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It's almost the defition. A tensor $T_{ab}$ of rank $2$ is symmetric if, and only if, $T_{ab}=T_{ba}$, and antisymmetric if, and only if, $T_{ab}=-T_{ba}$. So from this definition you can easily check that this decomposition indeed yields a symmetric and antisymmetric part. Edit: Let $S_{bc}=\dfrac{1}{2}\left(A_{bc}+A_{cb}\right)$. Then ...


0

I think the answer to this is basically yes. You do have to be careful, because in general, you can't interpret inner products in relativity as measures of whether something is "orthogonal" to something else in the Euclidean sense. E.g., a lightlike vector has a zero inner product with itself. This is because the metric isn't the Euclidean metric. However, ...


1

To address the first question you can consider $$ \nabla_av^b := v^b ,_a$$ since $$ \nabla : \Gamma(E) \rightarrow \Gamma(E\otimes T^*M)$$ where E is any section(e.g. the vector field in question) and contracting it with the tangent vector field of the curve you get $$t^av^b,_a = 0 $$ this is similar to contracting a vector field with a dual vector $$ ...


2

Yes, you're exactly right, $\omega_b {C^b}_{ac} = {C^b}_{ac} \omega_b$. In general, the order of factors doesn't matter in a tensor expression like this. This is Wald, so technically you're supposed to think of those expressions as using abstract index notation instead of involving components in any particular basis, but it's also valid to think of them as ...



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