intersection number - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:35:56Z http://mathoverflow.net/feeds/question/71067 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71067/intersection-number intersection number Enchanted 2011-07-23T13:58:15Z 2011-07-23T21:58:16Z <p>I vaguely recall the following fact that I'd like to use in my research. It should be easy to see that this holds (if it does) but I can't seem to prove it.</p> <p>Let $p:X\longrightarrow S$ be a (regular) arithmetic surface over a Dedekind scheme $S$.</p> <p>Let $P:S\longrightarrow X$ be a section and let $\omega$ be a non-zero rational section of $\omega_{X/S}$. Let $K_{X} = \mathrm{div}(\omega)$ be the canonical divisor defined by $\omega$. (A better notation for $K_X$ would be $K_{X/S}$, maybe.)</p> <p>By definition, the intersection number $(K_X,P)$ is defined as $$\sum_{s } i_s (K_{X}, P) \log( \mathrm{card}( k (s)) ),$$ where the sum runs over the closed points of $S$ and $$i_s(K_{X},P) = \sum_{x} i_x(K_{X}, P) [k(x):k(s)],$$ where the sum runs over the closed points of $X_s$ and $i_x$ denotes the intersection number at $x$.</p> <p>Now, I wonder if the following equality is trivial to see.</p> <p>Write $\omega = df$ for some rational function $f\in K(X)$. (We assume this to be possible.) Do we have that </p> <p>$$(K_{X}, P) = \sum_{s} \mathrm{ord}_s(P^\ast\omega) \log(\mathrm{card}(k(s)))?$$</p> <p>To see this, it suffices to prove the following equality: $$\mathrm{ord}_s(P^\ast \omega) =i_s (K_{X}, P) .$$ Does this hold?</p> http://mathoverflow.net/questions/71067/intersection-number/71076#71076 Answer by Qing Liu for intersection number Qing Liu 2011-07-23T18:48:16Z 2011-07-23T21:58:16Z <p>This is correct if $P(S)$ is not contained in the support of $\mathrm{div}(\omega)$. It comes essentially from the definition of $i_x(K_X, P)$. You don't need $\omega$ to be an exact differential from. However the intersection number depends on the choice of $\omega$ (as well as the Weil divisor $K_X$). You can check this by yourself by multiplying $\omega$ by a non-zero constant in $K(S)$ and see the effect on the total intersection number. If $P(S)$ is contained in the support of $K_X$, then you can't define $i_x(K_X, P)$. </p> <p><b>EDIT</b>. Let me add some more details. Denote by $K(X)$ the field of rational functions on $X$, viewed as a constant sheaf on $X$. Then $\omega\in \omega_{X/S}\otimes K(X)$. Hence $\omega\cdot\omega_{X/S}^{\vee}$ is a subsheaf of $K(X)$, hence equal (not only isomorphic) to $O_X(-D)$ for some Cartier divisor $D$ on $X$. We have $$\omega_{X/S}=\omega\cdot O_X(D).$$ A straightforward local computation shows that $\mathrm{div}(\omega)=D$ as Cartier divisors. Let us identify $P$ with $P(S)$. Let $I\subset O_X$ be the ideal sheaf defining $P$ in $X$. As $P$ is not contained in the support of $D$, $D|_P$ is a well defined Cartier divisor on $P$. Namely, if a local equation of $D$ at some point $x\in P$ is given by $f_x\in K(X)$, then we can write $f_x=a/b$ with $a, b\notin I_x$ (here we use the fact that $O_{X,x}$ is a UFD). Then a local equation of $D$ restricted to $P$ is $\bar{a}/\bar{b}$ where $\bar{c}$ means the image of $c$ in $O_{X,x}/I$. </p> <p>The above equality restricted to $P$ reads $$P^{*}(\omega_{X/S})=P^{*}(\omega) \cdot O_P(D|_P).$$ So $P^{*}(\omega)$ is a rational section of </p> <p>$P^{*}\omega_{X/S}$ and its divisor on $S$ is $D|_P$. </p> <p>To get an intersection number independent of the choice of a rational section $\omega$, you have to use Arakelov intersection theory. </p>