Let $\{S_a\}_{a \in A}$ be a partition of embedded connected submanifolds of a manifold of $M$ where each $S_a$ has the same dimension, say $k$.
Is it true that this a foliation of $M$?
I guess so:
By assumption: $M = \coprod_{a \in A} S_a$ and if $p \in M$, there is a unique $a \in A$ such that $p \in S_a$. Because $S_a$ is an embedded submanifold, there is a chart $(U, \phi)$ of $M$ such that $\phi(U \cap S_a) = \mathbb{R}^k \times \{0\}_{n-k}$.
My doubt comes from the following: my book claim that the following is not a foliation of $\mathbb{R}^2$:
$$\{(c,y): y > 0\}_{c \in \mathbb{R}} \cup \{(c,y): y < 0\}_{c \in \mathbb{R}}\cup \{(x,0) : x \in \mathbb{R}\}$$
Clearly, this is a partition of $\mathbb{R}^2$ with $1$-dimensional submanifolds. So it satisfies the above, so there is a contradicting lurking somewhere?
What am I missing?
Best Answer
A $p$-dimensional foliation of an $n$-dimensional manifold $M$ is a decomposition of $M$ into a union of disjoint connected submanifolds $\{L_{\alpha}\}_{\alpha \in A}$, called the leaves of the foliation, with the following property: for every point $p \in M$, there is a coordinate chart $(U, (x^1, \dots, x^n))$ with $p \in U$ such that for each leaf $L_{\alpha}$, the components of $U \cap L_{\alpha}$ are given by the equations $x^{p+1} =$ constant, $\dots$ , $x^n =$ constant.
The non-example is not a foliation because there are no such charts for points on the $x$-axis. To see this, consider a point of the form $(c, 0)$. Suppose there is a chart $(U, (x^1, x^2))$ with $(c, 0) \in U$ such that for every leaf $L_{\alpha}$, the connected components of $U\cap L_{\alpha}$ are given by $x^2 =$ constant. Consider the leaf $L_{\alpha} = \{(c, y) \mid y > 0\}$. There is $\varepsilon > 0$ such that one component of $U\cap L_{\alpha}$ is $\{(c, y) \mid 0 < y < \varepsilon\}$, so there is a constant $k$ such that $\{(c, y) \mid 0 < y < \varepsilon\}$ is given by the equation $x^2 = k$. By continuity, $x^2(c, 0) = k$ which is a contradiction as $(c, 0) \not\in U\cap L_{\alpha}$.