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Let $(M,g)$ be a Riemannian manifold, endowed with the Levi-Civita connexion $\nabla$ induced by $g$. By the very definition of the Levi-Civita connexion $\nabla$, we indeed know that $\nabla g=0$, i.e., the (total) covariant derivative of the metric tensor vanishes. Now assume that $G$ is another metric tensor field on $M$, such that it satisfies $\nabla G=0$, that is, the (total) covariant derivative of this alternative metric tensor vanishes, as well.

I have realised that trivially $G=k g$, where $k\in\mathbb{R}^+$, is a solution for $\nabla G=0$. However, I am interested to know whether this is the most general case, or else if there are other alternative metric tensors $G$ with vanishing covariant derivatives (with respect to the Levi-Civita connextion induced by $g$), which are not positive multiples of the original metric tensor $g$.

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    $\begingroup$ In Euclidean space, there are many. For instance both $dx^2 +dy^2$ and $dx^2 +2dy^2$ are parallel with respect to the same connection. More generally, given any metric tensor $g$ we can apply the transformation $Ag$ where $A$ is a symmetric positive definite matrix to get another parallel metric. $\endgroup$
    – Gabe K
    Jun 18, 2022 at 13:45
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    $\begingroup$ @GabeK Interesting! That's true since in a flat Riemannian manifold, we always have chart in which the connexion components $\Gamma^i_{jk}$ of $\nabla$ vanish. So the equation $\nabla G=0$ is reduced locally to $\partial{G_{ij}}/\partial{x^k}=0$ which results in $G_{ij}$ to be constants, hence your correct examples. $\endgroup$
    – Mohamad Emami
    Jun 18, 2022 at 14:13

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In an irreducible Riemannian manifold $(M,g)$, every symmetric tensor that satisfies $\nabla^{g}T=0$ must be of the form $T=kg$ for some constant $k$. See Theorem 10.3.2 in Chapter 10 of Peter Peterson's Riemannian geometry, 3rd Edition. It is shown there how to use this fact to conclude that irreducible symmetric spaces are in fact Einstein manifolds.

In a reducible manifold, this mustn't be the case: given $(M,g)$ simply connected and complete, a theorem of de Rham tells us that it can be written as a product $(M,g)=(M_{1}\times...\times M_{n},g_{1}+...+g_{n})$, where the sum of the metric tensors is orthogonal, and each $(M_{i},g_{i})$ is an irreducible space. Then, one can define $T=k^{i}g_{i}$ for any real constants $k^{i}$, which may be different from one another. Thus $T$ is not a multiplication of $g$ by a factor. If $(M,g)$ is not simply-connected or not complete, then this simply becomes a local argument, where $M$ and $M_{i}$ are replaced with local simply connected neighborhoods.

The example above in the comments is a private case of this: a locally-flat space can always be locally decomposed into an orthogonal Riemannian products of $\mathbb{R}$.

I suspect that the other direction is also true, namely that every $\nabla^{g}T=0$ must be locally of this form. But I have no reference for this nor verified this myself.

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    $\begingroup$ The global de Rham Theorem requires not only the hypothesis of simple connectivity but also that the metric be complete. The local de Rham Theorem does not need either hypothesis. $\endgroup$
    – Robert Bryant
    Jun 18, 2022 at 14:14
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    $\begingroup$ Thank you, I forgot completeness. I fixed in the body of the answer. $\endgroup$
    – MySheperd
    Jun 18, 2022 at 14:19
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    $\begingroup$ Yes, and I think that on a general manifold locally every such tensor is of the form $G=\sum_{i} k_{i}g_{i}$ for orthogonal metrics $g_{i}$ such that $g=\sum_{i}g_{i}$. I think this is correct, but I am not sure. $\endgroup$
    – MySheperd
    Jun 18, 2022 at 20:20
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    $\begingroup$ @MySheperd: Minor correction: it is "Petersen", not "Peterson". $\endgroup$
    – Alex M.
    Jun 19, 2022 at 10:27
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    $\begingroup$ @Alex M. Thanks, sorry for any other typo that might come up :) $\endgroup$
    – MySheperd
    Jun 19, 2022 at 11:38

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