The convex hull of any nonempty subset of the n + 1 points that define an n-simplex is called a face of the simplex.

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In the case of triangle 0-faces (vertices) is 3, which is reasonable, a triangle has 3 vertices, and 1-faces (edges) could be interpreted as that a triangle has 3 edges, the question is

what does 2-faces = 1 mean?

in the case of tetrahedron, 2-faces = 4 and 3-faces = 1, where do they come from?

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    $\begingroup$ The 2-face of the triangle is the whole thing. 2-faces are faces that have 3 vertices. So there are 4 triangular faces of the tetrahedron. $\endgroup$ Jul 28 '19 at 13:57

An $n$-face is just a $k$-face where $n=k$.

So what's a $k$-face?

Following the definition of a face in your first sentence, for any $k=0,...,n$, and for any $k+1$ element subset of the given set of $n+1$ points, the convex hull of that subset is a face of the given $n$-simplex. Furthermore, if I apply the definition of a simplex (unstated in your post), that face is itself a $k$-simplex.

So the sensible definition is that a $k$-face is a face that is a $k$-simplex.

Then the question arises: how many $k$-faces are there in an $n$-simplex? This is answered here, but I'll spell it out in a bit more detail. This number is the same as the number of $k+1$ element subsets of an $n+1$ element set, which is the same as the number of combinations of $n+1$ things taken $k+1$ at a time, which is given by the formula $$\frac{(n+1)!}{(k+1)! \, ((n+1)-(k+1))!} = \frac{(n+1)!}{(k+1)! \, (n-k)!} $$

For example, for $n=3$ and $k=1$, in the given set of $n+1=4$ points there are exactly six subsets with $k+1=2$ points. This comes from the formula for combinations of $4$ things taken $2$ at a time: $$\frac{4!}{2! (4-2)!}=6 $$ Therefore a $3$-simplex has exactly six $1$-faces.

When $k=n$, there is just one $n+1$ element subset of the given set of $n+1$ points, namely the whole set. And so there is just one $n$-face of the given $n$-simplex, namely the whole thing.


The idea stems - much more general than the above explanation for the simplices, which then indeed is related to the Pascal triangle, aka binomial coefficients - from the generalisation of the Euler formula.

For convex polyhedra you probably know the form $V-E+F=2$, i.e. the number of vertices ($V$) minus the number of edges ($E$) plus the number of faces ($F$) equals 2. If you would use quite generally $f_k$ to be the number of $k$-faces ($k$-dimensional elements), you would get for some $d$-dimensional polytope the dimensionally generalised formula $$1-(-1)^d=\sum_{k=0}^{d-1}(-1)^k f_k$$ This very formula could even become much nicer, if we'd define $f_{-1}=1$ (kind of saying there is a unique nullitope, an empty element, associated to the polytope, which furthermore is incident to all vertices) and we also would define $f_d=1$ as well (kind of saying there would be a unique $d$-volume, the polytope itself). For then we can simply write $$0=\sum_{k=-1}^d(-1)^k f_k$$ A different explanation would occur in the context of Hasse diagram. Here you represent each symmetry inequivalent type of elements (e.g. differentiating between triangles and squares etc., within the polygonal faces) by a node. and you connect those nodes via the incidence relations. Again, when adding those extremal elements $f_{-1}$ and $f_d$, then those Hasse diagrams get the nice topology of what also becomes known an antitegum(*). Within a Hasse diagram every subelement gets represented by the sub diagram between that elemental representing node and the nullitopal node. Furthermore any polytopal vertex-, edge-, etc. -figure's Hasse diagram is given by the subdiagram between the elementals representing node and the bulk- or $d$-volume node.

(*) In 3D the antitegum is the dual of an antiprism.

--- rk


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