I would assume that complex polytopes (according to Shephard and Coxeter) too could be alternated (snubbed), rectified, truncated, etc. - possibly given some further restrictions.

However I have currently no plan on how that might work in details. Could someone explain such operation actions, when being applied to those?

  • $\begingroup$ For context, you could explain how these operations work on real polytopes, and where the difficulty is in changing from real to complex. $\endgroup$
    – mr_e_man
    May 2, 2023 at 18:48
  • 1
    $\begingroup$ For real space polytopes their application is obvious, e.g. the alternation of a cube results in the vertex inscribed tetrahedron, the truncation of a cube results in the truncated cube, and the rectification of the cube results in the cuboctahedron. I.e. facets themselves become alternated, truncated, or rectified in turn, and some further facets get added at half / all former vertices... - However, a complex-1-dimensional element ("edge") will be already real-2-dimensional, dyadicity is lost, complex edges might have n vertices. $\endgroup$ May 2, 2023 at 19:18
  • $\begingroup$ Wikipedia says something about truncated complex polytopes. -- I haven't figured out the meaning of a complex Coxeter-Dynkin diagram. $\endgroup$
    – mr_e_man
    May 2, 2023 at 22:00
  • $\begingroup$ I do have a description of truncation in terms of abstract polytopes. It would take a bit of work to verify that it's a valid operation, when the dyadic property is dropped. Then to map the abstract polytope into complex space, all that needs to be done is to replace each vertex (of the original complex polytope) with a hyperplane, in a way that preserves symmetry. The other new elements are gotten by intersection with the hyperplane. $\endgroup$
    – mr_e_man
    May 2, 2023 at 22:03
  • $\begingroup$ Wikipedia's definitions say the symmetry transformations must be unitary. But the given matrices are not unitary. ...? $\endgroup$
    – mr_e_man
    May 2, 2023 at 22:23

4 Answers 4


I still have no clear $general$ solution for my own question. But still I have solved it partially for one familly of complex polygons. This partial answer then might serve as a basis for other and then more complete answers to come.

Consider first $_p\{4\}_2\equiv x_p\text-4\text-o_2$ (where the first notation is that being used by Coxeter, and the latter is just applying the inline CD notation for real space polytopes onto the former, i.e. $x$ denotes the ringed node of a CD, thence representing an "edge" type, and $o$ denotes a non-ringed node). That one is known to denote that complex polygon which has $p^2$ vertices and $2p$ (complex) $p$-edges. It well can be visualized by means of the 4D realspace polychoron which is the $p$-gonal duoprism $x\text-p\text-o\ \ x\text-p\text-o$.

In fact the latter has for incidence matrix $$\begin{array}{c|c|cc|ccc|cc} .\ \ \ \ \ .\ \ .\ \ \ \ \ . & p^2 & 2 & 2 & 1 & 4 & 1 & 2 & 2 \\ \hline x\ \ \ \ \ .\ \ .\ \ \ \ \ . & 2 & p^2 & * & 1 & 2 & 0 & 2 & 1 \\ .\ \ \ \ \ .\ \ x\ \ \ \ \ . & 2 & * & p^2 & 0 & 2 & 1 & 1 & 2 \\ \hline x\text-p\text-o\ \ .\ \ \ \ \ . & p & p & 0 & p & * & * & 2 & 0 \\ x\ \ \ \ \ .\ \ x\ \ \ \ \ . & 4 & 2 & 2 & * & p^2 & * & 1 & 1 \\ .\ \ \ \ \ .\ \ x\text-p\text-o & p & 0 & p & * & * & p & 0 & 2 \\ \hline x\text-p\text-o\ \ x\ \ \ \ \ . & 2p & 2p & p & 2 & p & 0 & p & * \\ x\ \ \ \ \ .\ \ x\text-p\text-o & 2p & p & 2p & 0 & p & 2 & * & p \\ \end{array}$$ According to $\mathbb{C}\equiv\mathbb{R}^2$ would in a complex polytope however only occur even-dimensional real space elements. In here those ones, to be selected, are just the real space vertices (all $p^2$), then representing the complex space vertices in turn, and the realspace $p$-gons (all $2p$), then representing the complex space $p$-edges. (All other elements get discarded.)

Therefore the incidence matrix of that complex polygon $_p\{4\}_2\equiv x_p\text-4\text-o_2$ just happens to become $$\begin{array}{c|c} p^2 & 2 \\ \hline p & 2p \\ \end{array}$$

Next consider the complex polygon $_2\{4\}_p\equiv x_2\text-4\text-o_p= o_p\text-4\text-x_2$. This one is known to be derived from the 4D real space $p$-gonal duotegum instead.

As the latter is just the dual of the just mentioned duoprism, its incidence matrix can be obtained from the former by a 180° rotation $$\begin{array}{cc|ccc|cc|c} p & * & 2 & p & 0 & 2p & p & 2p \\ * & p & 0 & p & 2 & p & 2p & 2p \\ \hline 2 & 0 & p & * & * & p & 0 & p \\ 1 & 1 & * & p^2 & * & 2 & 2 & 4 \\ 0 & 2 & * & * & p & 0 & p & p \\ \hline 2 & 1 & 1 & 2 & 0 & p^2 & * & 2 \\ 1 & 2 & 0 & 2 & 1 & * & p^2 & 2 \\ \hline 2 & 2 & 1 & 4 & 1 & 2 & 2 & p^2 \\ \end{array}$$

Within this complex polygon however the vertices clearly are 0-dimensional, independent of its complex or real consideration. The edges however are are $2$-fold only, that is the according $2$-gons will become degenerate and are just real edges again. Within this 4D embedding those can be found as the lacing ones, i.e. the ones connecting vertices of the x-y subspace polygon to the vertices of the z-w subspace polygon.

Accordingly the incidence matrix of $_2\{4\}_p\equiv x_2\text-4\text-o_p= o_p\text-4\text-x_2$ becomes $$\begin{array}{c|c} 2p & p \\ \hline 2 & p^2 \\ \end{array}$$

Note, that $o_p\text-4\text-x_2$ already is nothing but the rectified version of $x_p\text-4\text-o_2$ (and vice versa), at least when merely looking at these (complex) CDs. And indeed, while $x_p\text-4\text-o_2$ shows up $p$-fold edges (each such complex edge has $p$ vertices) and at each vertex there are exactly $2$ such edges, these counts are exactly reversed for $o_p\text-4\text-x_2$, having just $2$-fold edges, but $p$ of them per vertex.

Thus, when considering $x_p\text-4\text-x_2$, i.e. the truncation of either of those in the mutually other direction, it becomes clear that it would use both, the former $p$-fold edges $and$ those $2$-fold edges as well. Thus, when starting e.g. from $x_p\text-4\text-o_2$, each former vertex has to be delated into such a $2$-edge, that is the former vertex count ($p^2$) has to be doubled. Conversely, when starting with $o_p\text-4\text-x_2$ instead, that former vertex count ($2p$) would have to be multiplied by $p$ instead because it gets streched out into a (real space) $p$-gon each. Thus either way the new vertex count will be $2p^2$. In fact, the total incidence matrix here becomes $$\begin{array}{c|cc} 2p^2 & 1 & 1 \\ \hline p & 2p & * \\ 2 & * & p^2 \\ \end{array}$$

--- rk


Here's a decription of truncation in terms of abstract polytopes. But of course I won't require the dyadic property.

Let $P$ be an $n$-polytope (with $n\geq1$), and let $P_k$ be the set of rank $k$ elements of $P$, for $-1\leq k\leq n$. The least and greatest elements are $l\in P_{-1}$ and $g\in P_n$. Construct the truncate $P'$ as follows:

$$ P' = (P - P_0) \sqcup \left( \bigsqcup_{v \in P_0} ([v, g] - \{ v \}) \right) $$

(Explanation: $P-P_0$ means to remove the vertices from $P$. And $[v,g]=\{x\in P:v\leq x\leq g\}$, also denoted $g/v$, is the vertex figure at $v$. And $\sqcup$ is the disjoint union; it makes copies of sets instead of using the original sets, in case they share some elements. I'll use $x_0\in P'$ to denote an element of $P-P_0$, and I'll use $x_v\in P'$ to denote an element of $[v,g]-\{v\}$, copied from $x\in P$.)

Define the partial order on $P'$ in terms of the partial order on $P$, as follows:

  • $x_0\leq'y_0$ if and only if $x\leq y$.
  • $x_v\leq'y_v$ if and only if $x\leq y$.
  • $x_0\leq'y_v$ never (except $x=l$).
  • $x_v\leq'y_0$ if and only if $x\leq y$.
  • $x_v\leq'y_w$ never (where $v$ and $w$ are two different vertices).

The ranking on $P-P_0$ in $P'$ is the same as in $P$. The ranking on each $[v,g]-\{v\}$ in $P'$ is $1$ less than in $P$. The least and greatest elements in $P'$ are $l_0$ and $g_0$. It's straightforward to verify that this relation $(\leq')$ is in fact a partial order, given that $(\leq)$ is. And $P'$ is strongly connected.

I visualized this construction by drawing a 3D Hasse diagram, with a 2D diagram of $P$ in the $y,z$ plane at $x=0$, and diagrams of its vertex figures at $x=1$, all in a row. The vertex figures are $1$ lower than where they came from in $P$; the corresponding elements are connected by diagonal lines between $x=0$ and $x=1$. And the least element $l$ is connected to the vertex figures by diagonal lines going the other way.

Now suppose each element of $P_k$, in the abstract polytope, is mapped to a $k$-dimensional affine subspace of $\mathbb C^n$, such that $(\leq)$ becomes $(\subseteq)$. Then we need to do the same for $P'$.

The subspaces for $P-P_0$ in $P'$ can be the same subspaces as in $P$. The subspaces for a vertex figure $[v,g]-\{v\}$ in $P'$ are uniquely determined by $g_v$, since $x_v\subseteq g_v$ and $x_v\subseteq x_0$ together with dimensionality imply $x_v=g_v\cap x_0$. We can choose a fairly arbitrary subspace $g_v$ for each vertex $v$, subject to non-degeneracy (the subspaces don't contain things they're not supposed to). But that's not accounting for symmetry.

If $P$ is regular (or just isogonal) (and finite), then it has a definite centre (the unique point fixed by all symmetry transformations), which we may then call $\vec0\in\mathbb C^n$, so that points become vectors. I assume the transformations are unitary (they preserve the complex inner product $\vec v^\dagger\vec w$). A vertex $\vec v\in P_0$ determines an $(n-1)$-dimensional subspace; in fact a different possible subspace for each scalar $c\in\mathbb C$:

$$ g_v = \{ \vec x \in \mathbb C^n : \vec v^\dagger \vec x = c \} $$

To maintain symmetry, $c$ should be the same for all $\vec v$. So we have a $1$-complex-dimensional family, or a $2$-real-dimensional family, of possible truncates $P'$. For a real polytope we would choose one truncate that has uniform edge lengths; for a complex polytope we might choose $c$ such that the "edges" of $P'$ all have the same circumradius, for example.

  • $\begingroup$ Very good consideration! - I however propably would prefer scaling complex n-edges such that they become regular unit-edged real polygons. This "normation" then might allow to represent mD "uniform" complex polytopes as (according elemental subsets of) 2mD uniform real space polytopes, simply because these complex n-edges thus always get represented by regular unit-edged n-gonal real space polygons each. (Or a simple unit edge for n=2.) $\endgroup$ May 5, 2023 at 12:39
  • $\begingroup$ @Dr.RichardKlitzing - That is reasonable. But a complex edge has a unique circumradius, regardless of any real interpretation. An edge with $5$ vertices could be interpreted as either $\{5\}$ or $\{5/2\}$, which have different real edge lengths but the same circumradius. $\endgroup$
    – mr_e_man
    May 8, 2023 at 20:53
  • $\begingroup$ That mentioned disambiguation however only provides 2 quite separate scalings for 5-edges, but at least no continuity of various scalings. $\endgroup$ May 9, 2023 at 19:13

All of the Wythoff Mirror-edge constructions, and "alternation" also works.

The Wythoff operators are simply the Kepler names applied to the Stott markings they produce. But truncation, cantellation and cantitruncation are real operators.

All of the usual tricks one does on the Coxeter Dynkin diagrams still apply: that is you can find the count and type of surtope produced by any subset of nodes. The laws of bisecting angles between identical nodes still works.

Truncation of a polygon will cause new edges to appear at the vertex, the vertex then on more intense truncation works directly to the centre of the polygon.

A rectified polygon is just the dual. It's more with polyhedra and higher (which have a real image in 6d), that rectification works.

Two polygon-edges can meet only at one point. This is because in hopf fibruation, two points define a 2space, and if C is not part of the 2-space defined by A, B, then any line containing C can only cross AB at no more than one point.

In a polyhedron, like 3{3}3{3}3, two faces can share an edge, which means that two 4spaces will cross at a two-space. In a rectified polyhedron, like 3{4}2x{3}2, the faces are the duals of the face of the regulars at each end, so you have 2{4}3 = triangle tegum, the edges of it are dyads, shared in common with the triangles.

Alternation works. It's not mod 2 stuff, but one vertex per edge. Alternated P{4}2 simply generates a large polyteelon P, just as the alternated square just produces the diagonal. One can see that P{4}2 has P such diagonals.

One should note that the vertices of the edges can be numbered 0 to x-1, and that the alternation selects vertices whose sum is 0 mod x.

When prismatic elements arise, as in 6x{4}2{3}2x gives a prism face at the middle node, the prisms are still cartesian products of the the edges. Here the representation is a hexagonal prism, the top and bottom being 6teela, and the vertical edges are 2teela. The squares of the hexagonal prism play no part. The hexagonal prism is then attached by sharing the whole 6teelon with the 6{4}2, and the 2teela with the triangle 2{3}2. This is cantellation or runcination [the same in 3d].

The complex symmetries do not appear to have a rotation group. The mirrors are n-2 spaces that "rotate" space around it, to form successive images, but these have no alternating parity like euclidean mirrors do. The operation of several mirrors can leave the mirror planes undisturbed, and cause a clifford rotation (where everything rotates around the centre), but i have never seen anyone making a deal of it.

  • $\begingroup$ At least your proposal that "All of the Wythoff Mirror-edge constructions [...] works" is definitely wrong! Consider the simple complex polytope $x_p\{2\}x_r$ (the reducible one), which then might get requested within s.th. like $x_2\{4\}o_3\{3\}x_3$. Let $s:=p+r\ge2p$ then we get for $h=2(1/p+2/q+1/r-1)=2pr/(p+r)=2p-2p^2/s$, i.e. $s|2p^2$ additionally! That is, the latter Wythoffian complex polyhedron would not be allowed, simply because there $s=2+3=5$ does not divide $2p^2=8$. $\endgroup$ May 21, 2023 at 9:50
  • $\begingroup$ Where $x_p$ and $x_r$ are both greater than 2, then $x_p\{2\}x_r$ does not exist. $\endgroup$ May 21, 2023 at 10:13
  • $\begingroup$ First of all: why shall something like $x_p\{2\}x_r$ not exist if $p,r>2$? And further more my example of application was about $p=2$ and $r=3$, the to be required face $x_2\ .\ x_3$ of an hypothetical $x_2\{4\}o_3\{3\}x_3$. - However for those values of that complex face polygon we would derive $h=2/(1/p+2/q+1/r-1)=2/(1/2+2/2+1/3-1)=12/5$, which clearly is not integral. $\endgroup$ May 21, 2023 at 17:19
  • $\begingroup$ What is wrong with 12/5? $h$ is a petrie polygon, not just an integer. You are confused by someone misnaming everything. $\endgroup$ May 22, 2023 at 9:54

Just found out myself that Shephard himself kind of was on this track, when introducing (e.g. within his paper on UNITARY GROUPS GENERATED BY REFLECTIONS) what Coxeter later added into his second edition of his REGULAR COMPLEX POLYTOPES as "almost regular" complex polytopes, which both of them described by some awkward triangular Coxeter-Dynkin diagram with an included number (which in fact either of them defines differently) and possibly some further legs: those in fact just seem to be snubs of the regular ones!

--- rk


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