# Is a symmetric, parallel (0,2)-tensor a metric?

I'm interested in affinely connected spaces, on which a metric is not necessarily defined, i.e. $$(\mathcal{M},\Gamma)$$. Since (as a physicist) my goal is to consider a generalized model of gravity, I restrict myself to the space of possible connections compatible with a symmetry group, by requiring that the Lie derivative of the connection---along the vectors associated with the generators of the symmetry group---vanishes: $$\mathcal{L}_\xi \Gamma^a_{bc} = \xi^m \partial_m \Gamma^a_{bc} - \Gamma^m_{bc} \partial_m \xi^a + \Gamma^a_{mc} \partial_b \xi^m + \Gamma^a_{bm} \partial_c \xi^m + \frac{\partial^2 \xi^a}{\partial x^b \partial x^c} = 0.$$

In this space I've been able of define a symmetric tensor of type $$\binom{0}{2}$$, $$T_{ab}$$, which is parallel under the action of the torsion-free connection compatible with the symmetries (as defined above), i.e. $$\nabla^{\Gamma} T = 0$$.

In General relativity, the metric, $$g$$, is a symmetric $$\binom{0}{2}$$-tensor, that is required to satisfy the metricity condition, $$\nabla g = 0$$, i.e. it is parallel, under the (Levi-Civita) connection.

## Question(s)

Given the similarities a natural question is: Can the tensor $$T_{ab}$$ be interpreted like a metric? Is there a theorem ensuring that a parallel tensor of $$\binom{0}{2}$$-type is a metric?

## Side note

A while ago, I found a set of notes about holonomy groups saying that (if I remember well) the existence of a parallel $$\binom{0}{2}$$-tensor was equivalent to restricting the holonomy group to $$O(N)$$---where $$N = \dim(\mathcal{M})$$.

It sound to me like the action of the parallel transportation preserves the length of the vectors. Am I right?

Does it mean that starting from an affinely connected space I've end up in a Riemannian space? If so, Where did I make the transition? Since I've pulled the tensor $$T$$ out from a hat!!!

• What if $T=0$? You need to check that $T$ is nondegenerate, i.e. that the determinant of $T_{ab}$ is not zero. This nonzeroness is invariant, even though the determinant is not a scalar, as the determinant is a tensor in a tensor power of the volume form bundle. – Ben McKay Dec 7 '18 at 9:41
• Also, you might want to know if $T_{ab}$ is positive definite, or some other signature, to see if you are studying a Riemannian, Lorentzian, or more general pseudo-Riemannian manifold. – Ben McKay Dec 7 '18 at 9:43

You don't need parallelism: a symmetric tensor of type $$\binom{0}{2}$$ is positive definite if and only if it is a Riemannian metric, and has signature $$(1,n-1)$$ (where $$n$$ is the dimension of your manifold) if and only if it is a Lorentzian metric, and has signature $$(p,q)$$ if and only if it is a $$(p,q)$$ pseudo-Riemannian metric. But it is also true that an affine connection has holonomy contained in the orthogonal group $$O(p,q)$$ if and only if there is a pseudo-Riemannian metric of signature $$(p,q)$$ parallel for that affine connection.