# Determinant of an exact sequence sheafs

• Sep 30th 2008, 01:26 PM
cduston
Determinant of an exact sequence sheafs
Hey everyone, I'm studying the Adjunction formula and part of the proof is confusing me. I have two exact sequences on a subvariety Y of codimension 1 in variety X:

$0\to\mathcal{I}_Y\to\mathcal{O}_X\to\mathcal{O}_Y\ to 0$

where the $\mathcal{O}_X$ are structure sheafs and (I think) $\mathcal{O}_{X}(Y)$ are divisorial sheafs. We also have

$0\to T_Y \to T_{X}\mid Y \to N_{X\mid Y}\to 0$

Where these are tangent and normal bundles. The next line says "Taking determinants gives:"

$det T_X \mid Y=det T_Y\otimes N_{X\mid Y}$

which is the Adjunction formula. So can someone give me some insight into how one thinks about taking the determinants of something like an exact sequence?

Thanks!
• Sep 30th 2008, 07:53 PM
NonCommAlg
Quote:

Originally Posted by cduston
Hey everyone, I'm studying the Adjunction formula and part of the proof is confusing me. I have two exact sequences on a subvariety Y of codimension 1 in variety X:

$0\to\mathcal{I}_Y\to\mathcal{O}_X\to\mathcal{O}_Y\ to 0$

where the $\mathcal{O}_X$ are structure sheafs and (I think) $\mathcal{O}_{X}(Y)$ are divisorial sheafs. We also have

$0\to T_Y \to T_{X}\mid Y \to N_{X\mid Y}\to 0$

Where these are tangent and normal bundles. The next line says "Taking determinants gives:"

$det T_X \mid Y=det T_Y\otimes N_{X\mid Y}$

which is the Adjunction formula. So can someone give me some insight into how one thinks about taking the determinants of something like an exact sequence?

Thanks!

determinant here probably means the top exterior power, i.e. if $V$ is a vector space of dimension $n,$ then: $\det V = \bigwedge^n V.$ the reason that i think this might be the case is that

we know that if we have an exact sequence of vector spaces $0 \longrightarrow U \longrightarrow V \longrightarrow W \longrightarrow 0,$ where $\dim U=m, \ \dim W = n,$ then $\bigwedge^{m+n}V=\bigwedge^mU \otimes \bigwedge^n W.$ see if this helps!