Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $P$ be a polyhedron in $\mathbb{R}^n$ and $\omega \in \mathbb{R}^n$, viewed as a linear functional

$\text{face}_{\omega}= \{ u \in P : \omega\cdot u \geq \omega\cdot v\mbox{ for all }v \in P \}$.

How does this definition match with general notion of the face of a polyhedron?

share|improve this question
2  
This is the generalized notion of a "face", in which (in three dimensions) vertices, edges, and faces are all included. For example, consider the unit cube $[0,1]^3$. Try $\omega = (1,1,1)$, $\omega = (1,1,0)$, and $\omega = (1,0,0)$ and see what "faces" you get. –  Rahul Mar 4 '12 at 13:13
    
Well, I understand that but how do I know that every face according to this generalized definition is also a "face" according to the old definition. –  Mohan Mar 4 '12 at 13:21
    
It's a "face" in the old sense when $\omega$ is normal to one of the faces in the old sense. A choice of $\omega$ is a linear function on $\mathbb{R}^n$ which restricts to a linear function on the polyhedron. The face of $\omega$ is the locus of points of the polyhedron maximizing the function (exists by compactness yadda yadda), which is going to be one of the lower-dimensional cells. –  Neal Mar 4 '12 at 13:41
    
*at least one of (only one, if the polyhedron is convex). –  Neal Mar 4 '12 at 13:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.