# Nontrivial second order parallel symmetric tensor field on a general Riemannian manifold？

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?

• 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 $\Bbb R^n$. Dec 16 '18 at 3:18

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.

• Thanks a lot for your answer. @Travis Dec 18 '18 at 7:01
• You're welcome, I hope you found it helpful. Dec 18 '18 at 7:11
• 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 Dec 18 '18 at 7:19