For a $k$-dimensional subspace $E \subseteq T_p M$, $\exists$ a codimension $k$ submanifold $N \subseteq M$ with $T_p M = E \oplus T_p N$?

Question

Let $M$ be a manifold and $p \in M$. Let $E \subseteq T_p M$ be a subspace of dimension $k$. By using the Frobenius theorem I have learned that there exists a codimension $k$ submanifold $N \subseteq M$ with $T_p M = E \oplus T_p N$. But without the Frobenius theorem can we show this statement?

Answer 1

The Frobenius Theorem is stronger: It states that for a smooth manifold $M$, any involutive distribution $\mathscr{D}\subseteq TM$ is integrable.

That is, if for all $X,Y\in \Gamma(M, \mathscr{D})$, $[X,Y]\in \Gamma(M,\mathscr{D})$, then there is a submanifold $N\subseteq M$ such that $T_pN=\mathscr{D}_p$ for each $p\in M$.

As you said, with a little work we can deduce the statement that you propose from Frobenius' Theorem, but there is no need as we can do this via elementary tools. Here is a sketch of how. In specifying $E\subseteq T_pM$, we can choose a linear complement $F\subseteq T_pM$, i.e. such that $E\oplus F=T_pM$. So, we really just want to find a $N$ with $T_pN=F$. Take local coordinates $(U,x^1,\ldots, x^n)$ centered at $p$ - i.e. $p$ is identified with $0\in \mathbb{R}^n$. We can choose these coordinates so that $E=\operatorname{span}(\frac{\partial}{\partial x^1}|_p,\ldots, \frac{\partial}{\partial x^k}|_p).$ Then the submanifold ($N\subseteq U$) by $x^1=\cdots=x^k=0$ has tangent space at $p$ specified by $\operatorname{span}(\frac{\partial}{\partial x^{k+1}}|_p,\ldots, \frac{\partial}{\partial x^n}|_p).$ It follows that in local coordinates, $T_pM=E\oplus T_pN$, and hence the result is also true in the manifold.