Suppose I have the following 1-dimensional skeleton of a tetrahedron:
I assumed that it would have 4 1-dimensional holes but it turns out that the $1^{st}$ Betti number is 3.
i.e. the nullity of the 1-combinatorial Laplacian is 3.
However, I don't see why this is the case. Shouldn't the holes be where the faces should be? Can someone care to explain?
There is a relation linking these cycles. Namely, \begin{align*}\partial [0,1,3]+\partial[1,2,3]=&[1,3]-[0,3]+[0,1]+[2,3]-[1,3]+[1,2]\\ =&[0,1]+[2,3]+[1,2]\end{align*} and \begin{align*}\partial[0,1,2]+\partial[0,2,3]=&[1,2]-[0,2]+[0,1]+[2,3]-[0,3]+[0,2]\\ =&[0,1]+[2,3]+[1,2].\end{align*} Thus, the dimension is at most $4-1=3$.