Let $M$ be a Riemannian manifold, $E \to M$ a vector bundle and $\nabla : \Gamma(E) \to \Gamma(T^*M \otimes E)$ a connection on $E$.
I'm a bit confused on the definition of flatness. The word is used a bit vaguely here and there.
Lee calls a Riemannian manifold flat if and only if the curvature tensor $R $ vanishes identically.
Some authors call the connection flat if $R$ vanishes identically.
Is this just a poor choice of words or am I misunderstanding something here? To be clear, I'm trying to understand the definition of a flat connection, but I can't find a good resource anywhere. Someone also defined flat connection by $\nabla^2 = 0$, but this is different than $R = 0$ right? If $F$ is a $(k,l)$-tensor field, then $\nabla^2F$ is a $(k,l+2)$-tensor field and I don't see how $\nabla^2_{X,Y}F= \nabla_X(\nabla_YF) – \nabla_{\nabla_XY}F$ is related to the curvature.
Any clarification for this would be appreciated.
Best Answer
Curvature is a property of a connection. When talking about connections on tangent bundles, we abuse language slightly and speak of curvature of the underlying manifold. See here for some more remarks about why we omit the distinction in language about objects ‘living on $M$’ vs ‘living on $TM$’.
Next, given any linear connection $\nabla$ in a vector bundle $(E,\pi,M)$, one can construct a whole bunch of operators $d_{\nabla}$, called the exterior covariant derivative (relative to $\nabla$) which take $E$-valued $k$-forms on $M$ to $E$-valued $(k+1)$-forms on $M$, i.e $d_{\nabla}:\Omega^k(M;E)\to \Omega^{k+1}(M;E)$. This is 95% analogous to how given our usual definition of $df$ for smooth functions, we get a unique extension (after imposing some conditions) of $d$ to higher order forms.
Now, with this technology, for any section $\psi$ of $E$, we can show that $d_{\nabla}^2\psi$ is not the zero section, but rather $R\cdot_{\text{ev}}\psi$, where the $\cdot_{\text{ev}}$ denotes the evaluation pairing $\text{End}(E)\oplus E\to E$. Or more explicitly, for each $x\in M$, and each $h_x,k_x\in T_xM$, we have that \begin{align} (d_{\nabla}^2\psi)_x(h_x,k_x)&=R_x(h_x,k_x)[\psi(x)]. \end{align} Or if you don’t like to go to the pointwise level, we can stay at the field level and say that for all vector fields $X,Y$ on $M$, \begin{align} (d_{\nabla}^2\psi)(X,Y)&=R(X,Y)[\psi]=\nabla_X\nabla_Y\psi-\nabla_Y\nabla_X\psi-\nabla_{[X,Y]}\psi. \end{align} Therefore, vanishing of $R$ is equivalent to $d_{\nabla}^2=0$ (i.e every section $\psi$ of $E$ gets sent to the zero $E$-valued $2$-form on $M$).
You should note that on a general vector bundle $(E,\pi,M)$ with a linear connection $\nabla$, it doesn’t even make sense to speak of iterated covariant derivatives $\nabla\nabla$. Why? Because if $\psi$ is a section of $E$, then $\nabla\psi$ is an $E$-valued $1$-form on $M$, or equivalently, a section of $T^*M\otimes E$. So, if I want to take a covariant derivative of $\nabla\psi$, then I would need a connection $\nabla^{T^*M\otimes E}$ on the larger bundle $T^*M\otimes E$… and in general just having a connection on $E$ does not give me a connection on $T^*M\otimes E$. You would have to either arbitrarily fix a connection on $TM$, then consider the associated connection on $T^*M\otimes E$. Or, if your vector bundle was $E=TM$ to begin with, then of course $T^*M\otimes E=T^*M\otimes TM$ is again a tensor bundle of $M$, so we have a naturally induced connection.
In any case, the iterated $\nabla\nabla\psi$ is not necessarily the same as $d_{\nabla}^2\psi$ (which is always defined). Indeed, let’s just stick to the tangent bundle $TM$. Then for any torsion-free and flat connection $\nabla$ (e.g think of $\Bbb{R}^n$ with the usual flat connection) and any smooth function $f$, we have that $\nabla\nabla f=\nabla(df)$ is called the Hessian of $f$, and it is a symmetric $(0,2)$ tensor field on $M$, and not necessarily $0$. On the other hand, $d_{\nabla}^2f=d^2f=0$.
Having said this, there are some authors who use the notation $\nabla^2$ to mean not the iterated $\nabla\nabla$, but rather $d_{\nabla}^2$. Notationally, I find this confusing, but it is what it is.