Decomposing polyhedral cones into "direct sums" and a polynomial

Question

This question consists of two parts. I'm not breaking it up into two separate ones because posing the second question would essentially require me two rewrite the first one. Also, to some extent, the second question makes sense only if there is a proper and well-known answer to the first one.

Part 1

For convenience sake I'll only be concerned with convex polyhedral cones with vertex at the origin. (More precisely: whose face of minimal dimension contains the origin, to take cones containing whole affine subspaces into account.)

Given a pair of such cones, one in $\mathbb{R}^m$ and one in $\mathbb{R}^n$ one may consider their "direct sum" -- their Minkowski sum in $\mathbb{R}^{m+n}$. What I'm interested in is the other direction. It may so happen that a cone decomposes into a direct sum, i.e. the containing space decomposes into the sum of two independent subspaces in such a way that our cone is the Minkowski sum of its intersections with the subspaces.

Given a decomposition of a cone, one may now proceed to decompose the components and so on, obtaining a decompositon of our cone into indecomposable ones. It is natural to expect some uniqueness result to hold for this decomposition. What I'm looking for is the correct, concise and (if applicable) well-established formulation of this statement. (And a reference, of course.)

Remark. It looks as if the face lattice of a direct sum of cones is the product of their face lattices. Is there such a decomposition theorem for (geometric) lattices? And might it be that the cone's decomposition is somehow determined solely by the combinatorics of its face lattice?

Part 2

What made me stumble upon the above is the following. To each of my cones there is a polynomial associated of the form $$(1-t)^{d_1}(1-t^2)^{d_2}\ldots(1-t^k)^{d_k}.$$ The number $k$ is some number which is much smaller than the dimension of the cone.

After some examination I was able to deduce that $d_1$ is the number of indecomposable components (in the above sense) which are one dimensional lines (not rays!), i.e. it is the dimension of the cone's apex (face of minimal dimension). Furthermore, $d_2$ is the the number of components which are non-simplicial cones. (Any indecomposable cone is either a line, a ray or an indecomposable non-simplicial cone.)

Unfortunately, that's as far as I got right now. Does this polynomial look reminiscent of anything from lattice theory or convex geometry?

Clarification. The $d_i$ are defined in terms of some combinatorial data specific to the cones in consideration (actually, nothing more than the face cones of the Gelfand-Tsetlin polytope). What I'm trying to do is generalize this definition to arbitrary polyhedral cones.

Answer 1

The decomposition of polyhedral cones as mentioned in Part 1 is treated in my article Completions of fans, J. Geom. 100 (2011), 147--169. Following Igor's wish I will give a short version here.

(1) Consider a real vector space $V$ and a family $(A_i)_{i\in I}$ of polyhedral cones (called just cones in the following). The sum of vector spaces $\sum_{i\in I}\langle A_i\rangle$ is direct if and only if every $x\in\sum_{i\in I}A_i$ (Minkoswki sum) can be written uniquely in the form $x=\sum_{i\in I}x_i$ with $x_i\in A_i$ for every $i\in I$. In this case we say (by abuse of language) that the sum of cones $\sum_{i\in I}A_i$ is direct and denote it by $\bigoplus_{i\in I}A_i$.

(2) A cone $A$ is called decomposable if it is the direct sum of two cones different from $0$, and indecomposable otherwise. A decomposition of a cone $A$ is a set $Z$ of indecomposable cones different from $0$ such that $A=\bigoplus_{B\in Z}B$.

(3) Let $(A_i)_{i\in I}$ be a family of cones whose sum $A=\sum_{i\in I}A_i$ is direct. Concerning faces of $A$ we have the following (which can be checked readily on use of the definition of a face):

a) The faces of $A$ are precisely the cones of the form $\bigoplus_{i\in I}B_i$ where $B_i$ is a face of $A_i$ for every $i\in I$;

b) If $B_i$ is a face of $A_i$ for every $i\in I$, then $B_i=(\bigoplus_{j\in I}B_j)\cap A_i$ for every $i\in I$;

c) If $B$ is a face of $A$ then $B\cap A_i$ is a face of $A$ for every $i\in I$, and $B=\bigoplus_{i\in I}(B\cap A_i)$.

(It follows from this that there is an isomorphism of ordered sets between the ordered set of faces of $A$ and the ordered set the product of the ordered sets of faces of $A_i$ for $i\in I$.)

(4) Theorem: Every cone has a decomposition; a cone has a unique decomposition if and only if it contains no line or it is a line.

Sketch of proof: Let $A$ be a cone. Existence of a decomposition of $A$ follows by induction on the dimension of $A$. Choosing a basis $X$ of the greatest subspace $S$ contained in $A$ and a section $q$ of the canonical projection $p\colon V\rightarrow V/S$ we get $A=\sum_{x\in X}\langle x\rangle+q(p(A))$, and this sum is direct. Hence, uniqueness of a decomposition of $A$ implies that it contains no line or it is a line. Conversely, a line obviously has a unique decomposition. It remains to show that if $A$ contains no line then it has a unique decomposition. This is a bit technical and can be read at loc.cit.

(5) In loc.cit. the above results are proven with additional rationality hypotheses. Moreover, as an application one gets a nice proof of the fact that every fan has a simplicial subdivision with the same $1$-dimensional cones.

Answer 2

Here is a relatively simple proof (I hope) of the uniqueness in part 1: If a polyhedral cone $P$ contains no line, then it is generated by the finite set of its extreme rays $E = \{ \mathbb{R}v_i \mid i=1,\dotsc, p\}$. If the surrounding space $V$ decomposes into $V= V_1\oplus V_2$ such that $ P = (P\cap V_1) \oplus (P\cap V_2)$, then we have $E = (E\cap V_1 ) \cup (E\cap V_2) $ (from extremality). Conversely, if the last equality holds for some vector space decomposition $ V = V_1 \oplus V_2$, then $P$ decomposes accordingly. Now uniqueness of a decomposition is almost immediately (existence is clear): If $$ V = \bigoplus_i V_i = \bigoplus_j W_j \quad\text{with}\quad E = \bigcup_i (E\cap V_i) = \bigcup_j (E\cap W_j)$$ are two decompositions, then $ E = \bigcup_{i,j} (E\cap V_i\cap W_j)$ and thus $\{V_i\cap W_j\}$ also defines a decomposition of the cone. See Theorem 4 in this paper.