Let $L$ be a line bundle on a smooth algebraic variety $X$. Let $\sigma_i:U_i \times \mathbb{C} \to L_{|U_i} $ be its local trivializaations and $u_{ij}$ the transition functions satisfying $\sigma_j=u_{ij}\sigma_i$. Following Welters "Polarized Abelian Varieties and the Heat Equations" (Comp. Math 1983 - Lemma 1.16), let $L \in H^1(X,\Omega_X^1)$ be the class given by the cocyle {$\frac{du_{ij}}{u_{ij}}$} and consider the short exact sequence

$$ 0 \to \mathcal{O}_X \to Diff^{(1)}(L) \to T_X \to 0,\ \ \ \ \ \ (*) $$

where $ Diff^{(1)}(L) $ is the sheaf of 1st order differential operators on $L$ and $T_X$ is the tangent sheaf. The RHS map is just the symbol map of $L$. Welters claims that the coboundary map $\delta: H^0(X,T_X) \to H^1(X,\mathcal{O}_X)$ is given by $- \cup [L]$, i.e. minus the cup product by the class $[L]$.

On the other hand Atiyah (COMPLEX ANALYTIC CONNECTIONS IN FIBRE BUNDLES - 1957, Sect. 4) claims that the extension class of (*), i.e. the Atiyah class of $L$ in $H^1(X,\Omega_X^1)$, is given by $[L]$. But then the coboundary map should be given by $\cup [L]$ so there seems to be a contradiction. Does anybody see where the problem is?

  • 3
    $\begingroup$ The sign of the coboundary map depends very much on the chosen conventions. I am sure that if you analyze thoroughly the definitions in both papers you'll find out where the difference comes from. $\endgroup$ – abx Mar 14 at 2:23
  • $\begingroup$ @abx: yeah, that's what I thought as well. unfortunately, browsinf the paper I have found no trace of such change of convention. Maybe they simply tacitely assume opposite conventions? $\endgroup$ – IMeasy Mar 14 at 16:43

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.