the manifold of $ D=\{x,y,z,w \mid xw-yz \le 0\}$

Question

What geometric object $({{x,y,z,w}})$ is defined by the equation $xw-yz\le 0$ in $\mathbb R^4$? Using parametrics, can we describe this manifold? Is it possible to prove a diffeomorphism between this geometrical object and others?

Answer 1

The set $D$ is not a manifold; in fact, it's not even a manifold with boundary. It has nonempty interior (the set of all points where $xw-yz<0$), so if it were going to be a manifold with boundary it would have to be $4$-dimensional, and thus its boundary would have to be a $3$-manifold, which it's not. The interior, however, is a $4$-manifold, because it's an open subset of $\mathbb R^4$.

Here's how to see that $\partial D$ is not a $3$-manifold. It's the set where $xw = yz$, and I claim this set is actually a cone on a torus. By that I mean it is homeomorphic to the quotient of the space $\mathbb S^1\times\mathbb S^1 \times [0,\infty)$, with the entire set $\mathbb S^1\times \mathbb S^1 \times \{0\}$ collapsed to a point. Consider the map $F\colon \mathbb S^1\times\mathbb S^1 \times [0,\infty) \to \mathbb R^4$, given by $$ (x,y,z,w) = F(\theta,\phi,r) = (r\cos\phi + r \cos\theta, r \sin\theta+ r\sin\phi, r\sin\theta- r\sin\phi, r\cos\phi- r\cos\theta), $$ where $\theta$ and $\phi$ are interpreted mod $2\pi$, so they represent points on a circle. A straightforward computation shows that $F$ maps into $\partial D$ and collapses the entire set $\mathbb S^1\times \mathbb S^1 \times \{0\}$ to the origin. A little more work shows that its image is all of $\partial D$ and it doesn't make any other identifications, so it descends to the quotient to show that $\partial D$ is homeomorphic to the cone I mentioned above.

This cone is not a $3$-manifold. The proof that it's not is a little technical, but here goes. Every point in a $3$-manifold has a neighborhood $U$ homeomorphic to a $3$-ball, and therefore when the given point $p$ is removed from $U$, the punctured set $U\smallsetminus \{p\}$ is simply connected. But if $U$ is any neighborhood of the origin in $\partial D$, when you remove the origin you get a set homeomorphic to an open subset of $\mathbb S^1\times \mathbb S^1 \times (0,\infty)$, and containing a torus of the form $\mathbb S^1\times \mathbb S^1 \times \{\varepsilon\}$ for some small $\varepsilon$. The set $U\smallsetminus \{0\}$ retracts onto this torus. Since a retract of a simply connected space is simply connected, while the torus is not, this shows that no punctured neighborhood of the origin in $\partial D$ is simply connected. Thus $\partial D$ is not a $3$-manifold, and $D$ is not a manifold with boundary.