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This answer benefited from a discussion with I. Izmestiev.

Consider a $3$-dimensional simple polytope $P$. The indegrees of the vertices can only be $0$, $1$, $2$ or $3$. The lowest vertex has indegree $0$, the highest vertex has indegree $3$, and the other vertices have indegree $1$ or $2$. For an index-increasing realization of $P$, there is a plane that separates the indegree-$\le 1$ vertices and the indegree-$\ge 2$ vertices.

Moreover, every face of $P$ contains a vertex of indegree $\le 1$ (its highest vertex) and a vertex of indegree $\ge 2$ (its lowest vertex). Therefore, the existence of the required realization means that there must be a plane that cuts all the faces.

From this point of view, it becomes much clearer that index-increasing realization is very unlikely to exist even in dimension $3$. Counter-examples can be easily constructed, for example, by taking truncations.

This answer benefited from a discussion with I. Izmestiev.

Consider a $3$-dimensional simple polytope $P$. The indegrees of the vertices can only be $0$, $1$, $2$ or $3$. The lowest vertex has indegree $0$, the highest vertex has indegree $3$, and the other vertices have indegree $1$ or $2$. For an index-increasing realization of $P$, there is a plane that separates the indegree-$\le 1$ vertices and the indegree-$\ge 2$ vertices.

Moreover, every face of $P$ contains a vertex of indegree $\le 1$ (its lowest vertex) and a vertex of indegree $\ge 2$ (its highest vertex). Therefore, the existence of the required realization means that there must be a plane that cuts all the faces.

From this point of view, it becomes much clearer that index-increasing realization is very unlikely to exist even in dimension $3$. Counter-examples can be easily constructed, for example, by taking truncations.

This answer benefited from a discussion with I. Izmestiev.

Consider a $3$-dimensional simple polytope $P$. The indegrees of the vertices can only be $0$, $1$, $2$ or $3$. The lowest vertex has indegree $0$, the highest vertex has indegree $3$, and the other vertices have indegree $1$ or $2$. For an index-increasing realization of $P$, there is a plane that separates the indegree-$1$ vertices and the indegree-$2$ vertices.

Moreover, every face of $P$ contains a vertex of indegree $1$ and a vertex of indegree $2$. Therefore, the existence of the required realization means that there must be a plane that cuts all the faces.

From this point of view, it becomes much clearer that index-increasing realization is very unlikely to exist even in dimension $3$. Counter-examples can be easily constructed, for example, by taking truncations.

This answer benefited from a discussion with I. Izmestiev.

Consider a $3$-dimensional simple polytope $P$. The indegrees of the vertices can only be $0$, $1$, $2$ or $3$. The lowest vertex has indegree $0$, the highest vertex has indegree $3$, and the other vertices have indegree $1$ or $2$. For an index-increasing realization of $P$, there is a plane that separates the indegree-$\le 1$ vertices and the indegree-$\ge 2$ vertices.

Moreover, every face of $P$ contains a vertex of indegree $\le 1$ (its highest vertex) and a vertex of indegree $\ge 2$ (its lowest vertex). Therefore, the existence of the required realization means that there must be a plane that cuts all the faces.

From this point of view, it becomes much clearer that index-increasing realization is very unlikely to exist even in dimension $3$. Counter-examples can be easily constructed, for example, by taking truncations.

This answer benefited from a discussion with I. Izmestiev.

Consider a $3$-dimensional simple polytope $P$. The indegrees of the vertices can only be $0$, $1$, $2$ or $3$. The lowest vertex has indegree $0$, the highest vertex has indegree $3$, and the other vertices have indegree $1$ or $2$. For an index-increasing realization of $P$, there is a plane that separates the indegree-$1$ vertices and the indegree-$2$ vertices.

Moreover, every face of $P$ contains a vertex of indegree $1$ and a vertex of indegree $2$. Therefore, the existence of the required realization means that there must be a plane that cuts all the faces.

From this point of view, it becomes much clearer that index-increasing realization is very unlikely to exist even in dimension $3$.

This answer benefited from a discussion with I. Izmestiev.

Consider a $3$-dimensional simple polytope $P$. The indegrees of the vertices can only be $0$, $1$, $2$ or $3$. The lowest vertex has indegree $0$, the highest vertex has indegree $3$, and the other vertices have indegree $1$ or $2$. For an index-increasing realization of $P$, there is a plane that separates the indegree-$1$ vertices and the indegree-$2$ vertices.

Moreover, every face of $P$ contains a vertex of indegree $1$ and a vertex of indegree $2$. Therefore, the existence of the required realization means that there must be a plane that cuts all the faces.

From this point of view, it becomes much clearer that index-increasing realization is very unlikely to exist even in dimension $3$. Counter-examples can be easily constructed, for example, by taking truncations.