# Why do we call the Riemann curvature tensor the curvature of spacetime rather than the curvature tensor of its tangent bundle?

I was studying the mathematical description of gauge theories (in terms of bundle, connection, curvature,...) and something bothers me in the terminology when I compare it with general relativity.

In gauge theory, take a vector bundle $$E \rightarrow M$$ over a Riemannian Manifold $$(M,g)$$ and suppose you have a connection $$\nabla: \Gamma(E) \longrightarrow \Gamma(E) \otimes \Gamma(T^*M)$$. Then, you defined the curvature tensor $$F_\nabla$$ associated to your connection $$\nabla$$ as an $$End(E)$$-valued 2 form i.e $$F_\nabla \in \Omega^2(M,End(E))$$ $$\nabla \circ \nabla := F_\nabla$$ where you extend $$\nabla$$ on the $$p$$-form.

Equivalently you can defined the curvature tensor as a map $$F_\nabla : TM \times TM \times E \rightarrow E$$ such that $$F_\nabla(X,Y)s=\nabla_X\nabla_Y s - \nabla_Y\nabla_X s - \nabla_{[X,Y]}s \qquad X,Y \in \Gamma(TM), s \in \Gamma(E)$$ Then, $$F_\nabla$$ give you information on the curvature of your bundle i.e if you take a vector $$v \in E$$ and you do the parallel transport of $$v$$ on a closed path with respect to the connection $$\nabla$$, the difference between $$v$$ and its parallel transport $$v'$$ will be given by $$F_\nabla$$.

Then, you remark that if as vector bundle you take $$E=TM$$ and as connection you take the Levi-Civita connection $$\nabla=\nabla^{LC}$$ you see that your curvature tensor associated to $$\nabla$$ is exactly the Riemann curvature tensor i.e $$F_\nabla = R$$

Now my question:

In the literature, the Riemann curvature tensor is often called the curvature of space-time. But in view of the definition of the curvature tensor on a vector bundle, why is $$R$$ not rather the curvature tensor of the tangent bundle (associated with the Levi-civita connection)?

(My answer would be: I suppose that in fact the Riemann tensor corresponds to the curvature tensor of tangent bundle (associated with the Levi-Civita connection) but it also gives plenty (if not all) of information on the curvature of space-time (since tangent space is very linked to the manifold) and therefore it is called curvature of space-time (and therefore it would be a problem of terminology). I'm not sure of this answer that's why I wanted an outside opinion.)

• We also have the Euler characteristic of a vector bundle, and the Euler characteristic of the tangent bundle is just called the characteristic of the manifold. Commented Jan 20 at 9:31

My answer would be: I suppose that in fact the Riemann tensor corresponds to the curvature tensor of tangent bundle (associated with the Levi-Civita connection) but it also gives plenty (if not all) of information on the curvature of space-time (since tangent space is very linked to the manifold) and therefore it is called curvature of space-time (and therefore it would be a problem of terminology). I'm not sure of this answer that's why I wanted an outside opinion.

The answer is that when we talk about the intrinsic curvature of a manifold, what we're talking about is the curvature of its tangent bundle. That's what we mean when we talk about spacetime being curved.

In other words, intrinsic curvature is characterized by how tangent spaces to different points on the manifold are connected via parallel transport. If the mapping from the tangent space at $$p$$ to the tangent space at $$q$$ is path-independent, then the space is said to be (intrinsically) flat. Otherwise, it possesses non-zero intrinsic curvature which is quantified by the Riemann tensor.

• I have another question then : The curvature tensor of a vector bundle is called curvature because it encodes a sort of curvature (intuitively) of the bundle or is called curvature because it's defined by analogy with the Riemannian curvature tensor?
– eomp
Commented Jan 19 at 16:50
• @eomp I would imagine the latter. Given smooth manifold (say, 2D curved surface embedded in a 3D Euclidean space), the curvature form associated to its tangent bundle is so named because it encodes the curvature (in the colloquial sense) of the surface. But the same object can be associated to generic vector bundles, not just tangent bundles, and the name curvature form is inherited from the study of curved surfaces and spaces. But this is more speculation on my part than actual knowledge - you'd have to ask an expert in the history of differential geometry. Commented Jan 20 at 5:55

• The Riemann curvature $$R$$ “of $$M$$” is referring to the curvature of a linear connection $$\nabla$$ in the tangent bundle $$TM$$ (which is a vector bundle).
• Referring to $$R$$ as “curvature of $$M$$” is technically an abuse of language.

Here are some more remarks, and seeing as your question is terminology-related, let’s be ‘pedantic’ with the language.

• A connection is really a piece of data on a fiber bundle. We have principal connections in principal bundles, we usually speak of linear connections in vector bundles, etc. A connection is NOT a feature ascribed to a single manifold $$M$$; what we really mean is that we’re referring to a linear connection on the tangent vector bundle $$TM$$. And of course in more specific situations, we mean more specific things (on a semi-RIemannian manifold, we always use the Levi-Civita connection unless something is explicitly mentioned to the contrary).
• Likewise, curvature is now understood to mean the curvature of a connection on a certain bundle. So, again curvature is NOT a quantitative property that you can assign to a manifold, so strictly speaking, phrases like “curvature of $$M$$” or “curvature of spacetime” are not correct.

The reason why we refer to things as “living on $$M$$” rather than as objects on $$TM$$ is simply for linguistic simplicity: imagine all you do is study general relativity or some concrete blackhole spacetimes. Then you know that you’re working with a single metric $$g$$, and you’re only working with the Levi-Civita connection on the tangent bundle, and its curvature. That’s the only thing you’ll ever refer to, so you would rather not beat an already dead horse.

Also, as you said, the tangent bundle $$TM$$ is intrinsically associated to $$M$$, so any property defined for $$TM$$ it would be reasonable for you to say is a property of $$M$$ as well. Another reason is that back when differential geometry was still being developed, we really had only the idea of the tangent spaces (and normal spaces) lying around, and we didn’t have more fancy examples of vector or principal bundles formally laid out, so we didn’t need to distinguish the levels of structure (such as a smooth manifold vs semi-Riemannian manifold, or tangent bundle vs vector bundle vs principal bundles etc).

The curvature tensor of a vector bundle is called curvature because it encodes a sort of curvature (intuitively) of the bundle or is called curvature because it's defined by analogy with the Riemannian curvature tensor?

is that it is both. First, the definition of $$F_{\nabla}$$ (also denoted as $$R_{\nabla}$$ by many, including me, depending on context) is almost identical to the one for the Riemann curvature tensor. So by that symbolic similarity alone, and the reverence to the greatness of Riemann as a mathematician, we still call $$F_{\nabla}$$ the Riemann curvature tensor (though sometimes people might reserve the name Riemann for when $$E=TM$$ because that’s the subject that Riemann developed). As for why it’s called curvature, well we’re defining a new object by almost the exact same formula so it makes sense to refer to it by the same name.

Some slightly better reasons for still calling it “curvature”:

• it is the obstruction to the existence of local parallel frames (see this answer of mine). So, $$F_{\nabla}$$ determines whether or not we have local basis of sections $$\{e_1,\dots, e_k\}$$ of $$E$$ over $$U\subset M$$ such that $$\nabla(e_i)=0$$ identically on $$U$$. Why do we care about this? Because first of all a local frame of sections determines a vector bundle trivialization $$\pi^{-1}(U)\cong U\times \Bbb{R}^k$$. Furthermore having $$\nabla(e_i)=0$$ means that you managed to locally trivialize the vector bundle in such a manner that you have ‘unchanging directions’.
• as you already said: it determines whether or not the parallel-transport isomorphism along a loop (which is homotopic to a point) is the identity map (i.e whether or not parallel transport ‘twists/jumbles up’ the fibers).
• Great answer, thank you very much!
– eomp
Commented Jan 20 at 9:41

Given a spacetime $$(M,g)$$, let's say that spacetime is flat if and only if each point $$p\in M$$ has an open neighborhood $$U\subseteq M$$ on which there are coordinates $$(x^1,\dots,x^m)$$ such that in this coordinate system we have$$g|U=\varepsilon_1dx^1\otimes dx^1+\dots+\varepsilon_mdx^m\otimes dx^m,$$ where $$\varepsilon_i=\pm1$$. Then it is a standard theorem that $$(M,g)$$ is flat if and only if the Riemann curvature tensor vanishes, $$R=0$$. The curvature operator of some connection on some vector bundle has absolutely no bearing on the flatness of $$(M,g)$$.

EDIT: Regarding the comment. It is instructive to look at one possible version of the proof that $$R=0\Longrightarrow (M,g)$$ is flat.

Start off with an orthonormal coframe $$w^1,\dots,w^m\in T^\ast_p M$$ at some point in $$M$$. Let $$U$$ be a neighborhood of $$p$$ which is star-shaped with respect to some chart. Let us extended the coframe elements $$w^1,\dots,w^m$$ at $$p$$ into some coframe $$(\theta^1,\dots,\theta^m)$$ in $$U$$ through parallel transport with respect to some linear connection $$\nabla$$ on the cotangent bundle $$T^\ast M$$ along radial curves emanating from $$p$$. Then

1. the resulting fields $$\theta^i$$ are parallel in $$U$$ (i.e. $$\nabla\theta^i=0$$) if and only if the curvature of the connection vanishes, i.e. $$R=0$$;
2. the resulting fields $$\theta^i$$ are exact (i.e. $$\theta^i=du^i$$ for some functions $$u^i$$ on $$U$$) if and only if torsion vanishes, i.e. $$T=0$$ (assuming the preceding condition is also met);
3. the resulting fields $$\theta^i$$ will form an orthonormal coframe if and only if the connection is metric-compatible.

These conditions imply that in the region $$U$$ we have $$g=\sum_{i=1}^m\varepsilon_i du^i\otimes du^i$$, i.e. spacetime is flat.

So concluding:

1. In order to construct Cartesian coordinate systems, we need to carry covectors around, i.e. we need a connection on $$T^\ast M$$ (but those are equivalent to connections on $$TM$$), which explains why the (co)tangent bundle is privileged over other vector bundles (another point is that (co)tangent bundles are natural vector bundles, so they are functorially associated to a manifold; a "foreign" vector bundle requires additional info not contained in $$(M,g)$$).
2. A preliminary condition for the contruction of a Cartesian coordinate system is the flatness of the connection, but to ensure that the resulting coframing is both exact and orthonormal requires torsionlessness and metric-compatibility, which uniquely fixes the connection to be the Levi-Civita connection of $$g$$.
3. For this reason, only one specific connection, viz. the Levi-Civita connection on the tangent bundle carries full information about the curvature of $$(M,g)$$ itself.
• Thank you for your reply. Yes, it is why I said at the end " R [...] also gives plenty (if not all) of information on the curvature of space-time" and it is why I think we called it curvature of space-times. My problem/question is about the fact that R is defined as a curvature tensor on a vector bundle so it should give information also on the curvature of tangent bundle...but no one ever talks about that !
– eomp
Commented Jan 19 at 13:08

Kind of for the same reason we might say "the antena on my car" rather than "the antena on the roof of my car" ... Referring to the tangent bundle is just extra level of specifivity that is not needed ... the tensor always lives on [the tangent bundle of] the manifold, so we don't need to mention it ... just as the antena always lives on [the roof of] the car ... so we don't need to mention the roof, usually.

• FWIW, I've owned several cars (including one currently!) where the antenna "lives" on the hood. Commented Jan 20 at 21:47