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If I had two tensors of rank $3$ and wanted their inner product, what would it be? Also, how could I represent the process with indices and please explain that? Could someone demonstrate this with two specific rank three tensors, with the elements shown? I have no idea how to show that or how the unit components would work. Thank you!

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  • $\begingroup$ What do you mean by "inner product" in this case? What do you expect the result to be? A scalar? Something else? $\endgroup$
    – Muphrid
    May 25, 2013 at 5:57
  • $\begingroup$ For a vector space $V$ over a field $\mathbb{k}$, an inner product is a linear map $V \otimes V \to \mathbb{k}$, $v \otimes v' \mapsto (v,v')$. If $V$ has a basis $\{v_1, \dots, v_n\}$, then the inner product is specified by the $n \times n$ matrix of $a_{i,j} = (v_i, v_j) \in \mathbb{k}$. $\endgroup$ May 25, 2013 at 6:16

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I think what you want to do is to multiply each element of the first tensor by the corresponding element of the second tensor, and add the results. At least, this is the most useful way of taking inner products of vectors and matrices. In case of order 3 tensors ("rank" usually means something different), the formula would look like $$\sum_{ijk}a_{ijk}b_{ijk}$$ assuming the numbers are real. Numerical example: two tensors $2\times 2\times 2$; the first is filled with numbers $5$, the second has $b_{ijk}=i+j+k$. Then you compute $$a_{111}b_{111}=5\cdot 3=15$$ $$a_{112}b_{112}=5\cdot 4=20$$ $$\dots$$ $$a_{222}b_{222}=5\cdot 6=30$$ and add all these products together.


(That said, without knowing what you want to do with this inner product, I can't be sure that this approach does what you want.)

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