Is a linear vector field a geodesible vector field?

Question

I have already asked this question in MSE; I repeat it here at MO.

Assume that $A\in M_n(\mathbb{R})$ is a non-singular matrix.

Is the flow of linear vector field $X'=AX$ a geodesible flow on $\mathbb{R}^n \setminus \{0\}$? Namely, is there a Riemannian metric on $\mathbb{R}^n \setminus \{0\}$ such that the trajectories of the linear vector field are unparametrized geodesics?

Remark: For $n=2$ the answer is affirmative, as we explain below:

Fact: A linear vector field associated to a non singular $ 2 \times 2$ real matrix is a geodesible vector field on the punctured plane.

Proof: Let $A$ be an invertible matrix. We denote by $X$ the linear vector field associated to $A$. We consider two cases:

1. $A^2$ has no real eigenvalue.

2. $A^2$ has real eigenvalue.

Case 1) In this case the linear vector field $Y$ associated to matrix $A^{-1}$ is transverse to $X$ on the punctured plane and satisfies $[X,Y]=0$ this obviously implies that $X$ is a geodesible vector field.

Case 2) If $A^2$ has real eigenvalue then $A$ is similar to one of the following matrices:

$$\begin{pmatrix} a&0\\ 0& b \end{pmatrix}\;; \begin{pmatrix} a&\epsilon\\ 0& a \end{pmatrix} \;;\begin{pmatrix} 0&b\\ -b& 0 \end{pmatrix} $$ For the first matrix the closed one form $\psi=axdx+bydy$ satisfies $\psi(X)>0$.So $X$ is a geodesible vector field. For the second matrix the $1$-form $\psi=axdx+aydy$ satisfies $\psi(X)>0$. For the third matrix the vector field is geodesible because we have a foliation of punctured plane by closed curve.

The reason of geodesibility of case $1$ and three matrices in case $2$ is discussed in the following post which is essentially based on page 71 of "Geometry of foliations" by Philippe Tondeur, Proposition $6.7$ and $6.8$.

Finding a 1-form adapted to a smooth flow

Please see also this related post: Is every real matrix conjugate to a semi antisymmetric matrix?

Answer 1

This observation seems to be very easy, but it takes care of many examples. Suppose that $A$ corresponds to a contraction (that is, all eigenvalues are $< 1$ in absolute value). Decompose ${\Bbb R}^n \backslash 0$ to a product $S^{n-1} \times {\Bbb R}$ using the diffeomorphism flow $e^{tA}$ (put a point $x\in {\Bbb R}^n \backslash 0$ to $(t \in {\Bbb R}, s\in S^{n-1})$ such that $e^{tA}= z\in S^{n-1}\subset {\Bbb R}^n$; such $t$ and $z$ are unique when $A$ is everywhere transversal to a unit sphere; however, any contraction vector field is conjugate to a contraction which is transversal to the unit sphere). Locally, this gives a diffeomorphism ${\Bbb R}^n \backslash 0=S^{n-1} \times {\Bbb R}$ such that your vector field $A$ is the translation along ${\Bbb R}$.

For this argument you don't even need linearity: Poincare-Dulac classify the contraction vector fields, and for all of them this argument can be applied as well.

We did it for conformally Kahler metrics and complex vector spaces in https://arxiv.org/abs/1210.2080 , Theorem 2.24.