What Sets Delta-Complexes Apart from Simplicial Complexes?

[ForMyThunder]

What is the difference between a delta-complex and a simplicial complex? Hatcher's book says that simplicial complexes are uniquely determined by their vertices. Could someone clarify this? Thanks.

 

[quasar987]

Well, a simplicial complex can be specified as a set of vertices, together with the specification of which vertices are to make up a simplex. From this raw data, one can then construct a topological space by gluing simplices accordingly. If there is a homeomorphism btw this space and a space X, this is called a triangulation of X.

The main difference btw simplicial and delta complexes is that in simplicial complexes, there is the restriction that two simplices must intersect in a common face (or not at all), whereas delta complexes do not have this restriction. So a delta-triangulation on a space X will typically have less triangles than a triangulation, and is so it is easier to find one, and computations (such as the Euler characteristic or the homology) are easier to perform.

For instance, the minimal delta-triangulation on the 2-torus has only 2 2-triangles, 3 1-triangles, and 1 0-triangle. The minimal "simplicial triangulation" of the torus has... well, I don't know, but the most obvious (to me) triangulation of the 2-torus has 18 2-triangles alone. (http://rip94550.files.wordpress.com/2008/07/triangulation-18.png)

 

[micromass]

Hi ForMyThunder!

The standard n-simplex has vertices (1,0,0,...,0),(0,1,0,...,0),...,(0,0,0,...,1). Now, given a map $\sigma:\Delta^n\rightarrow X$ of our Delta-complex, then we can call

$$\sigma (1,0,0,...,0),~\sigma (0,1,0,...,0),...,\sigma (0,0,0,...,1)$$

are the vertices of these maps. A simplicial complex is such that no two maps $\sigma_\alpha$ and $\beta$ have the same set of vertices!

For example, consider the square [0,1]x[0,1]. Then the points (0,0),(1,0) and (0,1) form a triangle which is homeomorphic to $\Delta^2$, so take that as a first map. The points (1,0),(0,1) and (1,1) also determine a map. Continuing further gives us a simplicial complex, because every collection of points belongs to at most 1 map.

However, if we would take another map from $\Delta^2$ to the triangle (0,0), (1,0), (0,1) and adjoing it to our complex, then there would be two maps with vertices (0,0), (1,0) and (0,1). This would not form a simplicial complex.

Hope that helped!

 

[ForMyThunder]

Thanks! I understand now.