Difference between geometric simplicial complex and abstract simplicial complex

complex-geometrygeneral-topologygeometric-topologysimplex

I don't quite understand the difference between geometric simplicial complex and abstract simplicial complex. Both of their definition and the way you describe them sound almost the same to me.

For example, if there is a set $P = \{a, b, c\}$ then there can be a geometric complex $G = \{a, b, c, \{a, b\}, \{b, c\}, \{c, a\}, \{a, b, c\}\}$, right?

But I guess the abstract complex $A$ is also $\{a, b, c, \{a, b\}, \{b, c\}, \{c, a\}, \{a, b, c\}\}$, which is expressed in the same with $G$.

I was also advised that it doesn't need to have edges. Then how should I understand a simplex such as $<a, b>$?

Best Answer

Simply these have different definitions. I'm not sure why you call your $G$ a geometric simplicial complex, it is not.

A pair $(X,S)$ is called an abstract simplicial complex, if $X$ is a (nonempty) set and $S$ is a collection of subsets of $X$ such that if $P\in S$ and $A\subseteq P$ then $A\in S$ and all singletons belong to $S$. I will call elements of $S$ abstract simplicies (even though I don't think this is a standard nomenclature). Note that this is an abstract approach, we don't have edges, faces, etc. Although we can imagine that if say $X=\{1,2,3\}$ and $S=\big\{\{1\},\{2\},\{3\},\{1,2\},\{2,3\},\{1,3\}, \{1,2,3\}\big\}$ then $\{1,2\}$ is an "edge" connecting vertex $1$ to $2$, while $\{1,2,3\}$ is a "triangle" on $3$ vertices. This observation is actually a basis for geometric realization (more on that later).

(Geometric) simplicial complex on the other hand is a pair $(X,S)$ where $X$ is a topological space and $S$ is a collection of continuous functions $f:\Delta^n\to X$ (where $\Delta^n$ denotes the $n$-simplex, i.e. $n$ dimensional triangle) such that:

  1. $f$ restricted to the interior of $\Delta^n$ is injective
  2. Restriction of $f$ to one of the faces of $\Delta^n$ belongs to $S$ as well. Here a face of $\Delta^n$ is naturally identified with $\Delta^{n-1}$.
  3. A subset $U\subseteq X$ is open if and only if $f^{-1}(U)$ is open for all $f\in S$.

This definition comes from Allen Hatcher's Algebraic Topology. Note that $n$ is not constant here, it varies, in fact condition 2. says that if $f:\Delta^n\to X$ belongs to $S$ then we have some maps $\Delta^m\to X$ for any $0\leq m<n$ in $S$ as well.

$S$ is often called a simplicial structure on $X$. Here we already have edges, faces, etc.

In other words geometric simplicial complex is a concrete topological space divided into subspaces, each homeomorphic to a triangle (have a look at a very closely related concept of triangulation). For example a triangle on the plane, or two triangles on the plane glued at a single vertex. Or boundary of a triangle. But things that are not triangles are also simplicial complexes, for example a sphere or a real line. Many topological spaces you'll encounter actually have simplicial structure, but not all of them, e.g. rationals $\mathbb{Q}$ (or any other non-locally connected space).

So what about geometric realization? There is a way of constructing a geometric simplicial complex from an abstract simplicial complex $K$ in a "nice way" (meaning abstract simplex $\{a_1,\ldots,a_n\}$ corresponds to triangle on $n$ vertices, i.e. $n-1$ dimensional triangle). This is called the geometric realization of $K$. Roughly speaking for any abstract simplex $\{a_1,\ldots,a_n\}$ we just take an $n-1$-dimensional triangle, and then we glue those triangles at common faces (prescribed in the abstract definition). Of course except for the dimension $0$ which has no faces.


Here are few examples of a geometric realization of some abstract simplicial complex:

Example 1. Let $X=\{1\}$ and $S=\big\{\{1\}\big\}$. Then the geometric realization of $(X, S)$ is a $0$-dimensional triangle, i.e. a discrete single point.

Example 2. Let $X=\{1,2\}$ and $S=\big\{\{1\},\{2\}\big\}$. Then we have two $0$-dimensional triangles, so the resulting geometric realization is a discrete space on two points.

Example 3. Let $X=\{1,2\}$ and $S=\big\{\{1\},\{2\},\{1,2\}\big\}$. This is more interesting. We have two $0$-dimensional triangles, and one $1$-dimensional triangle: an interval. We glue that $1$-dimensional triangle to $0$-dimensional triangles on faces (vertices in that case). The resulting space is homeomorphic to a closed interval.

Example 4. Let $X=\{1,2,3\}$ with $S=\big\{\{1\},\{2\},\{3\},\{1,2\},\{2,3\},\{1,3\}\big\}$. Then we take three $0$-dimensional simplices (points) for each $\{x\}$, three $1$-dimensional simplices (closed intervals) for each $\{a,b\}$ and glue them at vertices. This produces a triangle without interior.

Example 5. Let $X=\{1,2,3\}$ with $S=\big\{\{1\},\{2\},\{3\},\{1,2\},\{2,3\},\{1,3\}, \{1,2,3\}\big\}$. This is similar to example 1, except we have additional $\{1,2,3\}$ abstract simplex. So we have one additional step: we generate a $2$-dimensional triangle and glue its faces with what we've already constructed in example 1. This gives as a triangle together with its interior.

Example 6. Let $X=\{1,2,3,4\}$ with $S=\big\{\{1\},\{2\},\{3\},\{4\},\{1,2\},\{2,3\},\{1,3\}, \{1,2,3\}, \{1,4\}\big\}$. Similarly to example 2 we have a full triangle, but additionaly vertex $\{4\}$ outside, which is glued to vertex $\{1\}$ via $1$-dimensional triangle: a straight line. So it is a full triangle together with a line segment going out of one of the vertices.

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