# Computing the curvature of a connection (a specific example)

I'm learning about the curvature of a vector bundle, and trying to compute it for a simple example. Let $\nabla$ be the connection on a trivial line bundle over $\mathbb{R}^2$ given by $\nabla s = ds$.

I know that curvature is given by $F(X,Y)(s) = \nabla_X\nabla_Ys - \nabla_{Y} \nabla_Xs - \nabla_{[X,Y]}s$.

I found that $\nabla_X\nabla_Ys = \nabla_X(ds(Y)) = \nabla(ds(Y))(X) = d(ds(Y))(X) = 0$, and similarly $\nabla_{Y} \nabla_Xs =0$.

How do I compute $\nabla_{[X,Y]}s$? I've written it as $(ds)[X,Y] = ds(XY)-ds(YX) = (XY)(s)-(YX)(s)$ but I'm not sure where to go from here.

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You might be mistaken about the meaning of $ds(X)$.

The space of vector fields acts on smooth functions by derivation. So in local coordinates, if we write $X=\sum_{i=1}^nX_i \partial_i$ where $X_i\in \mathcal{C}^\infty$, then we get $$(X\cdot s)(p)=\sum_{i=1}^nX_i(p)\dfrac{\partial s}{\partial x_i}(p)=ds_p(X(p)).$$

Let's see what happens on your bundle $\mathbb R^2\times \mathbb R\rightarrow \mathbb R^2$ and your connection $\nabla$. I will use the global coordinates denoted by $(x,y)$ on $\mathbb R^2$ and the corresponding basis of the tangent space $(\partial_x,\partial_y)$.

For every section $s$ of your bundle and for every $X,Y\in \mathcal{C}^\infty(\mathbb R^2,\mathbb R)$, $\nabla_{X\partial_x+Y\partial_y}s$ is just the map given by $$p\in \mathbb R^2\mapsto X(p)\dfrac{\partial s}{\partial x}(p)+Y(p)\dfrac{\partial s}{\partial y}(p).$$

Furthermore, since the curvature is a skew-symmetric bilinear map, to compute $F$ it suffices to compute $F(X\partial_x,Y\partial_y)$. Other terms can be deduced by linearity.

It is then an easy computation to see that $$\nabla_{X\partial_x}\nabla_{Y\partial_y}s-\nabla_{Y\partial_y}\nabla_{X\partial_x}s = X\dfrac{\partial Y}{\partial x}\dfrac{\partial s}{\partial y}-Y\dfrac{\partial X}{\partial y}\dfrac{\partial s}{\partial x}$$ which is in fact also equal to $\nabla_{[X\partial_x,Y\partial_y]}$.

It follows that the curvature of $\nabla$ vanishes.

In fact, this result tells you that the Lie bracket $[X,Y]$ of two vector fields is exactly the vector field associated to the derivation $\mathcal{L}_X\mathcal{L}_Y-\mathcal{L}_Y\mathcal{L}_X$.

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What do you mean by "vector field associated to the derivation $\mathcal{L}_X\mathcal{L}_Y-\mathcal{L}_Y\mathcal{L}_X$? Isn't simply $\mathcal{L}_XY=[X,Y]$ true? – gofvonx Aug 20 '13 at 14:21
Yes exactly. I just wanted to see $\mathcal{L}_X$ as an operator acting on smooth functions instead of acting on tensor fields for the sake of simplicity. – Bebop Aug 21 '13 at 18:47