Multiple covariant derivatives In Aubin's book, he defines the curvature tensor to be
$R^{i}_{kij} = \nabla_i \nabla_j Z^{i} - \nabla_{j} \nabla_{i} Z^{i}$. What is this supposed to mean? $Z^{i}$ is not even globally defined and how do you take its covariant derivative? He also defines $\Delta f = \nabla_{i} \nabla^{i} f$, which I struggle to see the meaning of as well.
 A: Personally I think it's a tragedy that no book explains how to use this kind of notation and we just have to learn it by guessing basically. But to spare you that suffering, here's an explanation I hope will help: you're right that $Z^i$ is not globally defined, but that doesn't matter. Here the author is only interested in working in local coordinates, and by the nature of a connection, that is harmless. What Aubin actually writes is that:
$$R_{kij}^{\ell}Z^k = \nabla_i \nabla_j Z^{\ell} - \nabla_{j}\nabla_{i}Z^{\ell}$$
Let's see what both sides of this equality mean. On the left side, we have a real number expressed in Einstein notation, and so on the right side we must have a real number as well. Actually, it's a bit more subtle than that and to say this would be skipping a few steps. On both sides of the above equality we have a real function. What this equality actually means is that once you fix a chart $x: U \to M$ around a point $q \in M$, it's true that:
$$(R_{kij}^{\ell}Z^k)(p) = (\nabla_i \nabla_j Z^{\ell} - \nabla_{j}\nabla_{i}Z^{\ell})(p)$$
for every $p \in M$, where here $\partial_i \doteq \frac{\partial}{\partial x^i}$ for convenience. Now, $Z^k$ is the function $Z^k: U \to \mathbb{R}$ defined by $Z^k(p) = \text{$k$-th component of $Z(p)$ in the basis $\{\partial_1, \cdots, \partial_n\}$}$. Similarly, $R_{kij}^{\ell}$ is the function $R_{kij}^{\ell}: U \to \mathbb{R}$ defined by $(R_{kij}^{\ell})(p) = \text{$\ell$-th component of $(R(\partial_i, \partial_j)\partial_k)(p)$ in the same basis}$. And $\nabla_i \nabla_j Z^{\ell}$ is the function defined by $$(\nabla_i \nabla_j Z^{\ell})(p) = \text{$\ell$-th component of $(\nabla_i \nabla_j Z)(p)$ in the same basis }$$
Of course we still have to check this equality is indeed valid. But this is clear: writing in this local chart $Z = Z^k \partial_{k}$, we get $R(\partial_i, \partial_j)Z = Z^k R_{kij}^{\ell} \partial_{\ell}$ almost by definition. Therefore:
$$Z^k R_{kij}^{\ell} = (R(\partial_i, \partial_j)Z)^{\ell} = (\nabla_i \nabla_j Z - \nabla_{j}\nabla_{i}Z)^{\ell} = \nabla_i \nabla_j Z^{\ell} - \nabla_{j}\nabla_{i}Z^{\ell}$$
as previously claimed.
As for the latter definition, here's what's happening in all this mess: in Ricci-calculus, working in local coordinates as before, if $X = X^{i} \partial_i$, we denote this just by writing $X = X^i$ sometimes. Now, for a fixed vector field $Y$, we can think of $\nabla Y$ as a $(1, 1)$ tensor field, given by $\Gamma(TM) \ni X \mapsto \nabla_{X} Y \in \Gamma(TM)$. We define the trace of a $(1,1)$ tensor $T$ as $\text{tr}(T): M \to \mathbb{R}$, $p \mapsto (\text{tr}(T))(p) = \displaystyle{\sum_{i} g(A(E_i), E_i)(p)}$, where $\{E_1, \cdots, E_n\}$ is an orthonormal basis of $T_p M$. I'll leave the job of checking this definition doesn't depend on the basis and how to make sense of $\nabla Y$ as a $(1, 1)$ tensor to you. Here are some more lazy notations in order to make sense of this: for a vector field $\eta = (\eta^1, \cdots, \eta^k)$ in local coordinates, we write:
$$\nabla_i \eta^k = \frac{\partial \eta^k}{\partial x^i} + \Gamma_{ij}^{k}\eta^j$$
where as before, this means the $k$-th component of $\nabla_i \eta$ in local coordinates.
We might even get lazy and write $\text{div}(X^i) = \nabla_i X^i, \text{div}(\eta^j) = \nabla_{i} \eta^i$. Letting $\text{grad}(f)$ be the vector field defined by $g(\text{grad}f, \cdot) = \text{d}f(\cdot)$, sometimes we also write $f^i = \text{grad}(f) = \nabla^{i}f$. Defining $\Delta f = \text{div}(\text{grad}(f))$ and being incredibly lazy, we then get the notation $\Delta f = \nabla_i \nabla^i f$.
