First I fix definitions:

A polytope in $V$ is a subset that is bounded and is the intersection of half-spaces of $V$. A half-space is defined with respect to an affine form, and is that part of $V$ where the affine form is nonnegative.

Let $P$ be a polytope, $P = \cap_{n=1}^{m}H_n+$, for half-spaces $H_i+$.

A face is defined as the intersection of a support hyperplane with $P$. Where a support hyperplane is one such that the polytope is contained in either the nonnegative part of the affine form that vanishes on it, or the nonpositive part of that affine form.

Let $F_1 = P\cap H_1$, and $F_2 = P\cap H_2$. Then $F_1 \cap F_2 = P\cap H_1 \cap H_2$.

For $F_1 \cap F_2$ to be a face, there must be a support hyperplane $H_3$ such that $(H_1 \cap H_2) \subseteq H_3$. Intuitively it seems like this should be the case, since the intersection of $H_1 and H_2$ is a set of lower dimension on the boundary of the polytope, so it seems plausible that one can find a support hyperplane containing it.

How can I see this more clearly/rigorously?


1 Answer 1


Any bounding element of a polytope is said to be a face. Vertices are 0-faces (0-dimensional faces), edges are 1-faces, polygons are 2-faces, etc.

Whenever you consider 2 distinct supporting affine hyperplanes, then their intersection indeed is an affine hyperplane with 1 dimension less.

But note that generally such an intersection of hyperplanes needs not to be a support of an according section of the polytope in turn. In fact it might be only a support of some stellation of that given polytope. The faces would need to be neighbouring in order to have the intersection of their spanned affine hyperplanes to be a support of some (lower dimensional) face as well.

--- rk

  • $\begingroup$ How can I see that the intersection of affine hyperplanes is an affine hyperplane? I suppose the corresponding result is true of subspaces but i'm still unsure. $\endgroup$
    – trynalearn
    Sep 23, 2018 at 6:51
  • $\begingroup$ Any dimensional restriction is given algebraically by a further equation. Thus considering the intersection of 2 subspaces of codimension 1, i.e. given by 1 equation each, you would obtain a subspace of codimension 2, i.e. having 2 defining equations. (Except those 2 equations are contradicting or equivalent, both of which should be ruled out first.) $\endgroup$ Sep 23, 2018 at 8:27

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