This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Distribution and Integral Flow
Previously, in §Vector Fields, we saw that for any smooth vector field \(X\) defined on a given manifold \(M\), there exists a sufficiently small \(\epsilon>0\) such that there is a curve \(\sigma:(-\epsilon,\epsilon)\rightarrow M\) satisfying
\[\sigma'(t)=X(\sigma(t)),\qquad \sigma(0)=p\tag{1}\]The image \(S\) of such a curve \(\sigma\) in \(M\) can be viewed as a submanifold of \(M\) containing the point \(p\).
On the other hand, equation (1) determines not only the image of \(\sigma\), but also the way it is parameterized. In contrast, the submanifold \(S\) is determined independently of the parameterization of \(\sigma\), so it is determined solely by the 1-dimensional subspace \(\span(X_p)\) of \(T_pM\), rather than by the vector \(X_p\) itself.
Definition 1 Let \(M\) be an \(m\)-dimensional manifold, and let \(k\) be an integer with \(1\leq k\leq m\). A function assigning to each \(p\in M\) a \(k\)-dimensional subspace \(\mathcal{D}(p)\) of \(T_pM\) is called a \(k\)-dimensional distribution.
A \(k\)-dimensional distribution \(\mathcal{D}\) is said to be \(C^\infty\) if for each \(p\in M\) there exists a suitable open neighborhood \(U\) and smooth vector fields \(X_1,\ldots, X_k\) defined on it such that for each \(x\in U\),
\[\mathcal{D}(x)=\span\{(X_1)_x,\ldots, (X_k)_x\}\]holds.
As we saw above, a vector field \(X\) defines a 1-dimensional distribution via the map \(p\mapsto\span(X_p)\subseteq T_pM\). Then the submanifold \(S\) is characterized by the condition
\[T_xS=\mathcal{D}(x)\qquad\text{for all $x$}\]We therefore make the following definition.
Definition 2 For a \(k\)-dimensional distribution \(\mathcal{D}\) defined on \(M\), a manifold \(S\) satisfying
\[T_xS=\mathcal{D}(x)\qquad\text{for all $x$}\]is called an integral manifold of \(\mathcal{D}\).
If for each \(p\in M\) there exists an integral manifold containing \(p\), then \(\mathcal{D}\) is called an integrable distribution.
Frobenius Theorem
The following theorem is well known.
Theorem 3 (Frobenius) Let \(\mathcal{D}\) be a \(k\)-dimensional distribution defined on a manifold \(M\). Then \(\mathcal{D}\) is integrable if and only if \([X,Y]\in\mathcal{D}\) holds for any \(X,Y\in\mathcal{D}\).
Moreover, for any \(k\)-dimensional involutive distribution the following hold:
- For each \(p\in M\), there exists an integral manifold of \(\mathcal{D}\) containing \(p\).
-
Furthermore, by choosing a coordinate system \((U,\varphi)\) centered at \(p\) appropriately, the slices defined by the equations
\[x_i=\text{constant},\qquad i>k\]can be made to be integral manifolds of \(\mathcal{D}\).
- Finally, if \(\Phi:N\rightarrow M\) is a connected integral manifold and \(\Phi(N)\subseteq U\), then \(\Phi(N)\) is contained in exactly one of the slices from (2).
A distribution satisfying the latter condition is called involutive. Therefore, the Frobenius theorem may be summarized by saying that a distribution \(\mathcal{D}\) is integrable if and only if it is involutive.
One direction of this theorem can be proved quite easily.
Lemma 4 Let \(\mathcal{D}\) be a smooth distribution defined on a manifold \(M\). If \(\mathcal{D}\) is integrable, then \(\mathcal{D}\) is an involutive distribution.
Proof
Let \(X,Y\in\mathcal{D}\) and pick a point \(p\in M\). We must show that \([X,Y]_p\in\mathcal{D}(p)\).
Since \(\mathcal{D}\) is an integrable distribution, there exists an integral submanifold \(\Phi:S\rightarrow M\) of \(\mathcal{D}\) containing \(p\). Let \(s\in S\) be a point satisfying \(\Phi(s)=p\). For any \(x\in S\),
\[d\Phi_x:T_xS\rightarrow\mathcal{D}(\Phi(x))\]is an isomorphism, so we can find two vector fields \(\tilde{X},\tilde{Y}\) satisfying
\[d\Phi_s(\tilde{X}_s)=X_p,\qquad d\Phi_s(\tilde{Y}_s)=Y_p.\]Then these vector fields are \(\Phi\)-related to \(X\) and \(Y\) respectively, so by §Lie Derivative, ⁋Proposition 9, \([\tilde{X},\tilde{Y}]\) is \(\Phi\)-related to \([X,Y]\). Therefore
\[[X,Y]_p=d\Phi_s([\tilde{X},\tilde{Y}]_s)\in\mathcal{D}(p)\]holds.
Thus, the difficult part of the proof of Theorem 3 is the converse direction. This proceeds by induction on the dimension \(k\) of the distribution.
Lemma 5 Let \(M\) be an \(m\)-dimensional manifold, \(p\in M\) a point, and \(X\) a vector field satisfying \(X_p\neq 0\). Then there exists a suitable coordinate system \((U,\varphi)\) containing \(p\), with \(\varphi=(x^1,\ldots, x^m)\), such that
\[X|_U=\frac{\partial}{\partial x^1}\bigg|_U\]holds.
Proof
Choose a coordinate system \((V,\tau)\) centered at \(p\), with \(\tau=(y^1,\ldots, y^m)\), such that
\[X_p=\frac{\partial}{\partial y^1}\bigg|_p\]holds. Without loss of generality, we may assume that \(V\) is sufficiently small so that for a suitable \(\epsilon>0\) the map
\[(-\epsilon,\epsilon)\times V\rightarrow M;\qquad(t,q)\mapsto X_t(q)\]is a well-defined \(C^\infty\) map. (§Vector Fields, ⁋Theorem 6) Moreover, if we choose \(\epsilon>0\) small enough that the inclusion
\[(-\epsilon,\epsilon)\times W\subseteq V,\qquad \text{$W$ is an open neighborhood of the origin in $\mathbb{R}^{d-1}$}\]holds, then the map
\[\sigma: (-\epsilon,\epsilon)\times W;\qquad (t,a^2,\ldots, a^d)\mapsto \phi^t(\tau^{-1}(0,a^2,\ldots, a^d))\]is well-defined. Now,
\[d\sigma\left(\frac{\partial}{\partial r^1}\bigg|_0\right)=\frac{\partial}{\partial y^1}\bigg|_p=X_p\neq 0,\qquad d\sigma\left(\frac{\partial}{\partial r^i}\bigg|_0\right)=\frac{\partial}{\partial y^i}\bigg|_p\]so \(\sigma\) is nonsingular at the origin, and hence \(\sigma^{-1}\) defines a coordinate map.
Proof of Theorem 3
Assume the theorem holds for all \((k-1)\)-dimensional distributions, and let \(\mathcal{D}\) be a \(k\)-dimensional distribution. For a point \(p\in M\), we may assume that \(\mathcal{D}\) is spanned by \(k\) vector fields \(X_1,\ldots, X_k\) in a neighborhood of \(p\). Now apply Lemma 5 to find a coordinate system \((V,\tau)\) centered at \(p\), with \(\tau=(y^1,\ldots, y^k)\), such that
\[X_1|_V=\frac{\partial}{\partial y^1}\]holds.
Define \(k\) vector fields \(Y_1,\ldots, Y_k\) by the formulas
\[Y_1=X_1,\qquad Y_i=X_i-(X_i(y^1))X_1\quad(i\geq 2).\]Since the \(X_i\) are independent, so are the \(Y_i\).
Now let \(S\) be the slice defined by \(y_1=0\). Then by restricting \(Y_2,\ldots, Y_k\) to \(S\) we obtain vector fields
\[Z_i=Y_i|_S \qquad (i\geq 2).\]Since
\[Z_i(y^1)=Y_i(y^1)=0\]holds, the \(Z_i\) are independent vector fields contained in the tangent space of \(S\). Hence they span a \((k-1)\)-dimensional distribution on \(S\).
To use the induction hypothesis, let us show that this distribution is involutive. That is, we must show that \([Z_i,Z_j]\in\span(Z_2,\ldots, Z_k)\) for any \(i,j\).
Consider the inclusion \(\iota:S\rightarrow M\). Then the \(Z_i\) are \(\iota\)-related to the \(Y_i\), so it suffices to show that \([Y_i,Y_j]\in\span(Y_2,\ldots, Y_k)\). Now
\[Y_i(y^1)=X_i(y^1)-X_i(y^1)X_1(y^1)=X_i(y^1)-X_i(y^1)=0\]holds for all \(i\), and therefore \([Y_i,Y_j]y^1=0\). From this we see that the \([Y_i,Y_j]\) actually belong to \(\span(Y_2,\ldots, Y_k)\).
Applying the second assertion of the theorem to the involutive distribution \(\span(Z_2,\ldots, Z_k)\) on \(S\), we can choose a coordinate system \((w^2,\ldots, w^d)\) centered at \(p\in S\) so that the slices obtained from the equations
\[w^i=\text{constant},\qquad i>k\]are integral submanifolds of \(\span(Z_2,\ldots, Z_k)\).
To complete the proofs of the first and second assertions, define \(k\) functions
\[x^1=y^1,\quad x^j=w^j\circ\pi\]where \(\pi:V\rightarrow S\) is the projection eliminating the \(y_1\) component. Then the \((x^i)\) are independent functions, so we know there exists a coordinate system \((U,\varphi)\) having them as component functions. This coordinate system then satisfies the second assertion: the slices defined by
\[x^i=\text{constant},\qquad i>k\]are integral manifolds of \(\mathcal{D}\). To see this, it suffices to show that \(Y_i(x^{k+j})=0\) for all \(x^{k+1},\ldots, x^m\).
First, from the definition of the \(x^i\) we know that \(\partial x^j/\partial y^1=\delta_{j1}\) holds, and therefore on \(U\) we have
\[Y_1=\frac{\partial}{\partial x^1}.\]For the remaining \(Y_2,\ldots, Y_k\), using the formula
\[\frac{\partial}{\partial x^1}Y_i(x^{k+j})=Y_1(Y_i(x^{k+j})=[Y_1,Y_i]x^{k+j}\]and applying
\[[Y_1,Y_i]=\sum_{l=1}^k c_{il}Y_l\]to the right-hand side from the involutivity condition on \(\mathcal{D}\), we find
\[\frac{\partial}{\partial x^1}(Y_i(x^{k+j}))=\sum_{l=2}^k c_{il}Y_l(x^{k+j}).\]Now for a fixed slice \(W\), the \(Y_i(x^{k+j})\) are functions of the single variable \(x^1\), so the above is a system of \(k-1\) linear ODEs and we can solve it.
The slices thus obtained meet \(S\cap U\) at exactly one point, and there
\[Y_i(x^{k+j})=Z_i(w^{k+j})=0\]holds, so the proofs of the first and second assertions are complete.
Finally, we must prove the third assertion. This time let \(\pi\) be the projection from \(\mathbb{R}^m\) onto the last \(m-k\) coordinates. Then the image of \(\mathcal{D}\) under \(d(\pi\circ\varphi)\) is \(0\), so
\[d(\pi\circ\varphi\circ\Phi)\equiv 0\]holds for any \(y\in N\). But since \(N\) is connected, \(\pi\circ\varphi\circ\Phi\) is constant, and hence \(\Phi(N)\) is contained in a single slice.
References
[War] Frank W. Warner. Foundations of Differentiable Manifolds and Lie Groups, Graduate texts in mathematics, Springer, 2013
[Lee] John M. Lee. Introduction to Smooth Manifolds, Graduate texts in mathematics, Springer, 2012
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