# Contraction with the metric vanishes

## Main Question or Discussion Point

This question is probably silly, but suppose I have a contraction of the form ##g_{\mu \nu} C^{\mu \nu} = 0## where ##C^{\mu \nu}## is a tensor* and ##g_{\mu \nu}## is the metric tensor. Can I say that it must vanish for any ##g_{\mu \nu}##, and since in the most general case all ##g_{\mu \nu}## are non zero, then necessarely ##C^{\mu \nu} = 0##?

*##C^{\mu \nu}## is a symmetric tensor.

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Ibix
Since ##g_{\mu\nu}## is symmetric, ##C^{\mu\nu}## being antisymmetric is enough, I think.

kent davidge
Since ##g_{\mu\nu}## is symmetric, ##C^{\mu\nu}## being antisymmetric is enough, I think.
Yes. However it turns out that my ##C^{\mu \nu}## is also symmetric (unfortunately!)

DrGreg
Gold Member
This question is probably silly, but suppose I have a contraction of the form ##g_{\mu \nu} C^{\mu \nu} = 0## where ##C^{\mu \nu}## is a tensor* and ##g_{\mu \nu}## is the metric tensor. Can I say that it must vanish for any ##g_{\mu \nu}##, and since in the most general case all ##g_{\mu \nu}## are non zero, then necessarely ##C^{\mu \nu} = 0##?

*##C^{\mu \nu}## is a symmetric tensor.
No. Counterexample: ##C^{\mu \nu} = K^\mu K^\nu ## where ##\mathbf{K}## is any (non-zero) null vector.

Ibix and kent davidge
haushofer