# This set is a manifold

let $S$ be the set of pairs $(x,y)$ where x,y are orthogonal unit vectors in $\mathbb R^3$. i am trying to show this is a topological manifold. for starters one needs to define a suitable topology on it. i was thinking let a set $U$ be open in $S$ iff $U \cap S^2$ (intersection with sphere) is open in $S^2$ in the subspace topology? am i going in the right direction? I'd really appreicate some help. thank you

this is really about topological manifold, sorry for not stating it earlier.

here is the definition:

a manifold is a second countable Hausdorff space that is locally Euclidean.

• $S$ is a subset of $(\mathbb{R}^3)^2$, so the obvious thing to do is to give it its natural topology as a subspace of that product. – Chris Eagle Mar 4 '13 at 16:39
• @ChrisEagle yeah right, not sure why i didn't think of that. thanks! – tom b. Mar 4 '13 at 16:41
• @user58260: It is $\mathbb R^3\times\mathbb R^3$ – mrs Mar 4 '13 at 16:43

Your set $S$ is a subset of $\mathbb R^6$, so give it the subspace topology. That ensures it's 2nd countable and Hausdorff.

To show it's a manifold, notice $S$ is the pairs $(x,y) \in \mathbb R^3 \times \mathbb R^3$ such that:

$$|x|^2 =1,\ \ |y|^2=1, \ \ x\cdot y = 0$$

This is the same as saying $S = f^{-1}(1,1,0)$ where $f(x,y) = (|x|^2, |y|^2, x \cdot y)$.

So the idea would be to show $(1,1,0)$ is a regular-value of $f$, then apply the preimage theorem as Qiaochu cites. The pre-image theorem is basically just the implicit function theorem from calculus, but re-cast in a convenient formalism for saying things are manifolds.

• thank you for this great answer. it means there is no getting around smooth maps? (since the preimage theorem also requires $f$ to be smooth) it is not clear to me that every subset of $\mathbb R^6$ with the subspace topology is second countable: what if the subspace is the set of all $x \in \mathbb R^6$ with $x_n \in \mathbb R \setminus \mathbb Q$? but maybe this is not what you're saying in your first sentence. – tom b. Mar 9 '13 at 7:53
• Of course you can avoid smooth maps in the argument, but why bother? You don't need to, so don't bother with the extra effort. I think you're thinking too hard about the second countable claim -- the proof is direct. A base for the ambient topology restricts to a base for the subspace topology. So if one is countable, the other is too. – Ryan Budney Mar 11 '13 at 5:14
• right, of course. thanks, – tom b. Mar 12 '13 at 19:16
• Where does Qiaochu cite the preimage theorem? – Ben Millwood Mar 12 '13 at 19:41
• Ben Millwood: These two threads were simultaneous math.stackexchange.com/questions/320700/… – Ryan Budney Mar 12 '13 at 21:22

It is a smooth manifold of dimension 3. Indeed : let $F : S^2 \times S^2 \rightarrow \mathbb{R}$ be defined by $F(x,y)=\sum_i x_i y_i$ (the standard inner product). $F$ is a smooth map on the smooth manifold $S^2 \times S^2$ and its derivative does not vanish so it is a submersion. Bu the submersion property, the set of its zeroes is a manifold of dimension $2*2-1=3$.

• what do i have to show if i just want to show it is a topological manifold with no smoothness intended? – tom b. Mar 4 '13 at 20:05
• thanks for the great answer btw – tom b. Mar 4 '13 at 20:06
• Your map $F$ can't be a submersion, since there aren't any submersions $S^2 \times S^2 \to \mathbb R$. Submersions are open maps, but the domain is compact and the real numbers are not. $F$ fails to be a submersion on vectors $(v,v)$. But that shows you how to correct your argument. – Ryan Budney Mar 5 '13 at 7:38
• if you edit your answer could you also add an explanation of what the submersion property is please? i tried to look it up but couldn't find it anywhere. thanks – tom b. Mar 5 '13 at 7:46
• @Ryan: a very nice observation (that I had missed)! – Georges Elencwajg Mar 7 '13 at 20:02