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Let $\pi:E\to M$ be a smooth vector bundle over a smooth manifold, with $\text{rank}(E)=\text{dim}(M)$. For a section $\sigma$ of $E$ with a zero at $p\in M$, define the degree of the zero at $p$ to be the topological degree of the induced map from a small sphere in $T_pM$ to a small sphere in $E_p$.

One motivation for studying degrees of zeros is that they contain information about the topology of $E$. I think the following is true, although I couldn't find a good reference:

Theorem 1 (Hopf index theorem). Suppose the zeroes of $\sigma$ are the isolated points $p_1, \ldots p_k$, with degrees $d_1,\ldots d_k$ respectively. Then the Euler class of $E$ is $\chi(E)=\sum_{i=1}^kd_i$.

With this as motivation, my first question, the one stated in the title, is roughly (see the Example for an idea of what I'm getting at, and feel free to suggest a sharper version):

Are there conditions on [say, the symbol of] a linear differential operator $D:E\to F$, such that [some constraint] is satisfied by degree of any zero $p\in M$ of any local solution $\sigma\in\Gamma(E)$ to the PDE $D\sigma=0$?

Example. If $M$ is a Riemann surface and $E$ a holomorphic line bundle over it, the kernel of the delbar operator $\overline{\partial}:E\to T^{0,1}M\otimes E$ is precisely the holomorphic sections of $E$. By complex analysis, zeroes of holomorphic functions have positive degree.

Theorem 1 then yields the standard result that if a line bundle admits a global holomorphic section then its Euler class (aka first Chern class) is nonnegative.


Here's an idea I had for trying to prove a theorem of the sort I ask for in Question 1. Recall the definition of the local ring of a zero $p\in M$ of a section of E:

Write $\mathcal{O}_p$ for the ring of germs of smooth functions about $p$.

Definition. Let $\sigma\in\Gamma(E)$ be a smooth section which vanishes at $p$. The local ring of the germ $[\sigma]_p$, denoted $Q([\sigma]_p)$, is the quotient $\mathcal{O}_p/([\sigma]_p)$, where $([\sigma]_p)$ is the ideal of $\mathcal{O}_p$ generated by "components of $\sigma$": $([\sigma]_p)=\ <\{[v(\sigma)]_p:v\text{ a nonvanishing section of }E^*\}> \ \subseteq \mathcal{O}_p$.

Theorem 2 (Eisenbud-Levine-Khimshiashvili). Suppose $p$ is a zero of $\sigma$, and the local ring $Q([\sigma]_p)$ is a finite-dimensional algebra over $\mathbb{R}$. Then there is a canonical quadratic form on $Q([\sigma]_p)$, such that the degree of the zero of $\sigma$ at $p$ can be calculated as this quadratic form's signature.

Because a system of PDE is precisely a constraint on the local behaviour of a section, it seems plausible that local rings of zeros of solutions of a PDE might have interesting properties.

Are there conditions on [say, the symbol of] a linear differential operator $D:E\to F$, such that [some constraint] is satisfied by the signature of the local ring $Q([\sigma]_p)$ of any zero $p\in M$ of any local solution $\sigma\in\Gamma(E)$ to the PDE $D\sigma=0$?

Example. As in the previous example, let $E$ be a holomorphic line bundle over a Riemann surface $M$. By manipulating the Cauchy-Riemann equations, one can (I think!) classify the possible local rings of zeroes of a holomorphic section, and show that all of them have positive signature.

Theorem 2 then yields an alternative proof of the quoted result that zeroes of holomorphic functions have positive degree.

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I'll just remark that your 'Theorem 1' is a classical result; I think (I don't have a copy with me so that I can check) that it is mentioned in Milnor and Stasheff ("Characteristic classes"), where they define the Euler class as a obstruction to finding a nonzero global section. (By the way, you need both $M$ and the vector bundle $E$ to be oriented before you can define the local degree of an isolated zero of a section.) I sketched a proof in my answer to mathoverflow.net/questions/84521 . –  Robert Bryant Mar 14 '12 at 11:44
    
Thanks, yes. I'll fix it if I edit the question again. –  macbeth Mar 14 '12 at 12:23
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1 Answer

This question is a bit too general, and I think that it goes beyond the principal symbol.There clearly are constraints. Take for example the Laplace operator $\Delta$ acting on sections of the trivial complex line bundle on the round sphere $S^2$. The sections of this bundle are smooth complex valued functions functions and the functions in the kernel are constant.

Now replace this operator with the operator

$$\Delta_n=\Delta-n(n+1),$$

where $n$ is a nonnegative integer. The operator $\Delta_n$ has the same principal symbol as $\Delta$. The quantity $n(n+1)$ is an eigenvalue of $\Delta$ and the eigenfunctions are the restrictions to $S^2$ of the degree $n$ homogeneous polynomials (with complex coefficients) in $3$-variables. The behavior of such a polynomial $P(x,y,z)$ in a neighborhood of North Pole on the $2$-sphere is equivalent to the behavior near zero of the polynomial map

$$\mathbb{R}^2\ni (s,t) \mapsto P(s,t,1)\in\mathbb{C}$$

Above, the affine plane $(s,t)\mapsto (s,t,1)\in\mathbb{R}^3$ is the (affine) tangent plane to the sphere at the North pole. The real and imaginary parts of this polynomial map can be any polynomials of degree $\leq n$.

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Thanks for the example! I think you are addressing a global rather than a local version of my question, but this can easily be modified. Germs of sections in the kernel of $\Delta$ are complex functions whose real and imaginary parts are harmonic. Germs of sections in the kernel of $\Delta -n(n+1)$ are -- well, I'm not sure; some infinite-dimensional space; but anyway, your answer shows that their behaviour to $n$-th order is arbitrary. Thus the sets of isomorphism classes of local rings of germs of solutions to these PDE are (presumably) not the same. –  macbeth Mar 20 '12 at 18:59
    
By the way, can you suggest a better criterion than the principal symbol? –  macbeth Mar 20 '12 at 19:03
    
You can view a partial differential equation of order $N$ as imposing a linear constraints on the $N$-th jet. Thus you need to take in consideration the full symbol. In any case it seems very hard to predict what kinds of constraints this induces on the Eisenbud-Khimshiashvili pairing. –  Liviu Nicolaescu Mar 20 '12 at 20:03
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