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Equation 1.2.10 of Wolf's "Spaces of constant curvature" is: $[fX,gY] = fg[X,Y]+f(Xg)Y -g(Yf)X$ where $X$ and $Y$ are vector fields, and $[-,-]$ is the Poisson bracket.

I assume $f$ and $g$ are real valued functions, so $Xg$ and $Yf$ are also real valued functions (at each point one takes the directional derivative of the function along the vector given by the vector field at that point) and that, for example, $fX$, is just pointwise multiplication of $f$ and the directional derivative along $X$. The proof is left unstated, presumably because it is obvious to anyone with a grip of the subject.

Unfortunately I am very rusty and getting lost in the detail.

By antisymmetry I can see it is enough to prove that $[fX, Y] =f[X,Y] -(Yf)X$.

In turn this comes down to showing that $Y(fX)= fYX +(Yf)X$, i.e. that $Y(fX)h= fYXh +(Yf)Xh$ for any real valued function $h$ defined on relevant portions of the manifold.

I think this is simply a consequence of the chain rule for differentiation, but am getting quite confused with notation and where the various derivatives are being taken. Can anybody help me out?

Elliot Yu
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Oldlag
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    Lie bracket, not Poisson bracket :) ... – Ted Shifrin Jun 21 '22 at 15:14
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    Must admit I am more familiar with the Poisson bracket acting on functions rather than vector fields, as in the Hamiltonian formulation of mechanics. Wolf 1967 calls [X,Y] a Poisson bracket, while Spivak (Comprehensive introduction to Differential Geometry, Theorem 10) shows that it is the same thing as the Lie derivative of Y by X. Maybe Lie bracket is a more modern name. – Oldlag Jun 22 '22 at 08:45

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You have correctly interpreted the meaning of $Xg$, $Yf$, and $fX$. The remaining pieces of notation to decipher are the combinations of differential operators like $YX$ and $Y(fX)$. The key is to know that when differential operators are written in a sequence, the implied operation is composition. Thus when $YX$ and $Y(fX)$ are applied to $h$, what we get are $Y(Xh)$ and $Y(fXh)$.

Now that we know what $Y(fX) = f(YX)+(Yf)X$ means, we can prove it using the product rule (rather than the chain rule). Applied to $h$, the identity we want is really $Y(f\cdot Xh) = f\cdot Y(Xh) + (Yf)\cdot(Xh)$. This is true by differentiating $f\cdot (Xh)$ with $Y$ and applying the product rule.

Elliot Yu
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  • Many thanks for your rapid and clear answer. Yes I had appreciated that the sequences of operators were compositions, and meant to say the product rule. But my expansion was sloppy. Next step for me is to go back to basics and prove the product rule for, as you concisely put it, differentiating a product of functions "with Y". – Oldlag Jun 22 '22 at 08:28