Lie derivative of Almost complex structure $J$ along a vector field $X$.

868 Views Asked by At

I'd like some guidance in computing the Lie derivative of an almost complex structure $J$ on a smooth closed Riemannian manifold $M$ in terms of the Levi-Civita connection $\nabla$. The formula I'm trying to obtain is the following $$ (\mathcal{L}_X J)(Y)=(\nabla_X J)Y-\nabla_{JY}X+J\nabla_YX $$ (Ex 2.1.1 Salamon-McDuff J-Holomorphic curves and Symplectic Topology). here is what I did.

I can think of $J$ as a $(1,1)$-tensor field $J(\alpha,Y)$ and by definition the Lie derivative is

$\begin{align}(\mathcal{L}_X J)(\alpha,Y) &= X(J(\alpha,Y))-J(\mathcal{L}_X\alpha,Y)-J(\alpha,\mathcal{L}_XY)\\ &= X(J(\alpha,Y))-J(\mathcal{L}_X\alpha,Y)+J(\alpha,[X,Y])\\ &= X(J(\alpha,Y))-J(\mathcal{L}_X\alpha,Y)+J(\alpha,\nabla_YX)-J(\alpha,\nabla_XY) \end{align}$

The term $+J(\alpha,\nabla_YX)$ somehow resemble $J\nabla_YX $ even though with this (1,1)-tensor I'm not sure. Can someone help me with these computations?

1

There are 1 best solutions below

0
On BEST ANSWER

I think you're overcomplicating it by introducing this $\alpha$. In general, if $\nabla$ and $\nabla'$ are linear connections in vector bundles $E \to M$ and $E' \to M$, there is an unique linear connection $\nabla''$ in the bundle ${\rm Hom}(E,E')$ such that the product rule holds: $$\nabla'_X(F\psi) = F(\nabla_X\psi) + (\nabla''_XF)\psi,\qquad \mbox{for all } X \in \mathfrak{X}(M),\mbox{ } F \in \Gamma({\rm Hom}(E,E')) \mbox{ and }\psi \in \Gamma(E),$$where $F\psi \in \Gamma(E')$ is defined by $(F\psi)_x \doteq F_x\psi_x$. From now all let's denote all the connections simply by $\nabla$. You have that $J$ is a section of ${\rm Hom}(TM,TM)$, and we use the connection in this hom-bundle induced by the Levi-Civita connection of $(M,g)$. So you have $$\begin{align} (\mathcal{L}_XJ)(Y) &= \mathcal{L}_X(JY) - J(\mathcal{L}_XY) \\ &= [X, JY] - J[X,Y] \\ &= \nabla_X(JY) - \nabla_{JY}X - J(\nabla_XY - \nabla_YX) \\ &= \nabla_X(JY) - J(\nabla_XY) - \nabla_{JY}X + J(\nabla_YX) \\ &= (\nabla_XJ)(Y) - \nabla_{JY}X + J(\nabla_YX),\end{align}$$as wanted. In the first equal sign I use that $\mathcal{L}_X$ is a derivation, in the second one the characterization $\mathcal{L}_X = [X,\cdot]$ (when acting on vector fields, such as $JY$), in the third one that $\nabla$ is torsion-free, in the fourth one I just reorder terms, and in the last one the characterization of the connection induced in ${\rm Hom}(TM,TM)$ applied for $J$ playing the role of $F$ in the first explanation.

Also note that we did not use that $J$ is an almost complex structure. The above relation then holds for every section of ${\rm Hom}(TM,TM)$.