Weitzenböck Identities

I asked this question at Maths Stack Exchange, but I haven't received any replies yet (I'm not sure how long I should wait before it is acceptable to ask here, assuming there is such a period of time).

The Wikipedia page for Weitzenböck identities is explicitly example based. I am looking for a reference which takes a more rigorous approach (as well as a discussion of the Bochner technique). In particular, I am interested in references which focus on these identities in complex geometry.

I have already consulted Griffiths & Harris which is mentioned in the article, but it only contains one example. Berger's A Panoramic View of Riemannian Geometry doesn't have much more.

My interest in Weitzenböck identities has been motivated by a question arising from the following theorem:

Let $X$ be a Kähler manifold and $E$ a hermitian holomorphic vector bundle with Chern connection $\nabla$. Then for the Laplacians $\Delta_{\bar{\partial}} = \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$, $\Delta_{\partial} = \partial\partial^* + \partial^*\partial$, we have $\Delta_{\bar{\partial}} = \Delta_{\partial} + [iF_{\nabla}, \Lambda]$.

Is $\Delta_{\bar{\partial}} = \Delta_{\partial} + [iF_{\nabla}, \Lambda]$ an example of a Weitzenböck identity?

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Demailly's book (www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf) and Besse's "Einstein manifolds" have good discussions of these things. The identity you write is called a Bochner-Kodaira-Nakano identity, which is a complex geometric version of a Weitzenböck identity. By any name, these things express the difference between two Laplacians. –  Gunnar Magnusson Sep 14 '12 at 12:52
This paper, although written from a Riemannian geometry perspective instead of a complex one, has a seemingly general definition of Weitzenb\"ock identity. It's a bit long though. arxiv.org/abs/math/0702031 –  Paul Reynolds Sep 14 '12 at 12:58
The Weitzenboeck technique is mostly used to show that some (first order, self-adjoint) differential operator $D$ (acting on sections of an hermitian bundle over a compact manifold $M$) has trivial kernel. Since the $L^2$ norm of $Ds$ is $\int_M (D^2s,s)$, the goal is to write $D^2$ as some "standard Laplacian" plus a 0-order scalar term, $D^2=\Delta +A$. If $A$ is pointwise positive, by the maximum principle for $\Delta$ you get that $Ds=0$ implies $s=0$. –  Peter Dalakov Sep 14 '12 at 13:30

Here is roughly the philosophy of the Weitzenbock technique. (Most of what follows is taken from Berline-Getzler-Vergne book.)

Suppose that $E_0,E_1\to M$ are vector bundles on an oriented Riemann manifolds $M$ equipped with hermitian metrics. Denote by $C^\infty(E_i)$ the space of smooth sections of $E_i$.

A symmetric 2nd order differential operator $L: C^\infty(E_0)\to C^\infty(E_0)$ is called a generalized Laplacian on $E_0$ if its principal symbol $\sigma_L$ coincides with the principal symbol of a Laplacian. Concretely this means the following.

For a smooth function $f\in C^\infty(M)$ denote by $M_f$ the linear operator $C^\infty(E_0)\to C^\infty( E_0)$ defined by the multiplication with $f$. Then $L$ is a generalized Laplacian if for any $f_0,f_1\in C^\infty(M)$ and any $u\in C^\infty(E_0)$ we have

$$[\; [\; L,M_{f_0}\; ], M_{f_1}\; ]u = -2g( df_0,df_1)\cdot u$$

where $[-,-]$ denotes the commutator of two operators. Equivalently, this means

$$[[L,M_{f_0}],M_{f_1}]=-2M_{g(df_0,df_1)}.$$

One can show that if $L$ is a generalized Laplacian on $E_0$, then there exists a connection $\nabla$ on $E_0$, compatible with the metric on $E_0$, and a symmetric endomorphism $W$ of $E_0$ such that

$$L =\nabla^*\nabla +W.$$

The classical Weitzenbock formulas give explicit descriptions to the Weitzenbock remainder $W$ and the connection $\nabla$.

Usually the generalized Laplacians are obtained through Dirac type operators which are first order differential operators $D: C^\infty(E_0)\to C^\infty(E_1)$ such that both operators $D^\ast D$ and $D D^\ast$ are generalized Laplacians on $E_0$ and respectively $E_1$. We can rewrite this in a compact form by using the operator

$$\mathscr{D}: C^\infty(E_0)\oplus C^\infty(E_1)\to C^\infty(E_0)\oplus C^\infty(E_1),$$

$$\mathscr{D}(u_0\oplus u_1)= (D^*u_1)\oplus (D u_0).$$

Then $D$ is Dirac type iff $\mathscr{D}^2$ is a generalized Laplacian.

The Weitzenbock remainders of $D^\ast D$ and $D D^\ast$ involve curvature terms. If the Weitzenbock remainder of $D^*D$ happens to be a positive endomorphism of $E_0$, then one can conclude that

$$\ker D=\ker D^\ast D=0.$$

The Hodge-Dolbeault operator

$$\frac{1}{\sqrt{2}}(\bar{\partial}+\bar{\partial}^*): \Omega^{0,even}(M)\to \Omega^{0,odd}(M)$$

on a Kahler manifold $M$ is a Dirac type operator. For more details and examples you can check Sec. 10.1 and Chap 11 of my lecture notes.

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Thank you for your wonderful answer. I am not sure how the Hodge-Dolbeault operator (call it $D$ and ignore the constant factor of $\frac{1}{\sqrt{2}}$) fits in with the Laplacians $\Delta_{\partial}$ and $\Delta_{\bar{\partial}}$. You say that you can obtain a generalised Laplacian using $DD^*$ or $D^*D$, but neither of these give $\Delta_{\partial}$ or $\Delta_{\bar{\partial}}$. Instead, $DD^* + D^*D = \Delta_{\partial} + \Delta_{\bar{\partial}}$. –  Michael Albanese Sep 16 '12 at 14:25
Also, in Chapter 11 of your notes, you say that a Dirac operator is one which squares to be a generalised Laplacian which (I'm pretty sure) does not agree with what you have written. Using this definition instead and setting $D_{1, 0} = \partial + \partial^*$ and $D_{0,1} = \bar{\partial} + \bar{\partial}^*$ we have $D_{1,0}^2 = \Delta_{\partial}$ and $D_{0,1}^2 = \Delta_{\bar{\partial}}$. Is that what you meant? These are just the complex anologues of your Example 11.1.5 on the Hodge-de Rham operator $d + d^*$. –  Michael Albanese Sep 16 '12 at 14:29
Look at $\mathscr{D}=\frac{1}{\sqrt{2}}(\partial+\bar{\partial}): \Omega^{0,*}\to\Omega^{0,*}$.Then $\mathscr{D}^2$ is a Laplacian. The operator $\mathscr{D}$ is an example of a *graded Dirac, i.e., first order symmetric $\matscr{D}:C^\infty(E_0\oplus E_1)\to C^\infty(E_0\oplus E_1)$ such that $\mathscr{D}^2$ is Laplacian and $\mathscr{D}$ which maps sections of $E_0$ to sections of $E_1$ and vice-versa. It is uniquely determined by its restriction to $C^\infty(E_0)$. Check definition 11.1.1 of my notes and section 11.2.2 for the special case of Kahler manifolds. –  Liviu Nicolaescu Sep 16 '12 at 23:28
Sorry if I am missing something, but on page 530 of your notes you define the Hodge-Dolbeault operator to be $\bar{\partial} + \bar{\partial}^*$ but above you define it to be $\mathscr{D} = \frac{1}{\sqrt{2}}(\partial + \bar{\partial})$. Furthermore, isn't $\mathscr{D}(\Omega^{0,even}(M)) \subseteq \Omega^{1,even}(M)\oplus\Omega^{0,odd}(M)$? –  Michael Albanese Sep 17 '12 at 0:50
You're right The correct form is as in my book $\partial+\partial^*$. One needs the $1/{\sqrt{2}}$ factor only to get a Dirac operator. I will correct the post. –  Liviu Nicolaescu Sep 17 '12 at 8:05

As far as references are concerned, you can read about the Bochner technique in H.Wu, "The Bochner technique in differential geometry". For some applications of the Lichnerowicz formula you can check Ch.3 of Berline, Getzler, Vergne, "Heat Kernels and Dirac operators", and Ch.3, 5 of T.Friedrich's "Dirac Operators in Riemannian geometry".

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I don't know of any precise definition of Weitzenbock identities, which is closely related to or also known as the Bochner technique. It is basically a way to write some invariantly defined second order linear differential operator $D$ on a vector bundle over a manifold (often a complex one) in two different ways. One way, usually written in terms of a coboundary operator and its adjoint, shows that the kernel of the operator is an interesting topological or holomorphic invariant (usually some kind of cohomology). The other way is in the form $D = P^*P + R$, where $P$ is a linear first order operator and $R$ is a naturally defined geometric tensor, usually called some kind of curvature. This allows the kernel of $D$ to be studied under suitable assumptions on $R$ (usually that it is positive or nonnegative definite) by studying the kernel of $P$.
Also, you get from one form of the operator to the other by commuting covariant derivatives, which is how the $0$-th order curvature term $R$ appears.