2

In my complex analysis class today, my professor was showing us some of the ways that $\mathbb{C}$ is different than $\mathbb{R}^2$. One of the examples that he used was that $\mathbb{C}$ is a one-dimensional vector space over $\mathbb{C}$ and that $\mathbb{R}^2$ is a two-dimensional vector space over $\mathbb{R}$. Both of these are perfectly obvious statements. Why is this considered to be an example of (one of the many) differences between $\mathbb{R}^2$ and $\mathbb{C}$?

To me is seems that we can make the argument that $\mathbb{C}$ is also a two-dimensional vector space over $\mathbb{R}$, a basis being the set $\left\lbrace 1, i \right\rbrace$.

Umberto P. made the same point as my professor did here: What's the difference between $\mathbb{R}^2$ and the complex plane? Nobody gave an explanation for why this is a valid point.

Just to be clear, my question is not asking about the differences between $\mathbb{R}^2$ and $\mathbb{C}$. Instead, I am asking why comparing the dimensions of these sets as vector spaces over two different fields is valid. Thanks!

Community
  • 1
Tyler Levasseur
  • 133

2 Answers2

2

I would say that $\mathbb C$ is a vector space over $\mathbb R$ and $\mathbb C$, but it is not true that $\mathbb R^2$ is a vector space over $\mathbb C$.

Of course $\mathbb R^2$ can be made into a vector space over $\mathbb C$, but this involves choosing a basis and defining extra structure which is additional to that which comes with $\mathbb R^2$.

Mark Bennet
  • 101,769
0

I think your professor's example is flawed and does not really bring out any differences between $\mathbb{C}$ and $\mathbb{R}^2$. Both can be 2-d vector spaces over $\mathbb{R}$ or 1-d vector spaces over $\mathbb{C}$.

(Readers, please correct me if I'm wrong about this.)

Eric Auld
  • 28,997
  • I don't think that $\mathbb{R}^2$ is a one-dimensional vector space over $\mathbb{C}$, since that no longer has a well-defined scalar multiplication. For instance, $i$ is not a valid scalar here. (Right?) – Tyler Levasseur Aug 27 '13 at 04:11
  • Why not define $i \cdot (r_1, r_2) : = (-r_2, r_1)$ just as in a complex number? – Eric Auld Aug 27 '13 at 04:12
  • Good point. That totally does work. Sorry about that. – Tyler Levasseur Aug 27 '13 at 04:15
  • 2
    @EricAuld Of course, since $\mathbb{R}^2$ and $\mathbb{C}$ are both two-dimensional $\mathbb{R}$-vector spaces, you can find some way to identify them. I think the question is whether such an identification is natural. I think one could argue that this still depends on an $\mathbb{R}$-basis ${1,i}$ of $\mathbb{C}$ and depending on your definition, this might not be given – Cocopuffs Aug 27 '13 at 04:17
  • @TylerLevasseur I think the point that Umberto made in the post you referenced was that $\mathbb{R}^2$ as a vector space over $\mathbb{R}$, and $\mathbb{C}$ as a vector space over $\mathbb{C}$, have different algebraic properties. There is no question of them being isomorphic, because they are over different fields, but I belive Umberto wanted to bring out some of the consequences of the vector space $\mathbb{C}$ having what you might describe as a "bigger field of scalars" (i.e. $\mathbb{C}$ itself) compared with $\mathbb{R}^2$ having only $\mathbb{R}$. – Eric Auld Aug 27 '13 at 04:20
  • @EricAuld So, just to make explicit the point that (I think) is being made, is that in $\mathbb{R}^2$, scalar multiplication is defined as $\alpha (r_1,r_2) = (\alpha r_1, \alpha r_2)$, which only works for $\alpha \in \mathbb{R}$. I suppose that would explain my confusion in the first comment, where I mistakenly supposed that $\mathbb{R}^2$ was not a valid vector space over $\mathbb{C}$. If we let $\alpha \in \mathbb{C}$, we have the usual multiplication on $\mathbb{C}$, which allows for $\mathbb{R}^2$ to be a vector space over $\mathbb{C}$, and hence the different consequences. Sound right? – Tyler Levasseur Aug 27 '13 at 04:47
  • Yes, I think that's right, except for one thing: you're comparing $\mathbb{R}^2$ over $\mathbb{C}$ against $\mathbb{R}^2$ over $\mathbb{R}$, whereas your teacher is comparing $\mathbb{C}$ over $\mathbb{C}$ against $\mathbb{R}^2$ over $\mathbb{R}$. This is basically the same thing as far as vector spaces goes, but of course $\mathbb{C}$ has some extra structure as a vector space, because it has a multiplicative structure defined on it. (So they are a complex algebra in addition to being a complex vector space.) (Of course we could give $\mathbb{R}^2$ the same multiplication.) – Eric Auld Aug 27 '13 at 16:02