Lemma 45.10.4. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $X$ be a nonempty smooth projective scheme over $k$ which is equidimensional of dimension $d$. We have
Proof. Equality on the right. We have $[\Delta ] \cdot [\Delta ] = \Delta _*(\Delta ^![\Delta ])$ (Chow Homology, Lemma 42.62.6). Since $\Delta _*$ preserves degrees of $0$-cycles it suffices to compute the degree of $\Delta ^![\Delta ]$. The class $\Delta ^![\Delta ]$ is given by capping $[\Delta ]$ with the top Chern class of the normal sheaf of $\Delta \subset X \times X$ (Chow Homology, Lemma 42.54.5). Since the conormal sheaf of $\Delta $ is $\Omega _{X/k}$ (Morphisms, Lemma 29.32.7) we see that the normal sheaf is equal to the tangent sheaf $\mathcal{T}_{X/k} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\Omega _{X/k}, \mathcal{O}_ X)$ as desired.
Equality on the left. By Lemma 45.10.3 we have
We have used Lemmas 45.9.6 and 45.9.1. Write $\gamma ([\Delta ]) = \sum e_{i, j} \otimes e'_{2d - i , j}$ as in Lemma 45.9.7. Recalling that $\Delta ^*$ is given by cup product (Remark 45.9.3) we obtain
as desired. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)