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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!!!

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    $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Commented Dec 7, 2018 at 9:41
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    $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Commented Dec 7, 2018 at 9:43

1 Answer 1

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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.

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  • $\begingroup$ Thank you for the quick and clear answer. Could you please recommend me bibliography to continue researching about your arguments? $\endgroup$
    – Dox
    Commented Dec 7, 2018 at 12:38
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    $\begingroup$ For the definition of metric, Hawking and Ellis, The Large Scale Structure of Space Time, p. 36. For holonomy reduction, D. D. Joyce, Riemannian Holonomy Groups and Calibrated Geometry, p. 33, prop. 2.5.2. $\endgroup$
    – Ben McKay
    Commented Dec 7, 2018 at 13:26

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