Show that if $T_i$ are the components of covariant vector T, then $S_{ij}=T_iT_j-T_jT_i$ are the components of a skew-symmetric covariant tensor S.

The question is whenever working with equations of components form in tensor analysis, does the order in which the vector/tensor components are placed in an inner/outer product matter? i.e. $g_{\alpha \beta} \frac{\partial x^\alpha}{\partial z^\mu}\frac{\partial x^\beta}{\partial z^v}= \frac{\partial x^\beta}{\partial z^v}g_{\alpha \beta}\frac{\partial x^\alpha}{\partial z^\mu}, ds^2=g_{ij}dx^idx^j=dx^jg_{ij}dx^i$ ?

I've always thought that since the components themselves are scalars, such an order should be irrelevant. However, if I was correct, then why don't the terms $T_iT_j-T_jT_i $ cancel to zero?

  • 4
    $\begingroup$ I'm pretty sure you're correct and the exercise is meant to involve two vectors $T_i,V_j$ and the combination $T_i V_j - T_j V_i$. $\endgroup$ – jacob1729 Jun 2 '19 at 22:04
  • $\begingroup$ @ jacob1729 Ok thanks. But are there any exceptions to this "irrelevance of order" rule and how would we know that it is not applicable e.g. link when clearly only the components of the vectors (scalars) are being referred to ? $\endgroup$ – Chern-Simons Jun 2 '19 at 22:13
  • $\begingroup$ In your link the differential line elements $dx^i$ are implicitly being multiplied by the anti-symmetric wedge product $a\wedge b = -b \wedge a$. I've seen that done in a few places before but really I find it confusing and would much rather they write the wedges explicitly. $\endgroup$ – jacob1729 Jun 2 '19 at 22:19
  • $\begingroup$ Well.. actually $S_{ij}=0$ is skew-symmetric and covariant $\endgroup$ – acarturk Jun 3 '19 at 20:31

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