Let $P:\Gamma(E)\rightarrow\Gamma(F)$ be an elliptic partial differential operator, with index $=0$ and closed image of codimension $=1$, between spaces $\Gamma(E)$ and $\Gamma(F)$ of smooth sections of vector bundles $E\rightarrow M$ and $F\rightarrow M$ on a compact Riemannian manifold $(M,g)$ without boundary.

Question: What is the elliptic operator's associated Green's operator?

More concretely, $(M,g)=(\mathbb{S}^m,g)$ be the unit $m$-sphere with constant curvature =1 metric $g$, so $\text{Ricc}(g)=g$. Also let $E=F=S^2\mathbb{S}^m$, the space of $2$-covariant tensors on $\mathbb{S}^m$. Consider the operator: \begin{align*} P:\Gamma(S^2\mathbb{S}^m)&\rightarrow\Gamma(S^2\mathbb{S}^m)\\ h&\mapsto Ph_{ij}=\frac{1}{2}g^{kl}(\nabla_i\nabla_jh_{kl}+\nabla_k\nabla_lh_{ij}). \end{align*} The symbol is: $$ \sigma_P(\xi)h_{ij}=\frac{1}{2}g^{kl}(\xi_i\xi_jh_{kl}+\xi_k\xi_lh_{ij}). $$ We can show that $P$ is elliptic with index $=0$ and closed image of codimension $=1$.

Question: What is the Green's operator of $P$?

Reference request: A good reference on Green's operators for elliptic partial differential operators would be welcomed.

Addition: There is mention of Green's operator on pages 157 & 158 of Hamilton's 1982 paper The Inverse Function Theorem of Nash and Moser. In the context of the above, it goes roughly as follows:

Choose finite dimensional vector spaces $N$ and $M$ and continuous linear maps $$ j:\Gamma(S^2\mathbb{S}^m)\rightarrow N\qquad\text{and}\qquad i:M\rightarrow\Gamma(S^2\mathbb{S}^m). $$ Define another map \begin{align*} L:\Gamma(S^2\mathbb{S}^m)\times M&\rightarrow\Gamma(S^2\mathbb{S}^m)\times N\\ (h,x)&\mapsto L(h,x)=(Ph+ix,jh), \end{align*} which is required to be invertible. Then Theorem 3.3.3. states that:

  1. The inverse map $$ L^{-1}:\Gamma(S^2\mathbb{S}^m)\times N\rightarrow\Gamma(S^2\mathbb{S}^m)\times M $$ is a smooth, tame, and linear.
  2. For each $k\in\Gamma(S^2\mathbb{S}^m)$ there is a unique $h\in\text{kernal}\,j$ such that $Ph-k\in\text{image}\,i$. The resulting map \begin{align*} G:\Gamma(S^2\mathbb{S}^m)&\rightarrow\text{kernal}\,j\\ k&\mapsto Gk=h \end{align*} is smooth, tame, and linear, and is called the Green's operator of $P$.

New question: I'm wondering how to adapt this construction to the map $P$ above. What would the finite-dimensional vector spaces $N$ and $M$, and the maps $j$ and $i$, be to make this construction work?

  • $\begingroup$ Your additional question is substantially different from the original, since you seem to be interested in the Nash-Moser inverse function theorem now. I suggest you ask it as a separate question. Also, you haven't made it clear how you intend to interpret the inverse (aka the Green operator) or a non injective/surjective differential operator. $\endgroup$ Dec 8, 2014 at 9:11
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    $\begingroup$ Basically I'm trying to verify a step in a paper by Hamilton, which essentially says "use the Green's operator" of the map $P$ above. One can only assume that by the Green's operator of $P$, Hamilton means the definition in his paper in the Nash-Moser theorem, which is written at about the same time as the paper I'm reading. $\endgroup$
    – mdg
    Dec 8, 2014 at 9:17
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    $\begingroup$ So I think the second question is the same as the first - I'd like to construct the Green's operator of $P$. $\endgroup$
    – mdg
    Dec 8, 2014 at 9:46
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    $\begingroup$ If you use the inclusion $i \colon \operatorname{ker} P \to \Gamma(\cdots)$ and the orthogonal projection $j \colon \Gamma(\cdots) \to \operatorname{coker} P$, then $G$ is precisely the "pseudoinverse" that I mentioned in my answer. Other choices of $i, j, M$ and $N$ simply parametrize the many different ways one can define an "inverse" of a non-invertible linear operator. That last issue has nothing in particular to do with elliptic operators or functional analysis. It is an issue of ordinary linear algebra. $\endgroup$ Dec 8, 2014 at 11:03
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    $\begingroup$ An implicit assumption in the question is that the base manifold $M$ should be compact, otherwise the operator cannot be Fredholm. Moreover, one only gets the tame estimates needed for Nash-Moser under this assumption. It is satisfied in the examples given, but it's not explicitly stated. $\endgroup$ Dec 8, 2014 at 17:39

4 Answers 4


If it exists, the inverse of an elliptic operator $P$ is its Green's operator. In general, an inverse does not exist, but a parametrix does. A parametrix is an operator $Q$ such that $PQ-I$ and $QP-I$ are compact operators. (I am assuming that $P$ is an elliptic operator on a closed manifold and acts between sections of vector bundles.) The existence of a parametrix is a basic application of pseudodifferential calculus, but it can also be deduced from elliptic estimates.

The existence of a parametrix implies that $P$ is Fredholm, thus the kernel and cokernel of $P$ are finite-dimensional. Using this, one can make $P$ become part of an invertible operator $$ \begin{pmatrix} P& R_-\\ R_+& 0\end{pmatrix}^{-1}= \begin{pmatrix}E&E_+\\ E_-&E_{-+}\end{pmatrix}, $$ where $R_-$ has finite-dimensional domain and $R_+$ finite-dimensional range. This setup is presented, with many useful applications given, by Sjöstrand and Zworski in their paper Elementary linear algebra for advanced spectral problems. See, in particular, the section 2.4 on analytic Fredholm theory. The operator $P$ is invertible if and only if the finite-dimensional operator $E_{-+}$ is, and $P^{-1}=E-E_+ E_{-+}^{-1}E_-$. In the context of Theorem 3.3.3 of Hamilton's paper on the inverse function theorem set $P=L(f)$, $R_+=j$, and $R_-=i$. He calls $G(f)=E$ the Green's operator, which it is when he is allowed to ``forget'' his spaces $M$ and $N$. The finite-dimensional spaces $M$ and $N$ can be chosen as, respectively, the cokernel and the kernel of $P$. It may be useful to choose them larger if it is desired to have them idependent of $f$ in $P=L(f)$.

Decomposing with respect to kernel and cokernel is standard in Fredholm theory. In Deane Yang's answer the decomposition is interior relative to the spaces while here, and in Hamilton's paper, it is exterior.

  • $\begingroup$ Nice reference! Among other things, it also discusses the pseudoinverse. $\endgroup$ Dec 9, 2014 at 12:27

Start with the linear problem: Let $K \subset \Gamma(E)$ denote the kernel of $P$ and $\hat{K} \subset \Gamma(F)$ the cokernel. Assume that there are inner products on the vector bundles $E$ and $F$ and let $K^\perp$ and $\hat{K}^\perp$ the orthogonal complements with respect to the induced $L^2$ inner product on $\Gamma(E)$ and $\Gamma(F)$.

Then the operator $\hat{P} = \hat\pi\circ P$ restricted to $K^\perp$ is 1-1 and onto. So there is an inverse operator $\hat{Q}: \hat{K}^\perp \rightarrow K^\perp$. It's easily confirmed that $\hat{Q}$ is linear. Extend it to a linear map $Q: \Gamma(F) \rightarrow \Gamma(E)$.

It is now straightforward to use a priori elliptic estimates (for example, using pseudodifferential operators to construct a parametrix as described by Igor Khavkine) to establish that $Q$ is a bounded operator between the appropriate Sobolev spaces. So $Q$ is the Green's operator.

Next, the smooth tame estimates, where $P$ and the inner products on $E$ and $F$ are defined in terms of a Riemannian metric. Here, things gets trickier. You can get these estimates from the pseudodifferential approach by estimating the constant in the a priori estimates in terms of the symbol of the operator. This is described in a different setting (strictly hyperbolic systems) in a Duke Math Journal paper by Bryant-Griffiths-Yang. It is however better, if possible, to find a proof of the a priori estimates without using pseudodifferential operators. My favorite approach (I'm sure there are others) is to use Moser iteration (not the same as Nash-Moser iteration). I'm sure there's an explanation of how to get smooth tame estimates out of this, but I'm not sure where. Maybe in Hamilton's original Ricci curvature paper. Or his even older harmonic maps paper.

  • $\begingroup$ But why do you need Nash-Moser? Usually for a nonlinear elliptic PDE, the standard Banach space implicit function theorem suffices. So smooth tame estimates are not needed at all. Just the usual a priori estimates for the linear elliptic PDE. $\endgroup$
    – Deane Yang
    Dec 8, 2014 at 17:45

I believe that the construction of the Green operator $G$ for $P$ is standard and involves two important steps. First, one has to construct a parametrix $Q$ for $P$ and then find a correction term that turns it into an actual inverse. In your case, the operator $P$ seems to be non-invertible, because of the non-trivial kernel and co-kernel. So you need to be explicit about the kind of "inverse" you want to consider.

A parametrix $Q$ can be constructed as a pseudodifferential operator with principal symbol equal to the point-wise inverse of the principal symbol $P$. A parametrix by definition satisfies a relation of the form $QP = 1 - R$ and $PQ = 1 - R'$, where $R$ and $R'$ are smoothing operators. Then, in the case that $P$ is invertible, one can define $G = (1-R)^{-1}Q$ or equivalently $G = Q(1-R')^{-1}$, where the inverses are defined by their Neumann series, e.g., $(1-R)^{-1} = 1 + R + R^2 + R^3 + \cdots$. The idea is that a smoothing operator can be considered a small perturbation of the identity operator, so that the Neumann series always converges. In the non-invertible case, a pseudo inverse can be defined by orthogonally projecting out the kernel from the domain and the cokernel from the codomain of $P$.

There are many sources to read about pseudo-differential operators on manifolds and parametrices of elliptic operators. I'm not an expert, so I don't know what the best one is. Here's a relevant hit from Google: lecture notes on Linear Analysis on Manifolds by P. Albin. See in particular Theorem 22 and so on.


This is not an answer but a long comment. There is the following result which does not cover your example. Let $M$ be a compact smooth manifold. Let $P$ be a linear differential second order elliptic operator with smooth coefficients on functions on $M$. Then there exists a Green function $G(x,y)$ on $M\times M$ such that

1) $G$ is smooth outside of the diagonal.

2) $G$ is bounded from below.

More precise statement and indication of the proof can be found in the Appendix to this paper http://arxiv.org/PS_cache/arxiv/pdf/1111/1111.0403v2.pdf

Closely related question was also asked on MO here Estimates on the Green function of an elliptic second order differential operator.


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