Hamming metric isometries
Let $C$ be a vector subspace of $\mathbb{F}_q^n$, and define the Hamming distance as follows: If $v, w \in C$, then $d(v,w) =$ number of non-$0$ coordinates of $v-w$, and $d(v) = d(v,0)$.
For a linear map $f:C \to D$ between two subspaces, we say $f$ is \textbf{distance-preserving} if $d(f(v))=d(v)$, for all $v \in C$, and f is \textbf{adjacency-preserving} if $d(f(v))=d(v)$, for all $v \in C$ with $d(v)=1$.
It is obvious that a distance-preserving map is adjacency-preserving, but is the converse true?
Answer
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Thank you very much! Do you maybe have a concrete counterexample, of the non-vacuous nature?
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Yes, for example, consider the maps that sends v = (x,y,z,...) to f(v) = (x+y+z+..., 0,...). Clearly, if d(v) = 1 then d(f(v)) = 1, too. But you can obviously have d(v) > 1, while you always have d(f(v)) = x+y+z+... = 1 or 0 (e.g., for y=-x nonzero and others zero).
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* (please ignore the last "x+y+z...", it should not be there.)
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thanks so much!
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you're welcome. (I've added this example to the main answer, and also mention that for n=1, the two conditions are (obviously) equivalent, so in that case you (trivially) do have the converse.)
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Hey Ribs, I saw your other two questions about Euclidean Lattices, but can't comment there. IDK what the "taker" has answered but it's obvious that you can't have such a gamma because for A=B you have Ax - Bx = o ... Feel free to ask me (IDK how...) before offering $50 for an uninteresting question!
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