TITLE: Give an explicit embedding from $\mathbb{R}P_2$ to $\mathbb{R}^4$
QUESTION [13 upvotes]: I have heard that the least dimension $m$ required for $\mathbb{R}P_2$ to be embedded in the Euclidean space is 4, thus I wanted to find an explicit formulae for it. I found two possible strategies, but is not sure that they'll work. 

Define $\phi([x_1,x_2,x_3])=(|x_1|,|x_2|,|x_3|,x_1x_2+x_2x_3+x_3x_1)$, where $[x_1,x_2,x_3]$ is the eq.class under quotient from $S^2$. I hope that would be an embedding, since the last is a symmetric polynomial which is equal for $(x_1,x_2,x_3)$ and $-(x_1,x_2,x_3)$. 
My second stragy is more geometric approach. Note that the projective plane deleted a circle is a Möbius band $M$. Thus if I could "paste" the boundary of a closed disk (via the fourth dimension) onto the boundary circle of a Möbius Band, then I'm done. But since I'm no good at "visualizing" four dimensions, I don't know exactly how to proceed. 

My question is:
1) is 1. an embedding or not? (Or give another more elegant imbedding)
2) is there a way to realize my second approach? or is it hopeless?
3) Is there a more systematic way of doing such embeddings? (Frankly, If our world is in $\mathbb{R}_2$, I would probably not even be able to imagine how to imbed the torus into $\mathbb{R}_3$)

REPLY [4 votes]: With regard to your 2nd strategy, you remark that you have trouble "visualizing" four dimensions.
One way to think about your 2nd strategy is as follows: you can immerse $\mathbb RP^2$ into $\mathbb R^3$, by gluing a disk and a Möbius strip together along their boudnaries.  The problem, as you know, is that you get self-intersections.  But now imagine actually carrying this out (say with rubber models of a disk and a Möbius strip), but plotting not just the resulting points in $\mathbb R^3$, but
the times at which you placed each of the points of your model in their final positions. (So now you are plotting points in $\mathbb R^4$, not $\mathbb R^3$.)
The points which intersect will arrive at their final positions at different times, and so you will obtain an embedding of $\mathbb R P^2$ in $\mathbb R^4$.
More generally speaking, the old cliche of time being the "fourth dimension" can actually be very helpful for visualizing certain constructions in $\mathbb R^4$: you imagine making some geometric construction in $\mathbb R^3$, but then also imagine the construction as a process taking place over time; this then lets you "see" a construction in $\mathbb R^4$.