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Some literatures show that for some special manifolds (e.g. space forms, Sasakian mfd. etc.), the second-order parallel symmetric tensor fields are constant multiple of the associated metric tensor. Are there nontrivial second order parallel symmetric tensor fields on a general Riemannian manifold?

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  • $\begingroup$ The claim is not true for Euclidean space $(M, \bar g)$ (the zero-curvature space form): Every symmetric $2$-form $A_x \in S^2 T_0^* \Bbb R^n$ extends to a parallel symmetric $2$-form $A$ on $\Bbb R^n$. $\endgroup$ Dec 16 '18 at 3:18
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In general, no, the only parallel symmetric $2$-tensors on a (connected) Riemannian manifold $(M, g)$ are scalar multiples of the metric: Given any other nonzero, parallel, symmetric $2$-tensor $A$ and any point $x \in M$, the holonomy $\operatorname{Hol}_x(g)$ of $g$ is contained in the stabilizer of $A_x$ in $O(g_x)$, which is a positive-codimension subgroup, but for the holonomy group of a generic metric is $O(n, \Bbb R)$ or $SO(\Bbb R)$.

Since positive definiteness of a symmetric $2$-tensor is an open condition, if $A$ is as in the paragraph, then at least for small $\epsilon$, $g + \epsilon A$ is also a metric parallel with respect to the Levi-Civita connection $\nabla^g$ of $g$, and by uniqueness, $\nabla^g$ is also the Levi-Civita connection of $g + \epsilon A$: We say that $g, g + \epsilon A$ are affinely equivalent.

A theorem of Eisenhart describes (locally) exactly when this happens (see p. 303 of the reference): If $A$ is a parallel symmetric $2$-tensor on $(M, g)$, we can write it locally as $A = \sum_i \alpha^i g_i$ for some local decomposition $(U, g\vert_U) = (U_1, g_1) \times \cdots \times (U_n, g_n)$ of $(M, g)$ into a Riemannian product (necessarily the coefficients $\alpha^i$ are constants). In particular, if $(M, g)$ does not locally admit a Riemannian product decomposition around some point, $A$ must be a multiple of $g$.

Conversely, we can use this description to give simple examples. For any nontrivial product manifold $(M, g) \times (N, h)$, every symmetric $2$-tensor of the form $\alpha g \oplus \beta h$ is parallel, but among these tensors only those for which $\alpha = \beta$ are multiples of the product metric $g \oplus h$.

One can say more. See:

L.P. Eisenhart, "Symmetric Tensors of the Second Order Whose First Covariant Derivatives are Zero," [pdf warning] Trans. Amer. Math. Soc. 25:2 (1923), 297-306.

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  • $\begingroup$ Thanks a lot for your answer. @Travis $\endgroup$
    – G. Zhao
    Dec 18 '18 at 7:01
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    $\begingroup$ You're welcome, I hope you found it helpful. $\endgroup$ Dec 18 '18 at 7:11
  • $\begingroup$ Your answer is very helpful. By the way, if you are free, would you like to have a look at my latest questions? Thanks in advance. @Travis $\endgroup$
    – G. Zhao
    Dec 18 '18 at 7:19

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