Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I hope someone can help me, although this question is rather specific.

I am reading John Roe's chapter on Getzler symbols in "Elliptic operators, topology and asymptotic methods" to understand the proof of the Atiyah-Singer index theorem.

  • So, for the differential operators on functions of $M$ (say, $D(M)$), there is a symbol map to the space of constant coefficient operators (regarding a cartesion chart) on $T_p M$, named $C(TM)$. (This map seems quite natural, because $C_m(TM)$ is isomorphic to $D_m(M)/D_{m-1}(M)$)
  • By the same principle, there is a symbol map from the Clifford algebra $Cl(TM)$ to the exterior algebra $\Lambda^\bullet TM$.

Now regarding differential operators on the spinor bundle $\Sigma M$, we have $D(\Sigma M)=D(M) \otimes Cl(TM)$.

And so I thought:

Looking at the above, I would consider the symbol map from $D(\Sigma M)$ to sections of $C(TM) \otimes \Lambda^\bullet TM$. But that is not the one constructed in the Getzer calculous.

Rather, the symbol map used there maps $D(\Sigma M)$ to sections of $P(TM) \otimes \Lambda^\bullet TM$, where $P(TM) = \mathbb{C}[TM]\otimes C(TM)$, the space of constant coefficient operators on $TM$ with polynomial coefficients.

And here is my question:

Why is that? I guess, this construction is just the ingenuity of the whole construction. But can someone give me a motivation? What does this symbol map do, what the one I would have thought of cannot do?

share|improve this question
    
Just for interest, if someone stumbles across this question later: The symbol map defined in Roe's book mentioned above is flawed in the sense that it does not have the properties needed for the proof. This can be fixed, however, and also in a quite natural way. So my question basically came up because of the flaw in the book. –  Kofi Nov 25 '11 at 21:42

1 Answer 1

up vote 3 down vote accepted

Actually you can imitate the Getzler calculus using your symbol map $D(\Sigma M) \to C(TM) \otimes \bigwedge TM$ and you will still prove an interesting theorem, just not the Atiyah-Singer index theorem. In fact the theorem you will prove is basically Weyl's asymptotic formula for the eigenvalues of the Laplacian. The problem is that your symbol map doesn't use the Clifford module structure of the fibers of the spinor bundle in any essential way.

The natural way to fix this problem is to change the symbol map so that Clifford multiplication by a tangent vector becomes a first order operator instead of a 0th order operator, and indeed this is a property possessed by the Getzler symbol. Note that this gives the Dirac operator $D = \sum_i c(\partial_i) \partial_i$ Gezler order 2. Since the local approach to the Atiyah-Singer index theorem is based on the heat equation (due to the McKean-Singer formula), it is crucial that $D^2$ also have Getzler order 2. It's not obvious how to construct a symbol calculus in which both $D$ and $D^2$ have order 2, and achieving this motivates many of the difficult aspects of the Getzler calculus.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.