Commutativity of scalar/vector product: $a\mathbf{v}=\mathbf{v}a$ for all $a \in F$ and $\mathbf{v} \in V$

Question

There are traditionally 8 axioms to check whether a set $V$ together with a field $F$ constitute a vector space. A common list of axioms can be found here.

Missing from the list, however, is a condition that \begin{equation} a\mathbf{v}=\mathbf{v}a \end{equation} which one would hope holds for all $a \in F$ and $\mathbf{v} \in V$.

It seems like this condition is omitted because it follows from the other axioms. But in the abstract sense, how can one be ensured that a set $V$ and field $F$, together with the usual 8 vector space axioms, ensure that $a\mathbf{v}=\mathbf{v}a$ for all $a \in F$ and $\mathbf{v} \in V$?

Answer 1

As Kamal says in the comments, a vector space only requires meaning for multiplication of a vector with a scalar on one particular side, (usually the left by convention). If you add additional structure to the vector space by giving meaning to products of the form $\vec{v}a$, that's fine, but it's not part of the underlying vector space structure. It's a new kind of multiplication on top of the vector space scalar-vector multiplication.

Now, if you want to include scalar multiplication on the right in such a way that it gives $V$ another, separate, vector space structure from the structure that uses scalar multiplication on the left, then you could ask if it's possible for these two kinds of multiplication to be different. That is, is it possible to have $k\times V\to V$ be the scalar multiplication for one vector space structure on $V$, and $V\times k\to V$ be the scalar multiplication for another vector space structure on $V$, and have $a\vec{v}\neq\vec{v}a$ for some $a$, $\vec{v}$.

This can indeed happen. One example of this is if $V=\mathbb{C}$, left multiplication is regular multiplication, and right multiplication is multiplication by the conjugate. That is, $a\vec{v}=\vec{a\cdot v}$, but $\vec{v}a=\vec{v\cdot\bar{a}}$. Clearly the left-multiplication is defining vector space structure on $\mathbb{C}$ over $\mathbb{C}$. But also, the right-multiplication is doing so, as you can check the axioms. And yet $i\vec{1}=\vec{i}\neq\vec{-i}=\vec{1}i$.

So to summarize, it's possible to have vector spaces where $a\vec{v}\neq\vec{v}a$.

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If the field is $\mathbb{Q}$ or $\mathbb{F}_p$, then actually $a\vec{v}=\vec{v}a$, essentially because every scalar is "generated" by $1$.

$$\begin{align} m\vec{v}\frac{n}{m}&=(\overbrace{1+\cdots+1}^{m})\vec{v}\frac{n}{m}\\ &=\left(\overbrace{\vec{v}\frac{n}{m}+\cdots+\vec{v}\frac{n}{m}}^{m}\right)\\ &=\vec{v}\left(\overbrace{\frac{n}{m}+\cdots+\frac{n}{m}}^{m}\right)\\ &=\vec{v}n\\ &=\vec{v}(\overbrace{1+\cdots+1}^{n})\\ &=\overbrace{\vec{v}+\cdots+\vec{v}}^{n}\\ &=n\vec{v}\\ \implies\vec{v}\frac{n}{m}&=\frac{n}{m}\vec{v} \end{align}$$

Then consider $\mathbb{R}$, or any of its subfields, where every scalar is the limit of some rational numbers. So if $V$ is over $\mathbb{R}$ or any of its subfields, and if $V$ has a topology, and if the scalar multiplication map is required to be continuous, then I think you can conclude that here as well, $a\vec{v}=\vec{v}a$.

Answer 2

A small note: for the question considered here, the relevant vector space axiom is:

$\alpha \cdot(\beta \cdot v) = (\alpha\beta)\cdot v$ which tells us the group $F^{\ast}$ acts (as a left action) on the set $V$.

It is natural to ask, why can't we define a right action by:

$v \cdot \alpha = \alpha \cdot v$?

Well, we can, but we would like a right-action to satisfy:

$(v\cdot \beta)\cdot \alpha = v\cdot(\beta\alpha) = (\beta\alpha)\cdot v$, and what we get is:

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

In a field, or even a commutative ring, this is not an issue, but if one's "action upon an abelian group" is via a non-commutative ring it becomes a problem.

In other words, any left-vector space can be turned into an isomorphic right-vector space over the same field- in the case where we have an extension field of a smaller field, these two are the same entity.

This doesn't contradict alex.jordan's example, as the right-action he exhibits is a distinct action from the "natural" one induced here.

So, for example, when speaking of $R$-modules, if $R$ is commutative, it is typical to "pick a side and stick with it", although there are times (such as when dealing with tensor products) when it is convenient to view an $R$-module, as an $R,R$-bimodule (for a commutative ring $R$).

With vector spaces $R = F$, any vector space is already an $F,F$-bimodule (using the natural "left=right" action), but as alex.jordan's example shows, not ALL $F,F$-bimodule structures arise in this way.

Answer 3

Lurking in the background here is the notion of a module, which is like a vector space except the scalars need only form a ring and not a field. In this more general setting, one makes distinctions between "left modules" and "right modules", depending on whether the module is equipped with a map $R \times M \rightarrow M$ or a map $M \times R \rightarrow M$.

Simple example: The quaternions $\mathbb{H}$ are almost, but not quite, a field because multiplication is non-commutative. And though we often speak of "quaternionic vector spaces", we have to remember that, even for $\mathbb{H}^n$, the natural left- and right-multiplications don't agree.