[proofplan] We compute the second-order part of the variation of the Ricci tensor in local coordinates, because the principal symbol only records the coefficients of the highest-order derivatives of the variation tensor $h$. The variation of the Christoffel symbols gives the second derivatives of $h$ appearing in $D\operatorname{Ric}_g(h)$; all curvature, Christoffel, and first-derivative terms are lower order and therefore do not enter the symbol. Replacing each covariant derivative $\nabla_i$ in the second-order part by the covector component $\xi_i$ gives the displayed principal symbol. Finally, substituting $h=\xi\otimes\eta+\eta\otimes\xi$ shows that these infinitesimal-diffeomorphism directions are annihilated by the symbol. [/proofplan]

[step:Compute the second-order part of the linearized Ricci tensor]Fix $p \in M$, and let $(U,x)$ be a smooth coordinate chart around $p$ with coordinate vector fields $\partial_{x_1},\dots,\partial_{x_n}$. Write $\Gamma_{ij,k}$ for the coefficient of $\partial_{x_k}$ in $\nabla_{\partial_{x_i}}\partial_{x_j}$. Let $h \in \Gamma(S^2T^*M)$ be a smooth symmetric $2$-tensor. We use the Levi-Civita connection coefficient formula in the chart $(U,x)$ and differentiate it with respect to the metric parameter. Equivalently, differentiating the Koszul formula for the Levi-Civita connection gives a tensorial first-variation formula for the connection; after expressing that formula in the chart $(U,x)$, the only first-order terms in $h$ are the three covariant derivatives displayed below. The terms produced by differentiating $(g_s^{-1})_{k\ell}$ contain no derivatives of $h$, and after the principal symbol freezes coefficients at $p$, all such coefficient-variation terms are lower order. For a smooth one-parameter family of metrics $g_s$ with $g_0=g$ and $\frac{d}{ds}\big|_{s=0}g_s=h$, the first variation of the Christoffel coefficients is \begin{align*} (D\Gamma_{ij,k})_g(h) = \frac{1}{2}\sum_{\ell=1}^n (g^{-1})_{k\ell} \left( \nabla_i h_{j\ell} + \nabla_j h_{i\ell} - \nabla_\ell h_{ij} \right) + \text{terms of order zero in } h. \end{align*} Here $\nabla_i h_{j\ell}$ denotes $(\nabla_{\partial_{x_i}}h)(\partial_{x_j},\partial_{x_\ell})$. By tracing the coordinate expression for the Riemann curvature tensor in the definition of the Ricci tensor, the Ricci tensor has local coordinate expression \begin{align*} \operatorname{Ric}(g)_{ij} = \sum_{k=1}^n \partial_{x_k}\Gamma_{ij,k} - \sum_{k=1}^n \partial_{x_j}\Gamma_{ik,k} + \text{terms quadratic in the Christoffel coefficients}. \end{align*} When this expression is linearized, the quadratic Christoffel terms contribute only zeroth-order and first-order terms in $h$. Replacing the exterior partial derivatives by covariant derivatives changes only lower-order terms, because the difference between $\partial_{x_i}$ and $\nabla_i$ on tensor components is multiplication by Christoffel coefficients. Define the two second-order tensor expressions $A(h)$ and $B(h)$ by \begin{align*} A(h)_{ij} &:= \frac{1}{2}\sum_{k,\ell=1}^n (g^{-1})_{k\ell} \nabla_k \left( \nabla_i h_{j\ell} + \nabla_j h_{i\ell} - \nabla_\ell h_{ij} \right), \end{align*} and \begin{align*} B(h)_{ij} &:= \frac{1}{2}\sum_{k,\ell=1}^n (g^{-1})_{k\ell} \nabla_j \left( \nabla_i h_{k\ell} + \nabla_k h_{i\ell} - \nabla_\ell h_{ik} \right). \end{align*} Hence the second-order part of $D\operatorname{Ric}_g(h)$ is \begin{align*} (D\operatorname{Ric})_g(h)_{ij} &\equiv A(h)_{ij}-B(h)_{ij}, \end{align*} where $\equiv$ denotes equality modulo terms of differential order at most one in $h$.[/step]

[guided]We only need the second-order part because the principal symbol ignores every term involving at most one derivative of $h$. The Christoffel symbols depend on first derivatives of the metric, so their variation contains first derivatives of $h$. Differentiating the varied Christoffel symbols inside the Ricci tensor is therefore the unique source of second derivatives of $h$. Fix $p \in M$ and a smooth coordinate chart $(U,x)$ around $p$. The notation $\Gamma_{ij,k}$ means the scalar function on $U$ defined by \begin{align*} \nabla_{\partial_{x_i}}\partial_{x_j}=\sum_{k=1}^n \Gamma_{ij,k}\partial_{x_k}. \end{align*} Let $g_s$ be a smooth family of metrics with $g_0=g$ and variation $h=\frac{d}{ds}\big|_{s=0}g_s$. Differentiating the Koszul formula for the Levi-Civita connection gives the tensorial variation formula for $(D\nabla)_g(h)$; in the chart $(U,x)$, its first-order part has coefficients \begin{align*} (D\Gamma_{ij,k})_g(h) = \frac{1}{2}\sum_{\ell=1}^n (g^{-1})_{k\ell} \left( \nabla_i h_{j\ell} + \nabla_j h_{i\ell} - \nabla_\ell h_{ij} \right) + \text{terms of order zero in } h. \end{align*} The displayed part is first order in $h$. The omitted part comes from differentiating inverse-metric and coefficient factors, contains no derivative of $h$, and after freezing coefficients at $p$ contributes only lower-order terms to the principal symbol. Tracing the coordinate expression for the curvature tensor gives the Ricci tensor formula \begin{align*} \operatorname{Ric}(g)_{ij} = \sum_{k=1}^n \partial_{x_k}\Gamma_{ij,k} - \sum_{k=1}^n \partial_{x_j}\Gamma_{ik,k} + \text{terms quadratic in the Christoffel coefficients}. \end{align*} The quadratic Christoffel terms contain only the metric and its first derivatives. Their linearization therefore contains at most first derivatives of $h$, so they are lower order for a second-order operator. Also, replacing $\partial_{x_i}$ by $\nabla_i$ changes tensor-component formulas by terms involving Christoffel coefficients multiplying lower derivatives of $h$; this again changes only lower-order terms. Thus the second-order part is obtained by differentiating only the first-order part of $D\Gamma$: \begin{align*} (D\operatorname{Ric})_g(h)_{ij} &\equiv \sum_{k=1}^n \nabla_k (D\Gamma_{ij,k})_g(h) - \sum_{k=1}^n \nabla_j (D\Gamma_{ik,k})_g(h). \end{align*} Substituting the first-variation formula for $(D\Gamma)_g(h)$ into this expression gives \begin{align*} (D\operatorname{Ric})_g(h)_{ij} &\equiv \frac{1}{2}\sum_{k,\ell=1}^n (g^{-1})_{k\ell}\nabla_k\left(\nabla_i h_{j\ell}+\nabla_j h_{i\ell}-\nabla_\ell h_{ij}\right) - \frac{1}{2}\sum_{k,\ell=1}^n (g^{-1})_{k\ell}\nabla_j\left(\nabla_i h_{k\ell}+\nabla_k h_{i\ell}-\nabla_\ell h_{ik}\right). \end{align*} The symbol $\equiv$ means equality after discarding terms with at most one derivative of $h$. This is exactly the [equivalence relation](/page/Equivalence%20Relation) relevant to the principal symbol.[/guided]

[step:Replace the second covariant derivatives by covector components]The operator $L_g=D(-2\operatorname{Ric})_g$ is $-2$ times the preceding linearization. The second-order part of $L_g(h)$ is therefore \begin{align*} L_g h_{ij} &\equiv -\sum_{k,\ell=1}^n (g^{-1})_{k\ell}\nabla_k\left(\nabla_i h_{j\ell}+\nabla_j h_{i\ell}-\nabla_\ell h_{ij}\right) + \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\nabla_j\left(\nabla_i h_{k\ell}+\nabla_k h_{i\ell}-\nabla_\ell h_{ik}\right). \end{align*} The principal symbol is obtained by replacing each occurrence of $\nabla_a\nabla_b h$ by multiplication by $\xi_a\xi_b h$ and freezing the coefficients at $p$. Hence \begin{align*} \sigma_\xi(L_g)(h)_{ij} &= -\sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_k\left(\xi_i h_{j\ell}+\xi_j h_{i\ell}-\xi_\ell h_{ij}\right) + \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_j\left(\xi_i h_{k\ell}+\xi_k h_{i\ell}-\xi_\ell h_{ik}\right). \end{align*} Using the symmetry $h_{ab}=h_{ba}$ and the definitions of $|\xi|_g^2$ and $\operatorname{tr}_g h$, this becomes \begin{align*} \sigma_\xi(L_g)(h)_{ij}=|\xi|_g^2 h_{ij}+\xi_i\xi_j\operatorname{tr}_g h-\sum_{k,\ell=1}^n \xi_i(g^{-1})_{k\ell}\xi_\ell h_{kj}-\sum_{k,\ell=1}^n \xi_j(g^{-1})_{k\ell}\xi_\ell h_{ki}. \end{align*} This is the asserted formula.[/step]

[guided]The operator under consideration is $L_g=D(-2\operatorname{Ric})_g$, so its second-order part is obtained by multiplying the second-order part of $D\operatorname{Ric}_g$ by $-2$. Thus \begin{align*} L_g h_{ij} &\equiv -\sum_{k,\ell=1}^n (g^{-1})_{k\ell}\nabla_k\left(\nabla_i h_{j\ell}+\nabla_j h_{i\ell}-\nabla_\ell h_{ij}\right) + \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\nabla_j\left(\nabla_i h_{k\ell}+\nabla_k h_{i\ell}-\nabla_\ell h_{ik}\right). \end{align*} To form the principal symbol at $p$ in the covector direction $\xi\in T_p^*M$, we freeze the coefficient functions $(g^{-1})_{k\ell}$ at $p$ and replace each second covariant derivative $\nabla_a\nabla_b h$ by multiplication by $\xi_a\xi_b h$. Lower-order terms do not contribute by definition of the principal symbol. This gives \begin{align*} \sigma_\xi(L_g)(h)_{ij} &= -\sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_k\left(\xi_i h_{j\ell}+\xi_j h_{i\ell}-\xi_\ell h_{ij}\right) + \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_j\left(\xi_i h_{k\ell}+\xi_k h_{i\ell}-\xi_\ell h_{ik}\right). \end{align*} Now use the symmetry $h_{ab}=h_{ba}$. The positive term in the first sum is \begin{align*} \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_k\xi_\ell h_{ij}=|\xi|_g^2h_{ij}, \end{align*} and the trace term in the second sum is \begin{align*} \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_j\xi_i h_{k\ell}=\xi_i\xi_j\operatorname{tr}_g h. \end{align*} Relabelling dummy indices in the two remaining contraction terms yields \begin{align*} \sigma_\xi(L_g)(h)_{ij}=|\xi|_g^2 h_{ij}+\xi_i\xi_j\operatorname{tr}_g h-\sum_{k,\ell=1}^n \xi_i(g^{-1})_{k\ell}\xi_\ell h_{kj}-\sum_{k,\ell=1}^n \xi_j(g^{-1})_{k\ell}\xi_\ell h_{ki}. \end{align*} This is precisely the displayed formula in the theorem statement.[/guided]

[step:Exhibit kernel directions generated by symmetric products with $\xi$]Assume now that $\xi \ne 0$. For any $\eta \in T_p^*M$, define $h_{\xi,\eta}\in S^2T_p^*M$ by \begin{align*} (h_{\xi,\eta})_{ij}:=\xi_i\eta_j+\xi_j\eta_i. \end{align*} Let \begin{align*} a:=\sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_k\eta_\ell. \end{align*} Then \begin{align*} \operatorname{tr}_g h_{\xi,\eta} = \sum_{k,\ell=1}^n (g^{-1})_{k\ell}(\xi_k\eta_\ell+\xi_\ell\eta_k) = 2a, \end{align*} and \begin{align*} \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_\ell (h_{\xi,\eta})_{kj}=\sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_\ell(\xi_k\eta_j+\xi_j\eta_k)=|\xi|_g^2\eta_j+a\xi_j. \end{align*} The analogous contraction with the index $i$ gives \begin{align*} \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_\ell (h_{\xi,\eta})_{ki} = |\xi|_g^2\eta_i+a\xi_i. \end{align*} Substituting these three identities into the symbol formula gives \begin{align*} \sigma_\xi(L_g)(h_{\xi,\eta})_{ij}=|\xi|_g^2(\xi_i\eta_j+\xi_j\eta_i)+2a\xi_i\xi_j-\xi_i(|\xi|_g^2\eta_j+a\xi_j)-\xi_j(|\xi|_g^2\eta_i+a\xi_i)=0. \end{align*} Thus $h_{\xi,\eta}\in \ker\sigma_\xi(L_g)$ for every $\eta\in T_p^*M$. Since $\xi\ne 0$, choosing $\eta=\xi$ gives \begin{align*} h_{\xi,\xi}=2\xi\otimes\xi\ne 0. \end{align*} Therefore $\ker\sigma_\xi(L_g)$ contains a non-zero element, so $\sigma_\xi(L_g)$ has non-trivial kernel for every non-zero $\xi\in T_p^*M$.[/step]

[guided]Fix a non-zero covector $\xi\in T_p^*M$. The kernel directions we test are the symmetric products of $\xi$ with another covector. For any $\eta\in T_p^*M$, define the symmetric tensor $h_{\xi,\eta}\in S^2T_p^*M$ by \begin{align*} (h_{\xi,\eta})_{ij}:=\xi_i\eta_j+\xi_j\eta_i. \end{align*} Define the scalar $a\in\mathbb{R}$ by \begin{align*} a:=\sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_k\eta_\ell. \end{align*} This scalar is the $g$-[inner product](/page/Inner%20Product) of the covectors $\xi$ and $\eta$ after raising one index with $g^{-1}$. First compute the trace of $h_{\xi,\eta}$. Since $g^{-1}$ is symmetric and $h_{\xi,\eta}$ is symmetric, \begin{align*} \operatorname{tr}_g h_{\xi,\eta} &= \sum_{k,\ell=1}^n (g^{-1})_{k\ell}(\xi_k\eta_\ell+\xi_\ell\eta_k) = 2a. \end{align*} Next compute the contraction appearing in the third term of the symbol. Expanding the definition of $h_{\xi,\eta}$ gives \begin{align*} \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_\ell (h_{\xi,\eta})_{kj} = \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_\ell(\xi_k\eta_j+\xi_j\eta_k). \end{align*} Separating the two summands and using the definition of $a$ gives \begin{align*} \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_\ell (h_{\xi,\eta})_{kj} = |\xi|_g^2\eta_j+a\xi_j. \end{align*} The same computation with $j$ replaced by $i$ gives \begin{align*} \sum_{k,\ell=1}^n (g^{-1})_{k\ell}\xi_\ell (h_{\xi,\eta})_{ki} &= |\xi|_g^2\eta_i+a\xi_i. \end{align*} Substituting these three identities into the principal-symbol formula yields \begin{align*} \sigma_\xi(L_g)(h_{\xi,\eta})_{ij} = |\xi|_g^2(\xi_i\eta_j+\xi_j\eta_i)+2a\xi_i\xi_j -\xi_i(|\xi|_g^2\eta_j+a\xi_j) -\xi_j(|\xi|_g^2\eta_i+a\xi_i). \end{align*} The two $|\xi|_g^2$ terms cancel the corresponding products $|\xi|_g^2\xi_i\eta_j$ and $|\xi|_g^2\xi_j\eta_i$, while the two $a$-terms cancel $2a\xi_i\xi_j$. Hence \begin{align*} \sigma_\xi(L_g)(h_{\xi,\eta})_{ij}=0. \end{align*} Thus every tensor $h_{\xi,\eta}$ lies in $\ker\sigma_\xi(L_g)$. To make this kernel element non-zero, choose $\eta=\xi$. Since $\xi\ne 0$, the tensor \begin{align*} h_{\xi,\xi}=2\xi\otimes\xi \end{align*} is non-zero in $S^2T_p^*M$. Hence $\ker\sigma_\xi(L_g)$ contains a non-zero element for every non-zero $\xi\in T_p^*M$.[/guided]