0

When we think of $\mathbb{C}$ we are really thinking of the set of complex numbers which geometrically could be visualized as a plane but what is $\mathbb{C}^2$ and can it be visualized ?

atifcppprogrammer
  • 5,852
  • 2
    $\Bbb C^2$ is $4$-dimensional as a real vector space. You can still visualize it but you have to learn to think as a sequence of slices. – Gregory Grant Jul 21 '17 at 17:05
  • 1
    You can conceptualize $\mathbb{C}^2$ as a "plane" where the axes are "lines" representing compressed versions of $\mathbb{C}$. Of course, since those "lines" are actually copies of $\mathbb{C}$, they are $2$-dimensional over $\mathbb{R}$, so while a point of $\mathbb{C}^2$ can be represented as as an ordered pair of two elements of $\mathbb{C}$, it can also be represented as an ordered pair of two points of $\mathbb{R}^2$, and hence also as an element of $\mathbb{R}^4$. – quasi Jul 21 '17 at 17:14

1 Answers1

4

It has four real dimensions so it's hard to visualize but you can think of it as a sequence of 3-dimensional slices.

But one has to be very careful not to confuse the geometry of $\Bbb C^2$ as $4$-dimensional real space and the algebra of $\Bbb C^2$ as a two-dimensional complex space. For example a line $y=2x+3$ in $\Bbb C^2$ is geometrically a $2$-dimensional plane. That might lead you to think that every plane in $\Bbb C^2$ is the graph of a line, but that's not the case.

Also, you can have two planes in four dimensions that intersect in a single point. In fact two "lines" in $\Bbb C^2$ cannot intersect in more than one point even though two geometric planes in four dimensions can intersect in a (real) line.

Gregory Grant
  • 14,874
  • +1. For the OP, another point in the geometry-vs.-algebra picture: the real numbers $\mathbb{R}$ are infinite-dimensional (in fact, uncountably-dimensional) over $\mathbb{Q}$. – Noah Schweber Jul 21 '17 at 17:09