I have a question about a simple polytope. I am worried that my question would be inappropriate for mathoverflow. So I am sorry that I am ignorant of combinatorics.

Let $\mathcal{P}$ be a simple convex polytope of dimension $n$ embedded in $\mathbb{R}^n$. Then any vector $v \in \mathbb{R}$ defines a height function $h_v$ of $\mathcal{P}$ such that $$ h_v(x) = \langle v, x \rangle$$ where $\langle \cdot, \cdot \rangle$ is the usual inner product in $\mathbb{R}^n$. Assume that $v$ is chosen not to be perpendicular to any face of $\mathcal{P}$. Then $h_v$ gives an orientation on each edge of $\mathcal{P}$ such that $$ \overrightarrow{pq} \Leftrightarrow h_v(p) < h_v(q)$$ for every edge $\overline{pq}$ connecting $p$ and $q$. With this orientation, we can define an index $\mathrm{ind}(p)$ (or, indegree of $p$ as a digraph) as a number of edges which end at p. And we call $h_v$ is `index-increasing' if $$\mathrm{ind}(p) < \mathrm{ind}(q) ~\mathrm{ implies} ~h_v(p) < h_v(q)$$.

Allowing to deform $\mathcal{P}$ continuously (preserving combinatorial structure and convexity, but a normal fan can vary), is it always possible to find a deformation of $\mathcal{P}$ and a vector $v \in \mathbb{R}^n$ such that $h_v$ is index-increasing?

I would really appriciate if you could give me any comment. Thank you.