I was recently considering how one could motivate studying abelian groups from the study of linear algebra, and I presumed that the multiplicative group of the complex numbers would be a good a example of an abelian group that is not a vector space. However, upon closer inspection it seems that perhaps it is a real vector space, with appropriate scalar multiplication.

In particular, consider the scalar multiplication $(\lambda, z) \mapsto z^{\lambda}$. If our "vector addition" is actually complex multiplication, then this gives a two dimensional real vector space. However, this behaves quite a bit differently than ordinary finite-dimensional real vector spaces, in the sense that there exists nonzero $\lambda$ and $z$ with $z^{\lambda} = 1$ (which is the "zero" vector).

More generally, for many complex $z$ (e.g. roots of unity) there exists multiple distinct $\lambda_{j}$ with $z^{\lambda_{j}} = z^{\lambda_{j'}}$ (e.g. for $n$-th roots, $\lambda_{j} = j \cdot n$).

In other words, at least on one subspace (the unit circle) the action does not seem to be faithful (sort of - on closer reflection this is not strictly true). This comes as a shock and suggests to me that perhaps this isn't a vector space and that there is something I am missing. It is my understanding that all finite dimensional vector spaces are isomorphic to $\mathbb{R}^{n}$ for appropriate $n$, which does not seem to be the case here.

Thus my question: What I am missing here? Is there something that prevents a vector space from having this curious behavior, or is this actually a valid real vector space?

What I've looked at so far: it seems the only mention of this structure on MSE is this answer which says that, restricted to the positive reals under multiplication, it gives a vector space.


What you're looking for is the concept of a module over a ring $R$, which is an abelian group (written additively) together with compatible scalar multiplication by elements in $R$. You recover the notion of an abelian group by letting $R=\mathbb Z$: an abelian group is the same thing as a $\mathbb Z$-module. Vector spaces are modules over a field (or division ring).

Vector spaces are extremely special modules. The backbone theorems of linear algebra are mostly false for modules over more general rings. Most modules don't have a basis, in the sense that it's almost never true that there is a subset such that all elements of the module are unique linear combinations of elements of the subset. In a sense, that's the most "boring" case.

If a module does have a basis, then it's called a free module, and even then there's no need for a submodule of a free module to itself be free. This is true for abelian groups, but over more general rings it is not.

A free module need not have a concept of "dimension" either. The size of a basis of a free module is called the "rank" of the module. In general, a free module of rank $n$ is isomorphic to $R^n$. However, the rank need not be unique! There are rings where $R^m\cong R^n$ for all $m, n\geq 1$. Rings for which the rank is unique are said to have the invariant basis number property.

Already with the simplest non free abelian groups, $\mathbb Z/n\mathbb Z$, we have nonzero elements $m$ of the module and nonzero ring elements $a$ such that $am=0$. Elements for which such an $a$ exists that is not a zero divisor are called torsion elements, and your example has lots of them (specifically, the roots of unity) . Even if a module is not free, it need not have torsion elements. For example, $\mathbb Q$ is a torsion-free $\mathbb Z$-module, but it is about as far from being a free module as you can get: any two free submodules of rank $1$ intersect each other nontrivially.

Your group can be stripped to become a $\mathbb Q$-vector space, and indeed an $\mathbb R$-vector space. The group is the direct product $\mathbb R^+\times S^1$ when you decompose it into magnitude and direction. $S^1$, the unit circle, is what is making it fail to be a vector space, because the roots of unity are torsion elements and vector spaces can't have torsion. $\mathbb R^+$ under multiplication is actually isomorphic to $\mathbb R$ under addition via the exponential map $t\mapsto e^t$, and multiplication by a scalar becomes exponentiation, so it is a vector space in this way.

  • $\begingroup$ Thanks for the answer! This particular construction doesn't seem like it forms a module, however, since $a \cdot (b \cdot v) = (ab) \cdot v$ doesn't always hold. But modules might actually be a good way to introduce abelian groups. Interesting thought! $\endgroup$ – Jacob Maibach Dec 23 '18 at 20:57
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    $\begingroup$ @Jacob Being an abelian group, it's a module over $\mathbb Z$, which is integer exponentiation. Since it has integer torsion, it can't possibly be a module over $\mathbb Q$, no matter which roots you end up choosing. $\endgroup$ – Matt Samuel Dec 23 '18 at 20:59
  • $\begingroup$ @Jacob However, if you take out imaginary numbers and negative numbers, you're left with the positive real numbers under multiplication. This is indeed a module with rational exponentiation. $\endgroup$ – Matt Samuel Dec 23 '18 at 21:02
  • $\begingroup$ Could you add/emphasize that point about why it can't be extended to a $\mathbb{Q}$-module in your answer? I think that is exactly what I was looking for. $\endgroup$ – Jacob Maibach Dec 23 '18 at 21:11
  • $\begingroup$ @Jacob See the edit. $\endgroup$ – Matt Samuel Dec 23 '18 at 21:17

Embarrassingly, the same question I linked has an answer explaining just this.

Namely, for $z = i$, note that $(z^{4})^{1/2}$ is not equal to $z^{(4 \cdot 1/2)} = z^{2}$. Why? Because $i^{4} = 1$ so $(i^{4})^{1/2} = 1^{1/2} =1$ which is quite distinct from $i^{2} = -1$. That is, the lack of uniqueness of square-roots (or roots more generally) in the complex numbers either prevents scalar multiplication from being well-defined, or it prevents it from satisfying the axiom that $a \cdot (b \cdot v) = (ab) \cdot v$ (for scalars $a, b$ and vector $v$).

Edit: Generalizing this gives a short proof that a vector space is torsion-free. Presume there exists a nonzero scalar $a$ and vector $v$ such that $a \cdot v = 0$. Then note that $$ a^{-1} \cdot (a \cdot v) = (a^{-1}a) \cdot v = v, $$ yet also $$ a^{-1} \cdot (a \cdot v) = a^{-1} \cdot 0 = 0. $$ Therefore $v = 0$. More generally, this shows that scaling must be bijective. That is, for distinct scalars $a$ and $b$, and vector $v$ with $a \cdot v = b \cdot v$, we have $(a - b) \cdot v = 0$ implying once again that $v = 0$.


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