The definition of a manifold with corners is analogous to that of a smooth manifold, except that, instead of being locally diffeomorphic to $\mathbb R^n$, is locally diffeomorphic to $[0,+\infty)^k\times \mathbb R^{n-k}$ where $0\leq k\leq n$.

The classical example is the quadrant $Q=[0,+\infty)^2\subset \mathbb R^2$, which has $(0,0)$ as corner point.

Consider $X := \mathbb R^2 \setminus \mathrm{int}(Q).$ Is $X$ a manifold with corners?

The only problematic point is the origin which clearly has to be a corner point. We would like to show that a neighbourhood of the origin in $X$ is diffeomorphic to $Q$.

The naive idea would be to consider a map $\varphi: X\to Q$, that informally closes $X$ like a book so that it becomes $Q$. This would give us an isotopy of $X$ over $Q$, but it does not seem to be a smooth map. Indeed, this isotopy at a certain time would also show that $X$ (and hence $Q$) is diffeomorphic to $[0,+\infty)\times \mathbb R$ which is a manifold with boundary (no corners).


1 Answer 1


No, the quadrant compliment is not a manifold with corners. Essentially, manifold corners have no notion of angle (i.e. smooth deformations may change the angles of a polygon), but they do have a notion of convexity (i.e. smooth deformations may not change a convex angle into a concave one).

To see this, note that there is no smooth curve $\gamma:(-\epsilon,\epsilon)\to[0,\infty)^2$ with $\gamma(0)=0$ and $\dot{\gamma}(0)\neq 0$. This property is local about the origin, and is preserved by smooth deformations, so it holds for all corner points of manifolds with corners. The origin in the quadrant complement $0\in\mathbb{R}^2\setminus(0,\infty)^2$ does not have this property, and is thus not a corner point.


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