1
$\begingroup$

I'm beginning to learn Abstract Algebra and my professor was talking of groups.

He said that Vector Spaces are Groups but not with Scalar Multiplication. In Vector Spaces

1.The operation of Addition if associative

2.It has an identity element namely the zero vector

3.It has an inverse, the negative of the vector.

and plus it is commutative A+B=B+A which makes it an abelian group.

But these conditions do not change with the inclusion of scalar multiplication then why does he say forget scalar multiplication.

Is there something I'm missing.

$\endgroup$
4
  • 4
    $\begingroup$ The writer is not disparaging scalar multiplication. They are simply remarking that a vector space is an abelian group under addition. Scalar multiplication is not (generally) a binary product on the underlying set so you can't really make a parallel statement about it. $\endgroup$
    – lulu
    Jul 30, 2021 at 16:46
  • $\begingroup$ Most simple examples of groups, like $\mathbb{Z}$ under addition, can be given extra structure (like by considering multiplication, or in your example scalar multiplication). This does not alter the fact that, deep down, you have a group. This group doesn't need to know about all the other fancy stuff - it forgets all that gaudy decoration, so you can too. It just cares about the one operation. $\endgroup$
    – user1729
    Jul 30, 2021 at 16:51
  • $\begingroup$ @lulu....I'm new to this...would you mind me asking...what do you mean when you say binary product? $\endgroup$
    – Orpheus
    Jul 30, 2021 at 16:56
  • 1
    $\begingroup$ A group is a set, $S$ together with a binary product...that is, a map $\mu: S\times S\to S$. That map must have certain properties. There must an identity element, i.e. some $e\in S$ with $\mu(e,s)=s=\mu(s,e)$ for all $s$, it must be associative, there must be inverses. Scalar multiplication on, say, a real vector space is a map from $\mathbb R\times V\to V$ satisfying certain properties. Very different. $\endgroup$
    – lulu
    Jul 30, 2021 at 17:09

1 Answer 1

3
$\begingroup$

It's not that scalar multiplication makes a vector space less of a group - rather, it makes a vector space more than just a group. There's a pair of general terms (coming from logic, specifically model theory) which are useful here and which I'll define informally:

  • A reduct of a structure is what you get by "forgetting" some of the original structure. For example, the integers with just addition is a reduct of the integers with both addition and multiplication.

  • An expansion is the opposite: we take a structure and "add" some additional operations/relations/"stuff" to it. "$\mathcal{A}$ is an expansion of $\mathcal{B}$" is the same as "$\mathcal{B}$ is a reduct of $\mathcal{A}$."

Expanding a structure doesn't change any of its existing properties, it just broadens the range of properties we consider. For example, distributivity doesn't make sense if all we have is addition, but once we also have multiplication it's a thing we can meaningfully consider.

Sometimes taking a reduct doesn't actually lose any information. For example, consider the real numbers with addition, multiplication, and the ordering. The ordering is in fact definable from addition and multiplication alone: $x\le y$ iff $x+z^2=y$ for some $z$. So if I "forget" the ordering, I haven't really lost any information. By contrast, the scalar multiplication in a vector space is generally not redundant. So a vector space really is more than "just" a group.

$\endgroup$
1
  • $\begingroup$ @GEdgar Oof, good catch - fixed! $\endgroup$ Jul 30, 2021 at 17:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .