If we have a differentiable manifold $M$ and a vector field $X$, with $d$ the exterior derivative and $\nabla$ an affine connection on $M$. If $f$ is a smooth function, what are the $(q,r)$ tensor field components for the following tensor fields:
$$ T = \nabla_X (df) $$
$$ U = d(L_X (df)) $$
Where $L_X$ is the Lie derivative.
I am struggling to understand how the affine connection and the exterior derivative effects the tensor field components.
Any help appreciated.
Best Answer
If $M$ is a manifold with a connexion $\nabla$, and if $f$ is a function on $M$, $X$ and $Y$ are vector fields on $M$.
First: $$ \left(\nabla_X\left(\mathrm{d}f\right)\right)Y = \nabla_X\left(\mathrm{d}f(Y)\right) - \mathrm{d}f\left(\nabla_XY\right) = \nabla_X\left(\nabla_Y f\right) - \nabla_XY \cdot f = X\cdot\left(Y\cdot f\right) - \nabla_XY\cdot f $$ is the Hessian of $f$ with respect to $\nabla$. It is usually denoted $\nabla^2_{X,Y}f$. It is a bilinear form. In the case of a torsion free connexion, for example in the riemannian case, it is symmetric in $X$ and $Y$.
In local coordinates, say $\mathrm{d}f = \frac{\partial f}{\partial x^i}\otimes\mathrm{d}x^i$, then $$ \nabla_X \mathrm{d}f = \nabla_X\left(\frac{\partial f}{\partial x^i} \right)\otimes \mathrm{d}x^i + \frac{\partial f}{\partial x^i}\otimes \nabla_X \mathrm{d}x^i $$ and you can express $\nabla_X\mathrm{d}x^i$ thanks to the Christoffel symbols and the components of $X$. Second: $$ d\circ L_X\circ d = d\circ \left( d\circ i_X + i_X\circ d\right)\circ d = d^2 \circ i_X \circ d + d\circ i_X \circ d^2 = 0 $$ because of the Cartan formula ($L_X = d\circ i_X + i_X \circ d$) and because $d^2=0$. Hence, $d\left(L_X\left(df\right)\right)=0$.