Vector space over a set which is not commutative

Question

I was wondering whether you could have a vector space over a set of elements $S$ which do not satisfy all of the Abelian group properties. That is since scalar multiplication and vector addition are just functions, could you define them in such a way that it compensates the lack of commutativeness/associativeness in the set $S$?

You can define such a structure. That won’t be a vector space though. — mathcounterexamples.net, May 04 '21 at 05:56
As soon as you have a group which admits scalar multiplication, it must be abelian (provided $1$ is a valid scalar). See here for instance. — Chris Grossack, May 04 '21 at 06:20

score 6 · Answer 1 · answered May 04 '21 at 06:33

6

As others say, it won't be a vector space since it does not satisfy the axioms. However, you can have similar structures with more relaxed axioms. One case that is commonly studied is a module over a ring.

Module at Wolfram

answered May 04 '21 at 06:33

badjohn

8,854

score 3 · Accepted Answer · answered May 04 '21 at 05:59

A vector space, by definition, is defined over a field, which is a set that satisfies the Abelian group properties.

So no, you cannot define a vector space over a set of elements which is not Abelian, because whatever you define will, by definition, not be a vector space.

Vector space over a set which is not commutative

2 Answers2