Let $V$ be a finite dimensional vector space over $\mathbb{R}$, and $C\subset V$ a convex cone of the form $C=\mathbb{R}_{\geq0}v_i$ for finitely many $v_i$'s in $V$. How can one describe the stabilizer of $C$ in $GL(V)$?

Here one naturally defines the stabilizer of $C$ to be $GL(C)$ consisting of elements $g\in GL(C)$ such that $gC=C$. Say with respect to a base $(e_i)$ of $V$ and some integer $1\leq r\leq d$ one writes $$C=\sum_{i=1}^{r}\mathbb{R}_{\geq0}e_i +\sum_{j>r}\mathbb{R}e_j$$ then $GL(C)$ is the set of $g=(g_{ij})\in GL_d(\mathbb{R})$ such that $g_{ij}\geq0$ if $i\leq r$ and that the same holds for $ g^{-1} $.

My questions are:

(1) how large could $GL(C)$ be? It is clear that if in the above case with $r=d$ in the expression of $C$ along a basis $e_i$, then using the Bruhat-Tits decomposition in $GL(V)$ one finds large open subset of $GL(C)$ preserving $C$. Can $GL(C)$ be recovered essentially this way by choosing suitable basis?

(2) It seems that to characterize the difference $d-r$ one only needs to find out the split tori contained in $GL(C)$, inspired by the Bruha-Tits decomposition along a suitable basis $(e_i)$. Is this alwas true that the $r-d$ serves as a rank function for $GL(C)$?

(3) Could there be any improvements if one replaces $GL(C)$ by the set of linear maps $a\in End(V)$ such that $aC\subset C$, which is a monoid instead of a group?


  • $\begingroup$ A few comments: in the definition of $GL(C)$, I think you mean it to consist "of elements $g\in GL(V)$" instead of $g\in GL(C)$. Now, if you have finitely many $v_i$, and WLOG assume that they are convexly independent, then I think (not completely sure) that your $GL(C)$ should be generated by linear maps that permutes the $v_i$'s up to a positive scaling factor, this would be a (fairly trivial) bound on the size of $GL(C)$. For question (3), the monoid set would certainly be larger, so I am not sure what you would mean by "improvements" in the question. $\endgroup$ Dec 17 '10 at 15:56
  • $\begingroup$ Perhaps you should have a look to J. Faraut, A. Korányi, Analysis on symmetric cones. Oxford University Press, 1994. $\endgroup$ Dec 17 '10 at 16:22
  • $\begingroup$ @Denis: it is not clear to me that the OP is working on an inner product space, nor that the cones are self-dual. But thanks for pointing out that book: I think I should take a look at it. $\endgroup$ Dec 17 '10 at 17:06

As Willie already said, you should look at the action of your group on the the extreme rays of C. Look at the kernel of this action, K. The group GL(C)/K is then finite (and bounds on its order can be derived from the fact that it will be a permutation action, that is realised in a subspace of certain dimension...)

Regarding $K$, there will be a partition of the set of extreme rays into components $I_1,\dots,I_k$, so that each basis of the linear span of $C$ consisting of extreme rays will have $\dim (V_{I_j})$ elements from $V_{I_j}$, the linear span of $I_j$. Then $K$ will induce the multiplication by positive scalars action on each $V_{I_j}$, and will be the direct product of these actions.

(Thanks to David Speyer for pointing the error in the original description of $K$).

  • $\begingroup$ This is the right idea, but the details are wrong. In $\mathbb{R}^4$, consider the cone spanned by $(1,0,0,0)$, $(0,1,0,0)$, $(0,1,1,0)$, $(0,1,0,1)$ and $(0,1,1,1)$. That's $5$ cones in $4$ space. Nonetheless, there is a continuous symmetry $(w,x,y,z) \mapsto (tw, x,y,z)$ which rescales the first ray and leaves the others constant. $\endgroup$ Dec 17 '10 at 18:25
  • 2
    $\begingroup$ The right statement is to take the vector matroid $M$ given by the rays of your cone and decompose $M$ into connected components. Then $\mathbb{R}^n$ will break up into a direct sum $\bigoplus V_I$, indexed by the connected components of $M$, so that, if a ray $\rho$ is in component $I$ then $\rho \in V_I$. Then $K$ will be the group of matrices which act by a scalar on each $V_I$. $\endgroup$ Dec 17 '10 at 18:29
  • $\begingroup$ In my first comment, "5 cones" should read "5 rays". $\endgroup$ Dec 17 '10 at 19:44
  • $\begingroup$ Oops, I stand corrected. What David says is that the rays can always be partitioned into what in matroid terminology is called (irreducible) components, $I$'s. Each $I$ spans the subspace $V_I$ so that $\sum_I \dim(V_I)$ equals the dimension of the span $U$ of $C$, and so each basis of $U$, consisting of rays (i.e. each maximal independent set of the matroid) will have $\dim(V_I)$ rays from each $I$. From this the description of $K$ given by David follows. $\endgroup$ Dec 18 '10 at 6:02

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