8
$\begingroup$

@BrianM.Scott 's answer to this question Q: 3-dimensional array suggests that there is no standard concept of symmetry for 3-, 4-, N-dimensional arrays, in constrast to the case for 2-D arrays, as in linear algebra for matrices. Are there alternative definitions of symmetry for higher-dimensional arrays? Are there specific definitions that are widely used in certain contexts, e.g. in tensor calculus?

(I don't have a specific need; I'm trying to help implement a symmetry test for a matrix library [core.matrix for the Clojure language]. Since the library allows higher-dimensional arrays, there's a question about whether there is a natural choice for what the symmetry test should return for higher-dimensional arrays.)

$\endgroup$

2 Answers 2

5
$\begingroup$

I would say an $N$-way array is symmetric if an element is the same if the indices is permuted, for all permutations.

So, an array $T$ with elements $t_{i_1, i_2, \dots, i_N}$ is symmetric if for all permutations $\sigma \in S_N$, where $S_N$ is the symmetric group on $\{1, 2, \dots, N\}$, we have $$t_{i_1, i_2, \dots, i_N} = t_{i_{\sigma 1}, i_{\sigma 2}, \dots, i_{\sigma N}}$$ for all elements $t_{i_1, i_2, \dots, i_N}$.

For $2 \times 2$ arrays this reduces to just swapping the two indices, leading to the equation $t_{i,j} = t_{j,i}$.

So, if we have a $2 \times 2 \times 2$ array $T$, we want the following equalities to hold: $$ \begin{align} t_{1,1,1} \\ t_{1,1,2} = t_{1,2,1} = t_{2,1,1} \\ t_{1,2,2} = t_{2,1,2} = t_{2,2,1} \\ t_{2,2,2} \end{align} $$

$\endgroup$
2
  • $\begingroup$ Very nice Calle. Makes sense, intuitively. I guess a generalization of both yours and @Shuchang's answers would be to define a symmetry concept by any (not necessarily proper) subset of the permutations possible given the number of dimensions. Then $T$ is symmetric iff for every element $e$, it's equal to each element whose indexes are one of the specified permutations of $e$'s indexes. Maybe this is implicit in Shuchang's answer. (Hmm, a further generalization: allow arbitrary functions from indexes to others. That doesn't feel like symmetry, but might for some specialized application.) $\endgroup$
    – Mars
    Dec 28, 2013 at 19:34
  • $\begingroup$ Sure, only using certain permutations would yield a partially symmetric array. $\endgroup$
    – Calle
    Dec 28, 2013 at 21:45
5
$\begingroup$

When talking about symmetry for finite dimensional arrays, we usually need a symmetry group and an index set on which the group actions. If every element remain invariant when its index varies under action of symmetry group, we call it symmetry. For example, for torsion-free connection $\Gamma^i_{kj}=\Gamma^i_{jk}$, we call it symmetric because there's an index set $\{j,k\}$ such that element remains invariant when indices in the set varies under symmetry group $S_2$.

In general, given a set of bases $e_1,\ldots,e_n$ for each dimension of array, array can be expressed as $A=A_{i_1,\ldots,i_n}e_1\otimes\cdots\otimes e_n$. This is tensor algebra $T(V)$. To give different symmetry sub-algebras, we shall specify a equivalent relation $\sim$ on bases $e_i$'s via non trivial subgroup of a symmetry group $S_n$ (say, $e_i\otimes e_j=e_j\otimes e_i$). Then the symmetry tensor algebra is precisely $S(V)=T(V)/\sim$.

$\endgroup$
2
  • $\begingroup$ Thanks Schuchang. I'm not sure I understand everything in the answer. It sounds as if for arrays of any dimension n > 2, there is more than one concept of symmetry, and that none of these different symmetry relations is preferred (unless the context imposes some reason to prefer one over the other). Am I thinking about this the right way? $\endgroup$
    – Mars
    Dec 23, 2013 at 4:19
  • $\begingroup$ @Mars Exactly. $T_{123}=T_{213}$ is symmetric, $T_{123}=T_{132}$ is also symmetric. It depends on how you specify. $\endgroup$
    – Shuchang
    Dec 23, 2013 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.