Boy's surface and normal vectors

Question

Denote by $X$ the image under the immersion $RP^2 \to R^3$ (that is, the Boy's surface). At each point $x \in X$ consider a small orthogonal segment that is centered at the point $x$. Denote by $S$ the set of all endpoints of these segments. Thus, we obtain a new surface $S$. What is this? I think it should be $S^2$ but I cannot prove this. This construction looks like orientable double cover but there is a problem: the Boy's surface has self-intersection points (the whole curve of such points), therefore for the natural map $p: S \to X$ the preimage $p^{-1} (x)$ for any self-intersection point $x$ consists of 4 points instead of 2. So, it is not a covering or even not a ramified covering with finite number of ramification points, so we cannot use Riemann-Hurwitz formula to find the Euler characteristic of $S$. Am I missing something? How can it be done?

Answer 1

Say your surface is parametrized by some smooth function $f(x,y)$ (this can be said because Boy's surface is a smooth immersion of $\mathbb RP^2$ into $\mathbb R^3$). Then you consider $$ g_\varepsilon(x,y) = f(x,y) + \varepsilon \cdot \frac{\partial_xf(x,y)\times \partial_yf(x,y)}{\|\partial_xf(x,y)\times \partial_yf(x,y)\|}=:f(x,y) + \varepsilon \cdot n(x,y), $$ so $g_\varepsilon$ is just a normal shift of $f$. Since $f$ is an immersion, the vector $n(x,y)$ is always defined and its length equals $1$ (and moreover, $n$ is smooth as well). Therefore you conclude that $ g_\varepsilon - f = \varepsilon n $ becomes arbitrarily small if $\varepsilon \to 0$. Regarding first derivatives you see that first derivatives of the difference between $g_\varepsilon - f$ are given by $\varepsilon\cdot \partial_xn$ and $\varepsilon \cdot \partial_yn$ respectively. Again they become arbitrarily small, provided $\varepsilon \to 0$ and you conclude that $g_\varepsilon$ and $f$ are $C^1$-close. From there you get that $g$ is again an immersion since immersions form an open set in the $C^1$-topology.