The regular polytopes are completely described by their Coxeter diagram.
Is there a way to compute their dihedral angles (angle between adjacent facets) from the combinatorial data in the diagram? Clearly, the dihedral angles are well-defined by the diagram only (after all, I can construct the geometric object from the diagram). But I am specifically asking about a not too complicated direct way (a formula maybe) to obtain these angles from the diagram.
Moreover, I am explicitly looking for an approach that does not presuppose the classification of regular polytopes (reflection groups, root systems, etc.). I want an "algorithm" that only takes in a valid Coxeter diagram, and works with that.
Update 1
Here is a different way to phrase it. If I have $n$ lineary independent unit vectors $v_1,...,v_n\in\Bbb S^{n-1}$, and all I know of them are numbers $m_{ij}\in\Bbb N,i,j\in\{1,...,n\},i\not=j$ so that
$$\langle v_i,v_j\rangle = \cos\Big(\frac{\pi}{m_{ij}}\Big),$$
then how to find the angle between $v_n$ and the hyperplane spanned by $v_1,...,v_{n-1}$. The dihedral angle is twice that angle.
Update 2
Let $\bar v_1,...,\bar v_{n-1}\in\Bbb R^n$ be the dual basis of $v_1,...,v_{n-1}$ in the span of these vectors.
Let $M=(v_1,...,v_{n-1})\in\Bbb R^{n\times (n-1)}$ is the matrix with the $v_i$ as columns, and equivalently, $\bar M\in\Bbb R^{n\times(n-1)}$ the matrix with the $\bar v_i$ as columns. Then
$$\cos \angle (\mathrm{span}(v_1,...,v_{n-1}),v_n)=\|M \bar M^\top\! v_n\|.$$
I think it should even be possible to show that this is the same as $\langle v_n,\bar v_{n-1}\rangle/\|\bar v_{n-1}\|$. Can any of this be nicely expressed using only the $m_{ij}$?
You might be interested in the according outline provided here. $-$ Essentially that one comes up with the formula asked for here (and several ones more, including their derivations, as well as providing various explicite numerical values):
$$\begin{array}{lll} \text{2D:} & \text{@-p-o} & \alpha(1,1')=\arccos\left(1-2\cdot\frac P4\right)\\ \text{3D:} & \text{@-p-o-q-o} & \alpha(2,2')=\arccos\left(1-2\cdot\frac{Q}{4-P}\right)\\ \text{4D:} & \text{@-p-o-q-o-r-o} & \alpha(3,3')=\arccos\left(1-2\cdot\frac{4-P}4\cdot\frac R{4-P-Q}\right)\\ \end{array}$$
where $P=4\cdot\cos^2\left(\frac{\pi}p\right)$, $Q=4\cdot\cos^2\left(\frac{\pi}q\right)$, and $R=4\cdot\cos^2\left(\frac{\pi}r\right)$.
--- rk