# Lifting quadratic forms on the cotangent bundle to higher level forms

Backround

In several complex variables, an essential tool is Hormander's machinery for solving the $\overline{\partial}$ problem with $L^2$ estimates.

If $\alpha$ is a $(p,q+1)$ form on a domain $\Omega \subset \mathbb{C}^n$, and one is attempting to solve $\overline{\partial} u = \alpha$ with a weight function $\phi$ one of the terms which must be estimated to get the method to work is

$$\int_\Omega \left[ \sum_{|I|=p,|J|=q} \sum_{j,k=1}^n \frac{\partial^2 \phi}{\partial z_j \partial \overline{z_k}} \alpha_{I,jJ}\overline{\alpha_{I,kJ}} \right]e^{-\phi} dV$$

where $I,J$ are multi-indices, $kJ$ means "stick the index $j$ to the front of $J$", and $\alpha_{I,K}$ is the coefficient of $dz^I\wedge d\overline{z}^K$

This is a term in the integration by parts identity called the "Morrey-Kohn-Hormander" identity.

Let's give the notation

$$H_\phi^{(p,q)}(\alpha,\alpha) = \sum_{|I|=p,|J|=q} \sum_{j,k=1}^n \frac{\partial^2 \phi}{\partial z_j \partial \overline{z_k}} \alpha_{I,jJ}\overline{\alpha_{I,kJ}}$$

$H_\phi^{(0,1)}$ is a quadratic form on $(0,1)$ forms which is called the "complex Hessian" of $\phi$. It seems like the rest of the $H_\phi^{(p,q)}$ are "built out" of the complex hessian, in the sense that the entries are all sums of entries from the complex hessian in a structured way.

Question

While the most satisfying explanation of this term might just be "it is what shows up when you integrate by parts", I am hoping for a more conceptual explanation. Is $H_\phi^{(p,q)}$ "natural" in any way? Is it the "unique way to lift a quadratic form on $\Lambda^{(0,1)}$ to $\Lambda^{(p,q)}$ such that..."?

I would also be interested in the corresponding question in the real case. Given a quadratic form $Q$ on $\Lambda^1(\mathbb{R^n})$, say $Q(dx^i,dx^j) = Q_{i,j}$ and define $Q^k$ on $\Lambda^k$ by

$$Q^k(\alpha,\alpha) = \sum_{|I|=k-1} \sum_{j,k=1}^n Q_{j,k} \alpha_{jI}\alpha_{kI}$$

does $Q^k$ have a nice characterization? Also, does anyone have any pointers to this stuff written out for manifolds? The formulas I am working with in the case of $\mathbb{C}^n$ or $\mathbb{R}^n$ do not look like they transform in a particularly nice way under coordinate transformations.

## 1 Answer

I think that the answer to your question(s), when everything is sorted out, is 'no'. What is actually going on in the complex case that you are concerned with is that there are two Hermitian forms in the problem, the 'background' (flat) Kähler form $K$, for which the $\mathrm{d}z^i$ are 'unitary', that would be written in the form $$K = \frac{\sqrt{-1}}2\ \left(\mathrm{d}z^1\wedge \mathrm{d}\bar z^1 + \cdots + \mathrm{d}z^n\wedge \mathrm{d}\bar z^n\right),$$ and the Hessian form of $\phi$, that would usually be written in the form $$H_\phi = \frac{\sqrt{-1}}2\ \left(\frac{\partial^2\phi}{\partial z^i\partial \bar z^j}\ \mathrm{d}z^i\wedge \mathrm{d}\bar z^j\right),$$ (N.B.: In the statement of your question, you are inconsistent about whether the coordinates $z$ are indexed with superscripts or subscripts. I think that this contributes to your confusion. Note that I am using the common convention that $\bar z^j$ means $\overline{z^j}$, since that looks better when typeset.)

Anyway, you'll notice that, in your formula for $H^{(p,q)}_\phi$, you are actually using $K$ to 'contract' most of the index pairs of $\alpha$, and using $H_\phi$ to contract one pair only. (N.B.: Check your summation in that formula; I think you should be summing over $J$ with $|J|=q{-}1$, not $|J|=q$.) A similar comment applies to your $Q^k$ (which also has problems with the superscript/subscript conventions).

It's the failure to recognize the implicit use of $K$ that is causing you to have problems figuring out why the formula doesn't transform correctly under change of variables. If you want to generalize the argument you are studying for domains in $\mathbb{C}^n$ to cover, say, domains in general Kähler manifolds, then you'll need to use the Kähler form to get your Hermitian inner product on $(p,q)$-forms.

At the level of linear algebra, you can see what is going on as follows: You are trying canonically to define an Hermitian form on the vector space $W = \Lambda^{(p,q)}(V)$ (the $(p,q)$-forms on a complex vector space $V$) by a linear expression in a Hermitian form $H$ on $V$ itself. This would amount to finding a linear map $$L: \Lambda^{(1,1)}(V)^{\mathbb{R}} \longrightarrow \Lambda^{(1,1)}(W)^{\mathbb{R}}\subset \Lambda^{(p,q)}(V^*)\otimes \Lambda^{(q,p)}(V^*)$$ that is equivariant with respect to the action of $\mathrm{GL}(V)$. However, $\Lambda^{(1,1)}(V)^{\mathbb{R}}$ is an irreducible $\mathrm{GL}(V)$-module, and it is easy to see that it doesn't occur as a constituent of $\Lambda^{(p,q)}(V^*)\otimes \Lambda^{(q,p)}(V^*)$, so there is no $\mathrm{GL}(V)$-equivariant mapping like this, other than $L=0$. On the other hand, once you fix a background nondegenerate Hermitian form $K$ on $V$, then you can ask whether there is such an injective mapping $L$ that is equivariant with respect to $\mathrm{U}(V,K)\subset\mathrm{GL}(V)$, the $K$-unitary transformations of $V$, and the answer to this is 'yes'.

A similar analysis applies in the real case to show that you actually have to use a 'background' inner product on the vector space $V$ to get $Q^k$ to be well-defined.

• Sorry, I didn't see this response until now! This is very helpful, but it will take a few days to digest. Your answer shows that I need a background Hermitian form to have my expression make sense. I sort of anticipated this (after all, any sort of higher derivative data really needs a metric or at least a connection hanging around to make sense). Could you speak to the importance of this particular form? I.e. given a nondegenerate hermitian form $K$, and another hermitian form $H$, then I want an abstract characterization of $H^k$ acting on $\Lambda^{(p,q)}$. – Steven Gubkin Feb 19 '15 at 17:16
• Also, I have asked a number of questions which are sort of similar to this on this site, and you have always given me excellent answers. Thank you! I really appreciate them. – Steven Gubkin Feb 19 '15 at 17:18