Boy's surface as an immersion of $\mathbb{RP}^2$

Drag the dot on the square — or play Snake on it — and watch what happens on the surface.

The square ($\mathbb{RP}^2$)

Drag the dot. Push it off an edge and it reappears at the mirrored spot on the opposite edge — that is the gluing rule. The hollow dot is the antipodal partner point.

Boy's surface (in $\mathbb{R}^3$)

Drag to rotate, scroll to zoom. The colours match the square, so you can see which patch goes where. Anything hidden behind the surface really is behind it.

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The square is $[-1,1]^2$ with $p \sim -p$ on the boundary: the polygon $abab$ presenting $\mathbb{RP}^2$.

It is carried to $\overline{\mathbb{D}}$ by the odd map $$(u,v) \;\longmapsto\; w \;=\; \frac{\max(|u|,|v|)}{\sqrt{u^2+v^2}}\,(u+iv),$$ and then to $\mathbb{R}^3$ by the Bryant–Kusner parametrization of Boy's surface: with $D = w^6 + \sqrt{5}\,w^3 - 1$, $$g_1 = -\tfrac{3}{2}\operatorname{Im}\!\left[\frac{w(1-w^4)}{D}\right], \quad g_2 = -\tfrac{3}{2}\operatorname{Re}\!\left[\frac{w(1+w^4)}{D}\right], \quad g_3 = \operatorname{Im}\!\left[\frac{1+w^6}{D}\right] - \tfrac{1}{2}, \quad P = \frac{(g_1,g_2,g_3)}{g_1^2+g_2^2+g_3^2}.$$ Since $P(w) = P(-1/\bar w)$, this descends to $\mathbb{RP}^2$.

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