# Hypersurfaces homeomorphic to a hyperplane

Let $$X \subset \Bbb R^n$$ be a non compact, smooth, connected, orientable, algebraic hypersurface such that $$H_i(X,\Bbb Z)=0$$ for all $$i\not=0$$. Is $$X$$ homeomorphic to a hyperplane?

First, some terminology: An $$m$$-dimensional manifold $$M$$ is called an (integer) homology sphere if its homology groups are isomorphic to that of the sphere $$S^m$$. A topological space $$X$$ is called acyclic if it has zero reduced homology groups. It is a nice exercise to show that the complement to a point in a homology sphere is acyclic. There are homology spheres which are not simply-connected, the first one was discovered by Poincare. Therefore, the complement to a point in such a homology sphere is not simply-connected either, hence, is not even homotopy-equivalent to $$R^n$$.
Next, some 3-dimensional homology spheres embed as smooth submanifolds of $$S^4$$, they bound domains in $$S^4$$ known as Mazur manifolds, see this Wikipedia article. Furthermore, every smooth compact hypersurface in $$R^n$$ is isotopic to a nonsingular real-algebraic subset of $$R^n$$; this argument is due to Seifert, here
for details and generalizations. Thus, there exist 3-dimensional nonsingular hypersurfaces in $$S^4$$ or in $$R^4$$, whatever you prefer, which are homology spheres that are not homeomorphic to $$S^3$$. Lastly, removing a point from such a hypersurface and using the stereographic projection, we obtain a smooth algebraic hypersurface in $$R^4$$ which is acyclic but not simply-connected. It is likely one can even obtain algebraic hypersurfaces which are contractible and are not homeomorphic to Euclidean spaces. Such examples exist if you do not insist on having a hypersurface, see my answer here.
• Thank you very much for your answer and the precision in it! But for the case of surfaces in $\Bbb R^3$, is it true that X is homeomorphic to a plane? – Dino Mar 7 '19 at 7:07