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In a nutshell: a proposition in Abraham's Foundations of Mechanics states that, given a vector space $E$ with basis $e_1,\ldots,e_n$, the space $T^r_s(E)$ of tensors of type $(r,s)$, the dual basis $\alpha^1,\ldots,\alpha^n$, we have that $$\{ e_{i_1}\otimes\dots\otimes e_{i_r}\otimes\alpha^{j_1}\otimes\dots\otimes\alpha^{j_s} : i_k,j_k = 1,\ldots,n\}$$ is a basis of $T^r_s(E)$.

I'm immediately confused as the tensor product is defined between tensors, and $e_{i_1},\ldots,e_{i_r}$ are vectors, not tensors. What is the meaning of the above set then?

The proposition (and its proof, in case that helps):

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Sam
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  • A vector is a $(1,0)$ tensor. – K.defaoite Aug 13 '23 at 18:10
  • @K.defaoite could you explain how? – Sam Aug 13 '23 at 18:13
  • It is very cheap to explain what $\otimes$ is by simply saying that it is nothing else than the Kronecker product of the vector components. Books that omit that pedagogical part should not be read. – Kurt G. Aug 13 '23 at 18:25
  • @KurtG. It's very useful for someone just learning about tensors. However, not every book is written with them in mind. "Should not be read first" is more correct IMO. – Arthur Aug 13 '23 at 18:52
  • @Arthur I agree. It was a bit of a provocation. The book has in mind as "basic" audience "mathematicians, physicists and engineers interested in geometrical methods in mechanics assuming, a background in calculus, linear algebra, some classical analysis and point set topology." When it was published in 1967 I was only two years old and cannot comment on the knowledge one could assume of that audience about $\otimes$ at that time. – Kurt G. Aug 13 '23 at 19:31
  • Correction: published 1978. – Kurt G. Aug 13 '23 at 19:39
  • @KurtG that seems relevant to know, but I'm still not sure what the meaning of $e_i$ in the tensor product of the excerpt means. – Sam Aug 13 '23 at 20:04
  • $e_i$ and $e_j$ are basis vectors of $E,.$ The author assumes dim$E=n$. So: please take $E=\mathbb R^n$ choose your favourite basis, and write out the Kronecker product $e_i\otimes e_j$ in components of those basis vectors. Best to edit the question to do so. We will then check. (This is a very important mission. Engineers in 1978 could not have been that much smarter than we are today. :) ) – Kurt G. Aug 13 '23 at 20:12
  • @Sam Where are you learning about tensors? Have you not seen something like this ? – K.defaoite Aug 14 '23 at 01:28

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$v \in E$ is naturally associated to $V \in T^1_0(E)$ via $V(\alpha) = \alpha(v)$ for all $\alpha \in E^*$, and $\alpha \in E^*$ is by definition an element of $T^0_1(E)$ since $v \mapsto \alpha(v)$ is a linear map that takes in a vector and produces a scalar.

The text is implicitly making these identifications and then applying the already defined tensor product of $T^r_s(E)$.

Nicholas Todoroff
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