# Give an explicit embedding from $\mathbb{R}P_2$ to $\mathbb{R}^4$

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.

1. 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 $-(x_1,x_2,x_3)$.

2. 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 $\mathbb{R}_3$)

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from en.wikipedia.org/wiki/… $f(x,y,z)=(xy,xz,y^2-z^2,2yz)$, $f:S^2\to\mathbb{R}^4$, satisfies $f(-x,-y,-z)=f(x,y,z)$ so factors through the real projective plane. –  yoyo Dec 16 '11 at 15:14
–  lentic catachresis Dec 16 '11 at 23:17

An example of an embedding (real-algebraic, hence real analytic, hence $C^\infty$, hence topological) of $\mathbb P^2(\mathbb R)$ into $\mathbb R^4$ is given by $$i: \mathbb P^2(\mathbb R)\to \mathbb R^4: [x:y:z]\mapsto [xy:xz:yz:x^2+2y^2+3z^2]$$

It is however impossible to embed $\mathbb P^2(\mathbb R)$ into $\mathbb R^3$: the simplest reason is that a closed smooth surface in $\mathbb R^3$ is orientable, whereas $\mathbb P^2(\mathbb R)$ is well-known not to be orientable.

As a consolation, it is possible to immerse non injectively $\mathbb P^2(\mathbb R)$ into $\mathbb R^3$: the image variety is called Boy's surface (no misogyny here: the name honors the mathematician Werner Boy)

Edit
As an answer to Michael's comment let me mention that according to a celebrated theorem of Whitney, every compact $n$-dimensional smoooth manifold ($n\gt1$) can be immersed into $\mathbb R^{2n-1}$ and embedded into $\mathbb R^{2n}$.
However for $\mathbb P^n(\mathbb R)$ you sometimes get better results and this has been the object of much research. For example $\mathbb P^{15}(\mathbb R)$ can be embedded into $\mathbb R^{24}$.
You will find many such results in an interesting table here .

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Hence I guess $\mathbb{P}^n(\mathbb{R})$ could be embedded similarly? Say, $[x_1,\dots,x_n]$ to $[x_1x_2,x_1x_3,\dots,{x_1}^2+\dots+n{x_n}^2]$, a total of $2n+1$ dimensions? Is $2n+1$ minimal here? Or is there a general conclusion for the minimal dimension required for an embedding? Thanks –  Michael Luo Dec 20 '11 at 9:09
Sorry, $(n-1)n/2+1$ dimensions. I am inclined to think $\mathbb{P}^n\mathbb{R}$ can be embedded into $\mathbb{R}^{n+1}$. –  Michael Luo Dec 20 '11 at 9:29

You can do #2 by what are called "Movie moves". Think of $\mathbb{R}^4 = \mathbb{R}^3 \times \mathbb{R}$. Then in level $\mathbb{R}^3\times 0$ embed the Möbius band. Smoothly propagate the boundary of the Möbius band to $\mathbb{R}^3\times \epsilon$. Now, the boundary of the Möbius band is a smooth unknot so we can insert the isotopy to the standard unknot in $\mathbb{R}^3\times [\epsilon,1-\epsilon]$. Now cap off the unknot inside of $\mathbb{R}^3\times [1-\epsilon,1]$.

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I think I got it. Thanks! –  Michael Luo Dec 20 '11 at 9:13

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 $\mathbb R^4$.

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Ah. I see your point! By assigning (continously) a time to parts of the surface and seeing it as a process rather than a concrete shape, one can eliminate the "intersections". This was very helpful! –  Michael Luo Dec 20 '11 at 9:03
@Michael: Dear Michael, I'm glad it helped. Regards, –  Matt E Dec 20 '11 at 13:15
An equally good fourth dimensional illustrator is hue. Draw a mobius strip which is green in the center turning to yellow on the edges, then attach a disk which is yellow on the edges turning to red in the center. The intersection points occupy the same physical space but have different color coordinates. Tada! –  Eric Jan 16 '12 at 8:26