Lie brackets

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Lie brackets of vector fields. Fields commute if and only if their flows commute.

A reference for this material is Section 9 of John M. Lee. Introduction to smooth manifolds. Second edition. Grad. Texts in Math., 218. Springer, New York, 2013. xvi+708pp. ISBN: 978-1-4419-9981-8.

Lie bracket Let $M$ be a smooth manifold, and $X,Y \in \mathfrak {X}(M)$. Then the map $[X,Y] : C^{\infty } (M) \to C^{\infty }(M)$ given by

\[ [X,Y] (f) : = XY f - YXf \]

is a global derivation. The vector field $[X,Y]$ is called the Lie bracket of $X$ and $Y$.

Solution:

For $f,g \in C^{\infty }(M)$, one has

\begin{align*} [X,Y] (fg) = & \, XY(fg) - YX (fg) \\ = & \, X( (Yf)g + f (Yg) ) - Y ( (Xf)g + f (Xg) ) \\ = & \, (XYf) g + (Yf )(Xg) + (Xf)(Yg) + f (XYg)\\ & \, \,\,\, - (YXf)g - (Xf )(Yg) - (Yf)(Xg) - f (YX g) \\ = & \, ( XY f - YXf ) g + f( XY g - YX g ) \\ = & \, ( [X,Y] f) g + f ([X,Y] g) . \end{align*}

Also, in a chart $(U, \varphi )$ one has

\[ X (x) = \sum _{i = 1} ^n X^i (x) \partial _i \vert _x , \hspace {1.5cm} Y (x) = \sum _{j = 1} ^n Y^j (x) \partial _j \vert _x . \]

Then

\begin{align*} [X,Y] f = & \, \sum _{i = 1 } ^n \sum _{j = 1} ^n [ X^i \partial _i ( Y^j \partial _j f ) - Y^j \partial _j ( X^i \partial _i f ) ]\\ = & \, \sum _{i = 1 } ^n \sum _{j = 1} ^n [ X^i (\partial _i Y^j) \partial _j f - Y^j (\partial _j X^i ) \partial _i f ] \\ & \,\,\, + \sum _{i = 1 } ^n \sum _{j = 1} ^n X^i Y^j (\partial _i \partial _j f - \partial _j \partial _i f)\\ = & \, \sum _{i = 1} ^n \left [ \sum _{j = 1 }^n ( X^j \partial _j Y^i - Y^j \partial _j X^i ) \right ] \partial _i f . \end{align*}

This shows that the $i$-th component of the vector field $[X,Y]$ is

\[ \sum _{j = 1 }^n ( X^j \partial _j Y^i - Y^j \partial _j X^i ) . \]

Properties of Lie bracket Let $f , g \in C^{\infty }(M)$ and $\mathfrak {X}(M)$, and $X,Y,Z \in \mathfrak {X}(M)$. Then

$[Y,X] = - [X,Y]$.
$[fX, gY] = fg [X,Y] + f(Xg)Y - g (Yf) X. $
$[X,[Y,Z]] + [Y, [Z,X] ] + [Z, [X,Y]] = 0$.

Solution:: Exercise.

Lie bracket is Lie derivative Let $X,Y \in \mathfrak {X}(M) $, $p \in M$, and $\Phi ^X : W \times (- \varepsilon , \varepsilon ) \to M$ the flow of $X$ around $p$. Then

\begin{equation}\label {eq:lie-derivative} [X,Y] (p) = \lim _{h \to 0 } \frac { (\Phi _{-h}^X )_{\ast } Y(\Phi _h^X(p)) - Y(p) }{h} . \end{equation}

Solution:

Fix $f \in C^{\infty }(M)$ and let $ g : ( - \delta , \delta ) \times (- \delta , \delta ) \to \mathbb {R} $ be given by

\begin{align*} g (s,t) : & = \, ( \Phi _{-t} ^X )_{\ast } Y (\Phi _s ^X (p)) (f) \\ & = \, Y( \Phi _s^X (p) ) ( f \circ \Phi _{-t} ^X ) \\ & = \, \frac {\partial }{ \partial u } \Big \vert _{u = 0} f ( \Phi _{-t}^X \Phi ^Y _u \Phi ^X _s (p)) , \end{align*}

where $\Phi ^Y_u$ is the flow of $Y$. We compute

\begin{align*} \frac {\partial g}{\partial s} (0,0) & = \frac {\partial }{\partial s} \Big \vert _{s = 0} Y ( \Phi ^X _s (p) ) (f) \\ & = \frac {\partial }{\partial s} \Big \vert _{s = 0} Yf (\Phi ^X _s (p )) \\ & = X Y f ( p ) \end{align*}

\begin{align*} \frac {\partial g}{\partial t} (0,0) & = \frac {\partial }{\partial t} \Big \vert _{t = 0}\frac {\partial }{\partial u} \Big \vert _{u = 0} f ( \Phi ^X _{-t} ( \Phi ^Y _u (p) )) \\ & = \frac {\partial }{\partial u} \Big \vert _{u = 0}\frac {\partial }{\partial t} \Big \vert _{t = 0} f ( \Phi ^X _{-t} ( \Phi ^Y _u (p) )) \\ & = \frac {\partial }{\partial u} \Big \vert _{u = 0} [ - Xf ( \Phi ^Y _u (p)) ] \\ & = - YXf (p) \end{align*}

Then by the chain rule, one has

\begin{align*} \frac {d}{dh} \Big \vert _{h= 0} g (h,h) & = \frac {\partial g}{\partial s} (0,0) + \frac {\partial g}{\partial t} (0,0) \\ & = X Y f ( p ) - YXf (p) \\ & = [X,Y] (p) f. \end{align*}

Finally, note that the right hand side of (eq:lie-derivative) applied to $f$ also equals $\frac {d}{dh} \Big \vert _{h= 0} g (h,h)$. Since $f$ was arbitrary, the result follows.

Commutators of flows Let $X,Y \in \mathfrak {X}(M)$, $p \in M$, and $\Phi ^X : W \times (-\varepsilon , \varepsilon ) \to M$, $\Phi ^Y : W \times (- \varepsilon , \varepsilon ) \to M$ their respective flows around $p$. Then the following are equivalent:

$[X,Y] = 0$ in a neighborhood of $p$.
$\Phi _s^X \Phi _t^Y = \Phi _t^Y \Phi _s^X $ for small enough $s$ and $t$ in a neighborhood of $p$.

Solution:

Assume $[X,Y] = 0$ in a neighborhood of $p$. By Proposition pro:lie-derivative, for $x$ near $p$, the vector

\[ ( \Phi ^X _{-t} )_{\ast } Y (\Phi _t^X (x)) \in T_xM \]

does not depend on $t$ for small $t$. Hence

\[ ( \Phi ^X _{t} )_{\ast } Y(x) = Y ( \Phi ^X_t (x) ) . \]

Let $\gamma (s) : = \Phi ^Y_s (x)$. Then

\begin{align*} \frac {\partial }{\partial s} \Big \vert _{s = 0 } \left [ \Phi ^X_t ( \gamma (s) ) \right ] & = (\Phi ^X_t)_{\ast } \gamma ' (s) \\ & = (\Phi ^X_t)_{\ast } Y (\gamma (s) ) \\ & = Y ( \Phi ^X_t (\gamma (s))) . \end{align*}

This implies that $s \mapsto \Phi ^X_t (\gamma (s) )$ is the flow line of $Y$ starting at $\Phi ^X_t (x)$. Therefore

\[ \Phi ^Y_s ( \Phi _t ^X (x) ) = \Phi ^X_t ( \gamma (s) ) = \Phi ^X_t ( \Phi _s ^Y (x) ). \]

Now assume $\Phi _s ^X \Phi _t ^Y = \Phi _t^Y \Phi _s ^X$ for small enough $s$ and $t$ in a neighborhood of $p$ and let $x$ in such neighborhood. For $f \in C^{\infty }(M)$, one then has

\begin{align*} XY f(x) & = X(x) (Yf) \\ & = \frac {\partial }{\partial s} \Big \vert _{s = 0} Yf ( \Phi ^X_s (x) ) \\ & = \frac {\partial }{\partial s} \Big \vert _{s = 0} \frac {\partial }{\partial t} \Big \vert _{t = 0} f ( \Phi ^Y _t (\Phi ^X _s (x) ) ) \\ & = \frac {\partial }{\partial t} \Big \vert _{t = 0} \frac {\partial }{\partial s} \Big \vert _{s = 0} f ( \Phi ^X _s (\Phi ^Y _t (x) ) ) \\ & = \frac {\partial }{\partial t} \Big \vert _{t = 0} Xf ( \Phi ^Y_t (x) ) \\ & = Y(x) (Xf) \\ & = YX f (x). \end{align*}

This shows $[X,Y](x) = 0$.

Let $f: M \to N$ smooth, $X, Y \in \mathfrak {X}(M)$, and $V, W \in \mathfrak {X}(N)$ such that $X$ and $V$ are $f$-related, and $Y$ and $W$ are $f$-related. Then $[X, Y]$ and $[V, W]$ are $f$-related.

Solution:

Let $\alpha \in C^{\infty }(N)$. Then

\begin{align*} XY ( \alpha \circ f ) & = X ( ( W \alpha ) \circ f ) \\ & = ( VW \alpha ) \circ f , \end{align*}

\begin{align*} YX ( \alpha \circ f ) & = Y ( ( V \alpha ) \circ f ) \\ & = ( WV \alpha ) \circ f . \end{align*}

Therefore

\[ [X,Y] (\alpha \circ f) = ( [V,W] \, \alpha ) \circ f . \]

Since $\alpha $ was arbitrary, we deduce $[X,Y]$ and $[V,Y]$ are $f$-related.