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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2013

Paper 49

For a smooth <function>, <pullback of a smooth function> is composition. The <chain rule> gives
$$
\boxed{\mathcal L_Xf=\left.\frac{d}{ds}\right|_0f(\phi_s(p))=X(f).}
$$
For a <vector field> $Y$, use any local coordinates. To first order,
$$
\phi_s^\mu(x)=x^\mu+sX^\mu(x)+O(s^2),\qquad
(d\phi_s)^{-1\mu}{}_{\nu}=\delta^\mu{}_{\nu}-s\partial_\nu X^\mu+O(s^2).
$$
Multiplying this inverse differential by $Y(\phi_s(x))$ gives
$$
(\phi_s^*Y)^\mu=Y^\mu+s\bigl(X^\nu\partial_\nu Y^\mu-Y^\nu\partial_\nu X^\mu\bigr)+O(s^2).
$$
Thus
$$
\boxed{\mathcal L_XY=[X,Y],\qquad
[X,Y]^\mu=X^\nu\partial_\nu Y^\mu-Y^\nu\partial_\nu X^\mu.}
$$
This is the <Lie bracket of vector fields>, whose action on a function is $X(Yf)-Y(Xf)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 49 / 1 / c / Solution

Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 49 / 1 / c / Solution

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Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 49 / 1 / c / Solution

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