I Vanishing of Contraction with Metric Tensor

[kent davidge]

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.

 

[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]

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.

 

[haushofer]

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.

From a Linear Algebra point of view, you take the trace. A traceless matrix is not necessarily the zero matrix.
Also, do the counting: $g_{\mu \nu} C^{\mu \nu} = 0$ puts one constraint on your tensor $C^{\mu \nu}$, while in D dimensions a general tensor $C^{\mu \nu}$ has $D^2$ components. The condition that $C^{\mu \nu} = C^{[\mu \nu]}$, i.e. $C^{\mu \nu}$ is antisymmetric, means $C^{(\mu \nu)} = 0$, which are $\frac{1}{2}D(D+1)$ constraints. So $g_{\mu \nu} C^{\mu \nu} = 0$ cannot imply that $C^{(\mu \nu)} = 0$. Of course, $C^{(\mu \nu)} = 0$ does imply that $g_{\mu \nu} C^{\mu \nu} = g_{\mu \nu} C^{(\mu \nu)} = 0$; you get a linear combination of zeroes.

So, to answer your question: no, most definitely not.