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This is a re-editing of a prerviously posted question:

Let $(M,g)$ be a Riemannian manifold. Let $C:TM\to TM$ be symmetric positive definite. Define the metric $$ (X,Y)_C = (X,CY)_g. $$ Denote by $\nabla^g$ and $\nabla^C$ the Levi-Civita connections of $g$ and $C$ and by $d^{\nabla^g}$ and $d^{\nabla^C}$ the corresponding exterior covariant derivatives. Denote by $R^g$ and $R^C$ the corresponding curvature operators (i.e., the $(2,2)$-tensors which are two forms with values in $\wedge^2TM$).

We denote by $K_g(TM)$ and $K_C(TM)$ the bundles of algebraic curvatures with respect to $g$ and $C$, i.e., the sub-bundles of $End(\wedge^2TM)$ of tensors satisfying an algebraic Bianchi identity (which depends on the metric).

I am looking for a "simple" relation between the curvature operators $R^g$ and $R^C$.

The first observation is that those two tensors do not satisfy the same symmetries: one is a section of $K_g(TM)$ and the other is a section of $K_C(TM)$. However, $(C^{-1})^*R^C$ is a section of $K_g(TM)$. Moreover (unless I am wrong), both $R^g$ and $(C^{-1})^*R^C$ satisfy the same differential Bianchi identity, $$ d^{\nabla^g} R^g = d^{\nabla^g} (C^{-1})^*R^C = 0. $$

Thus, the natural object to estimate is the difference $$ R^g -(C^{-1})^*R^C, $$ which is a section of $K_g(TM)$ satisfying a differential Bianchi identity. Since the curvature difference depends on two derivatives of $C$, the natural suspect for this difference is $$ d^{\nabla^g}(d^{\nabla^g} C)^T, $$ which as easily verified is of the desired type.

The question is whether it is true that $$ R^g -(C^{-1})^*R^C = d^{\nabla^g}(d^{\nabla^g} C)^T $$ (possibly up to a constant)?

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    $\begingroup$ It’s the top order terms which are linear. There are however a first order nonlinear terms, too. So it’s the linearization of the curvature difference. $\endgroup$
    – Deane Yang
    Mar 21, 2019 at 14:00
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    $\begingroup$ The difference between the connections is a tensor which can be written in terms of $C$ and its covariant derivative. Putting this into the definition of the curvature tensor and expanding everything out will give you the formula for the curvature difference. You’ll recognize the second order terms. $\endgroup$
    – Deane Yang
    Mar 21, 2019 at 14:27
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    $\begingroup$ Thanks. Note that nothing here is linear in $C$, since the connection $\nabla$ is the connection associated with $C$. Obviously, the problem I posed is dual to using the $g$-connection, however differentiating $C^{-1}$ rather than $C$. $\endgroup$
    – Raz Kupferman
    Mar 21, 2019 at 16:37
  • $\begingroup$ Yes. Good point. $\endgroup$
    – Deane Yang
    Mar 21, 2019 at 17:40

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