# faces of a simple polytope

Question: Why is the face of a simple polytope a simple polytope?

My reasoning thus far:

I only have to show that this is true for any facet $F$ of $P$. Now let's take $v$ a vertices of $F$ and we show that $v$ is in $d-1$ facets of $F$.

Now take $F'$ a facet of $F$ containing $v$. Then by the diamond property, there exists a unique facet $G$ of $P$ containing $v$ such that $F'$ is in $G$ and $F$. Therefore the number of facets of $F$ containing $v$ is smaller or equal to $d-1$ (since there are only $d-1$ facets of $P$ containing $v$ other than $F$).

However I don't see how I can show that it cannot be strictly smaller than $d-1$

• I'm not sure I understand your remark. The definition states that for any $v$ it must be in $d$ facets. But why would that imply that by taking a facet $F'$ of $P$ any vertice $v'$ of that facet is still in $d-1$ facets of $F'$ i.e. $d-1$ ridges of $P$? PS: sorry to the previous comment. It seems I deleted it without knowing it. Apr 10, 2017 at 18:28
• Do you know about vertex figures? It seems like you should be able to bound below by $d-1$ using the vertex figure at $v$. If not -- what is your definition of a polytope (or what book are you following), or do you have any other properties of face lattices? Apr 11, 2017 at 0:45
• Good one! Indeed I have a result that says that if $P$ is simple then its vertex figure for any vertex is a simplex. however I haven't got a proof for this result. So I'll have a look at that now :-) Apr 11, 2017 at 8:24

1) If $P$ is simple, $Fig^v(P)$ is a simplex
2) The order set $(F^v,\subset)$ is isomorphic to the face lattice of $Fig^v(P)$ ($F^v$ is the set of faces of $P$ containing $v$)
By 2) we have that the face lattice of $F$ in the polytope $F^v$ is isomorphic to the face lattice of a facet of $Fig^v(P)$. By 1) it is the face lattice of a simplex. Therefore $F$ in $F^v$ is a simplex and it has d-1 facets who all include $v$. However it is possible that there are over facets of $F$ in the polytope $P$ who contain $v$. So we know that there are at least $d-1$ facets of $F$ who contain $v$.