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Let $d_i\in\mathbb N$, $I_i:=\{1,\ldots,d_i\}$ and $u\in\mathbb R^{d_1}\otimes\mathbb R^{d_2}\otimes\mathbb R^{d_3}$. It's somehow clear to me that we may regard $u$ as a three-dimensional array (see Corollary 3 below) and hence consider its entry $u_{i_1i_2i_3}$ at the index $(i_1,i_2,i_3)\in I_1\times I_2\times I_3$. By fixing the middle index $i_2\in I_2$, we may define $$v^{(i_2)}_{i_1i_3}:=u_{i_1i_2i_3}\;\;\;\text{for }(i_1,i_3)\in I_1\times I_3$$ and I guess we can again consider $v^{(i_2)}$ as being an element of $\mathbb R^{d_1}\otimes\mathbb R^{d_3}$.

However, I'm really struggling to see that all the identifications involved are really legitimate. I'm used to think of tensor products in a more abstract fashion (see my definition of the tensor product below$^1$).

Is the process of "fixing the middle index", which is effectively a (surjective?) transformation from $\mathbb R^{d_1}\otimes\mathbb R^{d_2}\otimes\mathbb R^{d_3}$ to $\mathbb R^{d_1}\otimes\mathbb R^{d_3}$, a special case of a more general result of somehow "folding" a tensor product space $E_1\otimes E_2\otimes E_3$ to $E_1\otimes E_3$?

Remark: Tensors $u$ of order 3 like this are so-called cores in the tensor-train decomposition which I'm trying to understand.


$^1$ If $E_i$ is a $\mathbb R$-vector space, I'm defining $$(x_1\otimes x_2)(B):=B(x_1,x_2)\;\;\;\text{for }B\in\mathcal B(E_1\times E_2)\text{ and }x_i\in E_i,$$ where $\mathcal B(E_1\times E_2)$ is the space of bilinear forms on $E_1\times E_2$, and $$E_1\otimes E_2:=\operatorname{span}\{x_1\otimes x_2:E_i\in E_i\}\subseteq{\mathcal B(E_1\times E_2)}^\ast.$$

Lemma 1: If $I$ is a finite nonempty set, $(e_i)_{i\in I}$ is the standard basis of $\mathbb R^I$ and $E$ is a $\mathbb R$-vector space, then the linear extension $\iota_1$ of $$a\otimes x\mapsto(a_ix)_{i\in I}\tag1$$ is an isomorphism between $\mathbb R^I\otimes E$ and $E^I$.

Lemma 2: If $I_i$ is a finite nonempty set, then $$\iota_2:\left(\mathbb R^{I_2}\right)^{I_1}\to\mathbb R^{I_1\times I_2}\;,\;\;\;a\mapsto\left(a(i_1)(i_2)\right)_{(i_1,\:i_2)\in I_1\times I_2}$$ is an isomorphism.

Corollary 3: If $k\in\mathbb N$ and $d_1,\ldots,d_k\in\mathbb N$, then $$\bigotimes_{i=1}^k\mathbb R^{d_i}\cong\mathbb R^{d_1\times\cdots\times d_k}\tag2.$$ Proof: This follows by induction from Lemma 1 and Lemma 2.

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Yes, mapping $A \otimes B \otimes C$ to $A \otimes C$ by choosing a coordinate of $B$ (equivalently, basis element of dual space $B^*$) is indeed a special case of choosing any element of $B^*$, considering it as a map $B \to k$ (where $k$ is the field), and then extending that map to $A \otimes B \otimes C \to A \otimes k \otimes C \cong A \otimes C$.

Your $v^{(i_2)}$ is sometimes called a slice, or more specifically a $2$-slice, since the index in the $2$nd position was fixed. This map, that sends a tensor to its $i_2$'th $2$-slice, is surjective: any matrix arises as the $2$-slice of a tensor given by sticking that matrix into the appropriate entries, and filling the rest with any old thing (zeros, whatever).

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  • $\begingroup$ Thank you for your answer. The isomorphism is $(a\otimes b\otimes c)(b^\ast):=b^\ast(b)(a\otimes b)$, right? Is there an established notation which lets me denote the evaluation of $(a\otimes b\otimes c)$ at a basis functional in a more convenient way? The introduction of the notation $v^{(i_2)}_{i_1i_3}$ is quite annoying. $\endgroup$ – 0xbadf00d May 21 at 4:55
  • $\begingroup$ I'm not sure of any really great notation. Sometimes it's called a "contraction" (contracting by $b^*$) and they use a symbol like $\neg$ ($b^* \neg a \otimes b \otimes c = b^*(b)(a \otimes c)$), but I suspect that's far from universal. I think there are different conventions and different notations in geometry, physics, engineering, representation theory... :-( One problem with saying "oh we'll just write $b^*$ for the map contracting tensors by $b^*$ in the 2nd entry" is, what if $A=B=C$, so $b^*$ could act in any position? Ugh. Maybe someone else has a good idea, but I don't know, sorry. $\endgroup$ – Zach Teitler May 21 at 5:01
  • $\begingroup$ The notion of contraction that I'm aware of is also called the vector-valued trace. It is the mapping $$(A\otimes B^\ast)\otimes(B\otimes C)\to A\otimes C\;,\;\;\;(a\otimes b^\ast)\otimes(b\otimes c)\mapsto b^\ast(b)a\otimes c.$$ Is this what you mean? $\endgroup$ – 0xbadf00d May 21 at 5:07
  • $\begingroup$ Yeah, pretty much that. I guess I was thinking primarily of the situation where you have a fixed element $b^* \in B^*$, yielding a map $A \otimes B \otimes C \to A \otimes C$; but that's just a restriction or specialization of what you wrote. I suppose the identification behind this is $\operatorname{Hom}(B^*, \operatorname{Hom}(A \otimes B \otimes C, A \otimes C)) \cong \operatorname{Hom}((A \otimes B^*) \otimes (B \otimes C), A \otimes C)$. Your vector-valued trace is an element of the right-hand Hom. My version of contraction is a particular map in the left-hand Hom. $\endgroup$ – Zach Teitler May 21 at 5:24
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    $\begingroup$ Okay, denoting the linearization of $$(E_1⊗ E_2^\ast)\times(E_2⊗ E_3)\to E_1⊗ E_3\;,\;\;\;(x_1⊗φ_2)\times(x_2⊗ x_3)\mapsto\langleφ_2,x_2\rangle_{E_2}x_1⊗ x_3$$ by $\operatorname{tr}_{E_2}$, we may note that for any $φ_2\in E_2^\ast$, the linear extension $\iota_{φ_2}$ of $$x_1⊗ x_2⊗ x_3\to x_1⊗φ_2⊗ x_2⊗ x_3$$ is an embedding of $E_1⊗ E_2⊗ E_3$ into $(E_1⊗ E_2^\ast)⊗(E_2⊗ E_3)$ and hence we could define $\operatorname{slice}_{φ_2}:=\operatorname{tr}_{E_2}\circ\:\iota_{φ_2}$. By choosing $φ_2$ to be a basis functional, we can express $v^{(i_2)}_{i_1i_3}$ in terms of this slice operator. $\endgroup$ – 0xbadf00d May 21 at 11:40

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