Assume that there exists a first order elliptic operator $D$ acting on functions from $\mathbb{R}^n$ to some vector space $V$. What can we conclude about $V$?

For example, is the dimension of $V$ always greater than $[n/2]$? Is $D$ necessarily some kind of modified Dirac operator? Can we somehow construct a pointwise Clifford module structure on $V$?

To sum it up, I would like to know if there is any truth to the statement "all first order elliptic operators are basically Dirac operators".


2 Answers 2


Here is a simple way of producing first order elliptic operators $\newcommand{\bR}{\mathbb{R}}$ $C^\infty(\mathbb{R}^n, W)\to C^\infty(\bR^n, W)$ with constant coefficients. Denote by $L(W)$ the space of linear operators $W\to W$. Consider a map $\newcommand{\si}{\sigma}$

$$\si :\bR^n\to L(W), $$

such that $\si(x)$ is invertible for any $x\in\mathbb{R}^n\setminus 0$. Denote by $e_1,\dotsc, e_n$ the canonical basis of $\bR^n$. Now set $\si_k=\si(e_k)$ and define $\newcommand{\pa}{\partial}$

$$ D=\sum_{k=1}^n \si_k \pa_k: C^\infty(\mathbb{R}^n, W)\to C^\infty(\bR^n, W). $$

This is an elliptic operator because its symbol is the map $\si: \bR^n\to L(W)$, up to a multiplication by $\sqrt{-1}$. Let me point two things.

  • All elliptic first order p.d.o.'s with constant coefficients are obtained in this fashion and
  • The symbol at a point $p \in\bR^n$ of any first order p.d.o. $C^\infty(\mathbb{R}^n, W)\to C^\infty(\bR^n, W)$ is a map $\si_p:\bR^n\to L(W)$ with the above property.

The problem is then to find the lowest dimensional vector space $W$ such that there exists a linear map $\si: \bR^n\to L(W)$ with the property that $\si(x)$ is invertible for any $x\in\mathbb{R}^n\setminus 0$.If $W$ is real, the minimal dimension is computed by the so called Radon-Hurwitz numbers. For details, I refer to the book Topological Geometry, by I.R. Porteous. At the end of Chapter 13 of the book you can find a discussion on these numbers and the above problem. In any case, the solution is not simple, it is related to the problem of vector fields on spheres and was elucidated by J.F. Adams using $K$-theoretic techniques. Atiyah has a discussion where he explains that the index theorem for Dirac operators implies the index operators for arbitrary operators. I'll add the reference once I find it. I have posted on MO a similar question myself a while ago.

Note. In the paper Topology of elliptic operators, Proc. Sympo.Pure. Math., vol. 16, Amer. Math. Soc 1970, p 101-119, Atiyah essentially answers your first question in the complex case, again referring to the work of Adams.

In the paper The index of elliptic operators, Colloquium Lectures, Amer. Math. Soc. Dallas, 1973 Atiyah gives a brief explanation why the index theorem for Dirac operators implies the index theorem in general. Essentially, it has to do with the Bott periodicity theorem. You can find both papers in the 3rd volume of Atiyah's Collected Works.

  • $\begingroup$ Why is any elliptic pdo of order 1 of this form? Same problem as in my answer, I would say. $\endgroup$ Nov 20, 2013 at 20:59
  • $\begingroup$ The symbol of the operator, up to a factor of $\sqrt{-1}$ is $\si(\xi)=\sum_k\xi_k\sigma_k$. By construction this is invertible if $\xi\neq 0$. Also, it need not be a Dirac operator. $\endgroup$ Nov 20, 2013 at 21:08

Yes. Let $D$ be a first order elliptic operator from $V$-valued functions to $W$-valued functions. Consider the operator

$$ P = \begin{pmatrix} 0 & D^{\ast} \\ D & 0 \end{pmatrix} $$

acting on the vector bundle $V \oplus W$. The operators $P$ is formally selfadjoint and graded. Let $p(x, \xi)$ be the principal symbol, which is selfadjoint in the sense that $p(x,\xi)^{\ast}=p(x,\xi)$. Consider the square of the symbol $p(x,\xi)^2$, which is selfadjoint and positive definite (since it is the square of a selfadjoint linear map). Thus $p(x,\xi)^2 = q(x,\xi)$, with $q$ quadratic and positive definite in $\xi$. Define a new Riemann metric on $\mathbb{R}^n$ by $|v|^2 := q(x,v^{\sharp})$, when $v \in T_x \mathbb{R}^n$ and $\sharp$ is the musical isomorphism. You find out that the symbol of $P$ defines a Clifford structure on the trivial bundle $V \oplus W$, but with a possibly different metric on $\mathbb{R}^n$ [Here is a gap, see the comments below]. Then $P$ is a Dirac type operator, except that it might not be selfadjoint with respect to the new metric, but at least its symbol is self-adjoint, and so $P$ is selfadjoint up to order $0$ operators.

In particular, the vector space $V \oplus W$ will have the structure of a graded Clifford module for $Cl^n$. Or, $V$ will be an ungraded module over $Cl^{n-1}$. This puts restrictions on $\dim (V)=\dim (W)$. If $n=2m+1$ is odd, then $Cl^{2m} = Mat(2^m, \mathbb{C})$, and $\dim (V)$ is a multiple of $2^m$. If $n=2m$ is even, then $Cl^{2m-1} = Mat(2^{m-1}, \mathbb{C})\oplus Mat(2^{m-1}, \mathbb{C})$, and $\dim (V)$ is a multiple of $2^{m-1}$.

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    $\begingroup$ But, $p^2=q$ is a priori an endomorphism of $V\oplus W$ that doesn't have to be skalar and does not define a Riemannian metric in general. Or am I missing something? $\endgroup$ Nov 20, 2013 at 19:59
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    $\begingroup$ The above construction has a problem that Kofi pointed out, namely $p(x,\xi)^2$ is a symmetric, positive definite endomorphism of $W$. $\endgroup$ Nov 20, 2013 at 20:12

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