0

$\vee$ denotes the wedge product of the two manifolds. Let's say $p$ is the point connecting the two spheres. I would say an open neighborhood of this point is homeomorphic to an open subset of $\mathbb{R}^{n+2}$ an so $S^n \vee S^n$ could not be an $n$-dimensional manifold, but I am not entirely sure about it.

If $S^n \vee S^n$ is not a smooth manifold is it anyway possible to compute its de-Rham-cohomology?

NicAG
  • 661
  • 4
    The $n-$th homology group of $S^n \vee S^n$ is $\mathbb Z \oplus \mathbb Z$ which is impossible for a connected $n-$manifold. The $n-$th homology of a connected $n-$manifold is always $0$ or $\mathbb Z$. – Noel Lundström Apr 18 '20 at 00:06
  • A good way to test your guess is to think about the case $n=1$. – Eric Wofsey Apr 18 '20 at 01:37
  • 1
    Near-duplicate: https://math.stackexchange.com/questions/2694888/is-it-true-that-the-wedge-of-two-manifolds-is-not-a-manifold – Eric Wofsey Apr 18 '20 at 01:38
  • Its not even a topological manifold, never mind a smooth one. Look at figure-8, and try to see why local-Euclideaness fails at the point connecting the two circles. – feynhat Apr 18 '20 at 04:40
  • It’s fine if $n=0$. – John Palmieri Apr 18 '20 at 05:45

1 Answers1

1

Let $S^n_1$ and $S^n_2$ be disjoint copies of $S^n$ with basepoints $x_1$ and $x_2$ respectively, let $q\colon S^n_1 \sqcup S^n_2 \to S^n \vee S^n$ be the quotient map, and suppose $x_0 = \{x_1, x_2\}\in S^n \vee S^n$ is the wedge point. I claim there is no euclidean open set containing $x_0$, and will sketch a proof.

The definition of the quotient topology on $S^n \vee S^n$ is that $U$ is open iff $q^{-1}(U)$ is open, so if $x_0\in U$ then $q^{-1}(U)$ must be the disjoint union of non-empty open sets $U_1\subset S^n_1, U_2\subset S^n_2$ containing $x_1$ and $x_2$ respectively. In particular $U = U_1 \vee U_2$.

Now you have to argue that an open set in $S^n \vee S^n$ of the form $U_1 \vee U_2$ cannot be homeomorphic to an open $n$-ball when $n>0$ (the wedge of $0$-spheres is still a $0$-manifold, as John Palmieri points out in their comment to your question). Hint: consider the number of connected components of $U\setminus x_0$.

William
  • 9,310