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I have a question about tensors. Okay, you could say that this question would fits more properly on Math.StackExchange but, here it's more cozy. Anyway:

A vector, as we all learned in first year, could be written in this form: \begin{equation} \textbf{V} = \sum A_{i} \textbf{e}_{i} \end{equation}

Okay, the thing begins more "rich" with linear algebra when, due to the formalism, one could now tell the difference between a contravariant vector and a covariant vector. Respectively:

\begin{equation} \textbf{V} = \sum A^{i} \textbf{e}_{i} , \textbf{V} = \sum A_{i} \textbf{e}^{i} \end{equation}

If you are quite curious you may reach the study of tensors. Following the same "idea" one could now write a tensor (of rank 3) with the same "shape" of a vector. A contravariant tensor and a covariant tensor. Respectively:

\begin{equation} \textbf{T} = \sum \sum \sum T^{ijk} \textbf{e}_{i}\textbf{e}_{j}\textbf{e}_{k},\\ \textbf{T} = \sum \sum \sum T_{ijk} \textbf{e}^{i}\textbf{e}^{j}\textbf{e}^{k} \end{equation}

Question:

The first thing that we learn when begin to talk about tensors, in a linear algebra perspective, is the tensor product. Because of that I know that what I've wrote above is, in fact:

\begin{equation} \textbf{T} = \sum \sum \sum T^{ijk} \textbf{e}_{i}\otimes \textbf{e}_{j}\otimes \textbf{e}_{k} \end{equation}

The point is, I still did not grasp the notion that $\textbf{e}_{i}\otimes \textbf{e}_{j}$ could be a fundamental way to think about inner products for example, or even what $\textbf{e}_{i}\otimes \textbf{e}_{j}\otimes \textbf{e}_{k}$ really really means in this concept; can someone help with this?. I mean "producing tensors" in the form as above.

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  • $\begingroup$ Hi Countto10. So, I'm Brazillian, and most of my sources are in portuguese. But, books like: Marion (classical mechanics), Hartle (general relativity); Greub (linear algebra) and Hoffman-Kunze (Linear algebra) are my main sources. $\endgroup$ – M.N.Raia Jul 2 '17 at 12:50
  • $\begingroup$ I think one way to think is that they "eat" eachother. Ex: $[e_1\otimes e_2](e^1\otimes e^2)=e^1(e_1)\cdot e^2(e_2)$. Ex: $[e^1\otimes e^2](e_1\otimes e_2)=e^1(e_1)\cdot e^2(e_2)$ $\endgroup$ – Emil Jul 2 '17 at 16:22
  • $\begingroup$ I also believe that contraction is when you just feed one of them. Ex: $C(2,1)(a\otimes b\otimes c,d)= b^i(d_i)a\otimes c$ (note: haven't verified this. also used a makeshift notation since I haven't found the real notation). $\endgroup$ – Emil Jul 2 '17 at 16:34
  • $\begingroup$ Also, the components are $T^{ijk}=\mathbf{T}(e^i\otimes e^j \otimes e^k)$. The $e_i\otimes e_j \otimes e_k$ are there to handle all other cases (since T is multilinear you can let them handle the other possible arguments for T). Unless I am mistaken. $\endgroup$ – Emil Jul 2 '17 at 16:49
  • $\begingroup$ One more thought, maybe the contraction example above should have been written like $C(2,4)(a\otimes b\otimes c\otimes d)$, to stay in tensor-land, so to speak.. $\endgroup$ – Emil Jul 2 '17 at 20:09
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due to the formalism, one could now tell the difference between a contravariant vector and a covariant vector

This is incorrect: covariant (respectively contravariant) does not refer to the vector, rather to the transformation properties of the components of such vector along some basis.

A vector $\mathbf{v}$ is an element of a vector space $V$. A co-vector is a map from a vector space into a field $\alpha\colon V\to\mathbb{F}$ and as such an element of the dual vector space $V^*$. A tensor of type $(r,s)$ is a map $\tau\colon V^r\times {V^*}^s\to\mathbb{F}$.

Given the above, a tensor product of two tensors $T\colon A\to X$, $S\colon B\to Y$ is another linear map defined as $$ T\otimes S\colon A\otimes B\to X\otimes Y $$ such that $(T\otimes S)(a\otimes b)=T(a)\otimes S(b)$ however you choose $a\in A, b \in B$.

By linearity you can construct tensor product of any types.

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