If $U\in\bigotimes_{k=1}^p\mathbb R^{n_k}$ and $U^{(k)}\in\mathbb R^{r_{k-1}}\otimes\mathbb R^{n_k}\otimes\mathbb R^{r_k}$ with$^1$ $$U_{i_1,\:\ldots\:,i_p}=\sum_{j_0=1}^{r_0}\cdots\sum_{j_p=1}^{r_p}U^{(1)}_{j_0i_1j_1}\cdots U^{(p)}_{j_{p-1}i_pj_p}\tag1,$$ then $(1)$ is called a *tensor-train decomposition* of $U$. $(1)$ has to be understood with respect to the isomorphisms $\bigotimes_{k=1}^p\mathbb R^{n_k}\cong\mathbb R^{n_1\times\cdots\times n_p}$ and $\mathbb R^{r_{k-1}}\otimes\mathbb R^{n_k}\otimes\mathbb R^{r_k}\cong\mathbb R^{r_{k-1}\times n_k\times r_k}$.

Is there a generalization of this decomposition type to general $\mathbb R$-vector spaces (or Banach/Hilbert spaces) $E_i,F_j$? Something like $U\in\bigotimes_{k=1}^p E_k$, $U^{(k)}\in F_{k-1}\times E_k\times F_k$, $$U=\iota\bigotimes_{k=1}^pU^{(k)}\tag2,$$ where some identification/transformation $\iota$ is applied to the right-hand side?

I'm quite sure that it should be possible to write $(1)$ in the form of $(2)$ using a suitable tensor contraction$^2$. I'm strongly interested in a way to express $(1)$ without the identification with matrices.

**EDIT 1**: Clearly, we have the canonical isomorphisms, \begin{equation}\begin{split}\Xi_k:\mathbb R^{r_{k-1}\times n_k\times r_k}&\to\mathbb R^{r_{k-1}}\otimes\mathbb R^{n_k}\otimes\mathbb R^{r_k},\\\left(a_{j_{k-1}i_kj_k}\right)&\mapsto\sum_{j_{k-1}=1}^{r_{k-1}}\sum_{i_k=1}^{n_k}\sum_{j_k=1}^{r_k}a_{j_{k-1}i_kj_k}\left(f^{(k-1)}_{r_{k-1}}\otimes e^{(k)}_{i_k}\otimes f^{(k)}_{j_k}\right),\end{split}\end{equation} where $\left(e^{(k)}_1,\ldots,e^{(k)}_{n_k}\right)$ and $\left(f^{(k)}_1,\ldots,f^{(k)}_{r_k}\right)$ denote the standard bases of $\mathbb R^{n_k}$ and $\mathbb R^{r_k}$, respectively, and $$\Lambda_k:\mathcal L(\mathbb R^{n_k},\mathbb R^{r_{k-1}}\otimes\mathbb R^{r_k})\to\mathbb R^{r_{k-1}}\otimes\mathbb R^{n_k}\otimes\mathbb R^{r_k}\;,\;\;\;A\mapsto\sum_{i_k=1}^{n_k}e^{(k)}_i\otimes Ae^{(k)}_i.$$ Now, it's easy to see that $$\Lambda_k^{-1}\left(U^{(k)}\right)=\sum_{i_k=1}^{n_k}\left\langle\;\cdot\;,e^{(k)}_{i_k}\right\rangle\operatorname{tr}_{21}\left(U^{(k)}\otimes e^{(k)}_{i_k}\right)\tag3.$$ So, $$\Lambda_k^{-1}\left(U^{(k)}\right)(x_k)=\operatorname{tr}_{21}\left(U^{(k)}\otimes x_k\right)\;\;\;\text{for all }x_k\in\mathbb R^{n_k}.\tag4$$ Does this allow us to rewrite $(1)$?

**EDIT 2**: I guess the natural generalization is the following: Let $E_i$ be a $\mathbb R$-vector space, $I$ be a finite nonempty set and $U\in\bigotimes_{i\in I}E_i$. Let $\langle\;\cdot\;,\;\cdot\;\rangle_{\bigotimes_{i\in I}E_i,\:\bigotimes_{i\in I}E_i^\ast}$ denote the canonical duality pairing between $\bigotimes_{i\in I}E_i$ and $\bigotimes_{i\in I}E_i^\ast$. Then we may look for a factorization of $$\left\langle U,\bigotimes_{i\in I}\varphi_i\right\rangle_{\bigotimes_{i\in I}E_i,\:\bigotimes_{i\in I}E_i^\ast},\tag5$$ where $\varphi_i\in E_i^\ast$ for $i\in I$. In the case of $E_i=\mathbb R^{n_i}$ we obtain the matrix entry $U_{i_1,\:\ldots\:,i_p}$ by inserting for $\varphi_{i_k}$ the basis functional corresponding to the $i_k$th standard basis vector of $\mathbb R^{n_k}$. Now, I guess the cores are naturally assumed to be some $U^{(i)}\in F_{i-1}^\ast\otimes E_i\otimes F_i$, where $F_i$ is another $\mathbb R$-vector space. I guess we need to build the trace $$\operatorname{tr}\left(\bigotimes_{i\in I}\varphi_i\otimes U^{(i)}\right)\in\bigotimes_{i\in I}F_{i-1}^\ast\otimes F_i\subseteq\bigotimes_{i\in I}\mathcal L(F_{i-1},F_i).$$ Now I think the operators induced by the embedded on the right-hand side only need to be concatenated (which might be express in terms of the tensor constraction as well; see my related question Tensor contraction (vector-valued trace) on $\bigotimes_{i=1}^k\mathcal L(E_{i-1},E_i)$) and we need to assume that $\dim F_0=\dim F_p=1$.

It would be great if someone could put these pieces together to formulate a generalization.

$^1$ $r_0=r_p=1$.

$^2$ Let $I,J\subseteq\mathbb N$ be finite and nonempty and $E_i,F_j$ be $\mathbb R$-vector spaces for $(i,j)\in I\times J$. If $I_0\subseteq I$ and $J_0\subseteq J$ with $|I_0|=|J_0|\ge1$, $\phi:I_0\to J_0$ denotes the canonical bijection and $$E_i=F_{\phi(i)}^\ast\;\;\;\text{for all }i\in I_0\tag6,$$ then the *tensor contraction (or vector-valued trace) with respect to $(I_0,J_0)$* is given by \begin{equation}\begin{split}\operatorname{tr}_{I_0J_0}:&\bigotimes_{k\in I}E_k\otimes\bigotimes_{l\in J}F_l\to\bigotimes_{k\in I\setminus I_0}E_k\otimes\bigotimes_{l\in J\setminus J_0}F_l,\\&\bigotimes_{k\in I}x_k\otimes\bigotimes_{l\in J}y_l\mapsto\prod_{i\in I_0}\left\langle x_i,y_{\phi(i)}\right\rangle_{F_{\phi(i)}}\bigotimes_{k\in I\setminus I_0}x_k\otimes\bigotimes_{l\in J\setminus J_0}y_l.\end{split}\end{equation} If $(i,j)\in I\times J$, then $\operatorname{tr}_{ij}:=\operatorname{tr}_{\{i\}\{j\}}$.