$Hamming metric isometries$

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$Hamming metric isometries$

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?

Linear Algebra Algebra

Ribs

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The converse is not (always) true. In particular, the adjacency-preserving condition may be vacuously satisfied (so that all maps have this property), but of course not all maps are distance-preserving, for example the zero map never is, unless $C = \{o\}$.

As very simple counter-example, consider a subspace C that does not contain any vector with $d(v) = 1$, for example $C$ equal to the linear span of the single vector (1, 1, 0, ..., 0): here, all nonzero vectors in $C$ have two nonzero components, so there's no $v \in C$ with $d(v) = 1$.

In that case, the adjacency-preserving condition is vacuously true, i.e., all linear maps defined on $C$ (into whatever other vector space) do have this property.

But of course not all are distance-preserving -- for example the zero map isn't (since $f(v)=o$, you have $d(f(v)) = 0$ for all vectors $v$, but $d(v) > 0$ for all nonzero vectors.

For a less trivial (counter) example, consider the maps that sends vectors $v = (x,y,...,z)$ to $f(v) = (x+y+...+z, 0,..., 0)$. Clearly, if $d(v) = 1$ (i.e., $v$ has just one nonzero component) then $d(f(v)) = 1$, too. But you can obviously have $d(v) > 1$ (unless $n=1$, in which case the two conditions are indeed trivially equivalent), while you always have $d(f(v)) = 1$ or 0 (e.g., for $y = -x$ and others zero).

M F H

345

Ribs

0

Thank you very much! Do you maybe have a concrete counterexample, of the non-vacuous nature?
- M F H
  
  0
  
  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).
- M F H
  
  0
  
  * (please ignore the last "x+y+z...", it should not be there.)
Ribs

0

thanks so much!
- M F H
  
  0
  
  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.)
M F H

0

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!

$Hamming metric isometries$

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