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?

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    $\begingroup$ In the case of pure Riemannian geometry (i.e. caring only about the Levi-Civita connection), the "natural tensors" are all contractions of the metric and covariant derivatives of the curvature. I think you can make this rigorous in some categorical sense, but it's certainly true if we take the path of studying the metric in normal coordinates - see e.g. this article. For an arbitrary metric connection I'm not sure what happens - perhaps they're all contractions of covariant derivatives of $g$, $R$ and $T$? $\endgroup$ Commented Jul 5, 2017 at 11:06
  • $\begingroup$ My link above is dead. The article it pointed to is Guarrera, D.T., Johnson, N.G., Wolfe, H.F. (2002): The Taylor Expansion of a Riemannian Metric $\endgroup$ Commented May 26, 2019 at 23:55

2 Answers 2


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.

  • $\begingroup$ Could you explain how to derive $d^{\nabla}R=0$ with the expressions above? I can't bring the $(d^{\nabla}R)(X,Y,Z,W)$ to the same form as the usually represented one: $0 = (\nabla_{X}R)(\cdot, W, Y,Z) + \text{cycl} X\rightarrow Y\rightarrow Z$, as using the definition above I'm always getting extra terms, resulting from covariant differentiation of the type: $\nabla_{X}(R(\cdot, W, Y,Z))$ $\endgroup$
    – K.T.
    Commented Apr 22, 2023 at 21:59
  • 1
    $\begingroup$ See pages 4 and 5. $\endgroup$
    – Ivo Terek
    Commented Apr 22, 2023 at 22:57

The Cotton tensor is one such example. I also saw 3rd order Codazzi tensor being used ("Handbook of Differential Geometry", Volume 1, p. 933).


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