# Is the complement of a quadrant a manifold with corners?

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).

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.