When is the union of embedded smooth manifolds a smooth manifold? - MathOverflow

When is the union of embedded smooth manifolds a smooth manifold?

2011-10-21T01:52:30Z

Suppose we have k embeddings of one single smooth manifold into one other, such that the intersections are manifolds,too. What are sufficient conditions, such that the union of those embeddings is a smooth submanifold in the image manifold?

To be more precise: Suppose we have a manifold M and a manifold N (both smooth). Moreover we have k-many embeddings $f_1: M \rightarrow N$ , $\ldots$, $f_k: M \rightarrow N$ such that the intersections of $f_j(M) \cap f_i(M) \neq \emptyset$ are in general not empty but it is required that $f_j(M) \cap f_i(M)$ is itself a submanifold of $N$. Is it possible at all that the union $\cup f_j(M)$ for all $1 \leq j \leq k$ is a submanifold of N?

Of course it is if all $f_j$'s are equal, but suppose some of them are not. Then in general the union is not a submanifold as we can see for example if we embedded $\mathbb{R}^2$ into $\mathbb{R}^3$ by $f_1(x_1,x_2):=(0,x_1,x_2)$ and $f_2(x_1,x_2):=(x_1,x_2,0)$ and $f_3(x_1,x_2):=(x_1,0,x_3)$. Then the intersections are lines and hence are manifolds by themself, but the union of the images of the $f_j$'s is not a manifold.

The question is: Are there conditions under which the union is a submanifold or not?

Of course one sufficient condition is that there is a $i$ such that $f_i(M)=\cup f_j(M)$ $i \neq j$. So the more interesting situation is, when we have $f_i(M)\neq \cup f_j(M)$ for all $i \neq j$.

Answer by Mark Grant for When is the union of embedded smooth manifolds a smooth manifold?

2011-10-21T09:13:44Z

For simplicity, I will assume that all manifolds are connected and closed.

An obvious sufficient condition is that the for each pair of embeddings, their images are either disjoint or equal.

To see that this is also necessary, suppose that the two embeddings $f\colon M \hookrightarrow N,\, f'\colon M' \hookrightarrow N$ of copies of $M$ intersect in a submanifold $Q$. Note that $Q = f^{-1}(M') \subseteq M$ can be viewed as a closed submanifold of $M$. This implies that either $Q=M=M'$, or $Q$ has dimension less than that of $M$. In this latter case, the union $f(M)\cup f'(M')$ will not be a submanifold (each point of $Q$ is a double-point singularity of the immersion $f\sqcup f'$).

The same argument extends to disconnected manifolds and shows that the images of each connected component must be pairwise equal or disjoint.

Answer by Anton Lukyanenko for When is the union of embedded smooth manifolds a smooth manifold?

2011-10-21T14:18:13Z

The union will usually not be a manifold. Here are three examples:

1) x (two intersecting lines)

2) 8 (two tangent circles)

3) The topologist's sine curve, the union of two disjoint embeddings of R.

Based on these, I'm guessing there's no local condition that guarantees the union is a manifold. You probably can't do much better than requiring the manifolds to be closed and disjoint.

Answer by Ryan Budney for When is the union of embedded smooth manifolds a smooth manifold?

2011-10-21T18:00:52Z

To expand on my comment, say for all $i$, $M_i$ is an $m$-dimensional submanifold of $N$. So in your question, $M_i = f_i(M)$. But the $M_i$'s need not be diffeomorphic for the answer to hold, below.

Further assume that $\overline{M_i} \cap M_j = \overline{M_i \cap M_j} \cap M_j$ for all pairs $i \neq j$, and $M_i \cap M_j$ is also an $m$-dimensional submanifold of $N$ for all $i \neq j$.

Then I claim $\cup_i M_i$ is an $m$-dimensional submanifold of $N$. The proof is fairly mechanical, basically the closure condition rules out the "topologist's sine curve" example Anton Lukayenko gives. It ensures that for any point in the union, a chart neighbourhood for an $M_i$ can be restricted to a chart neighbourhood of any overlapping $M_j$ because $M_j$ can't approach from a transverse direction. The condition that intersections all are $m$-dimensional rules out the figure-8 case, etc.