# Complex Numbers $\stackrel{?}{=} \mathbb{R}^ 2$

Suppose we have a vector field over real numbers $\mathbb R^2$. In additon to vector field proporties define inner product $(x,y) = x_1\cdot y_1 + x_2\cdot y_2$, where $x_1,x_2,y_1,y_2$ are real numbers.This structure is called $2$-dimensional Euclidean Space. ( call it $\text{E-2}$ ).

I want to define complex number field from $\text{E-2}$. I have to satify $5$ addition axiom, $5$ multiplication axiom and a distribution axiom. The problem is although addition axioms are trivial, for multiplication I have to define; $xy = (ac-bd,ad+bc)$ where $x = ( a,b )$, $y = (c,d)$ as an extra axiom.

My questions are:

1. Is there any natural way to define complex numbers without such additional axioms, by using inner product ?
2. By defining complex numbers from $\text{E-2}$, there is no need to use inner product between two numbers, this bothers me as well, am I missing something ?
• You mean (trivial) vector bundle, not vector field. – WillO Aug 18 '15 at 12:49
• In the Euclidean plane there is no preferred direction, but in the complex plane the real line is special (it's invariant under complex conjugation). So you can't express complex multiplication (in a coordinate-free way) using the inner product only. – Hans Lundmark Aug 18 '15 at 13:45
• You should probably make a distinction between your inner product notation and your ordered pair notation. – Cameron Buie Aug 20 '15 at 22:31
• You may be interested in this. See pages 7 and 8 for a "twisting" approach to constructing division algebras from $\Bbb R$. See page 11 for the Clifford Algebra/inner product way to construct division algebras from $\Bbb R$. – pjs36 Aug 21 '15 at 0:08

I will denote the (standard) inner product by $$\langle\cdot,\cdot\rangle$$ and consider the elements of $$\Bbb R^2$$ as column vectors. Put $$A=\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}$$ and $$B=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}.$$

We define an operation $$\otimes$$ on $$\Bbb R^2$$ by $$\vec w\otimes\vec z:=\langle\vec w,A\vec z\rangle\vec e_1+ \langle\vec w,BA\vec z\rangle\vec e_2,$$ where $$\vec e_1=\begin{bmatrix}1\\0\end{bmatrix}\text{ and }\vec e_2=\begin{bmatrix}0\\1\end{bmatrix}.$$

It's rather contrived, but it does the job. Note that left-multiplication by $$A$$ is the operation of complex conjugation, that $$\vec e_1$$ is the $$\otimes$$-identity, and that left-multiplication by $$B$$ is the same as $$\otimes$$-multiplication by $$\vec e_2,$$ which serves as our "$$i$$".

Let me see if I can shed some light on the connection between inner product and complex multiplication (and how I developed the peculiar operation above).

First, I observed that, if there was some function $$f:\Bbb C^2\to\Bbb R$$ satisfying $$f(a+bi,c+di)=ac+bd$$ for all $$a,b,c,d\in\Bbb R$$ (that is, if $$f$$ acts like our inner product), then $$w\overline z=(w_1+iw_2)(z_1-iz_2)=w_1z_1+w_2z_2+i(w_2z_1-w_1z_2)=f(w,z)+i(w_2z_1-w_1z_2)$$ and similarly $$\overline wz=f(w,z)+i(w_1z_2-w_2z_1),$$ whence we find that $$f(w,z)=\frac12\left(w\overline z+\overline wz\right).$$ Readily, then, we have that $$f(w,\overline z)=\frac12\left(wz+\overline w\overline z\right),$$ and (a bit less obviously) that $$f(w,i\overline z)=\frac12\left(-iwz+i\overline w\overline z\right)=\frac1{2i}\left(wz-\overline w\overline z\right).$$ Therefore we have that $$f(w,\overline z)\cdot 1+f(w,i\overline z)\cdot i=wz.\tag{\star}$$ All that remains, then, is to decide what the equivalents to $$1$$ and $$i$$ would be in $$\Bbb R^2$$ (which was fairly natural), and how to express conjugation and multiplication by $$i$$ as linear transformations on $$\Bbb R^2$$ (not too tricky).

Added: We have some freedom in the representatives we can choose, but we can't just pick any two vectors. Let's call our vector representatives of $$1$$ and $$i$$ by the names $$\vec 1$$ and $$\vec i,$$ respectively. Let's first find some necessary conditions for them.

First of all, note that we obviously need $$\vec 1$$ and $$\vec i$$ to be linearly independent. Assume for the moment that there exist $$2\times 2$$ real matrices $$C$$ and $$R$$ corresponding to complex conjugation and multiplication by $$i$$ (the latter of which rotates the plane). In particular, this means that the following hold:

• $$C\vec 1=\vec 1$$
• $$R\vec 1=\vec i$$

By our work in the above section--in particular, by translating $$(\star)$$ into the desired terms--we find that we need $$\langle\vec w,C\vec z\rangle\vec 1+\langle\vec w,RC\vec z\rangle\vec i=\vec w\otimes\vec z\tag{\heartsuit}$$ for all $$\vec w,\vec z\in\Bbb R^2.$$

Now, since $$\vec 1$$ needs to be our $$\otimes$$-identity, then in particular, $$\vec 1=\vec1\otimes\vec1=\langle\vec 1,C\vec 1\rangle\vec1+\langle\vec1,RC\vec1\rangle\vec i=\langle\vec1,\vec1\rangle\vec1+\langle\vec1,R\vec1\rangle\vec i=\langle\vec1,\vec1\rangle\vec1+\langle\vec1,\vec i\rangle\vec i,$$ whence linear independence shows us that $$\langle\vec 1,\vec1\rangle=1$$ and $$\langle\vec1,\vec i\rangle=0.$$ Another property we need is that $$\vec1=-\vec i\otimes\vec i,$$ from which we can further deduce that $$\langle\vec i,\vec i\rangle=1.$$ Therefore, we require that $$\left\{\vec1,\vec i\right\}$$ be an orthonormal basis for $$\Bbb R^2.$$

From this, we can conclude that $$\vec1=\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}$$ for some $$\theta\in\Bbb R.$$ Now, having chosen such a vector, we have two choices for $$\vec i.$$ (Can you see why, and what they are?) This might seem a bit alarming, but many constructions of $$\Bbb C$$ run into precisely this same problem, as discussed for example here and here. Regardless, it turns out that we must have either $$R=B$$ or $$R=-B$$ (with $$B$$ as given earlier), which depends on $$\vec1$$ and on which of the two options we choose for $$\vec i.$$

One final observation I will make (omitting the proof) is that we require $$C=\begin{bmatrix}\cos2\theta & \sin2\theta\\\sin2\theta & -\cos2\theta\end{bmatrix}.$$ In the case that $$\theta$$ is an integer multiple of $$2\pi$$--that is, that $$\vec1=\vec e_1$$--we have $$C=A.$$

I leave it to you to verify that the necessary conditions derived above are also sufficient to make $$\otimes$$ behave on $$\Bbb R^2$$ exactly the way we want complex multiplication to work. Observe, though, that there are uncountably-many different ways that we could define $$\otimes,$$ depending on our choice of $$\theta$$ and subsequent choice of $$\vec i.$$ The definition I gave above is about as close to canonical as you can get.