# Connections, The Kähler Condition, and Curvature – IV. Curvature

Unlike the case of partial differentiation for functions, in general, for sections of a vector bundle, partial differentiations for different directions do not commute. The failure of the commutativity is measured by the curvature of the connection. The commutativity of partial differentiation of a function in two different directions is equivalent of the composite of two exterior differentiations applied to functions. We do this for a real manifold ${M}$ of real dimension ${m}$ and a smooth vector bundle ${V}$ of rank ${r}$ over ${M}$ which can be a complex vector bundle or a real one. Take a local smooth basis ${e_\alpha}$ ${(1 \le \alpha \le r)}$ of ${V}$. We apply ${D}$ twice to ${e_\wedge}$ and get a local section ${DDe_\alpha}$ of ${V \otimes T_M^* \otimes T_M^*}$, where ${T_M^*}$ is the dual of the tangent bundle ${T_M}$ of ${M}$. By skew-symmmetrizing ${DD e_\alpha}$ in the two arguments for the tangent vectors of ${M}$, we get a local ${V}$-valued ${2}$-form ${D_\wedge D e_\alpha}$. We can express ${D_\wedge D e_\alpha}$ in terms of the connection ${\omega}$ and its exteiror derivative ${d\omega}$ as follows. We use the column vector ${e}$ with components ${e_\alpha}$.

$\displaystyle D_\wedge D e = D_\wedge(\omega e) = d\omega\,e - \omega \wedge De$

$\displaystyle = d\omega\,e - \omega \wedge \omega\,e = (d\omega - \omega \wedge \omega)\,e.$

We define the curvature to be ${d\omega - \omega \wedge \omega}$ and denote it by ${\Omega}$. It is an ${\text{End}(V)}$-valued ${2}$-form on ${M}$, because if ${f}$ is a local smooth matrix-valued function, then

$\displaystyle D_\wedge D(fe) = D_\wedge(df\,e + f\,De)$

$\displaystyle = ddf\,e - df \wedge De + df \wedge De + fD \wedge De = f\, D\wedge De.$

Let us look at the curvature tensor ${\Omega}$ is local coordinates. With respect to a local frame ${e_\alpha}$ ${(1 \le \alpha \le r)}$, we have

$\displaystyle \Omega_\alpha^\beta = \Omega_{\alpha i j}^\beta dx^i \wedge dx^j.$

Using the Christoffel symbol ${\Gamma_{\alpha i}^\beta}$ of the connection ${\omega_\alpha^\beta}$, we have

$\displaystyle \Omega_{\alpha i j}^\beta dx^i \wedge dx^j = d(\Gamma_{\alpha i}^\beta dx^i) - (\Gamma_{\alpha i}^\gamma dx^i) \wedge (\Gamma_{\gamma j}^\beta dx^j)$

and

$\displaystyle 2\Omega_{\alpha ij}^\beta = \partial_i \Gamma_{\alpha j}^\beta - \partial_j \Gamma_{\alpha i}^\beta + \Gamma_{\alpha i}^\gamma \Gamma_{\gamma j}^\beta - \Gamma_{\alpha j}^\gamma \Gamma_{\gamma i}^\beta.$

The curvature tensor ${\Omega}$ satisfies a Bianchi identity. The Bianchi identity is

$\displaystyle D_\wedge \Omega = d\Omega + \Omega \wedge \omega - \omega \wedge \Omega$

$\displaystyle =d(d\omega - \omega \wedge \omega) + (d\omega - \omega \wedge \omega) \wedge \omega - \omega \wedge (d\omega - \omega \wedge \omega)$

$\displaystyle -d\omega \wedge \omega + \omega \wedge d\omega + d\omega \wedge \omega - \omega \wedge \omega \wedge \omega - \omega \wedge d\omega + \omega \wedge \omega \wedge \omega = 0.$

In local coordinates, this says that

$\displaystyle \nabla_i \Omega_{\alpha jk}^\beta + \nabla_j \Omega_{\alpha ki}^\beta + \nabla_k \Omega_{\alpha i j}^\beta = 0.$

We would like to remark that two partial covariant differentiations in general do not commute, and the noncommutativity gives rise to the curvature. However, the skew-symmetrization of the results of three partial covariant differentiations becomes zero, which is the Bianchi identity. We can also define the curvature in invariant formulation. The curvature is defined to measure the failure of the commutativity of partial covariant differentiation. So for tangent vector fields ${X}$, ${Y}$ and a smooth local section ${s}$ of ${V}$, we should consider ${\nabla_X \nabla_Y s - \nabla_Y \nabla_X s}$. However, when ${X}$ and ${Y}$ are arbitrary tangent vector fields instead of tangent vector fields defined by coordinate functions, we do not expect to have commutativity of differentiations along ${X}$ and ${Y}$ even in the case of functions, because for a smooth function ${f}$, in general, ${X(Yf) - Y(Xf)}$ is nonzero and is equal to the derivative ${[X, Y]f}$ of ${f}$ along the direction of the Lie bracket ${[X, Y]}$ of ${X}$ and ${Y}$. So we have to make accommodation for that and consider

$\displaystyle R(X, Y)s = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s.$

For smooth functions ${\varphi}$, ${\psi}$, ${\theta}$, we have

$\displaystyle R(\varphi X, \psi Y,)(\theta s) = \varphi \psi \theta\, R(X, Y) s,$

because of the following computations.

$\displaystyle \nabla_{\psi Y}(\theta s) = \psi(Y\theta)s + \psi\theta\nabla_Ys.$

$\displaystyle \nabla_{\varphi X}(\nabla_{\psi Y}(\theta s)) = \varphi\nabla_X(\psi(Y\theta)s + \psi\theta\nabla_Ys)$

$\displaystyle = \varphi(X\psi)(Y\theta)s + \varphi\psi(XY\theta)s + \varphi\psi(Y\theta)\nabla_Xs$

$\displaystyle +\varphi(X\psi)\theta\nabla_Ys + \varphi\psi(X\theta)\nabla_Ys + \varphi\psi\theta\nabla_X\nabla_Ys.$

$\displaystyle [\varphi X, \psi Y] = \varphi(X\psi)Y - \psi(Y\theta)X + \varphi\psi[X, Y].$

$\displaystyle \nabla_{[\varphi X, \psi Y]}(\theta s) = \varphi(X \psi)(Y\theta)s - \psi(Y \varphi)(X \theta)s + \varphi\psi([X, Y]\theta)s$

$\displaystyle +\varphi(X\psi)\theta\nabla_Ys - \psi(Y\varphi)\theta\nabla_X s + \varphi\psi\theta\nabla_{[X, Y]}s.$

$\displaystyle \nabla_{\varphi X}(\nabla_{\psi Y}(\theta s)) - \nabla_{\psi Y}(\nabla_{\varphi X}(\theta s)) - \nabla_{[\varphi X, \psi Y]}(\theta s)$

$\displaystyle =\varphi\psi\theta(\nabla_X\nabla_Y s - \nabla_Y\nabla_X s - \nabla_{[X, Y]}s).$

So the value of ${R(X, Y)s}$ at a point ${P}$ depends only on the values of ${X}$, ${Y}$, ${s}$ at the point ${P}$. For

$\displaystyle X = {\partial\over{\partial x^i}}, \text{ }Y = {\partial\over{\partial x^j}}, \text{ }s = e_\alpha,$

it follows from definitions that

$\displaystyle R(X, Y)s = 2\Omega_{\alpha i j}^\beta e_\beta.$

So the curvature can be defined in an invariant way by

$\displaystyle R(X, Y)s = \nabla_X\nabla_Ys - \nabla_Y\nabla_Xs - \nabla_{[X, Y]}s.$

Let us consider the case when ${M}$ is a complex manifold of complex dimension ${n}$ and ${V}$ is a holomorphic vector bundle. Suppose we have a Hermitian metric ${H}$ along the fibers of ${V}$ and the connection ${\omega}$ is a complex metric connection. When we choose a local holomorphic basis ${e_\alpha}$ ${(1 \le \alpha \le r)}$, we have

$\displaystyle \Omega = d\omega - \omega \wedge \omega$

$\displaystyle = d((\partial H)H^{-1}) - (\partial H)H^{-1} \wedge (\partial H)H^{-1}$

$\displaystyle =(\partial + \overline{\partial})((\partial H)H^{-1}) - (\partial H)H^{-1} \wedge (\partial H)H^{-1}$

$\displaystyle =(\partial H)H^{-1} \wedge (\partial H) H^{-1} + \overline{\partial}((\partial H)H^{-1}) - (\partial H)H^{-1} \wedge (\partial H)H^{-1}$

$\displaystyle =\overline{\partial}((\partial H)H^{-1}) = \overline{\partial}\omega.$

This shows that ${\Omega}$ is actually an ${\text{End}(V)}$-valued ${(1, 1)}$-form on ${M}$, and it simply equals ${\overline{\partial}\omega}$ when expressed in terms of a local holomorphic basis of ${V}$. The Bianchi identity for ${\Omega}$ takes on a simpler form. In terms of local coordinates, the Bianchi identity is

$\displaystyle \nabla_k F_{\alpha i\overline{j}}^\beta + \nabla_i F_{\alpha \overline{j} k}^\beta + \nabla_{\overline{j}}F_{\alpha k i}^\beta = 0.$

Since the curvature tensor is of type ${(1, 1)}$, we simply have

$\displaystyle \nabla_k F_{\alpha i \overline{j}}^\beta = \nabla_i F_{\alpha k \overline{j}}^\beta.$

Analogously, we have also

$\displaystyle \nabla_{\overline{k}}F_{\alpha i \overline{j}}^\beta = \nabla_{\overline{j}} F_{\alpha i \overline{k}}^\beta.$

We denote by ${\text{Tr}\,\Omega}$ the trace of ${\Omega}$ so that ${\text{Tr}\,\Omega}$ is a ${(1, 1)}$-form on ${M}$. The ${(1, 1)}$-form ${\text{Tr}\,\Omega}$ is the curvature form of the determinant bundle ${\det V}$ of ${V}$ with the metric ${\det H}$ induced from the metric ${H}$ of ${V}$, because with respect to a local holomorphic basis,

$\displaystyle \partial\,\log\det H = H^{\alpha\overline{\beta}}\partial H_{\alpha\overline{\beta}} = \text{Tr}((\partial H)H^{-1})$

and

$\displaystyle \overline{\partial}((\partial(\det H))(\det H)^{-1}) = \text{Tr}(\overline{\partial}((\partial H)H^{-1})) = \text{Tr}\,\Omega.$

Now we come back to the real case. Consider the real manifold ${M}$ of real dimension ${m}$ and the tangent vector bundle ${T_M}$ with a Riemannian metric. Then the curvature tensor is given by

$\displaystyle \Omega_{khi}^\ell = \partial_i\Gamma_{kj}^\ell - \partial_j \Gamma_{ki}^\ell + \Gamma_{ki}^p \Gamma_{pj}^\ell - \Gamma_{kj}^p \Gamma_{pi}^\ell.$

In this case, we have another Bianchi identity. The earlier one is usually referred to as the second Bianchi identity. The earlier one is usually referred to as the second Bianchi identity, and the one we are going to discuss is called the first Bianchi identity. At one point, choose a coordinate system so all vanish at that point. Then at the point, we simply have

$\displaystyle \Omega_{kij}^\ell = \partial_i \Gamma_{kj}^\ell - \partial_j \Gamma_{ki}^\ell,$

from which it follows that

$\displaystyle \Omega_{kij}^\ell + \Omega_{ijk}^\ell + \Omega_{jki}^\ell = 0,$

because of the symmetry of ${\Omega_{kj}^\ell}$ in ${k}$ and ${j}$. This is the first Bianchi identity. The first Bianchi identity is a consequence of the torsion-free condition. We can see this more clearly in the following way. Suppose ${e_i}$ ${(1 \le i \le m)}$ is its dual frame. Let ${\omega_i^j}$ with

$\displaystyle De_i = \omega_i^j e_j$

be the matrix valued ${1}$-form of the connection. The torsion-free condition is

$\displaystyle d\omega^i + \omega_j^i \wedge \omega^j = 0.$

Taking exterior derivative of both sides and using the original equation to get rid of ${d\omega^i}$, we get

$\displaystyle 0 = d\omega_j^i \wedge \omega^j - \omega_j^i \wedge d \omega^j = d\omega_j^i \wedge \omega^j - \omega_j^i \wedge d\omega^j$

$\displaystyle = d\omega_j^i \wedge \omega^j + \omega_j^i \wedge \omega_k^j \wedge \omega^k = \Omega_j^i \wedge \omega^j.$

When we write

$\displaystyle \Omega_j^i = \Omega_{jk\ell}^i \omega^k \wedge \omega^\ell,$

we have

$\displaystyle \Omega_{jk\ell}^i + \Omega_{k\ell j}^i + \Omega_{\ell jk}^i = 0,$

which is the first Bianchi identity. There is another symmetry we want. Let

$\displaystyle R_{k\ell i j} = 2g_{\ell p}\Omega_{kij}^\ell.$

From

$\displaystyle \Omega_{jk}^i = {1\over2}g^{i\ell}(\partial_jg_{k\ell} + \partial_kg_{j\ell} - \partial_\ell g_{jk}),$

we have

$\displaystyle R_{k \ell i j} = \partial_i \Gamma_{\ell,\,kj} - \partial_j \Gamma_{\ell,\,ki}$

$\displaystyle = {1\over2}\partial_i(\partial_j g_{k\ell} + \partial_k g_{j \ell} - \partial_\ell g_{jk}) - {1\over2}\partial_j(\partial_i g_{k\ell} + \partial_k g_{i \ell} - \partial_\ell g_{ik})$

$\displaystyle = {1\over2}(\partial_i \partial_k g_{j \ell} + \partial_j \partial_\ell g_{ik} - \partial_i \partial_\ell g_{jk} - \partial_i \partial_\ell g_{ik} - \partial_i \partial_\ell g_{jk} - \partial_j \partial_k g_{i\ell}).$

Thus,

$\displaystyle R_{k\ell ij} = R_{ijk \ell}.$

We consider now the Kähler case. From

$\displaystyle 2\Omega_{kij}^\ell = \partial_i \Gamma_{kj}^\ell - \partial_j \Gamma_{ki}^\ell,$

and the fact that the only nonvanishing components of the Christoffel symbols are of type ${\Gamma_{\beta\gamma}^\alpha}$ and ${\Gamma_{\overline{\beta}\overline{\gamma}}^{\overline{\alpha}}}$, we conclude that when ${k}$ and ${\ell}$ are of different type, ${\Gamma_{kij}^\ell}$ must vanish. So ${R_{k\ell i j}}$ vanishes unless ${K}$ and ${\ell}$ are of different type. We can also see this from the fact that the curvature of the complex metric connection is of type ${(1, 1)}$. By the above symmetry, we conclude that the only nonzero components of the curvature tensor are of type ${R_{\alpha\overline{\beta}\gamma\overline{\delta}}}$, ${R_{\overline{\alpha}\beta\gamma\overline{\delta}}}$, ${R_{\alpha\overline{\beta}\overline{\gamma}\delta}}$, ${R_{\overline{\alpha}\beta\overline{\gamma}\delta}}$.