# Friday, March 05 :::{.remark} Recall that we set up a differential complex, whose objects were vector bundles and differentials were differential operators (i.e. linear combinations of partial derivatives) in local trivializations. We pulled back to tangent bundles (?) and defined the *symbol* of an operator, and saw that when taking the symbol complex of the deRham complex. the sequence of maps was given by wedging against a tautological one-form. This was an *elliptic complex* because the maps became wedging with a covector. ::: :::{.example title="of an elliptic complex"} Let $X\in \Mfd_\CC$ and $\bundle{E}\to X \in \VectBundle_\CC$ be holomorphic. There is a resolution \[ 0 \to \bundle{E} \mapsvia{i} \bundle{E} \tensor A^{0, 0} \mapsvia{\delbar} \bundle{E} \tensor A^{0, 1} \mapsvia{\delbar} \cdots .\] What is the symbol complex? Consider the projection $\pi: T\dual X\to X$, and use pullbacks to get a sequence \[ 0 \to \pi^* \bundle{E} \tensor A^{0, 0} \mapsvia{\sigma( \delbar )} \pi^* \bundle{E} \tensor A^{0, 1} \mapsvia{\sigma( \delbar )} \cdots .\] Here the symbol \( \sigma(\delbar) \) replace $\dd{}{t \conj{z}_i}$ with the corresponding function on $T\dual X$, say $\conj{y}_i$. Then \( \sigma( \delbar) = \sum_i \conj{y}_i \, d\conj{z}_i \wedge (\wait) = \conj{ \alpha } \wedge (\wait) \). As before, at a point $(p, \alpha)$ where \( \alpha\neq 0 \) in \( T\dual X \), we get \[ 0 \to \bundle{E}_p \mapsvia{\conj{ \alpha} \wedge (\wait)} \bundle{E}_p \tensor \Wedge^{0, 1}_p X \mapsvia{\conj{ \alpha } \wedge (\wait)} \bundle{E}_p \tensor \Wedge^{0, 2} X \to \cdots ,\] which is an exact sequence of vector spaces. So \( ( \bundle{E} \tensor A^{0, p}, \delbar ) \) is an elliptic complex. ::: :::{.slogan} The symbol being exact is approximately the top-order part being nowhere-vanishing. ::: :::{.remark} The next theorem computes the cohomology of an elliptic complex using Chern and Todd classes. ::: :::{.theorem title="Atiyah-Singer Index Theorem"} If \( ( \complex{\bundle{E}}, d) \) is an elliptic complex of smooth vector bundles on a compact oriented $X\in \Mfd^n_\RR$, then \[ \chi(\complex{ \bundle{E} }, d) = \sum (-1)^i \dim \qty{\ker d^i \over \im d^{i-1} } = (-1)^{\dim(X) \choose 2} \int_X {\ch \over \eul}( \complex{\bundle{E}} ) \td(TX \tensor_\RR \CC) .\] ::: :::{.remark} Here we define \( \ch( \complex{\bundle{E}} ) \da \sum_i (-1)^i \ch( \bundle{E}^i ) \). What does it mean to divide by the Euler class? Let \( \ts{ x_i, -x_i } \) be the Chern roots of the complexified tangent bundle \( TX\tensor \CC \), then \( \eul(X) \da \prod x_i \) is the product where we pick one of each of the Chern roots from each of the pairs. The preferred sign to choose is the one for which \( \int_X \prod x_i = \chi_\Top(X) \). Dividing just means to take the Chern character, then if it's divisible by $\prod x_i$, we do so. We have \[ \td(TX\tensor \CC) = \prod_i \qty{x_i \over 1 - e^{-x_i}} \qty{-x_i \over 1 - e^{-x_i}} .\] Thus \[ {\td(TX\tensor \CC) \over \eul(X) } = \prod_i {1\over x_i} \qty{x_i \over 1 - e^{-x_i}} \qty{-x_i \over 1 - e^{-x_i}} ,\] but note that this doesn't necessarily make sense. However, all all computations we'll see, there will be enough cancellation to make this well-defined. ::: :::{.exercise title="Chern character of the de Rham complex"} $\ch( \complex{\Omega}X \tensor \CC) = \prod_i (1-e^{x_i}) (1 - e^{-x_i})$ for $X\in \Mfd_\RR^{2n}$ even dimensional. ::: :::{.example title="?"} Supposing $X\in \Mfd_\RR^2$ is a genus $g$ surface, we have \[ \OO \to \Omega^1\tensor \CC \to \Omega^2 \tensor \CC ,\] and \( \ch(\complex{ \Omega }) = \ch( \OO) - \ch( \Omega^1 \tensor \CC) + \ch(\Omega^2\tensor \CC) \). The Chern roots of $TX \tensor \CC$ are \( \ts{ x_i, -x_i } \), which come in pairs. So \[ \ch( \complex{\Omega} ) = 1 - e^{x_i} - e^{x_i} + e^{-x_i + x_i} = (1 - e^{-x_i})( 1 - e^{x_i} ) .\] From the theorem, we're supposed to have \[ \chi( \complex{\Omega}, d) &= (-1)^{n(n-1) \over 2} \int_X {\prod_i (1 - e^{-x_i})( 1 - e^{x_i} ) \over \prod_{i=1}^n x_i } \prod_i \qty{x_i \over 1 - e^{-x_i}} \qty{-x_i \over 1 - e^{-x_i}} \\ &= (-1)^{n(n-1) \over 2} \int_X \prod_{i=1}^n (-x_i)\\ & = \int_X \prod_i x_i \\ &= \chi_\Top(X) && \text{C-G-B} .\] Letting $d=\dim X = 2n$, we have \[ (-1)^n (-1)^{d(d-1) \over 2} = (-1)^n (-1)^{n(2n-1)} = (-1)^2n = 1 . \] ::: :::{.example title="?"} We can prove HRR using this theorem: we have \[ \chi(X, \bundle{E} ) = \chi( \bundle{E} \tensor A^{0, \wait}, \delbar ) \equalsbecause{\text{ASIT}} \int_X { \ch(\bundle{E} \tensor A^{0, \wait} ) \over \eul(X) } \td(TX \tensor_R \CC) .\] We have $\ch( \bundle{E} \tensor A^{0, \wait} ) = \ch(\bundle{E}) \ch( A^{0, \wait} )$ where \( \ch(A^{0, 1}) = \sum_I (-1)^i \ch(\Wedge^i A^{0, 1} ) \). The Chern roots of - $TX$ are \( \ts{ x_i } \) - $A^{1, 0} = T\dual X$ are \( \ts{ -x_i } \) - $A^{0, 1}$ are \( \ts{ -x_i } \) So we obtain \[ \chi( \bundle{E} ) &= (-1)^n \int_X {\prod ( 1- e^{x_i}) \over \prod x_i} \prod_i \qty{x_i \over 1 - e^{-x_i}} \qty{-x_i \over 1 - e^{-x_i}} \\ &= \int_X \ch( \bundle{E} ) \prod_i {x_i \over 1-e^{-x_i} } \\ &= \int_X \ch( \bundle{E} ) \td(TX) ,\] which is HRR. :::