I present three different ways I've seen the Riemann curvature written:

  1. $R(X,Y)Z=D_XD_YZ-D_YD_XZ-D_{[X,Y]}Z$

  2. $R(e_c,e_d)e_b=D_{e_c}D_{e_d}e_b-D_{e_d}D_{e_c}e_b-D_{[e_c,e_d]}e_b$.

  3. $R^{\rho}_{\space \space\sigma \mu \nu}=dx^{\rho}(R(\partial_{\mu},\partial_{\nu})\partial_{\sigma})$

However I'm not told much more and I'm struggling to make sense of when we would need to use one or the other.

I'm assuming the $e_a$ in 2. are basis vectors, e.g. $e_a=\frac{\partial}{\partial x^1}$? So the second is just the curvature applied to basis vectors? But why would we want to do that? Also, doesn't $[e_c,e_d]=0$?

And with 3. I'm just completely lost. How do you relate $R(X,Y)Z$ and $R^{\rho}_{\space \space \sigma \mu \nu}$?

(This is quite an open question on the understanding of what's the intrinsic difference between these three and when we would need to use each in simple calculations. Or even a question about coordinate expressions of tensors in general.)

  • $\begingroup$ Concerning the commutator $[e_c,e_d]$, no, generally, the frame need not commute. However, if $[e_c,e_d]=0$ then there exists a coordinate system for which the coordinate derivations reproduce the frame locally. On the other hand, if $[e_c,e_d] \neq 0$ then there is no such coordinate system. $\endgroup$ Apr 4, 2014 at 1:33
  • $\begingroup$ @user13223423, Regarding the commutator: As noted below, $R : \Gamma(M) \times \Gamma(M) \times \Gamma(M) \to \Gamma(M)$ is multilinear over $C^{\infty}$ functions on $M$. If you naively tried to define $R$ to just check the commutativity of $D_X$ and $D_Y$ (i.e. $R(X, Y) = \left[ D_X, D_Y \right ]$), you would end up with an expression that is not multilinear over $C^{\infty}$ functions. In this case, you don't have the nice tensorial properties (i.e. $R$ wouldn't be induced by a (2, 1) tensor field). If you hang around Riemannian Geometry long enough, you start to notice . . . (cont.) $\endgroup$
    – THW
    Apr 4, 2014 at 16:15
  • 1
    $\begingroup$ . . . that the Lie bracket/commutator keeps reappearing. It is almost always the case (at least to my knowledge) that when the Lie bracket/commutator appears it serves the purpose of making sure that the resulting operation is multilinear over $C^\infty$ functions. This was always something a mystery to me at first. A nice reference for the role of the Lie bracket in Riemannian Geometry is Peter Peterson's text, where the motivation for the use/role of the Lie bracket is emphasized a bit more (at least to me) than in other texts. $\endgroup$
    – THW
    Apr 4, 2014 at 16:22
  • $\begingroup$ Really great tips. You're right, things like this will only become completely transparent to me if I continue using them day in day out $\endgroup$
    – Phibert
    Apr 5, 2014 at 11:23

1 Answer 1


There is probably someone more qualified to answer this, but I will take a crack at it.

Regarding 1.):

They are viewing $R : \Gamma\left(M\right) \times \Gamma\left(M\right) \times \Gamma\left(M\right) \to \Gamma \left(M\right)$, where $\Gamma\left(M\right)$ is the set of $C^{\infty}$ vector fields on the Riemannian manifold $M$. This is very much an invariant definition of $R$ as it makes no appeal to coordinates, but rather only to objects that are invariantly associated to $M$ and its Riemannian metric: the Lie bracket of vector fields and the covariant derivative/connection.

The key observation here (in my opinion) is that $R$ is multilinear on the level of $C^{\infty}$ functions. I believe that most people refer to this as tensorial and it tells us that $R$ is induced by a $(3, 1)$ tensor (As a side note, applying the so called musical isomorphisms coming from the metric, one can obtain a corresponding $(4, 0)$ tensor field.)

Regarding 2.):

The answer is "Yes, this is evaluation of $R$ on the triple of vector fields $e_{a}, e_b, e_c$, where (presumably) $e_1$, $e_2, \ldots e_n$ constitue a frame field on $M$. The key observation (and the connection with part 1.) and part 3.)) is the following: $R\left(e_a, e_b\right)e_c$ is a vector field on $M$. This means that we can express $R \left(e_a, e_b\right)e_c$ in terms if the frame field $e_{1}, \ldots e_{n}$ as

$$R \left(e_a, e_b\right)e_c = A^{i}e_{i},$$ where the $A^{i}$ are $C^{\infty}$ functions on $M$. (Note that we are making use of the summation convention on repeated indices.) This leads us naturally to the connection between 2.) and 1.),3.) . . .

Regarding 3.):

Since $R : \Gamma\left(M\right) \times \Gamma\left(M\right) \times \Gamma\left(M\right) \to \Gamma \left(M\right)$, then in a coordinate frame field, we obtain the following

$$R \left(\partial_a, \partial_b\right)\partial_c = A^{i}\partial_{i}.$$ Note that I have replaced the frame field $e_1, e_2, \ldots , e_n$ with the coordinate frame field $\partial_1, \partial_2, \ldots \partial_n$ and that while the form of the expression is the same, the coefficient functions $A^{i}$ depend on the frame used.

In order to find the coefficient functions $A^{i}$, we need to pair the vector field $R \left(\partial_a, \partial_b\right)\partial_c$ with the dual basis elements $dx^{\rho}$. By doing so, we obtain

\begin{align*} dx^{\rho}\left(R \left(\partial_a, \partial_b\right)\partial_c\right) &= dx^{\rho}\left(A^{i}\partial_{i}\right)\\ &= A^{i} dx^{\rho}\left(\partial_{i}\right)\\ &=A^{i}\delta^{\rho}_{i}\\ &= A^{\rho}. \end{align*}

Thus, the vector field $R \left(\partial_a, \partial_b\right)\partial_c$ is expressed with respect to the frame field as

$$ R \left(\partial_a, \partial_b\right)\partial_c = dx^{\rho}\left(R \left(\partial_a, \partial_b\right)\partial_c\right)\,\partial_{\rho}.$$

But this brings us full circle and expresses $R$ as a (3, 1) tensor field with respect to the coordinate frame field as

$$R^{\rho}_{a b c} dx^{a} \otimes dx^{b} \otimes dx^{c} \otimes \partial_{\rho}$$

where $R^{\rho}_{a b c}$ is the $\rho^{th}$ component of the vector field $R\left(\partial_a, \partial_b\right) \partial c$ with respect to the the coordinate frame.

Finally, regarding the issue of how to relate $R(X, Y)Z$ with $R^{\rho}_{a b c}$, we will express $X, Y$ and $Z$ in coordinates relative to the frame field and then rely on the multi-linearity of $R$ over $C$^{\infty}$ functions. Observe that we have

\begin{align*} X &= X^{i}\partial_{i}\\ Y &= Y^{j}\partial_{j}\\ Z &= Z^{k}\partial_{k}, \end{align*} where indices on the coefficient functions $X^{i}, Y^{j}, Z^{k}$run from 1 to n. Evaluating $R(X, Y)Z$, we obtain \begin{align*} R(X, Y)Z &= R\left(X^{i}\partial_{i}, Y^{j}\partial_{j}\right)Z^{k}\partial_{k} &\hspace{.5in} \textrm{(Sub)}\\ &= X^{i}Y^{j}Z^{k}R\left(\partial_{i}, \partial_{j}\right)\partial_{k} &\hspace{.5in} \textrm{(Mult-linearity)}\\ &= X^{i}Y^{j}Z^{k}R^{\rho}_{ijk}\partial_{\rho} &\hspace{.5in} \textrm{(Defn. of $R^{\rho}_{ijk}$)} \end{align*}

Note: In the discussion of 3.), one could equally well use any frame field $e_{1}, \ldots , e_{n}$ and its corresponding dual frame field $\theta^{1}, \ldots, \theta^{n}$.


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