# If we can imagine $\mathbb{C}$ as a 2-dimensional Euclidean Space, then can we imagine $\mathbb{C}^2$?

It is easy to think of $\mathbb{C}^2$ as an ordered pair. I just wonder if it is possible to put $\mathbb{C}^2$ into illustration, since $\mathbb{C}$ has taken the role of two dimensional Euclidean Space.

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It depends whether you see $\mathbb{C}$ as a one-dimensional vector space over itself or as a bidimensional vector space over $\mathbb{R}$. In the first case you can try to see $\mathbb{C}^2$ as a plane but it won't be accurate, only a sketch. In the second case it would be impossible. –  Marra Mar 5 '13 at 1:29

In the same way that you can think of $\mathbb C$ as a two-dimensional object (but note that it would be inaccurate to view it as being exactly two dimensional Euclidean space since $\mathbb C$ has much more structure), so is possible to view $\mathbb C^2$ as a four dimensional object. For the same reason that $\mathbb C$ is not the same as Euclidean two-dimensional space, the resulting four dimensional object is not the same as $\mathbb R ^4$. Algebraically, this object is related to notions that generalize fields (i.e., the division ring of quaternions). It seems that you wish to understand $\mathbb C^2$ geometrically, for which it would help if you can visualize things in four dimensions. It's actually a bit easier than one might initially think, but certainly requires getting used to. Try thinking of the four dimensional simplex, and draw some of its projections onto three dimensional space. I hope this answers your question.
Yes, this construction worked out in detail produces normed division algebras up to a point. The four-dimensional normed division algebra is the ring $\mathbb{H}$ of quarternions. You lose something along the way, though: the quarternions are not commutative. If you go one step further and create a normed division algebra out of $\mathbb{H}^2$, you get the octonions $\mathbb{O}$ which are not even associative. John Baez has a great article about this available here.