# Is a linear vector field a geodesible vector field?

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 puntured 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 foliation " by Philip Toender, Propsition $6.7$ and $6.8$

• What in the second case $a(xdx+ydy)$ gives the geodesic metric? If it is true it seems that $a(x_1dx_1+\cdots x_ndx_n)$ gives the geodesic metric of the Jordan blocks matrix. – yaoliding Oct 5 '17 at 1:09
• @yaoliding thank you very much for your comment. According to your comment it seems the answer to my question is affirmative. For n=2 we have $a(xdx+ydy).((ax+\epsilon y)\partial_x +ay \partial_y)) =a(x^2+y^2)+\epsilon xy >0$. – Ali Taghavi Oct 5 '17 at 7:05