# the inverse of determinant line bundle?

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?

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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
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

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!
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