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Triangulation of a topological space is its representation as the realization of a simplicial complex.

Suppose $X$ is a topological space. Then its triangulation is the realization $|K|$ of a simplicial complex $K$ along with a homeomorphism $h:|K|\rightarrow X$.

Example. Suppose a is a 1-simplex $$a = v_0v_1.$$ Then its faces are $v_0$ and $v_1$.

Example. The simplest cell complex representation of the circle - one 0-cell and one 1-cell - is not a simplicial complex.

For a given space, its triangulation isn't unique: any further subdivision of the above simplicial complex will be a triangulation of the circle.

Example. The familiar representation of the cylinder isn't a triangulation simply because the 2-cell is a square not triangle. Cutting it in half diagonally doesn't make it a triangulation because a new 2-cell α is glued to itself. Adding more edges does the job.

Exercise. Find a triangulation of the torus. Solution:

Exercise. Find a triangulation for each of the main surfaces.

Once all the cells are simplices, the triangulation can be found via the so-called barycentric subdivision: every simplex get a new vertex inside and all possible faces are added as well.

For the 3-simplex above, one:

  1. keeps the 4 original vertices,
  2. adds 6 new vertices on each edge,
  3. adds 4 new vertices on each face, and
  4. adds 1 new vertex inside.

Then one adds many new edges and faces.

Exercise. Find a triangulation for the cube.