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EDIT: I thought on rephrasing the question in another way:

I have been working recently with a tensor that satisfies

$A_{ijkl}=A_{i+b,j+b,k+b,l+b}$ $\forall$ $i,j,k,l$ $\in$ Z

$$dist(i,j,k,l)\leq M$$

where all indices are meant to be integers (also b with $b\geq 0$), and dis(i,j,k,l) is the distance between all pair of indices, so $|i-j|\leq M$, $|i-k|\leq M$ etc... with 6 total distances. Because of this symmetry, it is said that one can just fix one of the values of the indices, say $i=0,1,..,b-1$ and generate the other elements from the symmetry relation above.

I am struggling to see this, as for example, considering the case of a matrix $C_{i+b,j+b}=C_{i,j}$ it is clear to me how can one do this: one just calculates values for a single row, say $i=0$ and since $C_{ij}=0\forall |i-j|>M$, one is left with $2M+1$ independent terms. Then, the rest of matrix elements can be derived by using the symmetry relation:

$$C_{i+b,j+b}=C_{i,j}$$

However, for the $A$ tensor above I am having difficulties to see this and how one could in principle recover all the missing elements of the set, if I set $i=0$ and calculate for the other indices. How can one get, for example $A_{1,1,2,3}$ if we only now those terms for $i=0$ ( that is, we know $A_{0,jkl}$ only ) for the case $b=1$?

Thanks !!

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  • $\begingroup$ It is not true; you are only allowing simultaneous addition to all indices at once, not each index separately. Think again about matrices (which, in your setting, are I suppose of infinite dimension). Draw matrix entry $C_{ij}$ at position $(i,j)$. Your symmetry allows you to equate entries as you move by a vector $(r,r)$ in the plane, not by $(0,r)$ or $(r,0)$. $\endgroup$
    – Ben McKay
    Commented May 18, 2020 at 12:48
  • $\begingroup$ I forgot to add that the tensor must satisfy $dist(i,j,k,l)\leq N$. This is also confusing me, as what is meant here by dist(i,j,k,l) is meant by the pair combinations of indices, so $|i-j|\leq N$ , $|i-k|\leq N$. $|i-l|\leq N$ ... and so on. Also, the matrix is required to satisfy the symmetry above with $|i-j| \leq N$. In that case, you can fix one of the matrix indices $i$ and compute a single row; the rest follows by symmery but this is what I don't see in the tensor case $\endgroup$
    – Zarathustra
    Commented May 18, 2020 at 13:00
  • $\begingroup$ So, you want to count the set of integer $4$-tuples $(i,j,k,l)$ with $|i|,|j|,|k|,|l|,|i-j|,\dotsc,|k-l|\leq N$, modulo the equivalence relation $(i,j,k,l)\sim(i+b,j+b,k+b,l+b)$? $\endgroup$
    – Zach Teitler
    Commented May 18, 2020 at 15:50
  • $\begingroup$ Yes, but what is confusing me is the notation $dist(i,j,k,l)$, is this supposed to represent pair-wise distances between the integers? For the case of a matrix, this would mean that elements of distance $|i-j|>N$ away from the diagonal are zero. $\endgroup$
    – Zarathustra
    Commented May 18, 2020 at 16:02

1 Answer 1

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Because of this symmetry, it is said that one can just fix one of the values of the indices, say $i=0,1,..,b-1$ and generate the other elements from the symmetry relation above.

The issue is very simple. For each integer $i’$ there exists $i $ defined above such that $i’=i+tb$ for some integer $t$. Then $$A_{i’j’k’l’}=A_{i+tb,(j’-tb)+tb, (k’-tb)+tb,(l’-tb)+tb}= A_{i,j’-tb, k’-tb,l’-tb}.$$

Remark that the addition of $tb$ to each index keeps the distances between the indices.

for example $A_{1,1,2,3}$ if we only now those terms for $i=0$ ( that is, we know $A_{0,jkl}$ only ) for the case $b=1$?

$A_{1,1,2,3}=A_{0+b,0+b,1+b,2+b}= A_{0,0,1,2}$.

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    $\begingroup$ Yes indeed, many thanks for the reply! $\endgroup$
    – Zarathustra
    Commented May 22, 2020 at 11:09

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