In paper of Hantzsche (1938) there is proof that boundary of tubular neighborhood of $RP^2$ in $R^4$ (denote it $M$) is $S^3/Q_8$. I was trying to read this work but I don't know German, so this is difficult.

I Hantzsche's paper there is following formula $(\sin u \cos \frac{v}{2}, \sin u \sin \frac{v}{2}, \cos u \cos v, \cos u \sin v) $ on page 49. Is it parametrization of $\mathbb RP^2$ ? Earlier I read: "Dreht man in einer Projektiven ebene eine Gerade (das ist ein topologischer Kreis) um einen ihrer Punkte durch en Winkel $\pi$, so uberstreicht sie dabei die projektive Ebene vollstandig. Wir benutzen diese Eigenschaft der projektiven Ebene um sie in den R4 einzulagern. " which Google translate to "In a projective plane, if one turns a straight line (that is a topological circle) around one of its points through an angle $\pi$, then it completely passes over the projective plane. We use this property of the projective plane to store them in the $R^4$."

Another idea I have is to define $\mathbb RP^2$ in $\mathbb R^4$ as efect of gluing three rectangles. Next obtain triangulation (or cubification) of M by thickening each rectangle to 4-cube. The boundary will be built out of twelve 3-cubes. I don't know though how to define gluings of these 3-cubes.

  • $\begingroup$ I think you can do this by double covering your normal bundle by a bundle over $S^2$, and then the boundary is an $S^1$ bundle over $S^2$, and so is a quotient of $S^3$. Which quotient is completely determined by the Euler number. All this assumes the tubular neighborhood you started with was nontrivial, but I'm guessing that's true. $\endgroup$
    – Steve D
    Sep 9, 2018 at 12:26

1 Answer 1


I have gotten most of the way there. I still don't know two things (listed at the end). I did find a nice reference that talks about this though (in a more general setting): Lawson's Normal Bundles for an embedded $\mathbb{RP}^2$ in a Positive Definite 4-manifold (PDF).

So let $N$ be your tubular neighborhood, and assume it is non-trivial (not just $\mathbb{RP}^2\times\mathbb{R}^2$). According to Rochlin's Two-dimensional Submanifolds of Four-dimensional Manifolds (PDF), it has Euler number $\pm2$. It is double-covered by a disk bundle over $S^2$, which has Euler number $\pm4$. The boundary is a nontrivial $S^1$ bundle over $S^2$, which is thus a quotient of $S^3$. If we write this boundary as $M$, then ignoring orientation issues we have a 4-sheeted covering $S^3\rightarrow M$, and since $M$ double-covers $\partial N$, we have an 8-sheeted covering $S^3\rightarrow \partial N$.

We can thus write $\partial N$ as $S^3/G$, where $G$ is a group of order $8$ acting on $S^3$. We get this group because the covering $S^3\rightarrow M$ comes from a quotient of an action by a subgroup of $S^3$ [thought of as unit quaternions], and then we extend that by the $C_2$ action going from $M$ to $\partial N$.

The two questions I have not yet answered are:

  1. Why is the normal bundle nontrivial?
  2. Why is $G$ the quaternion group and not $C_8$?

Edit: The question has changed substantially since I added this answer. The OP is looking at a specific embedding, and asking about explicit generators in $\pi_1(\partial T)$, the fundamental group of the boundary of the tubular neighborhood. Perhaps that can be extracted from this answer (with effort). Anyway, I'll complete this answer, which shows for any embedding $\mathbb{RP}^2\rightarrow\mathbb{R}^4$, the boundary of a tubular neighborhood is homeomorphic to $S^3/Q_8$.

Answer to (1): @JasonDeVito gave a nice explanation in the comments below. It also follows from Rochlin's paper cited above, although it is not easy to tease it out!

Answer to (2): We already have that $M$ [the boundary of the disk bundle over $S^2$] is $S^3/\langle i\rangle$, treating $\langle i\rangle$ as a subgroup of $S^3$, the unit quaternions. Now if we think about $S^3$ as pairs of complex numbers and $S^2$ as $\mathbb{CP}^1$, then the Hopf fibration is $(z,w)\mapsto [z:w]$. In this setting, the $C_2$ action corresponding to the covering $S^2\rightarrow\mathbb{RP}^2$ sends $[z:w]$ to $[-\overline{w}:\overline{z}]$. Up in $S^3$, this is simply multiplication by $j$. [I should be more careful here since we're dealing with a nonabelian group: the action is on the left for both $i$ and $j$ multiplying $S^3$]. Thus we have a factorization of the covering $$ S^3\rightarrow S^3/\langle i,j\rangle\rightarrow \partial T$$

Since $S^3$ is an 8-sheeted covering of both those spaces, the second arrow is a 1-sheeted covering; that is, a homeomorphism.

  • $\begingroup$ Thank you. Quoted papers are interesting in 4-manifolds area. I am looking for more elementary proof - how generators of $\pi_1$ can be defined and why they are order 4. These circles should be order 2 in homology $H_1=\mathbb Z_2+\mathbb Z_2$. I will add more details tomorrow to my question. Regards, $\endgroup$ Sep 9, 2018 at 20:01
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    $\begingroup$ For 1. The normal bundle of any embedding $f:\mathbb{R}P^{n}\rightarrow \mathbb{R}^m$ is non-trivial if $n$ is not of the form $2^k - 1$, (e.g. $n=2$). For, if $n$ doesn't have this form, then according to Milnor-Stasheff's Characteristic classes, Corollary 4.6, there is at least one non-zero Stiefel-Whitney class of $\mathbb{R}P^n$, say $w_i(T\mathbb{R}P^n) \neq 0$. However, if the normal bundle $\nu$ to $\mathbb{R}P^n$ in $\mathbb{R}^m$ is trivial, then $0 = f^*(w_i(T\mathbb{R}^m)) = w_i(f^* T\mathbb{R}^m) = w_i(T\mathbb{R}P^n\oplus \nu) = w_i(T\mathbb{R}P^n)$, giving a contradiction. $\endgroup$ Sep 10, 2018 at 16:31
  • $\begingroup$ @JasonDeVito: thanks! I also see how to do 2, so I'll edit so that the answer is complete. $\endgroup$
    – Steve D
    Sep 10, 2018 at 17:01
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    $\begingroup$ Very nice! One question: What is an Euler number? If it's just another name for Euler class - don't you need to know that the bundle is orientable to compute it? $\endgroup$ Sep 11, 2018 at 1:59
  • $\begingroup$ @JasonDeVito: it's the torsion Euler number (see Rochlin's paper). I guess you can also just work with the double-cover (the disk bundle over the sphere). $\endgroup$
    – Steve D
    Sep 11, 2018 at 2:03

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