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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?

thanks

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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. –  Willie Wong Dec 17 '10 at 15:56
    
Perhaps you should have a look to J. Faraut, A. Korányi, Analysis on symmetric cones. Oxford University Press, 1994. –  Denis Serre Dec 17 '10 at 16:22
    
@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. –  Willie Wong Dec 17 '10 at 17:06
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up vote 2 down vote accepted

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 $K$).

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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. –  David Speyer Dec 17 '10 at 18:25
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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 $V_I$. –  David Speyer Dec 17 '10 at 18:29
    
In my first comment, "5 cones" should read "5 rays". –  David Speyer Dec 17 '10 at 19:44
    
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. –  Dima Pasechnik Dec 18 '10 at 6:02
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