# The Unapologetic Mathematician

## Integral Submanifolds

Given a $k$-dimensional distribution $\Delta$ on an $n$-dimensional manifold $M$, we say that a $k$-dimensional submanifold $\iota:N\hookrightarrow M$ is an “integral submanifold” of $\Delta$ if $\iota_*\mathcal{T}_pN=\Delta_{\iota(p)}$ for every $p\in N$. That is, if the subspace of $\mathcal{T}_{\iota(p)}M$ spanned by the images of vectors from $\mathcal{T}_pN$ is exactly $\Delta_p$.

This is a lot like an integral curve, with one slight distinction: in the case on an integral curve we also demand that the length of $c'(t)$ match that of $X_{c(t)}$, not just the direction (up to sign).

Now, if for every $p\in M$ there exists an integral submanifold $N(p)$ of $\Delta$ with $p\in N(p)$, then $\Delta$ is integrable. Indeed, let $X$ and $Y$ belong to $\Delta$. Since $\iota_{*q}:N(p)_q\to\Delta_{\iota(q)}$ is an isomorphism of vector spaces at every point, we can find $\tilde{X}$ and $\tilde{Y}$ that are $\iota$-related to $X$ and $Y$, respectively. That is, $X_{\iota(q)}=\iota_*\tilde{X}_q$ for all $q\in N$, and similarly for $Y$ and $\tilde{Y}$. But then we know that $[X,Y]_{\iota(q)}=\iota_*[\tilde{X},\tilde{Y}]_q$, and so $[X,Y]_{\iota(q)}\in\iota_*\mathcal{T}_qN=\Delta_{\iota(q)}$.

June 30, 2011 - Posted by | Differential Topology, Topology

1. [...] around such that on . Then the curve with coordinates and for all other is a one-dimensional integral submanifold of through [...]

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2. [...] a -dimensional distribution on , which we say is “induced” by , and that any connected integral submanifold of should be contained in a leaf of . It makes sense, then, that we should call a leaf of a [...]

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