How is the dimension of the simplicial complex defined?

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Let $K$ be set of finite number of simplexes in $\mathbb{R}^{n}.$ If these simplexes are nicely fited together, $K$ is called a simplicial complex.

So in this definition what is the $\text{dim}\ K$ is it the number of simplexes in $K$ or is it just $n$? For example what is the dimension of the tetrahedron? $\sigma^3=\langle p_0p_1p_2p_3\rangle$

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An $n$-dimensional simplex is the convex hull of $n+1$ many affine independent points. So e.g. a tetrahedron has four corners, so it is 3-dimensional.

The dimension of a simplicial complex is the maximum of the dimensions of its simplices.


Example. The tetrahedron $T$ (seen as a simplicial complex) does only consist of a single simplex (and its sides), which has dimension three. Hence, the tetrahedron is also 3-dimensional when considered as a complex.

The boundary $\partial T$ of the tretrahedron consists only of triangles (and their sides). Each triangle as three vertices, hence is of dimension two. Since all contained simplices are two dimensional, $\partial T$ is two dimensional, despite it is embedded into $\Bbb R^3$.

Example. All above examples are homogeneous complexes, i.e. all contained (maximal) simplices are of the same dimension. Consider the following complex:

Its (maximal) simplices are a triangle and a dangling edge. Since the triangle has dimension two, and the edge has dimension one, the triangle (with its greater dimension) determines the dimension of the complex: the complex is 2-dimensional.

Note:$\,$ A simplex of a complex is maximal if it is not the side of another simplex in the complex.

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The dimension of a simplicial complex is defined to be the greatest of the dimensions of the simplices it contains.

This means that the tetrahedron is a simplicial $3$-complex, while its boundary can be represented by a simplicial $2$-complex.