In the book of J. Lee introduction to Riemmannian manifold. It says that for a $(k,l)$ tensor $F$, the second covariant derivative $\nabla^2_{X,Y}F$ is linear over $C^\infty (M)$ in $Y$, while $\nabla_X(\nabla_Y F)$ is not.
I just wonder why $\nabla_X(\nabla_Y F)$ is not linear in $Y$. In the identity
$$
\nabla_X(\nabla_YF)=\nabla^2_{X,Y}F+\nabla_{(\nabla_XY)}F
$$
the first term on RHS is linear in $Y$. It seems that $\nabla_{(\nabla_XY)}F$ is also linear in $Y$ since
$$
\nabla_{(\nabla_X(Y_1+Y_2))}F=\nabla_{\nabla_XY_1}F+\nabla_{\nabla_XY_2}F.
$$
I think I was wrong at some point, but I can't see it.
Best Answer
Your misunderstanding stems from "linear over $C^\infty(M)$": all involved maps are linear over $\mathbb{R}$. It is the multiplication with a smooth function $f$ that no longer behaves as wanted for $\nabla_X \nabla_{Y}$. More precisely, we have $$\nabla_X \nabla_{fY} F = \nabla_X (f \cdot \nabla_Y F) = (\mathcal{L}_X f) \cdot \nabla_Y F + f \cdot \nabla_X \nabla_Y F, $$ where $\mathcal{L}$ denotes the usual Lie-derivative. In case of function linearity, we would only expect the final contribution! Note that the difference from "expected behaviour" consists in the first order differential operator $(\mathcal{L}_X f) \cdot \nabla_Y$, wherefore $\nabla_X \nabla_Y$ is of second order in $Y$. You should now do the computation for $\nabla^2_{X,fY}$ instead and verify $\mathcal{C}^\infty$-linearity in this case.