# Importance of studying the curvature of a covariant derivative on a vector bundle

Let $$\mathcal{E}$$ be a vector bundle over $$M$$. Let $$\nabla : \Gamma (M,\mathcal{E}) \rightarrow \Gamma(M,T^*M \otimes \mathcal{E})$$ be a covariant derivative on M. I read that the choice of a covariant derivative $$\nabla$$ on $$\mathcal{E}$$ allows us to split the tangent bundle $$T \mathcal{E}$$ into the vertical bundle and a horizontal bundle.

Denote $$\mathcal{R}$$ the curvature of $$\nabla$$, which is defined by $$\mathcal{R}(X,Y) =[\nabla_X, \nabla_Y]- \nabla_{[X,Y]},$$ Where $$X$$ and $$Y$$ are two vector fields.

I'm wondering why it is useful to consider the curvature of a covariant derivative on a vector bundle?

Developing @Wyatt's second point ("The curvature is exactly how much the connection fails to be flat.") a bit further:

Curvature provides a useful tool to topologically classify vector bundles in terms of Chern-Weil theory.

Think of vector bundles over compact manifolds, for simplicity a line bundle over the torus, the latter thought as a square obtained as quotient of $$\mathbb{R}^2$$ by a lattice. For the vector bundle to be smooth, the fibers need to be periodic in a suitable sense and this claim restricts the freedom in choosing how they "twist over the torus". A good notion of a quantitative measure of "how they twist" is the curvature.

In fact, integrating the curvature over the torus gives you a topological invariant, that is, an integer number (up to some constant) which only depends on the topology of the bundle and is the same for whichever covariant derivative you have chosen. Moreover, if two vector bundles can be smoothly deformed into one another, then these numbers must be same, or conversely, if they are not the same, then the bundles cannot be smoothly deformed into one another. (Google for Chern numbers if you are interested). So "how much curvature there must be in total" is a topological property of the bundle, where by choosing different covariant derivatives, you can "distribute the curvature" in different ways over the manifold.

Going on from there, you can study Chern-Weil theory which provides a large class of such topological invariants by (up to some details) integrating suitable polynomials of the curvature (which become non-trivial if the base manifold has a higher dimension than 2). You can of course pretend to forget the notion of curvature and express these polynomials of the curvature as polynomials of the covariant derivative, but this makes it much harder to characterize these polynomials and would not be the usual way of mathematics in general to structurize theories as good as possible.

• "For the vector bundle to be smooth, the fibers need to be periodic in a suitable sense and this claim restricts the freedom in choosing how they "twist over the torus". @nicrot000 could you please elaborate further in this statement, I didn't understand it ? For example what do you mean by " periodic" in this context and by "twist over the torus" ? Dec 3, 2022 at 15:57
• Okay, maybe that was a bit too handwavey... Take a line bundle $\pi:E\to\mathbb{T}^2$ over the 2-torus and define the quotient projection $$p:\mathbb{R}^2\to\mathbb{R}^2/\mathbb{Z}^2=\mathbb{T}^2.$$ Then $p^*E$ is a line bundle over $\mathbb{R}^2$ for which you define the fiber over a point $x\in\mathbb{R}^2$ to be the space $E_{p(x)}$. Since $p$ is $\mathbb{Z}^2$-periodic, so is $p^*E$, in the sense that for $x\in\mathbb{R}^2$ and $\gamma\in\mathbb{Z}^2$ you have $(p^*E)_{x+\gamma}=(p^*E)_x$. Dec 4, 2022 at 17:29
• Via $p$ you can think of the torus as the unit square in $\mathbb{R}^2$ with the boundary pieces identified appropriately. If you want to define something on $\mathbb{T}^2$ by defining it on the unit square it has to be periodic in the sense of "it has to coincide on the left and right and on the lower and upper boundary piece of the unit square". Dec 4, 2022 at 17:36
1. One very important example is general relativity, where the curvature is related to the stress-energy tensor on spacetime. Look up "Einstein Equation". There are other examples in gauge theory where the curvature corresponds to the field strength.

2. The curvature is exactly how much the connection fails to be flat. Flat connections also give integrable systems of differential equations. Curved connections do not, so we can view curvature as an obstruction to integrability. Look up "Maurer-Cartan Equation".