Interior Angles of Codimension 3 in a Convex Polytope

Let $$\mathcal{P}^n$$ be a convex polytope in $$\mathbb{E}^n$$, where $$n\geq3$$, of full dimension—so an intersection of closed half-spaces in $$\mathbb{E}^n$$ that is bounded and whose interior in $$\mathbb{E}^n$$ is non-empty. Let $$f$$ be an $$(n-3)$$-facet of $$\mathcal{P}^n$$. Let $$F_1,\ldots,F_k$$ be the $$(n-1)$$-facets that contain $$f$$. Then inside any $$F_i$$, $$f$$ is the intersection of two $$(n-2)$$-facets, so it makes sense to talk about the dihedral angle of $$f$$ inside $$F_i$$; label this by $$\alpha_i$$. Then define the interior angle of $$f$$ to be the sum $$\alpha_1+\ldots+\alpha_k$$. In case this sounds very complicated, when $$n=3$$ this is just the total angle around a vertex in a convex polyhedron.

Is it true that (given the assumption that $$\mathcal{P}^n$$ is convex) the interior angle of any $$(n-3)$$-facet of $$\mathcal{P}^n$$ is less than $$2\pi$$? I believe it is true but I'm struggling to see how to prove it. Any ideas, or counterexamples?

• You can find the proof for $n=3$ here: math.stackexchange.com/questions/1319912/…. $n=4$ is I think an easy corollary: just project down to three dimensions along the edge $f$, so that $f$ becomes a vertex of the projection. In fact, as far as I can see, you can in general project down to three dimensions from arbitrary $n$ so that the entire $(n-3)$-flat containing the facet $f$ goes to a single point, yielding your desired result. Nov 19, 2019 at 4:35
• Yep, that definitely works, and since I'm asking the question in arbitrary dimension I don't think my question counts as a duplicate. Could you post this comment as an asnwer and then I can mark the question as answered? Thanks. Nov 20, 2019 at 10:31

1 Answer

Posting my comment as requested: The affirmative answer to this question for the case $$n=3$$ is already provided at Convex polyhedron and its Gauß-curvature. The case $$n=4$$ is a direct corollary of this case: project down to three dimensions along the edge $$f,$$ so that $$f$$ becomes a vertex of the projection. In fact, you can in general project down to three dimensions from arbitrary $$n$$ so that the entire $$(n-3)-$$flat containing the facet $$f$$ goes to a single point, yielding the desired result.