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Let $V$ be a real, finite-dimensional vector space, and let $C\subset V$ be a closed convex cone, with nonempty interior, and such that $C\cap (-C)=\{0\} $. Let $u\in\operatorname{GL}(V) $ such that $u(C)=C$. Is $u$ diagonalisable [edit: i.e., diagonalizable over $\mathbf{C}$]?

Note that there are a number of hypotheses under which this holds (see for instance this paper), but I don't know whether it holds in general.

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  • $\begingroup$ One can take another point of view and wonder, given an invertible self-operator $T$, whether it preserves a cone (salient, nonempty interior). In dimension 2 it is easy to see which ones do (namely reflections and positive diagonalizable). In dimension 3, in view of the answer, the unipotent (identity + rank 2) one also does. In such an approach, one has to figure out what is the cone generated by one orbit $T^\mathbf{Z}v$, which does not sound completely obvious to me. $\endgroup$
    – YCor
    Commented Nov 25 at 11:28
  • $\begingroup$ The question is settled in the negative but I just mention a maybe relevant source of counter-examples, namely the so-called parabolic automorphisms of algebraic surfaces. The following paper of J. Grivaux gives a useful exposition: arxiv.org/pdf/1307.1771 $\endgroup$
    – Lazzaro Campeotti
    Commented Nov 25 at 12:01

2 Answers 2

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No, see Example 6.1 in Tam's and Schneider's paper On the Core of a Cone-Preserving Map. For completeness, I will write the counterexample here, but the details can be found in the paper.

Let $V = \mathbb{R}^3$ and consider the cone $C = \{(x,y,z) \in \mathbb{R}^3 \mid 2xz - y^2 \geq 0 \text{ and } x \geq 0 \}$. Then for $$ A = \begin{bmatrix} 1 & 1 & 1/2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}, $$ we have $AC = C$, but $A$ is not diagonalizable over $\mathbb{C}$. It is true, however, that any automorphism of a cone which has an eigenvector in the interior of the cone is diagonalizable over $\mathbb{C}$. In this case, the condition does not hold as the only eigenvector lies on the boundary.

References

Tam, Bit-Shun; Schneider, Hans, On the core of a cone-preserving map, Trans. Am. Math. Soc. 343, No. 2, 479-524 (1994). ZBL0826.15015.

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  • $\begingroup$ Is there an example with a polyhedral cone? $\endgroup$
    – Sam Hopkins
    Commented Nov 25 at 13:35
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    $\begingroup$ @SamHopkins no, then it will be diagonalizable over $\mathbb{C}$. See Corollary 3.4 and Theorem 3.5 of the mentioned paper. $\endgroup$
    – user527492
    Commented Nov 25 at 14:21
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Every $u\in\mathrm{SO}(3)$ preserves a lot of such cones ($u$ fixes a nonzero vector $\xi$ and hence round cones around $\mathbf{R}_+\xi$) but, up to a few exceptions, $u$ is not diagonalizable.

If instead you were asking about being diagonalizable over $\mathbf{C}$, isometries cannot any longer provide a counterexample but the slightly more elaborate example in the previous answer applies anyway.

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  • $\begingroup$ Thanks! I was indeed meaning diagonalizable over $\Bbb{C}$, I should have made it clear. Anyway, as you say the previous answer settles the question. $\endgroup$
    – abx
    Commented Nov 25 at 10:39

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