1) The definition of the Riemann curvature tensor does not include a metric. So, if we have a smooth manifold(not a Riemannian manifold), we can define the Riemannian curvature tensor for it by just giving it a connection(not the Levi-Civita connection). No metric is needed.
Now, if we also assign a metric to the smooth manifold, we can take traces of the Riemann curvature tensor and get the Ricci scalar. Does this imply that, for a Riemannian manifold(not just smooth), the relevant quantity that measures the curvature is the Ricci scalar and not the full Riemannian curvature tensor? Because to define the Riemann curvature tensor we need a connection(not a metric) and to define the Ricci scalar we also need a metric.

2) Lastly, since the Riemann curvature tensor depends on the connection and not the metric and the connection gives the way to parallel transport vectors, does it mean that parallel transporting the same vector along the same closed curve on two different Riemannian manifolds that correspond to the same smooth manifold (but we assign the same connection but different metric to each one), we will get the same angle of rotation for that vector at the end-point(which is also the starting-point) of the curve?

Note: When talking about a connection, I do not mean the Levi-Civita connection which comes from a metric. The question is about the difference in the role that the connection and metric play(independently) in the Riemann curvature tensor.

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    $\begingroup$ The curvature tensor is defined for any connection on any vector bundle. If you have a Riemannian manifold and you consider the tangent bundle as vector bundle you can consider the Riemann connection (aka the Levi-Cìvita connection which is unique given the metric) and consider the curvature (Riemannian curvature) of this connection. This is strongly related to the Riemannian metric. $\endgroup$ Nov 1, 2017 at 17:57
  • $\begingroup$ @WarlockOfFiretopMountain I agree with you, but consider the following: Say we have two smooth manifolds, $S^1$. Say we give each one a different metric so that the smooth manifolds become two different Riemannian manifolds, say the round sphere and an ellipsoid. We can then assign to each one of them their respective Levi-Civita connection. Even after giving them their corresponding L-C connections, won't their Riemann curvature tensor(but not their Ricci tensor and Ricci scalar) be the same since the Riemann curvature is, by definition, independent of the metric? $\endgroup$ Nov 1, 2017 at 18:08
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    $\begingroup$ The Riemann Curvature tensor is not independent of the metric in those cases. Locally the components $R_{k i j}^l$ are given in terms of the Christoffel symbols $\Gamma_{i j}^k$ (and derivatives) which since the connection is L-C are given in terms of the metric coefficients $g_{i j}$ (and derivatives) $\endgroup$ Nov 1, 2017 at 18:12
  • $\begingroup$ @WarlockOfFiretopMountain Sorry, you are right. Let me state what I trying to say more clearly. Again say we have two smooth manifolds $S^2$. We can assign to each one of them the same non-Levi-Civita connection. Then we assign to them a metric so that one of the is what we perceive as the round sphere and the other is the ellipsoid. Since they have the same connections, their Riemann curvature tensor is the same. But, since they have a different metric, their Ricci tensor and Ricci scalar is different. $\endgroup$ Nov 1, 2017 at 18:23
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    $\begingroup$ In this case also the Ricci tensor will not depend on the metric since it is defined as a contraction of two indexes one covariant and the other contravariant so you don't need a metric ( $R_{i j}=R_{k i j}^k$). The Scalar curvature indeed depends on the metric. General Relativity uses (as far as I know) only L-C connections induced by the metrics so you don't have these kind of problems. $\endgroup$ Nov 1, 2017 at 18:35

3 Answers 3


I think the point that should be understood is that there are three different levels of structure here: a Riemannian structure (a choice of metric) is richer than a choice of affine connection, which in turn is stronger than a smooth structure (a choice of atlas). A smooth structure lets you define the notion of differentiable curves and tangent vectors. A connection just lets you define the notion of parallel transport (and thus geodesics). Enriching this to a Riemannian structure adds the notions of length, angle, volume, etc.

Since a metric determines a unique compatible Levi-Civita connection, if you have a Riemannian structure then you get a connection for free; but the converse is not true - given an affine connection, there may be many different Riemannian metrics having it as their Levi-Civita connection. (Of course, some connections aren't Levi-Civita for any metric at all, for example those with torsion.)

Whenever we're talking about parallel transport, geodesics, etc. in the context of Riemannian geometry, we're always using the Levi-Civita connection. The Ricci and scalar curvatures are defined as metric contractions of the curvature of the Levi-Civita connection. If you're forming the same contractions with the curvature of a different connection, then these are not what we normally call Ricci/scalar curvatures, and they have dubious geometric significance. The only sense in which the Riemannian curvature is "independent of the metric" is that it only depends on the metric via the connection; i.e. we can write in some sense $R(g) = R(\nabla(g))$. The dependence of $\nabla$ on $g$ should not be forgotten! To answer your first question directly, I would say that the relevant curvature information for a Riemannian manifold is the curvature tensor along with the metric, from which you can form not only the scalar and Ricci curvatures, but also more interesting things like the sectional curvature and curvature operator.

Your second question is kind of getting at something, though it needs a bit of tweaking. (A smooth structure doesn't provide a distinguished choice of connection!) What is true is that if you have two different metrics with the same Levi-Civita connection, then parallel transporting the same vector along the same loop will give you the same resulting vector. (If you choose to use some non Levi-Civita connection to do this instead, then of course this is still true - in this case you're not even using the Riemannian structure in any way.) In general, however, the two metrics will measure a different angle between the same pair of vectors.

  • $\begingroup$ Great answer. Have a couple questions though: To clarify things here, are you saying that the Riemann curvature is always defined with the Levi-Civita connection? I understand that this is what happens when one studies Riemannian manifolds, but, in general, aren't we able to define it with an arbitrary affine connection? If so, I interpret your answer as follows: the Riemann curvature tensor(and Ricci tensor and Ricci scalar) only have clear geometrical meaning when defined through the Levi-Civita connection. Is my understanding correc? $\endgroup$ Nov 2, 2017 at 14:59
  • $\begingroup$ Also, you said that if we define the Ricci scalar with a different connection, then it has a dubious meaning. What meaning is that then? $\endgroup$ Nov 2, 2017 at 15:00
  • $\begingroup$ on your last paragraph: Could you please clarify a bit the point where you said that parallel transporting the same vector along the same loop with the same (L-C or not L-C(if I understood correctly)) connection we will get the same resulting vector but different angle in each Riemannian manifold? How can we get the same resulting vector but different angle if the original vector is also the same? [maybe you want to continue this in a chat room?] $\endgroup$ Nov 2, 2017 at 15:09
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    $\begingroup$ The terms "Riemannian curvature/Riemann tensor" are usually used only to refer to the curvature tensor of the Levi-Civita connection of a metric, yes; but this is just a point of terminology - the construction of the curvature tensor from the connection works in for a general connection. The curvature tensor of an arbitrary connection has clear geometric meaning; what I would say does not is the Ricci or scalar curvatures formed by contracting a metric with an unrelated connection. $\endgroup$ Nov 2, 2017 at 23:20
  • $\begingroup$ Remember that a metric is an inner product, and we define angles via the identity $\cos \theta(u,v) |u| |v| = \langle u,v \rangle$; so changing the metric will often change the measured angles. $\endgroup$ Nov 2, 2017 at 23:21

I think the word "curvature" causes a bit of confusion to you. The curvature of a connection is something related a the connection you put on a manifold that in principle do not depend on the metric you have put on the manifold. If you want to rediscover the intuitive concept of curvature (Gaussian curvature* for example) used in the study of surfaces embedded in $\mathbb{R}^3$ (which is a metric concept), you start with a metric $g$ and take the curvature tensor $R$ of the induced Levi-Civita connection of $g$. In this case $R$ contains all the geometric information you expect. For example from $R$, you can construct the Sectional curvature $K$. $K(X,Y)$ will give you the Gaussian curvature of an embedded surface generated geodesically by the vectors $X,Y$. It is a very geometric interpretation. Also the Ricci tensor is a sum of sectional curvatures and the Scalar curvature is another sum of sectional curvatures. So the curvature tensor $R$ will give you really geometric informations coming from the metric but you have to consider the L-C connection.

Consider for example $(S^2,g_1)$ and $(S^2,g_2)$ Riemannian where $g_1$ is the metric given by the standard embedding in $\mathbb{R^3}$ with the standard flat metric and $g_2$ is the metric pulled back from an ellipsoid embedded in $\mathbb{R^3}$. You have two L-C connections, $C_1$ and $C_2$. If you compute the curvature tensor for $C_2$ (or another random connection not related to $g_1$) it will not give the intuitive geometric informations relevant for $(S^2,g_1)$ wich is (iso)metrically a sphere, even if you consider the scalar curvature.

2) No, because if you parallel transport a two vectors along a curve using a connection not even compatible with the metric (compatible means that $\nabla_X \langle Y,Z\rangle = \langle \nabla_X Y , Z \rangle + \langle Y , \nabla_X Z \rangle$ notice also that the L-C connection is by definition compatible with the metric), you can get different angles.
Instead if you use a connection compatible with the metric the parallel transport preserves the scalar product (given by the metric).

*I do not know if you are familiar with the Gaussian curvature of surfaces embedded in $\mathbb{R}^3$ but it is a fantastic scalar that for example if positive tells that the surface locally resembles a sphere, if negative instead tells that it resembles a saddle (like the saddle of a torus for example), the cylinder instead and the plane have Gaussian curvature 0. Probably it is really what fit best the layman term "curvature".
The Gaussian curvature is a metric concept (because it depends only on the first fundamental form) and is nothing more than the Sectional curvature of the Levi-Civita connection associated to the metric obtained restricting the standard flat metric of $\mathbb{R}^3$ to the surface. If you for example give another connection to the surface and compute the sectional curvature of it, it would be like computing the Gaussian curvature of another embedding (well, provided that the connection you are using is the L-C of some other metric).

Answers to the comments

1) I don't know any geometric interpretation in this case. The only thing I can think is that the curvature is linked (as you pointed out) to the holonomy i.e. what happens when you parallel transport a vector along a loop. In general if the curvature tensor is not trivial then the vector can be different.

2) The fact that the vector can be different can be stated for example in terms of angles onece you consider an auxiliary Riemannian metric. But this is just to state it in terms of angles. The point is that the vector after a loop is different.

2b) they will not result in the same change of angles, the two Riemannian metrics could measure different angles, if the transported vector is different they will measure it but the effective angle measured can be different. (Notice that this is not the case I was dealing with when talking about compatibility, in that case I was considering the parallel transport of two vectors and studying how the angle between them changes, in this case we are studying a single vector and after a loop we see how it change)

Maybe you should take a look at some books about Holonomy groups, unfortunately I know almost nothing about them.

  • $\begingroup$ 1) For your answer for my 1st question: If the two connections, $C_1$ and $C_2$ are not Levi-Civita, then what is the geometric interpretation(even if it is not the intuitive interpretation that we get with the L-C connection)? $\endgroup$ Nov 2, 2017 at 15:29
  • $\begingroup$ 2) On your answer to my 2nd question: I agree with what you are saying(since it is true by definition) but this part of my question had to do with the following: nearly everyone who explains the intuition behind curvature uses the parallel transport of a vector along a closed loop and seeing the change in the angle as if it is the defining property/result of non-zero curvature. But, the definition of the Riemann curvature tensor does not use a metric(except through the connection if it is a L-C connection). ** $\endgroup$ Nov 2, 2017 at 15:30
  • $\begingroup$ **Continuing: So, it seems like an ellipsoid and a sphere(each one with its own metric), when given the same (non-Levi-Civita connection) would result in the same change in angle for a vector parallel transported along the same closed loop. So, how do you address that? $\endgroup$ Nov 2, 2017 at 15:30
  • $\begingroup$ Thanks for the very informative answer. To conclude with my questions, [following your edits (2) and (2b)]: I don't understand why would the angle be different(again, comparing the same vector parallel transported along the same loop on two manifolds with the same connection and different metrics) since the curvature tensor is the same. Isn't the curvature tensor that tells us how much will the final vector change its angle with respect to the initial vector?And since it depends on the connection and not the metric, shouldn't the vector change its angle in the same way in both cases(manifols)? $\endgroup$ Nov 2, 2017 at 17:36
  • $\begingroup$ Consider $\mathbb{R}^2$, you put some esoteric connection on its tangent bundle, then consider a vector $V_0\in T_0\mathbb{R}^2$, you parallel transport along a loop and you get $V_1\in T_0\mathbb{R}^2$. Suppose that $V_1\neq V_0$ (so the connection choosen has curvature tensor not trivial). Now you can still define two different scalar product on $\mathbb{R}^2$ in such a way that $\langle V_0,V_1\rangle_1 \neq \langle V_0,V_1\rangle_2 $. $\endgroup$ Nov 2, 2017 at 17:54

This question is somewhat old and has already received answers - one thing I feel the need to point out is that while the curvature tensor (of any connection) characterizes the parallel transport only, it is fairly easy to see in a somewhat intuitive manner why the Levi-Civita (LC) connection also characterizes metric properties.

Consider a manifold $M$ with metric $g$, and a linear connection $\nabla$. Suppose that

  1. $\nabla$ is flat, i.e. $R(X,Y)Z=0$;
  2. $\nabla$ is metric compatible, i.e. $\nabla g=0$;
  3. $\nabla$ is torsionless, i.e. $T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]=0$.

Consider now an orthonormal cobasis $\theta^1_p,...,\theta^n_p\in T^\ast_p M$ at $p\in M$. Suppose that $\mathscr U\subseteq M$ is an open neighborhood of $p$ with coordinate functions $x^1,...,x^n$ and that in this coordinate system $\mathscr U$ is star-shaped with respect to $p$. Parallel transport the cobasis elements $\theta^a$ from $p$ to all points of $\mathscr U$ along radial lines (with respect to this coordinate system of course), thus we obtain a coframe $\theta^1,...,\theta^n$ defined in the entirety of $\mathscr U$.

  1. The consequence of flatness is that the coframe is parallel, i.e. $\nabla\theta^a=0$ ($a=1,...,n$).
  2. The consequence of metric compatibility is that the coframe is orthonormal at each point of $\mathscr U$, since the parallel transport induced by a metric compatible connection is a linear isometry of the tangent (thus cotangent) spaces.
  3. The consequence of torsionlessness is that the coframe $\theta^a$ is holonomic. To see this let $\xi\in\Omega^1(\mathscr U)$ be a parallel 1-form on $\mathscr U$. Then its covariant derivative in components is $$ 0=\xi_{i;j}=\xi_{i,j}-\Gamma^k_{ji}\xi_k, $$ and due to torsionlessness the connection coefficients $\Gamma^k_{ij}$ are symmetric in the lower indices, thus $$ \xi_{i,j}-\xi_{j,i}=0. $$ Then by Poincaré's lemma there is a function $\phi\in C^\infty(\mathscr U)$ (the domains is star-shaped so Poincaré's lemma holds) such that $\xi=d\phi$. It follows that for the coframe there are functions $y^a$ on $\mathscr U$ such that $$ \theta^a=dy^a. $$

Putting these together gives that the metric is $$ g=\eta_{ab}\theta^a\otimes\theta^b=\eta_{ab}dy^a\otimes dy^b, $$ where $\eta_{ab}=\mathrm{diag}(-1,...,-1,1,...,1)$, therefore we obtain that the metric is flat.

For this result to hold, the connection $\nabla$ has to satisfy all three conditions, flatness, metric compatibility and torsionlessness. But we know from Riemannian geometry that a metric compatible and torsionless connection must uniquely be the LC connection of the metric.

We thus see that for a connection to characterize the flatness of the metric it must uniquely be the LC connection.


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