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Math Help - Determinant of an exact sequence sheafs

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by cduston View Post
    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!
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