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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?

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  • $\begingroup$ 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. $\endgroup$ Nov 19, 2019 at 4:35
  • $\begingroup$ 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. $\endgroup$
    – Tom Sharpe
    Nov 20, 2019 at 10:31

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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.

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