# Complex numbers as coordinates

Is it possible to have an n-dimensional geometry where each coordinate can be a complex number, or would it make no sense, i.e. lead to contradictions?

Spacetime, can be described as having 4 dimensions, and the time axis is in imaginary units.

That way the Pythagorean theorem can produce a negative hypotenuse for a right triangle where one leg is aligned to the time axis and has greater length in imaginary units than the other leg has in real units.

But what if all of the coordinates were complex numbers and could have both real and imaginary parts.

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Yes. This is called complex geometry. –  Zhen Lin Mar 6 '12 at 14:00
Depends on what properties you want. If you want a geometry with negative "lengths," I suspect that the real problem is that you'll get a geometry with zero lengths. That is, the distance between $x$ and $y$ can be zero when $x\neq y$. In general, complex geometry defines lengths as non-negative real numbers –  Thomas Andrews Mar 6 '12 at 14:06
The spacetime example doesn't really apply. Each of the space coordinates is real while the time coordinate is pure imaginary (in this definition) so it is still "four real dimensional" –  Ross Millikan Mar 6 '12 at 14:17

As a general rule, since the "length" in standard geometry is the square root of a computed value, and we'd like "length" to be well-defined, we don't allow complex lengths.

There is a neat way to generalize geometry on complex numbers. If $x=(x_1,x_2),y=(y_1,y_2)\in\mathbb R^2$, then we define $x\cdot y = x_1y_1+x_2y_2$. Then $\sqrt{x\cdot x}$ is the distance from $0$ to $x$.

When $x,y\in\mathbb C^2$, we change our definition of the dot product:

$$x\cdot y = x_1\overline{y_1} + x_2\overline{y_2}$$

Where, for $z\in\mathbb C$, $\overline z$ is the complex conjugate.

Then $y\cdot x = \overline{x\cdot y}$.

In particular, $x\cdot x$ is now again a positive real number, so we can take its positive square root to get the distance of $x$ from zero, $|x|=\sqrt{x\cdot x}$, and we have $|x|=0$ if and only if $x=0$. This distance happens to be exactly the same distance that we'd get if we considered $\mathbb C^2$ to be $\mathbb R^4$.

Another thing you lose in complex geometry is the notion of "between-ness." Given three distinct points on a real line, there is exactly one of them that is between the other two. There is no such notion of "between" for points of a complex line. Essentially, the complex line is not "linear."

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