Vector spaces under multiplication (o4-mini-high)

What are vector spaces under multiplication?

In the usual definition, a vector space $V$ over a field $K$ consists of

1. An abelian group $(V,+)$ (the “vector-addition”), with identity element $0$.

2. A scalar multiplication $K\times V\to V$, satisfying

   • $1\cdot v = v$,
   • $(\alpha\beta)\cdot v = \alpha\cdot(\beta\cdot v)$,
   • $\alpha\cdot(u+v) = \alpha\cdot u + \alpha\cdot v$,
   • $(\alpha+\beta)\cdot v = \alpha\cdot v + \beta\cdot v$.

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1. “Multiplicative” notation for vector spaces

Nothing deep changes if you write your abelian group multiplicatively instead of additively. Concretely, you replace

• the additive identity $0$ by a multiplicative identity, call it $1_V$,
• the “sum” $u+v$ by a product $u\cdot v$,
• the inverse $-v$ by the group-inverse $v^{-1}$.

Then the scalar axioms become:

1. Unit:

   $$ 1_K \cdot v = v . $$

2. Compatibility of scalars:

   $$ (\alpha\beta)\cdot v \;=\; \alpha\cdot(\beta\cdot v). $$

3. Distributes over the group product:

   $$ \alpha\cdot(u\cdot v) \;=\; (\alpha\cdot u)\,\cdot\,(\alpha\cdot v). $$

4. Respects field-addition via the group product:

   $$ (\alpha+\beta)\cdot v \;=\; (\alpha\cdot v)\,\cdot\,(\beta\cdot v). $$

Terminology. A set $V$ equipped with a (written-multiplicative) abelian group structure and a field-action satisfying the four rules above is sometimes called a multiplicative or exponential vector space over $K$. But it is completely equivalent (indeed isomorphic) to an ordinary (additive) vector space: you just rewrite $u\cdot v$ as “$u+v$” and $1_V$ as “0.”

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1.1 Example: $\Bbb R_{>0}$

• Take $V=\{\,x>0\}$ with multiplication as the group law.

• For any real scalar $t$, define

  $$ t\;\cdot\;x \;=\; x^t. $$

  Then one checks:

  • $1\cdot x = x^1 = x$.
  • $(s+t)\cdot x = x^{\,s+t} = x^s \cdot x^t = (s\cdot x)\,(t\cdot x)$.
  • $s\cdot(xy) = (xy)^s = x^s y^s = (s\cdot x)\,(s\cdot y)$.
  • $(st)\cdot x = x^{\,st} = s\cdot(x^t) = s\cdot(t\cdot x)$.

Thus $\bigl(\mathbb{R}_{>0},\times\bigr)$ with exponentiation is a vector space over $\mathbb{R}$, written multiplicatively; under the isomorphism $x\mapsto\ln x$ it becomes the usual additive space $(\mathbb{R},+)$.

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2. Vector spaces “with” multiplication of vectors ≠

Sometimes one asks: “What if I try to multiply two vectors to get another vector?” That additional operation (bilinear in the vectors) turns your space into an algebra rather than a “vector space under multiplication.” In other words:

• A plain vector space only has

  1. an addition of vectors, and
  2. a scalar multiplication by field elements.

• A (unital) algebra over $K$ is a vector space $V$ plus a map

  $$ V\times V \;\longrightarrow\; V,\quad (u,v)\mapsto u\,v, $$

  which is bilinear:

  $$ (\alpha u)\,v = \alpha\,(u\,v),\quad u\,(\alpha v)=\alpha\,(u\,v),\quad (u+u')\,v = u\,v + u'\,v,\quad\ldots $$

  Examples include:

  • The space of $n\times n$ matrices with ordinary matrix‐multiplication,
  • Polynomial rings $K[x]$ with polynomial multiplication,
  • Any group algebra, Clifford algebra, etc.

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Summary

• If by “vector spaces under multiplication” you mean writing the abelian group law multiplicatively, you get the same notion of vector space, with scalar action behaving like exponentiation (e.g.\ $\mathbb{R}_{>0}$ with $x^t$).
• If you mean equipping a vector space with a new “vector–vector” product, you’re describing an algebra over the field, which is a strictly richer structure.

Either argument establishes that the set of all finite sequences of natural numbers is countable.

Note: ChatGPT’s uses of \R, which rendered correctly in its own UI, were changed into \mathbb{R} here.