When is the union of embedded smooth manifolds a smooth manifold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:09:28Z http://mathoverflow.net/feeds/question/78733 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78733/when-is-the-union-of-embedded-smooth-manifolds-a-smooth-manifold When is the union of embedded smooth manifolds a smooth manifold? Mirco 2011-10-21T01:52:30Z 2011-10-21T18:00:52Z <p>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?</p> <p>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? </p> <p>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. </p> <p>The question is: Are there conditions under which the union is a submanifold or not? </p> <p>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 $i \neq j$.</p> http://mathoverflow.net/questions/78733/when-is-the-union-of-embedded-smooth-manifolds-a-smooth-manifold/78752#78752 Answer by Mark Grant for When is the union of embedded smooth manifolds a smooth manifold? Mark Grant 2011-10-21T09:13:44Z 2011-10-21T10:31:39Z <p>For simplicity, I will assume that all manifolds are connected and closed.</p> <p>An obvious sufficient condition is that the for each pair of embeddings, their images are either disjoint or equal.</p> <p>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 $f\sqcup f'$).</p> <p>The same argument extends to disconnected manifolds and shows that the images of each connected component must be pairwise equal or disjoint.</p> http://mathoverflow.net/questions/78733/when-is-the-union-of-embedded-smooth-manifolds-a-smooth-manifold/78769#78769 Answer by Anton Lukyanenko for When is the union of embedded smooth manifolds a smooth manifold? Anton Lukyanenko 2011-10-21T14:18:13Z 2011-10-21T14:18:13Z <p>The union will usually not be a manifold. Here are three examples:</p> <p>1) x (two intersecting lines)</p> <p>2) 8 (two tangent circles)</p> <p>3) The <a href="http://en.wikipedia.org/wiki/Topologist%2527s_sine_curve" rel="nofollow">topologist's sine curve</a>, the union of two disjoint embeddings of R.</p> <p>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.</p> http://mathoverflow.net/questions/78733/when-is-the-union-of-embedded-smooth-manifolds-a-smooth-manifold/78785#78785 Answer by Ryan Budney for When is the union of embedded smooth manifolds a smooth manifold? Ryan Budney 2011-10-21T18:00:52Z 2011-10-21T18:00:52Z <p>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.</p> <p>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 $i \neq j$. </p> <p>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. </p>