a}\Gamma(\hat{g})^a_{ij}$ is
\begin{align*}
\sum_{a,b=1}^n \hat{g}^{ab}
\bigl(
\partial_{x_a}\partial_{x_i}\hat{g}_{jb}
+
\partial_{x_a}\partial_{x_j}\hat{g}_{ib}
-
\partial_{x_a}\partial_{x_b}\hat{g}_{ij}
\bigr),
\end{align*}
and the genuine second-order part of $2\partial_{x_j}\sum_a\Gamma(\hat{g})^a_{ia}$ is
\begin{align*}
\sum_{a,b=1}^n \hat{g}^{ab}\partial_{x_j}\partial_{x_i}\hat{g}_{ab}.
\end{align*}
Define the lower-order bookkeeping term $A_{ij}$ to be the sum of all terms produced when one of the outer derivatives falls on an inverse coefficient $\hat{g}^{ab}$ in the two traced-Christoffel derivatives below. By the inverse-derivative identity above, $A_{ij}$ is smooth in $\hat{g}^{-1}$ and $\partial\hat{g}$ and contains no second derivatives of $\hat{g}$. Expanding those two traced-Christoffel derivatives gives the following identity, where $A_{ij}$ is already defined:
\begin{align*}
\partial_{x_i}\left(\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,j}\right)+\partial_{x_j}\left(\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,i}\right) = \frac{1}{2}\sum_{a,b=1}^n \hat{g}^{ab}\bigl(\partial_{x_i}\partial_{x_a}\hat{g}_{bj}+\partial_{x_i}\partial_{x_b}\hat{g}_{aj}-\partial_{x_i}\partial_{x_j}\hat{g}_{ab}\bigr)+\frac{1}{2}\sum_{a,b=1}^n \hat{g}^{ab}\bigl(\partial_{x_j}\partial_{x_a}\hat{g}_{bi}+\partial_{x_j}\partial_{x_b}\hat{g}_{ai}-\partial_{x_j}\partial_{x_i}\hat{g}_{ab}\bigr)+A_{ij}.
\end{align*}
Since $\hat{g}^{ab}=\hat{g}^{ba}$ and coordinate partial derivatives commute on the smooth coefficient functions $\hat{g}_{ij}$, the first two terms in each parenthesis combine to
\begin{align*}
\sum_{a,b=1}^n \hat{g}^{ab}
\bigl(
\partial_{x_a}\partial_{x_i}\hat{g}_{jb}
+
\partial_{x_a}\partial_{x_j}\hat{g}_{ib}
\bigr),
\end{align*}
while the two final traced terms combine to
\begin{align*}
-\sum_{a,b=1}^n \hat{g}^{ab}\partial_{x_i}\partial_{x_j}\hat{g}_{ab}.
\end{align*}
Therefore the difference of the displayed second-order Ricci terms equals
\begin{align*}
-\sum_{a,b=1}^n \hat{g}^{ab}\partial_{x_a}\partial_{x_b}\hat{g}_{ij}
+
\partial_{x_i}
\left(
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,j}
\right)
+
\partial_{x_j}
\left(
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,i}
\right)
-
A_{ij}.
\end{align*}
Define $R_{ij}$ to be the sum of $A_{ij}$ and all quadratic Christoffel terms and inverse-coefficient derivative terms produced in this computation after multiplying by $-1$. By the preceding identities, $R_{ij}$ is a smooth expression in $\hat{g}^{-1}$ and $\partial \hat{g}$. Multiplying by $-1$ for $-2\operatorname{Ric}(\hat{g})_{ij}$ and collecting those lower-order terms gives
\begin{align*}
-2\operatorname{Ric}(\hat{g})_{ij}
=
\sum_{a,b=1}^n \hat{g}^{ab}\partial_{x_a}\partial_{x_b}\hat{g}_{ij}
-
\partial_{x_i}
\left(
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,j}
\right)
-
\partial_{x_j}
\left(
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,i}
\right)
+
R_{ij},
\end{align*}
[guided]
The only point that requires care is deciding which pieces of the Ricci tensor are genuinely second-order in $\hat{g}$. The Christoffel symbols $\Gamma(\hat{g})^k_{ij}$ are built from $\hat{g}^{-1}$ and $\partial \hat{g}$. Therefore, after differentiating a Christoffel symbol, a second derivative of $\hat{g}$ appears only when the derivative lands on the $\partial \hat{g}$ factor. If the derivative lands on $\hat{g}^{-1}$, then the identity
\begin{align*}
\partial_{x_c}\hat{g}^{ab}
=
-\sum_{p,q=1}^n \hat{g}^{ap}\hat{g}^{bq}\partial_{x_c}\hat{g}_{pq}
\end{align*}
shows that the resulting term is still first-order in $\hat{g}$.
Using the coordinate formula
\begin{align*}
\operatorname{Ric}(\hat{g})_{ij}
=
\sum_{a=1}^n \partial_{x_a}\Gamma(\hat{g})^a_{ij}
-
\partial_{x_j}\Gamma(\hat{g})^a_{ia}
+
\sum_{a,b=1}^n
\Gamma(\hat{g})^a_{ij}\Gamma(\hat{g})^b_{ab}
-
\Gamma(\hat{g})^a_{ib}\Gamma(\hat{g})^b_{ja},
\end{align*}
the quadratic Christoffel terms contain only first derivatives of $\hat{g}$. Hence they are lower-order for parabolicity. Expanding only the differentiated Christoffel terms and grouping the genuine second derivatives yields
\begin{align*}
-2\operatorname{Ric}(\hat{g})_{ij}
=
\sum_{a,b=1}^n \hat{g}^{ab}\partial_{x_a}\partial_{x_b}\hat{g}_{ij}
-
\partial_{x_i}
\left(
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,j}
\right)
-
\partial_{x_j}
\left(
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,i}
\right)
+
R_{ij}.
\end{align*}
Here $R_{ij}$ denotes the collected lower-order expression; it is defined before the displayed formula as the sum of all non-principal terms produced by the expansion of $-2\operatorname{Ric}(\hat{g})_{ij}$. It is smooth in $\hat{g}^{-1}$ and $\partial \hat{g}$ because it is obtained by addition, multiplication, and contraction from those quantities, and inversion of a positive-definite matrix is smooth on the open cone of positive-definite matrices. This completes the Ricci-expansion step: the principal Laplacian term and the two gauge derivatives have been isolated, while every remaining term has been placed in $R_{ij}$.
[/guided]
[/step]
[step:Compute the Lie derivative contribution and cancel the gauge terms]
The Lie derivative of $\hat{g}$ along $W$ has coordinate form
\begin{align*}
(\mathcal{L}_W\hat{g})_{ij}
=
\partial_{x_i}W_j+\partial_{x_j}W_i
-
2\sum_{k,\ell=1}^n \Gamma(\hat{g})^k_{ij}\hat{g}_{k\ell}W^\ell.
\end{align*}
The final term contains only first derivatives of $\hat{g}$, because $\Gamma(\hat{g})^k_{ij}$ is first-order in $\hat{g}$ and $W^\ell$ is first-order in $\hat{g}$ plus a background term. Using the expression for $W_i$ from the first step, the second-order part of $\partial_{x_i}W_j+\partial_{x_j}W_i$ is
\begin{align*}
\partial_{x_i}
\left(
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,j}
\right)
+
\partial_{x_j}
\left(
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,i}
\right).
\end{align*}
Define $S_{ij}$ to be the sum of the remaining pieces in $(\mathcal{L}_W\hat{g})_{ij}$, namely the terms produced when derivatives fall on $\hat{g}^{ab}$, the terms coming from differentiating the background Christoffel contribution in $W_i$, and the final Christoffel product term. These pieces contain at most first derivatives of $\hat{g}$ and at most second derivatives of $\bar{g}$, so $S_{ij}$ is smooth in $\hat{g}^{-1}$, $\partial\hat{g}$, $\bar{g}^{-1}$, $\partial\bar{g}$, and $\partial^2\bar{g}$. Therefore
\begin{align*}
(\mathcal{L}_W\hat{g})_{ij}
=
\partial_{x_i}
\left(
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,j}
\right)
+
\partial_{x_j}
\left(
\sum_{a,b=1}^n \hat{g}^{ab}\Gamma(\hat{g})_{ab,i}
\right)
+
S_{ij},
\end{align*}
[/step]
[step:Combine the Ricci and DeTurck expansions into quasilinear form]
Adding the expansions of $-2\operatorname{Ric}(\hat{g})_{ij}$ and $(\mathcal{L}_W\hat{g})_{ij}$ cancels the two traced-Christoffel gauge derivatives. Hence the Ricci-DeTurck equation becomes
\begin{align*}
\partial_t\hat{g}_{ij}
=
\sum_{a,b=1}^n \hat{g}^{ab}\partial_{x_a}\partial_{x_b}\hat{g}_{ij}
+
Q_{ij}\bigl(\hat{g}^{-1},\partial \hat{g},\bar{g}^{-1},\partial \bar{g},\partial^2\bar{g}\bigr),
\end{align*}
where
\begin{align*}
Q_{ij}:=R_{ij}+S_{ij}.
\end{align*}
The expression $Q_{ij}$ is smooth in the displayed variables because it is formed from finitely many sums, products, contractions, and matrix inversions, and matrix inversion is smooth on the open cone of positive-definite symmetric matrices.
[/step]
[step:Identify the principal symbol and prove strict parabolicity]
Fix a point $(x,t)\in U\times[0,T]$ and a non-zero covector
\begin{align*}
\xi=\sum_{a=1}^n \xi_a\,dx_a\in T_x^*M.
\end{align*}
The principal symbol of the second-order operator
\begin{align*}
h_{ij}\mapsto \sum_{a,b=1}^n \hat{g}^{ab}\partial_{x_a}\partial_{x_b}h_{ij}
\end{align*}
on symmetric $2$-tensors is the [linear map](/page/Linear%20Map)
\begin{align*}
\sigma_{\hat{g}}(\xi): S^2T_x^*M&\to S^2T_x^*M
\end{align*}
given by
\begin{align*}
\sigma_{\hat{g}}(\xi)(h)_{ij}
=
\left(\sum_{a,b=1}^n \hat{g}^{ab}\xi_a\xi_b\right)h_{ij}.
\end{align*}
Since $\hat{g}(x,t)$ is positive definite, its inverse matrix $(\hat{g}^{ab}(x,t))$ is positive definite on $T_x^*M$. Therefore
\begin{align*}
\sum_{a,b=1}^n \hat{g}^{ab}(x,t)\xi_a\xi_b>0
\end{align*}
for every $\xi\neq 0$. Thus the principal symbol is a positive scalar multiple of the identity on $S^2T_x^*M$, so the system is strictly parabolic at every positive-definite metric.
Let $I\subset[0,T]$ be a time interval, and let $0<\lambda\leq\Lambda<\infty$ be constants such that the eigenvalues of $\hat{g}(x,t)$ relative to $\bar{g}(x)$ lie in $[\lambda,\Lambda]$ for every $(x,t)\in M\times I$. Then the eigenvalues of $\hat{g}^{-1}(x,t)$ relative to $\bar{g}^{-1}(x)$ lie in $[\Lambda^{-1},\lambda^{-1}]$. Hence, for every covector $\xi\in T_x^*M$ and every $(x,t)\in M\times I$,
\begin{align*}
\sum_{a,b=1}^n \hat{g}^{ab}(x,t)\xi_a\xi_b
\geq
\Lambda^{-1}\sum_{a,b=1}^n \bar{g}^{ab}(x)\xi_a\xi_b.
\end{align*}
This lower bound is uniform on $M\times I$ because the same constant $\Lambda^{-1}$ works at every point and time. Therefore the Ricci-DeTurck equation is uniformly strictly parabolic on every time interval on which the stated two-sided eigenvalue bounds hold uniformly in space and time.
[/step]
