Let $(M,g)$ be a Riemannian manifold of dimension at least $4$.

We consider the differential operator $$D:\Gamma(TM)\to \Gamma (TM)$$

with $$D(X)=\nabla \circ Div(X)$$.

The principal symbol $\delta (D): \pi^* TM \to \pi^* TM$, as a bundle morphism, has constant fiber wise rank $1$ where $\pi$ is the natural projection from the unit cotangent bundle $S^*(M)$ to $M$. So the image of this bundle morphism is a line bundle over $S^*(M)$. The first Chern class of the complexification of this line bundle determines a cohomology class in $H^2(S^*M)\simeq H^2 (M)$. So we obtain a cohomology class $\lambda(M) \in H^2 (M)$.

Does this cohomology class have an alternative formulation?Is there a name for this cohomology class? In dimension $4$, does $\int_M \lambda(M) \wedge \lambda (M) $ depend on the Riemanian metric $g$ on $M$?

The motivation for this question is the following:

I was thinking to find a concept weaker than ellipticity of differential operators. Then I consider the condition that the symbol has constant rank for all fibers.(The full rank is the same as elliptic concept). The only example I found was the above differential operator.

What are some other examples of this type:i.e the symbol of our differential operator is not invertible but has constant fiber wise rank? Are there some theories which consider this generalized elliptic operators?

  • $\begingroup$ How do you get the isomorphism $H^2(SM)\simeq H^2(M)$? $\endgroup$ – Omar Sep 14 '17 at 18:19
  • $\begingroup$ @Omar If I would be correct, the isomorphism comes from $H^*(SM)\simeq H^*(M) \otimes H^*(S^{n-1})$. Now I revise the question assuming that $M$ is at least $4$ dimensional. $\endgroup$ – Ali Taghavi Sep 14 '17 at 18:28
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    $\begingroup$ @AliTaghavi That isomorphism doesn't hold in general eg if M is a 2-sphere. Perhaps you mean to say that the isomorphism $H^2(SM) \simeq H^2(M)$ comes from the Gysin sequence if $n$ is at least $4$. $\endgroup$ – Danny Ruberman Sep 14 '17 at 20:56
  • $\begingroup$ @DannyRuberman Dear Prof. Ruberman. Thank you very much for this very helpfull comment. I was thinking for a possible fiber bundle analogy of Kuneth formula or for the application of Leary Hirsch Theorem for this spher bundle. I was not aware og Gysin sequence. Thanks for informing me of this sequence. $\endgroup$ – Ali Taghavi Sep 15 '17 at 20:54
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    $\begingroup$ @AliTaghavi The Leray Hirsch theorem has a hypothesis that there be a cohomology class on the sphere bu Nell that restricts nontrivially to the sphere fiber. This is more or less the same as saying that the Euler class vanishes. You can find the Gysin sequence I n Milnor -Stasheff; it's the long exact sequence for the pair (E,S) together with the Thom isomorphism theorem. $\endgroup$ – Danny Ruberman Sep 17 '17 at 12:43

Up to sign, the principal symbol of the operator in question is the projection onto the tautological line bundle in $\pi^*TM$. To see this, identify $TM$ with $T^*M$ using the Riemannian metric, and compute the symbol at $w\in S(TM)$. The symbol of the divergence at $w$ is $\langle-,w\rangle$. The symbol of the gradient is $w$ itself. The composition is the projection onto the trivial real line bundle $L\subset\pi^*TM$ over $S(TM)$ that is spanned by $w$ at $w\in S(TM)$. All characteristic classes of $L$ vanish.

A first order with constant rank symbol, extending your example, would be the total differential $d\colon\Omega^\bullet(M)\to\Omega^\bullet(M)$. If $M$ is a surface, it is not hard to see that the symbol class becomes trivial over $S(TM)$.

  • $\begingroup$ Thank you for this very interesting answer. I understand your answer via this example: The linear map$ \begin{pmatrix} { \epsilon_1}^2& \epsilon_1 \epsilon_2 \\ \epsilon_1 \epsilon_2& {\epsilon_2}^2 \end{pmatrix}$ is a (not necessarily orthonormal) idempotent whose image is the span of $(\epsilon_1, \epsilon_2)$, an element of tautological bundle. But just a question: How does a single vector field is an example of the constant rank situation? For example $\partial_x$ with symbol $\epsilon_1$ does not satisfy the rank 1 condition. Right? thanks for introducing CR geometry. $\endgroup$ – Ali Taghavi Sep 19 '17 at 14:39
  • $\begingroup$ You are right. Maybe the second part of my answer is rubbish. $\endgroup$ – Sebastian Goette Sep 19 '17 at 16:49
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    $\begingroup$ Note that in classical index theory, the symbol is used to create a compactly supported symbol class on $T^*M$. Compact support is important to treat the symbol in $K$-theory. But for this, one needs the symbol to be invertible outside a compact subset in $T^*M$. For your operators, you would need in addition a way to deal with kernel and cokernel of the symbol in order to define a topological index. $\endgroup$ – Sebastian Goette Sep 20 '17 at 6:36

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