I recently encountered a simple but confusing question. Suppose we have a set of points that is of the following form

$$\{(x_1,y_1,x_2,y_2)\in \mathbb{R}^4:x_1=y_1-1,x_2=y_2-1\}.$$ The above set gives us the Cartesian product of two lines, which should give us a 2-dimensional plane in the 4-dimensional space. However, it seems impossible for me to sketch this 2-dimensional plane in any of these 3-dimensional spaces: $x_1-y_1-x_2,$ $x_1-y_1-y_2,$$x_1-y_2-x_2,$ and $x_1-y_2-y_1.$

But since the plane is 2-dimensional, isn't it true that theoretically we should be able to sketch it in a 3-dimensional space? If this is really impossible, is there any useful way that we could visualize it?

Apologize if this question is not very rigorously stated or correctly tagged, and please feel free to edit it or add more details. Many thanks in advance for any help and references!


1 Answer 1


Well a plane is just — a plane. So you can even visualize it in $\Bbb R^2$ without any distortions. The plane is the set of points

$$\begin{align} E:\begin{pmatrix} y_1-1 \\ y_1 \\ y_2-1 \\ y_2\end{pmatrix} % = \begin{pmatrix} \\ \\ \\ \end{pmatrix} &= \begin{pmatrix} -1\\ 0 \\ -1 \\ 0 \end{pmatrix} + y_1 \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + y_2 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix} \\ & =: P + y_1 \vec v + y_2 \vec w \end{align}$$

Obviously, $\vec v$ and $\vec w$ are perpendicular and have length $\sqrt 2$. Thus, to map $E$ into the Euclidean plane, apply the following mappings:

  • Map $P$ to the origin of $\Bbb R^2$.

  • Map $\vec u$ to $(\sqrt{2}, 0)\in \Bbb R^2$. This will lie on the $y_1$-axis.

  • Map $\vec w$ to $(0, \sqrt{2})\in \Bbb R^2$. This will lie on the $y_2$-axis.

Thus, any point $P + y_1 \vec v + y_2 \vec w\in E$ will map isometrically$^1$ to $\sqrt{2}\,\dbinom{y_1}{y_2} \in \Bbb R^2$ and vice versa. Isomatrically means that when you have any two points $Q_1, Q_2\in E$, then their distance in $E$ is the same like the distance of their images in $\Bbb R^2$.

Notice that the choice of such mapping is by no means unique. For example you can pick any point of $E$ to serve as origin.

If you prefer to treat it as part of $\Bbb R^3$, just embed $\Bbb R^2$ in it any which way you like.

$^1$Isometrically means that lengths (and thus also angles) are preserved.

  • $\begingroup$ Oh I see! What a nice method! Now I understand the structure of this mapping, thank you so much for your detailed explanation! $\endgroup$
    – Chang
    Jul 8, 2022 at 16:13

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