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Suppose I have some set of quantities $T_{ji···m···l···k}$. If the particular contraction $T_{ji···m···m···k}$ of this set of quantities is tensorial in nature, then is it true the that original quantity is tensorial in nature?

I ask this question in relation to this question from Riley, Hobson, and Bence. I believe I need to use the above lemma to solve the problem. Bonus points if you can prove the lemma as a special case of the quotient rule, although I don't think that's possible.

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2 Answers 2

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I think that the answer is no:

Suppose that $T_{ij}=1$ for each basis and pair of indices, then $\sum_iT_{ii}$ is independent of the basis (i.e. a scalar), but $T$ is clearly not a tensor.

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You can think of tensors as multilinear map on some module $\mathscr{G}$ (call this Type-I tensors). You can also write ${T^{ab...}}_{pq...}=\sum A^aB^b\cdots P_pQ_q\cdots$ where each element ($A^a\in\mathscr{G}^a$, $B^b\in\mathscr{G}^b$ etc) (Type-II tensors). Using contraction of the form

$${T^{ab...}}_{pq...}X_aY_b\cdots U^pV^q\cdots=\sum (A^aX_a)(B^bY_b)\cdots (P_pU^p)(Q_qV^q)\cdots \in \mathscr{G}$$

we see that the above expression defines a $\mathscr{G}$ multi-linear map. Thus, for any type II tensor, there exists an unique type-I tensor. However, it doesn't guarantee that all type I tensors are obtainable from type II tensors in this way and that there is a 1-1 correspondence between these two definitions. OP's question can be reformulated as : can all type-I tensors be obtained from type-II tensors? It is only possible when $\mathscr{G}$ is totally reflexive, i.e. an isomorphism exists between module $\mathscr{G}^{\alpha}$ and its dual $\mathscr{G}_{\alpha}$. Say, we choose $\mathscr{G}$ to be $\mathscr{C}^{\infty}$ smooth functions on a manifold, but let $\mathscr{G}^{\alpha}$ to be space of $\mathscr{C}^0$ smooth vector fields. In this case we see that the dual space $\mathscr{G}_{\alpha}$ can only have zero element, as no other vector fields $f_{\alpha}$ when contracted with $\mathscr{C}^0$ vector field will give $\mathscr{C}^{\infty}$ smooth functions. Total reflexivity is guaranteed, if $\mathscr{G}^{\alpha}$ has a finite basis (at least locally). An elaborate proof of this proposition is discussed in sections 2.3 and 2.4 of Spinors and space-time, Volume-I.

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  • $\begingroup$ My knowledge of the varoius algebraic structures you mention here is unfortunately woefully inadequate, so I'm afraid I don't completely understand. Would you be able to add a more narrow addendum which simply states (without proof even) if it is true that $T_{ij}x_jx_k$ is a tensor if $T_{ij}x_jx_i$ (a contraction of the former) is (in the particular problem I mention)? $\endgroup$
    – EE18
    Dec 18, 2022 at 14:10
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    $\begingroup$ $T_{ij}x^ix^k$ can be a tensor provided (i) $x^j:\mathscr{G}_j\to\mathscr{G}$ is a linear map and $T_{ij}$ is a composition of such linear maps and (ii) $\mathscr{G}^i$ has finite basis. You may verify that these criterions are not satisfied in the example given by Filippo $\endgroup$
    – KP99
    Dec 18, 2022 at 16:04
  • $\begingroup$ How is the function$$x^i:\mathscr G_i\to\mathscr G$$defined? $\endgroup$
    – Filippo
    Dec 18, 2022 at 22:39
  • $\begingroup$ In particular, how is $\mathscr G_i$ - i.e. the domain of $x^i$ - defined? $\endgroup$
    – Filippo
    Dec 18, 2022 at 22:56
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    $\begingroup$ The type-I and type-II tensors. It's best if you look at those particular sections in spinors and space-time $\endgroup$
    – KP99
    Dec 20, 2022 at 11:37

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