Defining curvature in two ways

Question

Given a Riemannian manifold $(M,g)$ one often defines the Curvature of the Levi-Civita connection $\nabla$ as $$R(X,Y)=\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]}$$ for vector fields $X$ and $Y$. However, some authors define the curvature as $$R=\nabla \circ \nabla.$$ These two definitions should coincide, but I have some difficulties proving this. Does anyone here know a resource that goes over this?

Answer 1

Using the notation $\nabla$ for the second definition is something I absolutely abhor, because they’re really talking about the exterior covariant derivative.

Let $(E,\pi,M)$ be a vector bundle equipped with a linear connection $\nabla$. So, by definition $\nabla$ eats a section $\psi$ of $E$ and produces $\nabla_{(\cdot)}\psi$, which has an open slot where it can eat a tangent vector $h_x\in T_xM$ to output a vector $\nabla_{h_x}\psi\in E_x$; so there’s is a $1$-form aspect because it takes a tangent vector $h_x$ as input, and there’s a bundle-valued aspect. So, $\nabla$ is a certain map $\Omega^0(M;E)\to\Omega^1(M;E)$. We can generalize this gives us a bunch of operators $d_{\nabla}$ which map, for each $k\geq 0$, $\Omega^k(M;E)\to \Omega^{k+1}(M;E)$, i.e $E$-valued $k$-forms on $M$ to $E$-valued $(k+1)$-forms on $M$. The operators $d_{\nabla}$ are called the exterior covariant derivative (on $E$, relative to $\nabla$). The idea behind this is similar to how from the usual differential of a function, we get exterior derivatives of forms of all orders.

You don’t need to know the specifics of this yet, but all you need to know is that it obeys a similar type of product rule to the usual exterior derivative. As a result, something very cool happens at the level of second exterior covariant derivatives. Let $\psi\in\Omega^0(M;E)$ be any smooth section, and let $f\in C^{\infty}(M)$ any smooth function. Then, by the product rule \begin{align} d_{\nabla}(f\psi)&=df\wedge \psi+ (-1)^0f\cdot d_{\nabla}\psi= df\wedge \psi+f\cdot d_{\nabla}\psi \end{align} (actually since we’re at $k=0$, we can write this equality as $\nabla(f\psi)=df\cdot \psi+ f\cdot\nabla\psi$… or for the first term people often write $df\otimes \psi$ as well). But now let us differentiate again; this time we really need to use $d_{\nabla}$, we cannot use $\nabla$ anymore: \begin{align} d_{\nabla}^2(f\psi)&=d_{\nabla}(d_{\nabla}(f\psi))\\ &=d_{\nabla}(df\wedge \psi+ f\cdot d_{\nabla}\psi)\\ &=[d^2f\wedge \psi + (-1)^1df\wedge d_{\nabla}\psi]+ [df\wedge d_{\nabla}\psi+ (-1)^0f\cdot d_{\nabla}^2\psi]\\ &=f\cdot d_{\nabla}^2\psi, \end{align} where we have used that the first term involving $d^2f$ is $0$ because $d^2=0$ for usual forms. The second and third term cancel out, and so only the last term remains. Therefore, we have shown that the operator $d_{\nabla}^2:\Omega^0(M;E)\to\Omega^2(M;E)$ is $C^{\infty}(M)$-linear, i.e value of $d_{\nabla}^2\psi$ at a point $x$ depends only on the value of $\psi$ exactly at $x$, rather than a neighbourhood of $x$. Thus, by the “tensor characterization lemma” this is equivalent to a bundle morphism $R:\bigwedge^2(TM)\to \text{End}(E)$, or equivalently, to an $\text{End}(E)$-valued $2$-form on $M$. So, more explicitly, for any section $\psi\in \Omega^0(M;E)$ any $x\in M$, any $h_x,k_x\in T_xM$ we have that $(d_{\nabla}^2\psi)_x(h_x\wedge k_x)\in E_x$ is equal to the value of $R(h_x\wedge k_x)\in \text{End}(E_x)$ on $\psi(x)\in E_x$: \begin{align} (d_{\nabla}^2\psi)_x(h_x\wedge k_x)&=R(h_x\wedge k_x)[\psi(x)]\in E_x. \end{align} So, this is the more detailed explanation of the equation $R=d_{\nabla}^2$.

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Relating to the first formula.

Above, I simply said quickly that “we can generalize this to give a bunch of operators $d_{\nabla}$”, but I didn’t tell you how exactly. Well, one can define $d_{\nabla}$ to be the unique collection of operators which are linear, such that the correct product rule holds, and such that $d_{\nabla}\psi=\nabla\psi$ for all sections $\psi$ of $E$ (i.e for $E$-valued $0$-forms on $M$). This would be analogous to how almost everyone defines the usual exterior derivatives (it’s the unique operator such that it is linear, and $df$ is defined as usual for functions $f$, and such that $d^2=0$, and such that a suitable product rule holds).

Alternatively, one can equivalently define $d_{\nabla}$ by an explicit formula analogous to the Palais formula for usual differential forms; we simply replace the Lie derivative in the first summation by the covariant derivative $\nabla$. So, super explicitly, for each $\omega\in \Omega^k(M;E)$, we define $d_{\nabla}\omega$ by declaring it eats $(k+1)$ many vector fields $X_0,\dots, X_k$ on $M$, such that its value on these vector fields is \begin{align} (d_{\nabla}\omega)(X_0,\cdots, X_k)&:=\sum_{j=0}^p(-1)^j\nabla_{X_j}\bigg(\omega(X_0\wedge\cdots\wedge \widehat{X_j}\wedge\cdots\wedge X_k)\bigg)\\ &+\sum_{0\leq i<j\leq k}(-1)^{i+j}\omega([X_i,X_j]\wedge X_0\wedge\cdots\wedge\widehat{X_i}\wedge\cdots\wedge\widehat{X_j}\wedge\cdots\wedge X_k).\tag{$*$} \end{align} One has to prove a bunch of things about this formula, the first of which is that the RHS is alternating in the $X_j$’s, and that it is $C^{\infty}(M)$-linear with respect to each of the $X_j$’s. Once this check is made, we can be sure that the object $d_{\nabla}\omega$ really does lie in $\Omega^{k+1}(M;E)$, and so we have a well-defined map $d_{\nabla}:\Omega^k(M;E)\to \Omega^{k+1}(M;E)$.

To show the equivalence of these two approaches, it suffices to show that this explicit formula is also linear in $\omega$, and that it satisfies a correct product rule (and obviously when $k=0$, the RHS only has one term, namely $\nabla_{X_0}\omega$, which means $d_{\nabla}=\nabla$ for $k=0$). Checking these details is a little annoying to do, and I’m not going to do it here, but I’m sure they’re done in detail for the case of the usual exterior derivative $d$ in Spivak’s A Comprehensive Introduction to Differential Geometry, Vol I. For this vector-bundle case, you mimic the same algebraic steps, by appropriately replacing all instances of the Lie derivative $L_{X_j}\left(\omega(X_0\wedge\cdots\wedge \widehat{X_j}\wedge\cdots\wedge X_k)\right)\equiv X_j\left( \omega(X_0\wedge\cdots\wedge \widehat{X_j}\wedge\cdots\wedge X_k)\right)$ (i.e the vector field $X_j$ acting on a function) with the covariant derivative $\nabla_{X_j}$ acting on a section.

So, with the equivalence of these approaches in mind, we can proceed easily:

• for $k=0$, the formula is just saying that if $\psi\in \Omega^0(M;E)$, i.e a section of $E$, and $X$ is a vector field on $M$, then $(d_{\nabla}\psi)(X):=\nabla_X\psi$.
• for $k=1$, the formula is saying that for any $\omega\in \Omega^1(M;E)$ and vector fields $X,Y$ on $M$, we have $(d_{\nabla}\omega)(X\wedge Y):=\nabla_X\left(\omega(Y)\right)-\nabla_Y\left(\omega(X)\right)-\omega([X,Y])$.

So, if we want to calculate $d_{\nabla}^2\psi$ for $\psi\in \Omega^0(M;E)$, then applying these two formulas immediately gives us \begin{align} (d_{\nabla}^2\psi)(X\wedge Y)&=\nabla_X((d_{\nabla}\psi)(Y))-\nabla_Y((d_{\nabla}\psi)(X))-(d_{\nabla}\psi)([X,Y])\\ &=\nabla_X\nabla_Y\psi-\nabla_Y\nabla_X\psi-\nabla_{[X,Y]}\psi. \end{align} Since the LHS is by our second definition equal to $R(X\wedge Y)\cdot \psi$, we have thus proven the equivalence of the first and second definitions.

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Extra Remarks.

Now, specialize to a connection $\nabla$ in the tangent bundle $TM$. Here, we have a very special map, namely the identity map $I:TM\to TM$, which can be viewed as a $TM$-valued $1$-form on $M$. As such, we can compute its exterior covariant derivative $d_{\nabla}I$. This will be a $TM$-valued $2$-form on $M$. Its value on a pair of vector fields $X,Y$ on $M$ is (use the second bullet point formula for $k=1$ from above): \begin{align} (d_{\nabla}I)(X\wedge Y)&=\nabla_X\bigg(I(Y)\bigg)-\nabla_Y\bigg(I(X)\bigg)-I([X,Y])\\ &=\nabla_XY-\nabla_YX-[X,Y]. \end{align} You may recognize this as precisely being the value of the torsion $T(X\wedge Y)$. So, this proves that the torsion of a connection in the tangent bundle is the exterior-covariant derivative of the identity morphism: $T=d_{\nabla}I$, and it is thus a $TM$-valued $2$-form on $M$ (in components, $T^{a}_{\,bc}$, this is the reason for the skew-symmetry in the bottom two indices). Or of course, we can reverse perspective, and define the torsion to be $T:=d_{\nabla}I$ (which is actually what I prefer), and then deduce the above formula for its values on a pair of vector fields.

Here are some related MSE answers of mine you may find helpful if you’re going down the rabbit hole of bundle-valued forms

• What kind of object is the (second) exterior derivative of a moving point $R$ in $\mathbb R^n$? What does $dR\wedge\omega \mathbf e$ mean?
• Definition of flatness for a connection on a vector bundle
• Proving $d^\nabla( d^\nabla \omega) = F^\nabla \wedge \omega$.

The first link contains a summary of all the essential things one needs to know about bundle-valued forms (the proofs are almost line-by-line adaptations of those in the usual scalar-valued case).