Definition question in algebraic topology.

Question

Definition: The $p$th (de Rham) cohomology group is the quotient vector space $$H^p(U) = \frac{Ker(d:\Omega^{p}(U)\to \Omega^{p+1}(U)}{Im(d:\Omega^{p-1}(U)\to \Omega^{p}(U))}$$

where $U \in \tau_{\mathbb{R}^n}$, that is $U$ is an open set in $\mathbb{R}^n$.

I am wondering, because it says that $H^p(U) = 0$ for any $p <0$. So what happens if $p > n$ in the definition? For example, if $U \subset \mathbb{R}^2$ is an open set, then what is $H^1(U)$? Is it also zero?

My book also states the Mayer-Vietoris as an exact sequence of cohomology quotient vector spaces,

$$\dots\to H^p(U) \to H^{p}(U_1) \oplus H^{p}(U_2) \to H^{p}(U_1 \cap U_2) \to H^{p+1}(U) \to \dots$$ where $U = U_1 \cup U_2 \in \tau_{\mathbb{R}^n}$. Now does the "dot, dot, dot" imply it never ends in $0$?

Answer 1

If $p>n$, then $H^p(U)=0$ since $\Omega^p(U)=0$ (for open $U\subseteq \mathbb{R}^n$).

If $U\subseteq \mathbb{R}^2$ is an open set, then $H^1(U)$ depends on $U$ (but it is always finite dimensional and its dimension counts the number of "holes" in $U$, roughly speaking).

The $\dots$ just signifies that the sequence extends over all integer $p$. However, this doesn't imply that it never ends in $0$ - in fact, all but finitely many terms in this sequence are $0$ by the above reasoning. Also, you could have a sequence consisting entirely of $0$'s too if you choose $U$ and $V$ appropriately!

Let me know if you need clarification on any of these points and I would be very happy to do so!

Answer 2

The space $\Omega^p(U) = \Gamma(\bigwedge^p T^*U)$, and $\bigwedge^p T^*U = 0$ for $p > \dim T^*U = n$. Hence all such $\Omega^p(U)$ vanish. The Mayer-Vietoris sequence does eventually terminate in zero modules, but it's not a short exact sequence (in general); that term refers specifically to a three-term sequence $0 \to L \to M \to N \to 0$.