Exercise 8.29 (Properties of the Lie Bracket)

Question

Proposition 8.28 (Properties of the Lie Bracket). The Lie bracket satisfies the following identities for all $X, Y, Z \in \mathfrak{X}(M)$ :
\begin{enumerate}[label=(\alph*)]
\item BILINEARITY: For $a, b \in \mathbb{R}$,
$$
\begin{aligned}
{[a X+b Y, Z] } & =a[X, Z]+b[Y, Z], \
{[Z, a X+b Y] } & =a[Z, X]+b[Z, Y]
\end{aligned}
$$
\item ANTISYMMETRY:
$$
[X, Y]=-[Y, X]
$$
\item JACOBI IDENTITY:
$$
[X,[Y, Z]]+[Y,[Z, X]]+[Z,[X, Y]]=0 .
$$
\item For $f, g \in C^{\infty}(M)$,
$$
[f X, g Y]=f g[X, Y]+(f X g) Y-(g Y f) X
$$
\end{enumerate}

Elu Thingol · Accepted Answer

\paragraph{Bilinearity.} Let $a,b \in \mathbb{R}$. Then for any $f \in C^{\infty}(M)$, we have:
\begin{align*}
    [aX + bY,Z]f &= (aX+bY)Zf - Z(aX+bY)f\
    &= aXZf + bYZf - aZXf - bZYf\
    &= (aXZf - aZXf) + (bYZf - bZYf)\
    &= a(XZf - ZXf) + b(YZf - ZYf)\
    &= a[X,Z]f + b[Y,Z]f
\end{align*}
Similarly, $[Z,aX+bY] = a[Z,X] + b[Z,Y]$.

\paragraph{Antisymmetry.} For any $f \in C^{\infty}(M)$, we have:
\begin{align*}
    [X,Y]f = XYf - YXf = -(YXf -XYf) = [Y,X]f
\end{align*}

\paragraph{Jacobi identity.} For any $f \in C^{\infty}(M)$, we have:
$$
\begin{aligned}
& {[X,[Y, Z]] f+[Y,[Z, X]] f+[Z,[X, Y]] f } \
&= X[Y, Z] f-[Y, Z] X f+Y[Z, X] f \
&\quad -[Z, X] Y f+Z[X, Y] f-[X, Y] Z f \
&= X Y Z f-X Z Y f-Y Z X f+Z Y X f+Y Z X f-Y X Z f \
&\quad -Z X Y f+X Z Y f+Z X Y f-Z Y X f-X Y Z f+Y X Z f\
&=0.
\end{aligned}
$$

\paragraph{Lie property.} For any $f,g \in C^{\infty}(M)$ and for any $h \in C^{\infty}(M)$, we have: 
$$
\begin{aligned}
{[f X, g Y] h } & =((f X)(g Y)-(g Y)(f X)) h \
& =(f X)(g Y) h-(g Y)(f X) h \
& =f(X g Y h)+f(g X Y h)-g(Y f X h)-g(f Y X h) \
& =f(g X Y h)-g(f Y X h)+f(X g Y h)-g(Y f X h) \
& =f g(X Y h)-g f(Y X h)+f(X g Y h)-g(Y f X h)\
& =f g((X Y h)-(Y X h))+f(X g Y) h-g(Y f X) h \
& =f g[X, Y] h+f(X g)(Y h)-g(Y f)(X h).
\end{aligned}
$$

Exercise 8.29 (Properties of the Lie Bracket)

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Exercise 8.29 (Properties of the Lie Bracket)

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