Classification of simplicial complexes without boundary

Question

I want to know how to construct all finite two dimensional simplicial complexes without boundary. By without boundary I mean every edge in the complex is incident to an even number of faces. Hence, the entire complex is in the kernel of the boundary operator as defined by simplicial homology with coefficients over $\mathbb{Z}/2$.

The only examples of complexes without boundary I am aware of are manifolds and pinched surfaces. A pinched surface can be constructed from a manifold by identifying a set of vertices (or edges) to one another.

Are there other types of complexes without boundary? Can all complexes without boundary be constructed from manifolds and identifications?

Answer 1

Take three tetrahedron and glue them along one of their faces. With the obvious simplicial complex structure this is what you are interested in since every edge is incident to either 2 or 4 faces. This is not the result of taking a disjoint union of manifolds with a simplicial structure and identifying vertices with vertices and edges with edges because we can split in to two cases:

It is a quotient of multiple manifolds:

If this were a quotient of k connected manifolds, then we would be able to partition our 2-simplices into k disjoint sets each of which has the property that their sum is a cycle. These sets are just the faces of each connected manifold.

This clearly can't be the case here since any cycle with a face of one of the tetrahedron in it must have the entire tetrahedron.

It is the quotient of a single manifold:

This cannot be the case because if we look at the shared 2-simplex and pick an edge, we could find which face was adjacent to the shared simplex, along this edge, prior to gluing. Say it is from the first tetrahedron. We can see our manifold must contain a copy of this tetrahedron since away from the shared simplex, every point has a Euclidean neighborhood. However, this tetrahedron can't possibly be the initial manifold since we are missing many faces. But we run into a contradiction since no connected manifold could have a copy of $S^2$ in it as a proper subspace.

We conclude that there is no manifold with a simplicial complex structure such that it has a quotient retaining its simplicial complex structure which is the simplicial complex we constructed.

Answer 2

Here's the the classification theorem of closed surfaces. A $2d$ (connected, compact) surface without boundary is either orientable or non-orientable, and can always be associated with a genus $g\in \mathbb{N}$.

Orientable surfaces of genus $g$ can be constructed by taking the $2$-sphere $S^2$, cutting $g$ holes in it (i.e. removing disks homeomorphic to $D^2$), and then gluing a handle onto each hole by attaching their boundaries in the obvious way.

Non-orientable surfaces of genus $g$ can be constructed by taking the $2$-sphere $S^2$, cutting $g$ holes in it (i.e. removing disks homeomorphic to $D^2$), and then gluing a Möbius band onto each hole by attaching their boundaries in the obvious way.

Furthermore, according to the textbook Differential Topology by Bjørn Ian Dundas,

The reader might wonder what happens if we mix handles and Möbius bands, and it is a strange fact that if you glue $g$ handles and $h > 0$ Möbius bands you get the same as if you had glued $h + 2g$ Möbius bands! For instance, the projective plane with a handle attached is the same as the Klein bottle with a Möbius band glued onto it. But fortunately this is it; there are no more identifications among the surfaces.

Then this is a complete classification of all compact connected $2$-surfaces without boundary. If I recall correctly, every finite simplicial complex will be compact.