Proving a space of pairs of orthogonal unit vectors is a manifold.

I've been asked to prove that the set of pairs of unit vectors from $\mathbb R^3$, $(x, y)$ such that $x$ and $y$ are orthogonal is a manifold. The question isn't specific on which topology the set has, so I'm assuming the topology is either inherited from the metric on $\mathbb R^6$ or else is something equivalent.

My first instinct was to prove it directly. As a subspace of $\mathbb R^6$, Hausdorff and Second Countable are immediate that way, but I don't feel like I have a good enough grip on what neighborhoods look like in this space to exhibit a local homeomorphism at each point.

My other thought was showing the whole space is homeomorphic to a known manifold. I'm not sure I understand the space enough to do that correctly, but I've played with it, and it feels like it could be homeomorphic to $S^2 \times S^1$, because once I select the first vector, there is a circle's worth of second vectors to choose. But I'm not sure enough about the specifics on that to feel good about it.

Am I on the right track?

If you don't want to understand your space $S$ explicitly and only show that it is a manifold, it is actually easier to show that your set $S$ is not only a topological manifold but also a smooth manifold. The reason is that your space $S$ is naturally defined as a zero set of a function and for such sets, the implicit function theorem gives us a sufficient criterion for $S$ to be a manifold. Namely, let $F \colon \mathbb{R}^6 \rightarrow \mathbb{R}^3$ be defined by

$$F(x,y) = (\| x \|^2, \| y \|^2, \left< x, y \right>).$$

Then we have $S = F^{-1}(1,1,0)$. If $(1,1,0)$ is a regular value of $F$, the implicit function theorem shows that $S$ is a smooth manifold and in particular a topological manifold. This way you don't have to deal with constructing local charts and verifying that the transition maps are homeomorphisms.

$$dF|_{(x,y)} = \begin{pmatrix} 2x & 0_{1 \times 3} \\ 0_{1 \times 3} & 2y \\ y & x \end{pmatrix}.$$
Here, $x,y \in \mathbb{R}^3$ are row vectors. If $(x,y) \in S$ then $x,y$ are lineraly independent and so $dF|_{(x,y)}$ has full rank and hence $(1,1,0)$ is a regular value for $F$ and $S$ is a smooth manifold.
If you want to understand your space $S$ explicitly, it turns out that this space is homeomorphic/diffeomorphic to $\operatorname{SO}_3(\mathbb{R})$ which in turn is homeomorphic/diffeomorphic to $\mathbb{RP}^3$ (and not $S^2 \times S^1$). For details, see here.
• Thank you for your advice, but we're very early in the course and don't really have access to all the machinery of differential topology. We haven't defined a smooth manifold yet. We were just given the definition of a manifold, a handful of examples, and then asked to prove that this was one. We're also not expected to know any cohomology yet, and we haven't seen the definition of $SO_3(\mathbb R)$ yet (although I know what it is, and it's at least intutively clear to me that it's homeomorphic to my space). We have been given the definition of $\mathbb RP^3$ though. – Jordan Hardy Feb 1 '17 at 5:03