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!!!



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


