# Connections, The Kähler Condition, and Curvature – VIII. The Second Fundamental Form in the Complex Manifold Case

We now look at the case of holomorphic vector bundles over a complex manifold and discuss the concept of second fundamental form. Now, ${M}$ is a complex manifold of complex dimension ${n}$, and ${E}$ is a holomorphic vector bundle of rank ${r}$ over ${M}$. Assume that ${E}$ has a Hermitian metric. Let ${E'}$ be a holomorphic vector subbundle of ${E}$ of rank ${s}$.

We can choose a local unitary basis ${e_\alpha}$ ${(1 \le \alpha \le r)}$ of ${E}$ such that ${e_\alpha}$ ${(1 \le \alpha \le s)}$ belongs to ${E'}$. Let ${\omega'}$ be the connection of ${E'}$ induced by the complex metric connection ${\omega}$ of ${E}$. In other words,

$\displaystyle {\omega'}_\alpha^\beta = \omega_\alpha^\beta$

for ${1 \le \alpha \le s}$. Another way of describing the connection ${\omega'}$ is that the ${E'}$-covariant derivative ${D's}$ of a local section of ${s}$ of ${E'}$ is obtained by taking its ${E}$-covariant derivative ${Ds}$ and then projecting ${Ds}$ onto ${E'}$ by the orthogonal projection. This connection agrees with the complex metric connecion of ${E'}$ with respect to the metric induced from that of ${E}$. The reason is as follows. Firstly, it is easy to see that the ${E'}$-covariant derivative of a local holomorphic section of ${E'}$ along any ${(0, 1)}$ direction is zero from the above description of the ${E'}$-covaraint differentiation. Secondly, if ${s}$ is a section of ${E'}$ above a local curve of ${M}$ with zero ${E'}$-covaraint derivative along the curve, then its ${E}$-covariant derivative ${Ds}$ is a linear combination of ${e_\alpha}$ ${(s < \alpha \le r)}$ and must be perpendicular to ${s}$. As a consequence,

$\displaystyle d\langle s, s\rangle = \langle Ds, s \rangle + \langle s, Ds\rangle$

must vanish along the curve, and the length of ${s}$ is constant. Let ${Q}$ be the orthogonal complement of ${E'}$ in ${E}$. We give ${Q}$ the complex structure of the quotient bundle ${E/E'}$. The difference ${Ds - D's}$ of the ${E}$-covariant derivative of ${s}$ and the ${E'}$-covariant derivative of ${s}$ of ${s}$ is a 1-form with values in ${\text{Hom}(E', Q)}$. This ${\text{Hom}(E', Q)}$-valued ${1}$-form is called the second fundamental form of ${E'}$ in ${E}$, and we denote it by ${B}$. Let

$\displaystyle s = \sum_{\alpha = 1}^r s^\alpha s_\alpha$

be the representation of ${s}$ in terms of some local holomoprhic atlas ${e_\alpha}$. Then

$\displaystyle Ds = \sum_{\alpha=1}^r ds^\alpha e_\alpha + \sum_{\alpha=1}^r s^\alpha De_\alpha$

and

$\displaystyle D's = \sum_{\alpha=1}^r ds^\alpha e_\alpha + \sum_{\alpha=1}^r s^\alpha D'e_\alpha.$

Hence,

$\displaystyle Ds - D's = \sum_{\alpha=1}^r s^\alpha(De_\alpha - D'e_\alpha)$

is a ${Q}$-valued ${(1, 0)}$-form. Thus, the second fundamental form ${B}$ must be a ${\text{Hom}(E', Q)}$-valued ${(1, 0)}$-form, and we can write

$\displaystyle Ds = D's + Bs$

for a section ${s}$ of ${E'}$.

Let us now consider the case of the quotient bundle. Take a local holomorphic basis ${e_\alpha''}$ ${(s + 1 \le \alpha \le r)}$ of ${Q}$. We use the same notation ${e_\alpha''}$ to denote local sections of ${E}$ orthogonal to ${E'}$. The holomorphicity of basis ${e_\alpha''}$ means that for some sections ${e_\alpha'}$ ${(s + 1 \le \alpha \le r)}$ of ${E'}$, the sections ${e_\alpha' + e_\alpha''}$ ${(s + 1 \le \alpha \le r)}$ are holomorphic sections of ${E}$. We take also a local holomorphic basis ${e_\gamma}$ ${(1 \le \gamma \le s)}$ of ${E'}$. We have

$\displaystyle (De_\alpha'')^{(0, 1)} = \overline{\partial}e_\alpha'' = -\overline{\partial}e_\alpha'.$

When we write

$\displaystyle e_\alpha' = \sum_{\gamma=1}^s a_\alpha^\gamma e_\gamma$

in terms of the local holomorphic basis ${e_\gamma}$ ${(1 \le \gamma \le s)}$ of ${E'}$, we see that

$\displaystyle -\overline{\partial}e_\alpha' = -\sum_{\gamma = 1}^s (\overline{\partial}a_\alpha^\gamma)e_\gamma.$

Hence, ${-\overline{\partial}e_\alpha'}$ is an ${E}$-valued ${(0, 1)}$-form. Since from the definition of ${D}$ one has

$\displaystyle \left\langle (De_\alpha''^{(1, 0)}, e_\gamma \right\rangle = \partial \left\langle e_\alpha'', e_\gamma \right\rangle - \left\langle e_\alpha'', \overline{\partial}e_\gamma\right\rangle = 0,$

it follows that ${(De_\alpha'')^{(1, 0)}}$ is a ${Q}$-valued ${(1, 0)}$-form. So we can define a connection ${D''}$ on ${Q}$ by

$\displaystyle D''e_\alpha'' = De_\alpha'' + \overline{\partial}e_\alpha'.$

We claim that this connection is the complex metric connection of ${Q}$. It is a complex connection, because we have observed earlier that

$\displaystyle (De_\alpha'' + \overline{\partial}e_\alpha')^{(0, 1)} = 0$

and the ${(1, 0)}$-form ${(De_\alpha'')^{(1,0)}}$ has values in ${Q}$. It is also a metric connection, because

$\displaystyle \left\langle De_\alpha'' + \overline{\partial}e_\alpha', e_\beta''\right\rangle + \left\langle e_\alpha'', De_\beta'' + \overline{\partial}e_\beta'\right\rangle = \left\langle De_\alpha'', e_\beta''\right\rangle + \left\langle e_\alpha'', De_\beta''\right\rangle = d\left\langle e_\alpha'', e_\beta''\right\rangle.$

We write

$\displaystyle Ds'' - D''s'' = Cs''$

for sections ${s''}$ of ${Q}$. The operator ${C}$ is a ${\text{Hom}(Q, E')}$-valued ${(0, 1)}$-form given by

$\displaystyle Ce_\alpha'' = -\overline{\partial}e_\alpha' = -\sum_{\gamma = 1}^s (\overline{\partial} a_\alpha^\gamma)e_\gamma.$

We call ${C}$ the second fundamental form of the quotient bundle ${Q}$ of ${E}$.

Another more invariant way of representing the second fundamental form ${C}$ of the quotient bundle ${Q}$ of ${E}$ is the following. We have

$\displaystyle Ce_\alpha'' = \overline{\partial}e_\alpha''.$

The local basis ${e_\alpha''}$ ${(s + 1 \le \alpha \le r)}$ of the orthogonal complement of ${E'}$ in ${E}$ simply describes the monomorphism from ${Q}$ to ${E}$ which lifts ${Q}$ to the orthogonal complement of ${E'}$ in ${E}$. let us call this monomorphism ${\varphi}$. Then ${C}$ is simply ${\overline{\partial}\varphi}$. The entity ${\overline{\partial}\varphi}$ is a priori only a ${\text{Hom}(Q, E)}$-valued ${(0, 1)}$-form which is ${\overline{\partial}}$ exact. Our previous discussion shows that it is actually a ${\text{Hom}(Q, E')}$-valued ${(0, 1)}$-form. However, as a ${\text{Hom}(Q, E')}$-valued ${(0, 1)}$-form, in general it is only ${\overline{\partial}}$ closed and is not ${\overline{\partial}}$ exact. Suppose we have another Hermitian metric for ${E}$. Then we would have a different monomorphism ${\varphi'}$ from ${Q}$ to ${E'}$ and a different second fundamental form ${C'}$. The difference of ${\varphi}$ and ${\varphi'}$ is a homomorphism from ${Q}$ to ${E'}$, because ${\varphi}$ and ${\varphi'}$ are different liftings of ${Q}$ to ${E}$. So the difference ${C - C'}$ of the two second fundamental forms of ${Q}$ equals ${\overline{\partial}(\varphi - \varphi')}$, which is a ${\overline{\partial}}$ exact ${\text{Hom}(Q, E')}$-valued ${(0, 1)}$-form.

We want to relate ${C}$ to the second fundamental form ${B}$ of the subbundle ${E'}$ of ${E}$. For sections ${s'}$ and ${s''}$ of ${E'}$ and ${Q}$, respectively, we have

$\displaystyle 0 = d\left\langle s', s''\right\rangle = \left\langle Ds', s''\right\rangle + \left\langle s', Ds''\right\rangle$

$\displaystyle =\left\langle D's + Bs', s''\right\rangle + \left\langle s', D''S'' + Cs''\right\rangle$

$\displaystyle =\left\langle Bs', S''\right\rangle + \left\langle s', Cs''\right\rangle.$

Hence, ${C}$ is simply the negative of the adjoint ${B^*}$ of ${B}$ with respect to the Hermitian metrics of ${E'}$ and ${Q}$. So

$\displaystyle D = D'' - B^*.$

We now compute the curvatures of ${\Omega^E}$, ${\Omega^{E'}}$, and ${\Omega^Q}$. We choose a local orthonormal frame ${e_\alpha}$ ${(1 \le \alpha \le r)}$ so that ${e_\alpha}$ ${(1 \le \alpha \le s)}$ belongs to ${E'}$. Write

$\displaystyle De_\alpha = \sum_{\beta = 1}^r \omega_\alpha^\beta e_\beta.$

Since the connection is compatible with the metric by differentiating

$\displaystyle \langle e_\alpha, e_\beta \rangle = 0 \text{ or }1,$

we conclude that

$\displaystyle \omega_\alpha^\beta = -\overline{\omega_\beta^\alpha}.$

The second fundamental form ${B}$ of ${E'}$ is given by

$\displaystyle Be_\alpha = \sum_{\beta = s+ 1}^r \omega_\alpha^\beta e_\beta$

for ${1 \le \alpha \le s}$. From

$\displaystyle \Omega_\alpha^{E\text{ }\beta} = d\omega_\alpha^\beta - \sum_{\gamma=1}^r \omega_\alpha^\gamma \wedge \omega_\gamma^\beta$

and

$\displaystyle \Omega_\alpha^{E'\text{ }\beta} = d\omega_\alpha^\beta - \sum_{\gamma = 1}^s \omega_\alpha^\gamma \wedge \omega_\gamma^\beta,$

we have

$\displaystyle \Omega_\alpha^{E\text{ }\beta} = \Omega_\alpha^{E'\text{ }\beta} + \sum_{\gamma = s+1}^r \omega_\alpha^\gamma \wedge \overline{\omega_\beta^\gamma}.$

Thus,

$\displaystyle \Omega_\alpha^{E\text{ }\alpha}(X, \overline{X}) \ge \Omega_\alpha^{E'\text{ }\alpha}(X, \overline{X})$

(no summation over ${\alpha}$) for any vector ${X}$ of type ${(1, 0)}$, and equality holds if and only if

$\displaystyle B(X)e_\alpha = 0.$

In invariant formulation, we have

$\displaystyle \left\langle \Omega^E(X, \overline{X})\xi, \xi\right\rangle \ge \left\langle \Omega^{E'}(X, \overline{X})\xi, \xi\right\rangle$

for any ${\xi}$ in ${E'}$ and any vector ${X}$ of ${M}$ of type ${(1, 0)}$.

Let us now look at the case of the quotient bundle. We can identify the quotient bundle ${Q}$ as the subbundle of ${E}$ which is the orthogonal complement of ${E'}$ in ${E}$. We needed the complex structure of ${Q}$ only to define the connection of ${Q}$ as a complex metric connection. Once we get the connection of ${Q}$, we can ignore the complex structure of ${Q}$. The calculation of the curvature tensor depends only on the connection. So when it comes to comparing the curvature tensors, the computation of the quotient bundle case is the same as the subbundle case. There is however one difference. The second fundamental form of the subbundle is an endomoprhism valued ${(0, 1)}$-form. So when it comes to evaluating the exterior product of the second fundamental form and its complex conjugate transpose at ${(X, \overline{X})}$ for some ${(1, 0)}$-vector ${X}$, there is a sign difference between the quotient bundle case and the subbundle case. So we have

$\displaystyle \left\langle \Omega^E(X, \overline{X})\xi, \xi\right\rangle \le \left\langle \Omega^Q(X, \overline{X})\xi', \xi'\right\rangle$

for any ${\xi}$ in the orthogonal complement of ${E'}$ in ${E}$ and for any vector ${X}$ of ${M}$ of type ${(1, 0)}$, where ${\xi'}$ is the image of ${\xi}$ in ${Q}$.