# Symmetric tensor components

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 !!

• 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 $(r,0)$. May 18, 2020 at 12:48
• 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 May 18, 2020 at 13:00
• 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 $(i,j,k,l)\sim(i+b,j+b,k+b,l+b)$? May 18, 2020 at 15:50
• 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. May 18, 2020 at 16:02

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