Embedding of 3-manifold $S^3/Q_8$ in $\mathbb R^4$ 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.
 A: 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:


*

*Why is the normal bundle nontrivial?

*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.
