4
$\begingroup$

Does a symbolically elliptic sequence of differential operators have an analytic index? cohomology? For example, is there any concrete meaning of the Todd genus of an almost complex manifold in terms of the Dolbeault operator?

Consider a smooth manifold $M$ with vectors bundles $E_0$, $E_1$ and a linear differential operator $D\colon C^\infty(E_0)\to C^\infty(E_1)$ on their sections. The (principal) symbol $\sigma(D)$ of the degree $n$ operator is the leading term, ie, considered modulo operators of lower degrees. It can be interpreted as a vector bundle map from $E_0$ to $E_1$, not on $M$, but on its cotangent bundle, $\sigma(D)\in\Gamma(T^*M; Hom(E_0,E_1))$. Varying over the cotangent space at a particular point of $M$, it is homogeneous of degree $n$. Thus the symbol is zero along the zero section $M\subset T^*M$. The operator $D$ is elliptic if the symbol is an isomorphism of bundles away from the zero section.

An elliptic operator on a closed manifold is Fredholm, with finite dimensional kernel and cokernel. The index of a Fredholm operator is the difference between the dimensions of the kernel and cokernel. The index is a robust invariant. The index of a continuous family of Fredholm operators is constant. Two operators with the same symbol can be linearly interpolated, so they have the same index, so the index depends only on the symbol. Moreover, homotopy classes of elliptic symbols have constant indices (where a homotopy is a path of symbols all of which are elliptic, ie, isomorphisms). Determining the index from the homotopical data of the symbol is Gelfand’s index problem, solved by Atiyah—Singer, but that is not directly relevant to this question. For the purpose of this question, when I say analytic index I mean something concretely defined in terms of the operator, not produced by making arbitrary choices to introduce auxiliary operators.

Just as an exact chain complex generalizes an isomorphism, so does an elliptic complex generalize an elliptic operator. Let me define a composable sequence of differential operators $d_0,d_1\colon C^\infty(E_0)\to C^\infty(E_1)\to C^\infty(E_2)$ to be symbolically elliptic if their symbols form a chain complex of vector bundles on the cotangent bundle, exact away from the zero section. Is this a good concept?

The usual concept is an elliptic complex. While the ellipticity of an operator depends only on its symbol, for a sequence of composable operators to be an elliptic complex requires that it be symbolically elliptic and the non-symbolic condition that it be a chain complex, ie, that the operators compose to zero $d_1\circ d_0=0$. An elliptic complex, being a chain complex, has cohomology. Under the weaker condition of being symbolically elliptic, the sequence can be rolled up into a single elliptic operator $d_0+d_1^*\colon C^\infty(E_0\oplus E_2)\to C^\infty(E_2)$. Doing this uses the adjoint, defined by metrics on $M$ and metrics on the bundles. The sums of the even and odd cohomologies of an elliptic complex are naturally isomorphic to the kernel and cokernel of the single operator (eg, the Hodge theorem). In particular, they are finite dimensional. Moreover, the Euler characteristic of an elliptic complex is equal to the index of the associated elliptic operator.

A symbolically elliptic sequence of operators has an index that can be defined by making various choices. It can be perturbed to an elliptic complex. Or it can be folded up by metrics. But does the index have concrete analytic meaning about the original operators? In other words: can the kernel and cokernel of the rolled up operator be identified with something independent of choices? More weakly, if $D$ and $D’$ are the elliptic operators produced by different choices of metrics, are their kernels and cokernels canonically isomorphic? Are they even the same dimensions? (In the case of a length 2 complex, they have the same dimensions because the kernels are isomorphic to the kernel of $d_0$ plus the cokernel of $d_2$ and the dimension of the cokernels are fixed by the index.)

Similarly, an anti-self-adjoint elliptic operator $D=-D^*$ on a compact manifold has an (anti-self-adjoint) index in $\mathbb Z/2$, the parity of the dimension of the kernel. This index depends only on the symbol. An operator whose symbol is anti-self-adjoint can be deformed to a truly anti-self-adjoint operator with the same symbol, namely $\frac12(D-D^*)$. Can this index of this operator be interpreted in terms of the original operator?

$\endgroup$
  • $\begingroup$ An example of a symbolically skew-adjoint operator. For a 4$n$+1-dimensional holomorphic spin-c manifold the Dolbeault complex computing $H^*(X;\sqrt K)$ can be rolled up and made skew-adjoint, giving a Dolbeault semi-characteristic. But the Dolbeault operator for any line bundle topologically isomorphic to $\sqrt K$ has the same principle symbol. Does it have the same semi-characteristic $\sum\mathop{\textrm {dim}} H^{2k}(X;L)\in\mathbb Z/2$? Does this index tell you anything about the operator? $\endgroup$ – Ben Wieland Nov 20 '16 at 20:06
  • $\begingroup$ To answer my question, no it does not have the same semi-characteristic. But I think that this fails to be an example of a symbolically skew-adjoint operator. $\endgroup$ – Ben Wieland Nov 21 '16 at 18:04
  • $\begingroup$ And I shouldn't have said spin-c, which is a consequence of holomorphic. I meant with a choice of $\sqrt K$. $\endgroup$ – Ben Wieland Nov 21 '16 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.