What comes after the curvature tensor in "higher derivatives"? Let $(M,g)$ be a Riemannian Manifold and $\nabla$ be a metric compatible (not necessarily Levi-Civita) connection on $M$. The metric tensor itself is the 0-th order term in the covariant derivatives wrt to connection.
The next order is the torsion tensor, which is defined as $$T(X,Y) := \nabla_X Y - \nabla_Y X - [X,Y]$$, where $X,Y \in \mathfrak X(M)$ are vector fields on $M$. Now the torsion tensor is a 1st order term in the covariant derivatives wrt to the connection and furthermore expressed in the coordinates, $T$ has the derivatives of metric coefficients $g_{ij}$ wrt to the chosen coordinates.
The next order is the curvature tensor, defined as
$$R(X,Y)Z := \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z$$
for vector fields $X,Y,Z \in \mathfrak X(M)$ and the curvature tensor is clearly 2nd order in the covariant derivatives (apart from the last term) and expressed in the coordinates, it has the second derivatives of $g_{ij}$ wrt to the chosen coordinates.
In the usual differential geometry textbooks/lectures the sequence of higher order covariant derivatives ends here. What comes after the curvature tensor in the higher order derivatives in the metric coefficients and higher order in covariant derivatives? Can I systematically define higher order terms? If yes what are the meaning of these tensor fields?
 A: The Cotton tensor is one such example. I also saw 3rd order Codazzi tensor being used ("Handbook of Differential Geometry", Volume 1, p. 933). 
A: If $E \to M$ is a vector bundle equipped with a connection $\nabla$, one can generalize the exterior derivative ${\rm d}$ to an operator ${\rm d}^\nabla$ acting on $E$-valued forms, by mimicking the coordinate-free formula for ${\rm d}$: $$\begin{align} ({\rm d}^\nabla\omega)(X_0,\ldots,X_k) &= \sum_{i=0}^k (-1)^i \nabla_{X_i}\omega(X_0,\ldots, \hat{X_i},\ldots, X_k) \\ & \qquad + \sum_{0\leq i<j \leq k}(-1)^{i+j}\omega([X_i,X_j],X_0,\ldots, \hat{X_i},\ldots, \hat{X_j},\ldots X_k)\end{align}$$
This reduces to the usual exterior derivative when $E = M\times \Bbb R$ is the trivial line bundle and $\nabla$ is the standard flat connection. Now, if $X \in \mathfrak{X}(M)$ and $\psi \in \Gamma(E)$ is seen as an $E$-valued $0$-form, we have $$({\rm d}^\nabla\psi)(X) = \nabla_X\psi \quad \mbox{and} \quad ({\rm d}^\nabla({\rm d}^\nabla\psi))(X,Y) = R(X,Y)\psi.$$Want to generalize further? Sure, move on and compute $$({\rm d}^\nabla({\rm d}^\nabla({\rm d}^\nabla\psi)))(X,Y,Z) = R(X,Y)\nabla_Z\psi + R(Y,Z)\nabla_X\psi + R(Z,X)\nabla_Y\psi.$$This illustrates the general phenomenon that higher covariant exterior derivatives can be expressed in terms of $R$ itself. That's the reason why people don't go after such expressions. They just might naturally appear in computations. If $E = TM$, there is one special $TM$-valued $1$-form we can consider: the identity ${\rm Id}\colon TM \to TM$. With this definition, we have $({\rm d}^\nabla{\rm Id})(X,Y) = T(X,Y)$ and also $$({\rm d}^\nabla({\rm d}^\nabla{\rm Id}))(X,Y,Z) = ({\rm d}^\nabla T)(X,Y,Z) = R(X,Y)Z+R(Y,Z)X + R(Z,X)Y,$$which in particular says that the first Bianchi identity holds for torsion-free connections. The second Bianchi identity, in turn, is expressed by ${\rm d}^\nabla R = 0$. So if you know $R$, you know how to express any higher order derivatives you want.
