Given a covariant derivative, is there a metric tensor that has the covariant derivative as the metric connection?

Asked May 04 '21 at 08:24

Active May 04 '21 at 08:24

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In differential geometry or general relativity, we usually think of the metric tensor $g_{\mu \nu}$ first and then introduce the metric connection.

However, I wonder if we can go reverse.

That is, let $M$ be a smooth manifold equipped with some nontrivial affine connection $\nabla$. Then does there always exist a metric tensor $g_{\mu \nu}$ on $M$ that has $\nabla$ as its metric connection?

I tried to figure this out myself but it is more confusing than expected. Could anyone help me?

asked May 04 '21 at 08:24

Keith

7,673

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A first necessary condition would be that your covariant derivative is torsion-free. – Didier May 04 '21 at 08:34
Oh yes, that must be assumed of course. Then can we proceed? – Keith May 04 '21 at 08:36
Do you mean just being metric or do you also need torsion free? – Arctic Char May 04 '21 at 09:16
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@Keith: See https://mathoverflow.net/questions/54434/when-can-a-connection-induce-a-riemannian-metric-for-which-it-is-the-levi-civita – levap May 04 '21 at 09:19
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https://math.stackexchange.com/questions/789394/levi-civita-connection-compatible-with-riemaniann-and-pseudo-riemaniann-metric/789782#789782 – Moishe Kohan May 04 '21 at 18:39

Given a covariant derivative, is there a metric tensor that has the covariant derivative as the metric connection?

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