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I am having trouble understanding TAS. Let $(X,+)$ be a finite abelian group. A translation assiociation scheme is an association scheme $(X, \mathbf{R})$so that for all $(x,y) \in R_i \implies (x+z,y+z) \in R_i \quad \forall i \quad \forall z\in X$.

It is mentioned that if $(X, \mathbf{R})$ is a TAS and $X_i = \{x\in X: (x,0) \in R_i\}$ then $\{X_0,...,X_n\}$ forms a partition of $X$ and $R_i = \{ (x,y) \in X \times X: x-y \in X_i\}$.

  1. I dont really see why or what the point behind this statement is. Is it enough to have a Partition of $X$ (there are many ways to partition $X = \mathbb{Z_6}$ for example) and then the $R_i$ are given then? (because i know one symmetric association scheme with 6 points and 3 classes).

  2. It is said that the Hamming scheme $H(n,q)$ is also a TAS with $X = \mathbb{Z}_q^n$ and partition $X_i = \{ x \in X: wt_H(x) = i\}$ where $wt_H(x)$ counts the number of entries which are not $0$. I have seen already that it is an association scheme but why is it true if only the partition is given (or is there more to it) ?

azimut
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Youniz
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  • Please ask one question at a time. – Shaun Oct 15 '22 at 20:24
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    Is the definition of "translation association scheme" meant to say "... $(x,y) \in R_i \implies (x+z,y+z) \in R_i$ ..."? I am only guessing because I don't know what an association scheme is. If you have a chance, could you also define that term (or link to a standard definition?) – diracdeltafunk Oct 15 '22 at 21:30
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    @diracdeltafunk you are right i edited my question. A good introduction into the standard definition is given here https://encyclopediaofmath.org/wiki/Association_scheme – Youniz Oct 15 '22 at 21:51

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An association scheme is a partition $\{R_0,\ldots,R_d\}$ of $X \times X$ with certain properties. For a TAS, the partition can alternatively be given as $\{X_0,\ldots,X_d\}$ (with $X_i$ defined as in your question), which is a simpler description as it is a partition of $X$ (and not $X\times X$). The $R_i$ can be recovered as stated in your question.

For a TAS, the derived $R_i$'s must still satisfy the axioms of a translation scheme, of course. So for your first point, not all partitions of $\mathbb{Z}_6$ will give a TAS, but only very specific ones.

For your second point, in the moment that it is understood that the Hamming scheme is a TAS, it is natural to ask what the alternative partition $\{X_0,\ldots,X_d\}$ looks like. This is answered by the statement in your source.

azimut
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  • Thank you this cleared alot of things up. I have only one more question. To understand that the Hamming scheme is indeed a TAS it should be enough to see that the hamming distance is invariant under translation for all elements $z \in X$ right? – Youniz Oct 15 '22 at 23:24
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    @Youniz Yes, if you already have that the Hamming scheme is an association scheme, the only additional property needed for showing that it is a TAS is the translation invariance. – azimut Oct 16 '22 at 10:21