Help with a paper about tensors

Question

I came across something in a paper I am not able to understand jet. Unfortunately the author is kind of short with explanations. Maybe someone here can help me to understand this.

$M^d \in \mathfrak{R}^{n_1 \times n_2 \times \cdots \times n_d}$ is a $d$-$th$ order tensor with $n_1 \leq n_2 \leq \cdots \leq n_d$.

Rewrite $M^d$ as $(d-1)$-$th$ order tensor $M^{d-1}$ by combing its first and last components into one, and place the combined component into the last one in $M^{d-1}$, i.e.,

$$ M^d_{i_1 ,i_2 , ...,i_d} = M^{d-1}_{i_2 ,i_3 , ...,i_{d-1},(i_1 -1)n_d + i_d}, \forall 1 \leq i_1 \leq n_1, ...,1 \leq i_d \leq n_d$$

First I was thinking about tensor relaxation, but this reduces the rank by 2. And the hole place in last component wont make any sense to me. And if for all $1 \leq i_1 \leq n_1$ and $1 \leq i_d \leq n_d$ the combined component is placed in the last one this would mean by my understanding the last component is $n_1 n_d$ long.

Could someone maybe show me this by a random 4-$th$ order tensor, how to reduce it to a 3-$rd$ order tensor?

Thanks in advance for any help.

EDIT: The paper defines a tensor as follow

Consider the following multi-linear unction

$$F(x^1,x^2,...,x^d) = \underset{1 \leq i_1 \leq n_1, 1\leq i_2 \leq n_2 , ... , 1\leq i_d \leq n_d}{\sum}a_{i_1 i_2 ... i_d}x_{i_1}^1x_{i_2}^2\cdots x_{i_d}^d,$$

where $x^k \in \mathfrak{R}^{n_k}, k =1,2,...,d$. In the shorthand notation we shall denote $M = (a_{i_1 i_2 ... i_d}) \in \mathfrak{R}^{n_1 \times n_2 \times \cdots \times n_d}$ to b a $d$-th order tensor.

$\\$

EDIT 2: $\\$

I am still confused trying to imagine a tensors look by this method. Since $M^d \in \mathfrak{R}^{n_1 \times n_2 \times \cdots \times n_d}$ and $n_1 \leq n_2 \leq \cdots \leq n_d$. Let’s assume I have a $M_{2,3,4,5}$ tensor. The resulting tensor would be $M_{3,4,10}$.

But how to respahe a tensor like this:

$$ M_{:,:,1,1} = \left( \begin{array}{ccc} 1 & 2 & 3\\ 4 & 5 & 6 \end{array} \right) ; \ \ \ M_{:,:,2,1} = \left( \begin{array}{ccc} 7 & 8 & 9\\ 10 & 11 & 12 \end{array} \right) ; \ \ \ ... \\ M_{:,:,1,2} = \left( \begin{array}{ccc} 25 & 26 & 27\\ 28 & 29 & 30 \end{array} \right) ; \ \ \ M_{:,:,2,2} = \left( \begin{array}{ccc} 30 & 31 & 32\\ 33 & 34 & 35 \end{array} \right) ; \ \ \ ... \\ ... $$

into a $3 \times 4 \times 10$-Form? A part of the $M_{2,3,4,5}$ tensor is shown above. It's just counting up, the dots represent the rest.

Answer 1

if N is the new $3\times 4\times 10$ tensor we have $N_{1,1,1}=M_{1,1,1,1}\ N_{1,1,2}=M_{1,1,1,2}\ N_{1,1,3}=M_{1,1,1,3}\ N_{1,1,4}=M_{1,1,1,4}\ N_{1,1,5}=M_{1,1,1,5}\ N_{1,1,6}=M_{2,1,1,1}\ N_{1,1,7}=M_{2,1,1,2}\ N_{1,1,8}=M_{2,1,1,3}\ N_{1,1,9}=M_{2,1,1,4}\ N_{1,1,10}=M_{2,1,1,5}$
and
$N_{2,1,1}=M_{1,2,1,1}\ N_{2,1,2}=M_{1,2,1,2}\ N_{2,1,3}=M_{1,2,1,3}\ N_{2,1,4}=M_{1,2,1,4}\ N_{2,1,5}=M_{1,2,1,5}\ N_{2,1,6}=M_{2,2,1,1}\ N_{2,1,7}=M_{2,2,1,2}\ N_{2,1,8}=M_{2,2,1,3}\ N_{2,1,9}=M_{2,2,1,4}\ N_{2,1,10}=M_{2,2,1,5}$
and
$N_{3,1,1}=M_{1,3,1,1}\ N_{3,1,2}=M_{1,3,1,2}\ N_{3,1,3}=M_{1,3,1,3}\ N_{3,1,4}=M_{1,3,1,4}\ N_{3,1,5}=M_{1,3,1,5}\ N_{3,1,6}=M_{2,3,1,1}\ N_{3,1,7}=M_{2,3,1,2}\ N_{3,1,8}=M_{2,3,1,3}\ N_{3,1,9}=M_{2,3,1,4}\ N_{3,1,10}=M_{2,3,1,5}$
And I hope you can continue here (if you please you could raise my reputation by checking my answer as accepted)

Answer 2

Let $M_{ijkl}$ be a degree 4 tensor. Then $M_{jkl1}$ is a degree 3 tensor, just like $M_{jkl2},M_{jkl3}, \ldots,M_{jkln_l}$. Now glue these 3-tensors together by ranging them side by side (like in the matrix case), then we get a new tensor $N_{k,l,m}$ where $m$ takes the successive values $\underbrace{\overbrace{1,2,\ldots,n_l}^{n_l \text{ items}},\overbrace{n_l+1,n_l+2,\ldots,n_l+n_l}^{n_l \text{ items}},\ldots,\overbrace{n_l(n_i-1)+1,n_l(n_i-1)+2,\ldots,n_ln_i}^{n_l \text{ items}}}_{n_i\text{ groups of $n_l$ items}}.$
Whence the sequence $(i_1 -1)n_d + i_d$ in the formula of the paper. This is essensially a very old programming trick to manipulate tables in a programming language that only admit linear arrays: If $M_{ij}$ is a $4 \times 6$ matrix it would be converted in a length $24$ array $A$ where the element $M_{ij}$ is stored in $A_{6*(i-1)+j}$