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Is there a name for a mathematical space with a distance and ordering defined, and a subtraction operation defined, but no addition operation or scalar multiplication? Essentially this is like the real number line with no notion of an origin.

The concrete example I'm thinking of is datetimes; given two datetimes, we can subtract them to find an interval. We can add/subtract an interval to/from a datetime, or multiply an interval by a scalar, but we can't add two datetimes or multiply a datetime by a scalar.

So if d1 and d2 are datetimes, then d1 + (d1 - d2) has a well-defined meaning, but (d1 + d1) - d2 does not.

I'm also just realizing - the result of subtraction is not an element in the same space, so I guess this is a family of two sets $A$ and $B$, where

  • $a_i - a_j \in B$
  • $a_i \pm b_j \in A$
  • $b_i \pm b_j \in B$
  • $k \cdot b_i \in B$
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    $\begingroup$ Not sure what you are asking. What axioms would you want your subtraction to satisfy? Of course, you could take something like the integers and just "forget" about addition and think only about subtraction. Is that an example of what you want? $\endgroup$
    – lulu
    Dec 30, 2022 at 22:29
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    $\begingroup$ I would call this an "affine line". $\endgroup$ Dec 30, 2022 at 22:31

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This is called an "affine line", or more generally, an "affine space". Quoting from Wikipedia,

An affine space is a set $A$ together with a vector space $\overrightarrow{A}$, and a transitive and free action of the additive group of $\overrightarrow{A}$ on the set $A$.

In your example of datetimes, the set $A$ is the set of datetimes and $\overrightarrow{A}$ is the set of time intervals $B$. Wikipedia also quotes Marcel Berger:

An affine space is nothing more than a vector space whose origin we try to forget about, by adding translations to the linear maps.

The action of $\overrightarrow{A}$ on $A$ is transitive, which means that for every $a,b$ there is a $v \in \overrightarrow{A}$ so that $a+v = b$, where $+$ here indicates the action of $\overrightarrow{A}$ on $A$. Furthermore, since the action of $\overrightarrow{A}$ on $A$ is free, the $v$ here is unique. This allows us to define the subtraction $b-a:=v \in \overrightarrow{A}$ of any $a,b \in A$.

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