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From Wikipedia:

an n-simplex is an n-dimensional polytope which is the convex hull of its n + 1 vertices.

I was wondering if the definition is equivalent to say a simplex is synonym of a convex polytope?

Is simplex defined only for $\mathbb{R}^n$, not for other more general spaces?

Thanks and regards!

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4 Answers 4

up vote 8 down vote accepted

A simplex (in $\mathbb{R}^{n}$) is a special case of a convex polytope, namely one with exactly $n+1$ extremal points (the convex hull of $n+1$ points in general position - that is to say they don't lie in a $(n-1)$-dimensional affine subspace). An example of a convex polytope which isn't a simplex is given by any one of the platonic solids except the tetrahedron.

Yes, there are more general notions of simplices. In a geodesic metric space, for example, it makes sense to speak of convex hulls. In hyperbolic $n$-space or on the $n$-dimensional sphere one can then define a hyperbolic or spherical simplex as the convex hull $(n+1)$ points that don't lie in a totally geodesic submanifold (and that are not too far apart from each other in the case of a sphere).

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Thanks! I remember that convex hull makes sense for a subset of a vector space with ordered base field. Isn't it? Is a geodesic metric space such vector space? If not, how does it make sense to speak of convex hulls there? –  Tim May 1 '11 at 17:15
    
@Tim: In connection with vector spaces I've only seen convex hulls for real or complex vector spaces. A geodesic metric space is a metric space $(X,d)$ with the following property: For any two points $x,y$ there exists an isometry $\gamma: [0,d(x,y)] \to X$ such that $\gamma(0) = x$ and $\gamma(d(x,y)) = y$ (such an isometry is interpreted as a geodesic parameterized by arc length). A convex set $C$ is then a set such that for all $x,y \in C$ we have $\gamma([0,d(x,y)]) \subset C$ for all geodesics $\gamma$. The convex hull of $S$ is then the intersection of all convex sets containing $S$. –  t.b. May 1 '11 at 17:22
    
Thanks! So are these two unrelated definitions of convex hull? –  Tim May 1 '11 at 17:27
    
@Tim: No. For $\mathbb{R}^n$ they are equivalent. –  t.b. May 1 '11 at 17:28

The definition is a little unclear. In order for the convex hull of $n+1$ vertices in $\mathbb{R}^n$ to actually be $n$-dimensional, the vertices must be in general position: that is, they should not lie in a proper affine subspace of $\mathbb{R}^n$. For example, a square is a convex polytope which is the convex hull of $4$ vertices, but it is not a $3$-simplex: a $3$-simplex is a tetrahedron.

I suppose the answer to the last question is "no": when people talk about geometric simplices they are definitely referring to the ones in $\mathbb{R}^n$, although some people talk about abstract simplices in the sense of abstract simplicial complexes.

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Thanks! (1) So can I say a simplex is defined as a convex polytope whose vertices do not liey in a proper affine subspace of $\mathbb{R}^n$? (2) is it defined for more general space than $\mathbb{R}^n$? –  Tim May 1 '11 at 16:39
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@Qiaochu: The definition isn't incomplete (though it could be a bit clearer); I think the fact that they are in general position is covered by the statement that the convex hull is in fact an $n$-dimensional polytope. –  Zev Chonoles May 1 '11 at 16:39
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1) You need the condition that there are $n+1$ vertices, but yes. 2) A more general space than $\mathbb{R}^n$ will generally not have a notion of "convex hull." –  Qiaochu Yuan May 1 '11 at 16:40
    
@Zev: okay, that's fair. –  Qiaochu Yuan May 1 '11 at 16:41
    
@Qiaochu: Here is a book that discusses the abstract notion of a convex hull. amazon.com/Theory-Structures-North-Holland-Mathematical-Library/… –  Jay May 1 '11 at 23:57

I often work over simplices in $\mathbb{F}_q^d$ or other discrete spaces. Here, when we say a $k$-simplex, we mean a set of $(k+1)$-points spanning a $k$-dimensional subspace. I.e., for us, a $k$-simplex is just the set of vertices.

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Thanks! What is $\mathbb{F}_q^d$? –  Tim May 1 '11 at 17:25
    
Sorry. It is the $d$-dimensional vector space over the finite field with $q$ elements. As Qiaochu noted, there is no analog of "convex hull" in $\mathbb{F}_q^d$, which is why we simply refer to the simplex as the set of vertices. –  JavaMan May 1 '11 at 17:28

No, they are not equivalent. For example, an $n$-cube is a convex polytope but is not a simplex.

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Thanks! Why is the example not? What is the right definition? –  Tim May 1 '11 at 16:40
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It has too many vertices. –  Qiaochu Yuan May 1 '11 at 16:41
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@Tim Notice that every vertex in a simplex shares an edge (a 1-simplex) with every other vertex. –  yasmar May 1 '11 at 17:08

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