The Stacks project

Lemma 42.55.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a scheme locally of finite type over $S$. Let $\mathcal{E}$ be a locally free $\mathcal{O}_ X$-module of rank $r$. Then

\[ \prod \nolimits _{n = 0, \ldots , r} c(\wedge ^ n \mathcal{E})^{(-1)^ n} = 1 - (r - 1)! c_ r(\mathcal{E}) + \ldots \]

Proof. By the splitting principle we can turn this into a calculation in the polynomial ring on the Chern roots $x_1, \ldots , x_ r$ of $\mathcal{E}$. See Section 42.43. Observe that

\[ c(\wedge ^ n \mathcal{E}) = \prod \nolimits _{1 \leq i_1 < \ldots < i_ n \leq r} (1 + x_{i_1} + \ldots + x_{i_ n}) \]

Thus the logarithm of the left hand side of the equation in the lemma is

\[ - \sum \nolimits _{p \geq 1} \sum \nolimits _{n = 0}^ r \sum \nolimits _{1 \leq i_1 < \ldots < i_ n \leq r} \frac{(-1)^{p + n}}{p}(x_{i_1} + \ldots + x_{i_ n})^ p \]

Please notice the minus sign in front. However, we have

\[ \sum \nolimits _{p \geq 0} \sum \nolimits _{n = 0}^ r \sum \nolimits _{1 \leq i_1 < \ldots < i_ n \leq r} \frac{(-1)^{p + n}}{p!}(x_{i_1} + \ldots + x_{i_ n})^ p = \prod (1 - e^{-x_ i}) \]

Hence we see that the first nonzero term in our Chern class is in degree $r$ and equal to the predicted value. $\square$


Comments (0)


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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FEE. Beware of the difference between the letter 'O' and the digit '0'.