# Fundamental group of $Q:=\{[x,y,z,t,u,v] \in \mathbb{R}P^5 |x^2+y^2+z^2-t^2-u^2-v^2=0\}$ and a covering of a quadric

I was doing some exercises in order to prepare myself for the writing exam of General Topology. We didn't give enough attention to quadrics, so this exercise is giving me hard times, any tips about it?

I can almost see what could be happen with $$C$$, given that, heuristically speaking, between an elliptic hyperboloid and a torus there's a clear homomorphism. But $$Q$$ is completely exotic to me.

• Consider $$\,\,Q:=\{[x,y,z,t,u,v] \in \mathbb{R}P^5 | \,\,x^2+y^2+z^2-t^2-u^2-v^2=0\}$$. Using the map $$\pi: S^5 \to \mathbb{R}P^5\,\,$$ show that $$Q$$ is arc-connected and find its fundamental group $$\pi_1(Q)$$.
• Consider $$C :=\{[x,y,z,t] \in \mathbb{R}P^3 | \,\,x^2+y^2-z^2-t^2=0\}$$. Using the map $$\pi:S^3 \to \mathbb{R}P^3$$, show that the torus $$T^2$$ is a covering of $$C$$; and using the map $$\phi:S^1 \times S^1 \to S^1 \times S^1$$ where $$\phi(z,w)=(zw,z \bar{w})$$, show that the quadric $$C$$ is homeomorphic to $$T^2$$.

## 1 Answer

If you decompose $$\mathbb{R}^6$$ as $$\mathbb{R}^3\times\mathbb{R}^3$$, then solutions to $$x^2+y^2+z^2-t^2-u^2-v^2=0$$ can be thought of as pairs of $$\mathbb{R}^3$$ vectors of the same length. The vectors need to be nonzero if they are to correspond to points of $$\mathbb{R}\mathrm{P}^5$$, so we may scale the vectors to both have length $$1/\sqrt{2}$$, which puts the pair of vectors on $$S^5$$, and in particular on $$S^2\times S^2\subset S^5$$. The points of $$S^2\times S^2$$ then represent every point of $$Q$$ through the quotient map $$S^5\to \mathbb{R}\mathrm{P}^5$$.

The restriction of the quotient map to $$S^2\times S^2$$ is a connected double cover of $$Q$$, and since $$S^2\times S^2$$ is simply connected, we deduce $$\pi_1(Q)=\mathbb{Z}/2\mathbb{Z}$$.

For the set $$C$$, it is similar but instead it's pairs of two $$\mathbb{R}^2$$ vectors of the same length. It is the Clifford torus $$S^1\times S^1\subset S^3$$ that surjects onto $$C$$ through $$S^3\to \mathbb{R}\mathrm{P}^2$$ now. One can parameterize the Clifford torus by thinking of $$\mathbb{R}^4$$ as $$\mathbb{C}^2$$, then taking all $$(z,w)$$ with $$|z|=|w|=1$$, where we're thinking the 3-sphere as $$S^3=\{(z,w)\in\mathbb{C}^2:|z|^2+|w|^2=2\}$$. The quotient map identifies $$(z,w)\sim (-z,-w)$$. That is, each point is identified with it's 180 degree shift in each coordinate. If you're careful with constructing a fundamental domain, you can see the quotient is $$S^1\times S^1$$ as well. (I thought about $$S^1\times S^1$$ as $$\mathbb{R}^2$$ modulo the integer lattice. The quotient is a shift by $$(1/2,1/2)$$, and if you think about the integer lattice with the basis $$(1,0),(1,1)$$ then the $$(1,1)$$-circle is double covering a circle in the quotient.)