I have learned that, if we have a connected, oriented and compact n-dimensional manifold, the top de Rham cohomology is isomorphic to $$\mathbb{R}$$, i.e $$H_{\text{dR}}^n(M) \cong \mathbb{R}.$$ However, Poincaré's lemma states that for any star-shaped subset of $$\mathcal{U} \subset \mathbb{R}^n$$, it holds that $${H_{\text{dR}}^k(\mathcal{U}) = 0} \qquad \text{for all }k \neq 0.$$

Consider now the ball of radius $$1$$, this is evidently star shaped. Furthermore, it seems to me that it also suffices the properties in the first result. What is wrong in this reasoning?

• In just the topological setting, the idea is that because $\mathcal U$ is star-shaped, it is contractible – i.e. it can be continuously deformed to a point, and because (singular) cohomology is invariant under such continuous deformations, the (singular) cohomology of $\mathcal U$ must be that of a point, i.e. trivial for all $k \neq 0$. I suspect something similar is true in the de Rham setting? – Rylee Lyman Jan 24 at 16:20
• In your first statement you listed the property 'connected' twice. Is that a typo and is there a property missing? – quarague Jan 24 at 16:20
• one problem with the open ball is that although it is bounded, it is not compact? – Rylee Lyman Jan 24 at 16:22
• Yes, it had to be compact. Thanks! – Greg Jan 24 at 16:23
• There is an additional assumption needed to conclude $H^n(M)\cong \mathbb{R}$: $M$ must have no boundary. This fails in your ball example, if you take the closed ball. – Jason DeVito Jan 24 at 16:26

I know the statement in the form: "top de-Rham cohomology of compact, connected, oriented manifold without boundary is isomorphic to $$\mathbb{R}$$".