I'm trying to understand the equation following (2.15) on p.9 of Blau's Symplectic Geometry and Geometric Quantization.

For two vector fields $X,Y$ on a symplectic manifold $M$ we are told one statement of the tensorality of the Lie derivative is

$$i([X,Y]) = L(X)i(Y) - i(Y)L(X)$$

where $i(X)$ denotes inclusion into the first argument of the tensor being acted upon and $L(X)$ denotes the Lie derivative of the tensor being acted upon along the vector field $X$.

Here's my current understanding: from the linearity of tensors we have

$$i([X,Y])= i(XY-YX) = i(XY) - i(YX).$$

I would expect the next step somehow writes e.g. $i(XY)$ as $L(X)i(Y)$ but I cannot quite see the justification for this, and anyway wouldn't the result then be

$$i([X,Y]) = L(X)i(Y) - L(Y)i(X)~?$$


1 Answer 1


Well, the composition $XY$ of the vector fields $X$ and $Y$ may be an operator on $C^\infty(M)$, it is however not a vector field! Indeed, vector fields are first order linear partial differential operators, like, for example, the partial derivatives $\partial_x$ and $\partial_y$ on $C^\infty(\mathbf R^2)$. Their composition $\partial_x\partial_y$ is not first order. Your equation $$ i(XY-YX)=i(XY)-i(YX) $$ simply does not make sense since neither $i(XY)$ nor $i(YX)$ make sense.

The right way to derive the equation you want to understand is to consider a contraction $i(Y)T$ of a tensor $T$ as multplication $Y\cdot T$, and check that the Lie derivative of a contraction satisfies the Leibniz rule: $$ L_X(Y\cdot T)=L_X(Y)\cdot T+Y\cdot L_X(T), $$ which translates as $$ L_Xi(Y)=i([X,Y])+i(Y)L_X $$ since $L_XY=[X,Y]$. Voilà your formula!


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