Insight on difference between Euler characteristics of 2 manifolds: $\chi(U)-\chi(V)$?

Question

For the Euler characteristic, we have the inclusion-exclusion principle: $$\chi(U\cup V) = \chi(U)+\chi(V)-\chi(U \cap V),$$ and also the connected sum property: $$ \chi(U\#V) = \chi(U)+\chi(V)-\chi(S^n). $$

However, is there any relation or topological/differential-geometric interpretation for the difference between two Euler characteristics $$ \chi(U)-\chi(V)? $$ This is studied in usual set theory, but I lack an intuitive understanding of what the above could mean. For instance, such an understanding could be gotten for the $+$ case by using the first equation above: $\chi(U)+\chi(V)=\chi(U\cup V) + \chi(U \cap V)$. I also haven't found too many resources on the matter. Any advice?

Answer 1

An obvious observation is that for two CW complexes $X,Y$, the number $\chi (X) - \chi(Y)$ is an obstruction to the two complexes being homotopy equivalent. A fruitful thing to do is to ask if this can be realized geometrically.

Recall the basic constructions of the mapping cone and mapping cylinder of $f: X \rightarrow Y$, Hatcher chapter 0 will do if you are unfamiliar. We can use the inclusion-exclusion formula to obtain the equality $\chi(\operatorname{cone}(f))=\chi(\operatorname{cylinder}(f))+\chi(\operatorname{cone}(X))-\chi(X)$.

We know $\operatorname{cylinder}(f)\simeq Y$ and $\operatorname{cone}(X) \simeq *$, so we deduce that $\chi (\operatorname{cone}(f))-1=\chi(Y)-\chi(X)$. Notably, we recover the necessary condition that homotopy equivalent spaces have the same Euler characteristic by realizing that the cone on a homotopy equivalence is contractible.

In fact, studying the algebraic properties of the chain complexes of a mapping cone is extremely fruitful. One can view this difference as a primary obstruction to a map being a "simple" homotopy equivalence (because of course it should be a homotopy equivalence), and this is detectable on homology. By studying homology of covers we may detect when it is an actual homotopy equivalence, and by studying the chain complex itself we may finally find the last obstruction to being a simple equivalence. The relevant things to look up in regards to this are "Whitehead torsion" and "simple homotopy equivalences".

The reason I bring this up is that you specifically mention manifolds, and it turns out that studying invariants like this is how you prove things like the s-cobordism theorem or much more scary things like the parametrized stable s-cobordism theorem. You just have to move to more manifold like decompositions of manifolds (i.e. handle decompositions).

Answer 2

I don't know exactly what niceness hypotheses are required for this, but if $Y$ is, say, a finite CW complex and $X$ is a CW subcomplex of $Y$ then we should have

$$\chi_c(Y) - \chi_c(X) = \chi_c(Y \setminus X)$$

where $\chi_c$ is the compactly supported Euler characteristic, defined using cohomology with compact support $H_c^{\bullet}$. $\chi_c$ is not a homotopy invariant but besides that it behaves nicer in some ways, such as this one. We can equivalently write the above rleation as

$$\chi_c(Y) = \chi_c(X) + \chi_c(Y \setminus X).$$

Note that this is manifestly not true for the ordinary Euler characteristic!

Some $\chi_c$ examples:

• $\chi_c(X) = \chi(X)$ if $X$ is compact, since then compactly supported and ordinary cohomology coincide. Hence if $X$ and $Y$ are both compact above then we have $\chi(Y) - \chi(X) = \chi_c(Y \setminus X)$.
• $\chi_c(\mathbb{R}^n) = (-1)^n$ (this is a counterexample to homotopy invariance). This is because $\mathbb{R}$ is $[0, 1]$ minus two points, so $\chi_c(\mathbb{R}) = \chi_c([0, 1]) - 2 = -1$. It's still true that $\chi_c(X \times Y) = \chi_c(X) \times \chi_c(Y)$ so this determines the answer for $\mathbb{R}^n$. This can be used to explain the Euler characteristic of the sphere: we have $\chi(S^n) = \chi(\text{pt}) + \chi_c(\mathbb{R}^n) = 1 + (-1)^n$.
• The wedge of $2$ circles satisfies $\chi(S^1 \vee S^1) = \chi(\text{pt}) + \chi \left( (0, 1) \sqcup (0, 1) \right)$ (the result of deleting the wedge point) which gives $\chi(S^1 \vee S^1) = 1 - 2 = -1$ as expected.
• The torus $T^2$ satisfies $\chi(T^2) = \chi(S^1 \vee S^1) + \chi_c( (0, 1)^2) ) = -1 + 1 = 0$ as expected.
• The above argument generalizes to the following: if $X$ is a finite CW complex with $c_i$ different $i$-cells, then $\chi_c(X) = \sum (-1)^i c_i$ (with no compactness hypotheses).