Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am reading materials about the determinant defined by Knudsen-Mumford

http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&s1=103495&vfpref=html&r=11&mx-pid=437541

which assigns a graded line bundle to a perfect complex of locally free coherent $\mathcal{O}_X$-modules. Here, a graded line bundle is just a pair $(L,\alpha)$ where $L$ is a line bundle and $\alpha$ is a locally constant function $X\rightarrow \mathbb{Z}$. The tensor product of two graded line bundles are defined to be $$(L,\alpha)\otimes (M,\beta):=(L\otimes M,\alpha+\beta)$$ with an isomorphism $$\phi (L,\alpha)\otimes (M,\beta) \rightarrow (M,\beta)\otimes (L,\alpha) $$

which sends $l\otimes m$ to $(-1)^{\alpha\beta} m\otimes l$. It is said in Soule, Abramovich, Burnol and Kramer, 'Lectures on Arakelov Geometry' Chapter VI Section 1, that we can define $(L,\alpha)^{-1}=(L^{-1},-\alpha)$ (it is slightly different since in the book, $\alpha$ is defined to be mod 2) to be the inverse of $(L,\alpha)$ and the identity is $(\mathcal{O}_X,0)$.

My question is:

is the map from $(L,\alpha)$ tensor its inverse to the identity canonical? If so, how to explain the possible sign appearing in $(L,\alpha)\otimes(L^{-1},-\alpha)\rightarrow(L^{-1},-\alpha)\otimes(L,\alpha)$? And also how to explain the following diagram: $$\xymatrix{ (L,\alpha)\otimes(L^{-1},-\alpha) \ar[dr]\ar[dd]_{(-1)^{\alpha^2}} && (L,\alpha+1)\otimes(L^{-1},-\alpha-1) \ar[dl]\ar[dd]^{(-1)^{(\alpha+1)^2}}\\ &\mathcal(O)_X&\\ (L^{-1},-\alpha)\otimes(L,\alpha) \ar[ur] && (L^{-1},-\alpha-1)\otimes(L,\alpha+1) \ar[ul] }$$ where every arrow is an isomorphism?

share|improve this question
    
sorry for all the mess. I tried hard to make it look correct, but it did not go to the right track. Can someone please help me? –  Yujia Qiu Apr 29 '11 at 17:16
    
The problem seems to come from the ampersands (&). For some reason math including an ampersand seems not to be understood as math, and gets a big box around it. I don't know how to deal with xymatrix. –  Dan Ramras Apr 29 '11 at 17:58
3  
If anyone else edits, I think it will become Community Wiki. Charles, if you feel up to it, you can try removing the xymatrix, and just sticking the contents into a $3 \times 3$ array with arrows in separate cells. –  S. Carnahan Apr 29 '11 at 18:10
add comment

1 Answer

up vote 3 down vote accepted

A Picard groupoid is a symmetric monoidal category $G$ where all morphisms are isomorphisms and such that for any object $x\in G$, the functor $x\otimes-\colon G\rightarrow G$ is an equivalence of categories. Graded line bundles form a Picard groupoid. An inverse object to $x\in G$ is an object $x^\star\in G$ together with an isomorphism $\varphi\colon x\otimes x^\star\rightarrow e$, where $e\in G$ denotes the tensor unit. Every object has an essentially unique inverse object in the sense that, if $\bar{x}^\star$ is another one with $\bar\varphi\colon x\otimes \bar{x}^\star\rightarrow e$ then there is a unique isomorphism $\psi\colon x^\star\rightarrow \bar{x}^\star$ such that $\varphi=\bar{\varphi}(1_x\otimes \psi)$.

Inverse objects are canonical in this sense and they cannot be canonical in any other sense. Actually, you seem concerned about signs arising from the degree of, but even $L^{-1}$ is a choice!

share|improve this answer
    
Fernando, typo: phi bar instead of phi at the place where it is defined. –  Zoran Skoda Apr 30 '11 at 9:01
    
Thank you for you answer, Fernando. This seems to be a good reason. But isn't the dual of a line bundle $L$ canonically chosen to be $\operatorname{Hom}(L,\mathcal{O}_X)$? –  Yujia Qiu May 1 '11 at 17:38
    
@Zoran: Thanks! @Yijia: You may regard that choice as canonical in that context, that's true, but it could be misleading. –  Fernando Muro May 15 '11 at 14:34
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.