Consider $(M,g)$ a compact Riemannian manifold. When viewed as a second order (non-linear) differential operator $$ \text{Ric} : C^{\infty}(\text{Sym}^2_+T^*M) \to C^{\infty}(\text{Sym}^2T^*M), $$ the Ricci operator (taking a metric $g$ to its Ricci tensor $\text{Ric}(g)$) is seen to be non-elliptic. I'm reading the discussion about this in Chow and Knopf's The Ricci Flow: An Introduction (this is section 2 in chapter 3), where they claim that this non-ellipticity is due only to the invariance of the Ricci tensor under diffeomorphisms, i.e. $$ \text{Ric}(\varphi^*g) = \varphi^*\text{Ric}(g). $$ I'm trying to understand exactly to what extent is this statement true; is the non-ellipticity of $\text{Ric}$ equivalent to that invariance? Now what they first do is show that from this invariance, you can deduce that the kernel of the symbol $$ \sigma(D\text{Ric}_g(\xi)) : \text{Sym}^2T^*M \to \text{Sym}^2T^*M $$ has dimension at least $n = \dim M$ for $\xi \neq 0$. Basically, differentiating this invariance gives a complex $$ 0 \to T^*M \overset{\sigma(\delta_g^*)(\xi)}\longrightarrow \text{Sym}^2T^*M\overset{\sigma(D \text{Ric}_g)(\xi)}\longrightarrow \text{Sym}^2T^*M $$ for $\delta_g^*$ the $L^2$ adjoint of the divergence operator acting on $\text{Sym}^2T^*M$. So if we show that in fact $\ker(\sigma(D\text{Ric}_g))$ has dimension $n$ by showing that $\text{im}(\sigma(D\text{Ric}_g))$ has dimension $\frac12n(n-1)$, we will know that all the non-ellipticity is caused by this invariance.
Now conversely, to what extent does the non-ellipticity of $\text{Ric}$ implies its invariance under diffeomorphisms? I feel like Chow and Knopf's discussion is trying to adress this but I'm not completely satisfied with my understanding. What they do is show that the invariance wrt flows associated to vector fields implies the second contracted Bianchi identity $$ \nabla^jR_{ij} = \frac12\nabla_iR $$ which in turn implies an infinitesimal version of the invariance, namely $$ ((D\text{Ric}_g) \circ \delta_g^*)(\alpha) = \frac12\mathcal{L}_{\alpha^{\sharp}}(\text{Ric}_g) $$ for $\alpha$ a 1-form. They claim that this means the contracted second Bianchi identity is equivalent to the invariance of the Ricci tensor under diffeomorphisms but isn't this identity strictly weaker than the Bianchi identity?
In any case, they then express the contracted second Bianchi identity as $B_g(\text{Ric}(g)) = 0$ for $$ B_g : C^{\infty}(\text{Sym}^2T^*M) \to C^{\infty}(T^*M) $$ the Bianchi operator given by $B_g(h)_k = g^{ij}(\nabla_ih_{jk} - \frac12\nabla_kh_{ij})$. Then to show that the kernel of $\sigma(D\text{Ric}_g(\xi))$ is actually $n$-dimensional, they continue the sequence of symbols: $$ 0 \to T^*M \overset{\sigma(\delta_g^*)(\xi)}\longrightarrow \text{Sym}^2T^*M \overset{\sigma(D \text{Ric}_g)(\xi)}\longrightarrow \text{Sym}^2T^*M \overset{\sigma(B_g)(\xi)} \longrightarrow T^*M \to 0. $$ The contracted second Bianchi identity reading $B_g \circ \text{Ric}_g = 0$, it implies that $\text{im}(\sigma(D\text{Ric}_g)) \subset \ker(\sigma(B_g))$ and a little analysis of $\sigma(B_g)$ actually shows that this sequence is exact. But this means that $\dim(\text{im}(\sigma(D\text{Ric}))) = \frac12n(n-1)$ as wanted because then $\dim(\ker\sigma(B_g)(\xi)) = \frac12n(n-1)$.
Let's call the dimension of $\ker\sigma(D\text{Ric}_g)$ the number of degrees of non-ellipticity of $\text{Ric}_g$. I'd like to read the equivalence from this exact sequence as follows: The exactness at the first $\text{Sym}^2T^*M$ is a consequence of the invariance of $\text{Ric}$ under diffeomorphisms and contributes $n$ degrees of non-ellipticity to $\text{Ric}$. Conversely, if there are exactly $n$ degrees of non-ellipticity, this would imply the exactness at the second $\text{Sym}^2T^*M$ by counting dimensions, which (infinitesimally, at the level of symbols) is the contracted second Bianchi identity, which might imply the invariance under diffeomorphisms of $\text{Ric}$.
Is there any way to make this last argument rigorous? I really don't see how I can go from a statement on the symbol of the linearization of an operator to the analogous statement on the actual operator. Any vague ideas would be appreciated! If it helps, I'm reading this to try and really understand the idea behind deTurck's trick of "killing diffeomorphism invariance to make Ricci flow parabolic". I'd like to get a proof of the parabolicity of Ricci-deTurck flow without even having to calculate its symbol.
