I find myself confused with this problem:
"For each integer $n$ find a simplicial complex with Euler's characteristic $n$".
In the case of positive integers, the simplicial complex would be $n$ linearly independent vertices. Is this answer possible? And, for negative integers, I have been doing several cases where adding in an extra $0$-simplex and two $1$-simplices, I got negative integers as Euler's characteristic for example
- $K_2 = \{[v0, v1], [v1, v2], [v0, v2], [v1, v3], [v2, v3], [v0], [v1], [v2], [v3] \}$
- $K_3 = \{[v0, v1], [v1, v2], [v0, v2], [v1, v3], [v2, v3], [v1, v4], [v3, v4], [v0] , [v1], [v2], [v3], [v4]\}$
the $\chi(K_2) = -1$ and $\chi(K_3) = -2$ But I don't know if this is the way to solve the problem, or if it is only through a formula, I need help. Thank you
Once you have simplicial complexes with $\chi = -1,0,1$, you could in principle 'cheat' and take their disjoint unions, since simplicial complexes need not be connected. A point has $\chi = 1$, a circle has $\chi = 0$, and you seem to have found an example with $\chi = -1$ yourself, though I haven't checked what complex it is, though alternatively a figure-eight should work.
If your simplicial complexes have to be connected, you may instead wish to invoke wedge sums. Using the fact that $\chi(X \vee Y) = \chi(X) + \chi(Y) - 1$, you can get connected simplicial complexes with arbitrary Euler characteristic. For instance, the wedge of $n$ circles should have $\chi = 1 - n$, and if $X$ is a connected simplicial complex, then $\chi(X \vee S^2) = \chi(X) + 1$.