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A tensor exercise in a text reads: If $T_i$ are the components of a covariant vector $T$, show that $S_{ij}:=T_iT_j-T_jT_i$ is an order 2 covariant tensor $S$.

Am I missing something or is $S$ uniformly zero?

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Dear student: it would be helpful if you made explicit what text this is, as the source can make it easier to make sense of it (the tiny detail of whether this is a text intended for mathematicians or physicists would already be a useful piece of information!) – Mariano Suárez-Alvarez Jan 13 '12 at 21:57

No, but it is certainly antisymmetric. consider the two 1-tensors, $\partial_i$, and $A_j$, then look at $F_{ij}=\partial_i A_j - \partial_j A_i$. This has zero components, but they are not all zero... Maybe go through them component by component until that's clear :)

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You maybe speaking as a physicist (and then rules do not apply...) but «$\partial_i$» is not a tensor according to most definitions of what a tensor is (and there are a couple!) – Mariano Suárez-Alvarez Jan 13 '12 at 21:50
Wait, you seem to have misread the OP's question. The expression isn't $T_iU_j-T_jU_i$, it's $T_iT_j-T_jT_i$. If T is a tensor over the reals (as opposed to some kind of operator whose components might not commute), then S is zero, and the answer to the problem is that it's trivially true, since the zero tensor is indeed a tensor. – Ben Crowell Jan 14 '12 at 1:53
Ah, indeed, silly me. Is this the same Ben Crowell from PF? if so, cheers. – kηives Jan 14 '12 at 4:54

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