# Weitzenböck Identity for $\Delta_{\bar{\partial}_E}$

This question is related to this MO question and this MSE question.

Let $$E$$ be a hermitian holomorphic vector bundle over a hermitian manifold $$X$$. The bundle $$\bigwedge^{\bullet,\bullet}X\otimes E$$ has an induced hermitian metric. As $$E$$ is holomorphic, there is a Dolbeault operator $$\bar{\partial}_E$$ which acts on

$$\Omega^{\bullet,\bullet}(X, E) = \Gamma\left(X, \bigwedge\nolimits^{\!\bullet,\bullet}X\otimes E\right),$$

and it has adjoint $$\bar{\partial}_E^*$$. Both $$\bar{\partial}_E$$ and $$\bar{\partial}_E^*$$ are first order differential operators and

$$\Delta_{\bar{\partial}_E} = \bar{\partial}_E\bar{\partial}_E^* + \bar{\partial}_E^*\bar{\partial}_E$$

is a second order differential operator. Furthermore, $$2\Delta_{\bar{\partial}_E}$$ is a generalised Laplacian; that is,

$$\sigma_2(2\Delta_{\bar{\partial}_E})(\xi) = - \|\xi\|^2\operatorname{id}_{E_x}$$

where $$x = \pi(\xi)$$ and $$\pi : TX \to X$$ is the natural projection map.

Proposition 9.1.27 (Nicolaesu's Lectures on the Geometry of Manifolds)

Let $$L$$ be a generalised Laplacian on $$F$$. There is a unique metric connection $$\nabla$$ on $$F$$ and $$\mathfrak{R} \in \operatorname{End}(F)$$ such that

$$L = \nabla^*\nabla + \mathfrak{R}.$$

I want to know what the connection is for the generalised Laplacian $$2\Delta_{\bar{\partial}_E}$$ under the assumption that $$X$$ is Kähler. If I'm not mistaken, the following result is what I need.

Proposition 3.67 (Berline, Getzler, & Vergne's Heat Kernels and Dirac Operators)

Let $$\mathcal{W}$$ be a holomorphic vector bundle with hermitian metric on a Kähler manifold $$X$$. The tensor product of the Levi-Civita connection with the canonical connection of $$\mathcal{W}$$ is a Clifford connection on the Clifford module $$\bigwedge(T^{0,1}X)^*\otimes\mathcal{W}$$, with associated Dirac operator $$\sqrt{2}(\bar{\partial}_E + \bar{\partial}_E^*)$$.

First of all, $$\sqrt{2}(\bar{\partial}_E+\bar{\partial}_E^*)$$ is a Dirac operator for $$2\Delta_{\bar{\partial}_E}$$, i.e

$$\left(\sqrt{2}(\bar{\partial}_E+\bar{\partial}_E^*)\right)^2 = 2\Delta_{\bar{\partial}_E}.$$

If $$\mathcal{W} = \bigwedge(T^{1,0}X)^*\otimes E$$ where the first factor has the metric induced by the metric on $$X$$ and $$E$$ is equipped with a metric such that the tensor product metric agrees with the metric on $$\mathcal{W}$$. Then

$$\bigwedge(T^{0,1}X)^*\otimes\bigwedge(T^{1,0}X)^*\otimes E \cong \bigwedge\nolimits^{\!\bullet, \bullet}X\otimes E.$$

The connection on $$\bigwedge(T^{0,1}X)^*$$ is the Levi-Civita connection, and the connection on $$\bigwedge(T^{1,0}X)^*$$ is the Chern connection, but as the manifold is Kähler, the Chern connection coincides with the Levi-Civita connection. Therefore on

$$\left(\bigwedge(T^{0,1}X)^*\otimes\bigwedge(T^{1,0}X)^*\right)\otimes E \cong\bigwedge\nolimits^{\!\bullet, \bullet}X\otimes E$$

the connection is given by the Levi-Civita connection on the first factor and the Chern connection on the second factor.

1. Is this the metric connection from Proposition 9.1.27 for the generalised Laplacian $$2\Delta_{\bar{\partial}_E}$$?
2. If so, is it also the metric connection for the generalised Laplacian $$2\Delta_{\partial_E}$$?
3. Can any of this be generalised to the case where $$X$$ is just a hermitian manifold?

I'm still getting used to the whole language involved in this setup, so if it seems that I've missed something, please let me know.

Added later: Thanks to the two answers, several of my questions have been answered. If $$X$$ is a Kähler manifold, then on $$\Omega^{p,q}(X, E)$$ we have $$\Delta_{\bar{\partial}_E} = \nabla^*\nabla + \mathfrak{R}_1$$ and $$\Delta_{\partial_E} = \nabla^*\nabla + \mathfrak{R}_2$$ where, in both cases, $$\nabla$$ is the connection induced by the Chern connection. Combining these two decompositions, we have $$\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + (\mathfrak{R}_1 - \mathfrak{R}_2)$$. As $$\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + [iF_{\nabla}, \Lambda]$$ we see that $$\mathfrak{R}_1 - \mathfrak{R}_2 = [iF_{\nabla}, \Lambda]$$.

In the case that $$X$$ is a hermitian manifold, I am struggling to see how to generalise the above consideration. I know that $$\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + [iF_{\nabla}, \Lambda] + [\bar{\partial}_E, [L, \Lambda_{\bar{\partial}\omega}]] - [\partial_E, [L, \Lambda_{\partial\omega}]].$$ Combining the decompositions $$\Delta_{\bar{\partial}_E} = \nabla_1^*\nabla_1 + \mathfrak{R}_1$$ and $$\Delta_{\partial_E} = \nabla_2^*\nabla_2 + \mathfrak{R}_2$$, we have $$\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + (\nabla_1^*\nabla_1 - \nabla_2^*\nabla_2) + (\mathfrak{R}_1 - \mathfrak{R}_2).$$ Therefore $$(\nabla_1^*\nabla_1 - \nabla_2^*\nabla_2) + (\mathfrak{R}_1 - \mathfrak{R}_2) = [iF_{\nabla}, \Lambda] + [\bar{\partial}_E, [L, \Lambda_{\bar{\partial}\omega}]] - [\partial_E, [L, \Lambda_{\partial\omega}]].$$ What I can't deduce is the correspondence between the terms. On the right hand side, the first term is order zero, whilst the other two terms are order one. On the left hand side, the first bracket is at most order two, but I'm guessing they have the same principal part so it is actually order one. Is $$\nabla_1 \neq \nabla_2$$ or are they equal so that only $$\mathfrak{R}_1 - \mathfrak{R}_2$$ represents the right hand side. Surely this can't be the case as $$\mathfrak{R}_1$$ and $$\mathfrak{R}_2$$ are order zero, aren't they?

You can find all what you want to know in the chapter 1 of the book X.Ma and G. Marinescu:Holomorphic Morse Inequalities and Bergman Kernels.

In Kahler case, all the metric connection is the chern connection.

And for the general case(hermitian manifold), the connection is called Bismut connection (see section 1.2.3). And the Weitzenböck formula is in Theorem 1.4.7

For the relation of the two laplacian $\Delta_{\partial}$ and $\Delta_{\overline{\partial}}$ see Theorem 1.4.12 and Corollary 1.4.13.

• You say that for the general case, the connection is called the Bismut connection, but this can't be the metric connection for $2\Delta_{\bar{\partial}_E}$ and $2\Delta_{\partial_E}$. If it were, the difference $2\Delta_{\bar{\partial}_E} - 2\Delta_{\partial_E}$ would have order zero, but the torsion of the Chern connection is order one. So the metric connections for $2\Delta_{\bar{\partial}_E}$ and $2\Delta_{\bar{\partial}_E}$ must be different. Is one of them the Bismut connection, and the other something else? Aug 4, 2013 at 1:59
• the Bismut connection is only for $\Delta_{\bar{\partial}_E}$ considere as an operator acting on $\Lambda^{0,\cdot}T^*X$(if I remember correctly, but easy to write down for $\Lambda^{\cdot,\cdot}T^*X$). As you said, for $\Delta_{{\partial}_E}$, the connection will be different. As we have the Bochner-Kodaira-Nakano identity, the later is not so interesting...(of course, you can write it down) Sorry for the later answer, I am on vacation...
– shu
Aug 4, 2013 at 12:12

The answer for both questions 1, 2 is yes, the metric connection in Proposition 9.1.27 is the Levi-Civita induced connection if the background manifold is Kahler. When $X$ is only a Hermitian manifold, the result is not true. One such example is discussed in great detail in Bismut's paper

A local index theorem for non-Kahler manifolds, Math. Ann., 284(1989), 681-699.

• In the Kähler case, the metric connections for $2\Delta_{\bar{\partial}_E}$ and $2\Delta_{\partial_E}$ are the same. Is this still true in the hermitian case? Jul 24, 2013 at 14:54
• I don't know. Maybe The reference indicated by Shu has the answer. Jul 24, 2013 at 15:14
• @MichaelAlbanese, the two operators are not the same in the hermitian case. The difference of the two is samething related to the torion of chern connection. see theorem 1.4.12
– shu
Jul 24, 2013 at 15:26
• I know that the difference between the two covariant Laplacians must introduce a first order differential operator involving the torsion of the Chern connection because you must get the Bochner-Kodaira-Nakano identity. However, I haven't been able to find a direct statement or calculation of this form. Jul 28, 2013 at 10:11
• @MichaelAlbanese, the key point is using Bismut's connection, there is no first order differential operator. Have you checked the formula in Ma-Marinescu, they give the detailed calculations.
– shu
Jul 29, 2013 at 9:35