Is the set $x^2-y^2 = 0, z > y \geq0 $ a smooth manifold in $\mathbb{R}^3$?
I think that the answer is no, but I'm not really sure how to prove this as I'm having trouble visualizing how it looks. Any help would be appreciated!
2026-05-04 23:37:44.1777937864
Is the set $x^2-y^2 = 0, z > y \geq0 $ a smooth manifold in $\mathbb{R}^3$?
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$\newcommand{\Reals}{\mathbf{R}}$Hints on visualization:
The equation $0 = x^{2} - y^{2} = (x + y)(x - y)$ cuts out a pair of planes in $\Reals^{3}$ that intersect orthogonally along the $z$-axis.
The portion where $y \geq 0$ is the (generalized) cylinder parallel to the $z$-axis lying over the graph $y = |x|$, i.e., a "trough" comprising a pair of half-planes intersecting orthogonally.
The portion where $z > y \geq 0$ is the intersection of the trough with the open half-space $z > y$.
Can you take it from here?