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!!!