At the beginning a word of warning: this would be rather vague question: vague as it is, I'm not requiring a precise answer, rather some intuitive explanation.

In the flat case $M=\mathbb{R}^n$ there are some naturally constructed differential operators: the Laplace operator, the Hodge de-Rham operator $d+d^*$, the signature operator $d+d^*$ (using different grading), Dirac operator etc.

In the general case of arbitrary manifolds Dirac operator does not always exist: the manifold should be $spin^{c}$. On the contrary, for the Laplace operator to make sense no assumption on $M$ is required.

Suppose that I have some differential operator on $\mathbb{R}^n$: is there some way to answer immediately whether this operator makes sense on general manifolds or there will be some topological constraints for defining it on arbitrary manifold?

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    $\begingroup$ Its hard to say something at this level of generality. The way one usually constructs differential operators on manifolds is by bootstrapping. Meaning you first define a connection (preferably torsion free) on the tangent bundle, then pick a principal $G$-bundle and a connection on it. Now any bundle associated to that principal bundle will have a connection and any tensor product of these bundles with the tangent bundle or its relatives will have unique connection from leibniz rule. So now any principal symbol between bundles of the above type lifts uniquely to a differential operator. $\endgroup$ – Saal Hardali Oct 16 '17 at 21:00
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    $\begingroup$ Of course I forgot to say, any differential operator can be obtained as a sum of lifts of homogeneous symbols. $\endgroup$ – Saal Hardali Oct 17 '17 at 6:55
  • $\begingroup$ I do not recall seeing a claim the manifold should be $spin^{c}$. Maybe you meant it should have an associated spinor bundle, etc. $\endgroup$ – Bombyx mori Oct 18 '17 at 2:15
  • $\begingroup$ Bombyx mori I mean the a Dirac operator acting on (sections of) the bundle $S$ with the property that the Clifford multiplication acts irreductively on $S$. $\endgroup$ – truebaran Oct 18 '17 at 7:20
  • $\begingroup$ One way to give your question a more precise meaning is to think about operators which are intrinsically associated to some geometric structure on $M$ (which is true for the examples you mention). Indeed, in these examples, things boil down to the question of whether such a geometric structure can be defined on an arbitrary manifold (which is true for Riemannian structures but not for spin or spin$^c$-structures. Without the naturality involved, it seems very hard to give a reasonable meaning to the phrase "this operator makes sense on general manifolds". $\endgroup$ – Andreas Cap Oct 18 '17 at 9:02

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