The Unapologetic Mathematician

Mathematics for the interested outsider

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 […]

    Pingback by Integrable Distributions Have Integral Submanifolds « The Unapologetic Mathematician | June 30, 2011 | Reply

  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 […]

    Pingback by Foliations « The Unapologetic Mathematician | July 1, 2011 | Reply

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