the inverse of determinant line bundle? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T09:10:59Z http://mathoverflow.net/feeds/question/63446 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63446/the-inverse-of-determinant-line-bundle the inverse of determinant line bundle? Yujia Qiu 2011-04-29T17:11:35Z 2011-05-15T14:33:04Z <p>I am reading materials about the determinant defined by Knudsen-Mumford</p> <p><a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&amp;s1=103495&amp;vfpref=html&amp;r=11&amp;mx-pid=437541" rel="nofollow">http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&amp;s1=103495&amp;vfpref=html&amp;r=11&amp;mx-pid=437541</a></p> <p>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 <code>$$\phi (L,\alpha)\otimes (M,\beta) \rightarrow (M,\beta)\otimes (L,\alpha)$$</code></p> <p>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 $(\mathcal{O}_X,0)$.</p> <p>My question is:</p> <p>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: $$\xymatrix{ (L,\alpha)\otimes(L^{-1},-\alpha) \ar[dr]\ar[dd]_{(-1)^{\alpha^2}} &amp;&amp; (L,\alpha+1)\otimes(L^{-1},-\alpha-1) \ar[dl]\ar[dd]^{(-1)^{(\alpha+1)^2}}\ &amp;\mathcal(O)_X&amp;\ (L^{-1},-\alpha)\otimes(L,\alpha) \ar[ur] &amp;&amp; (L^{-1},-\alpha-1)\otimes(L,\alpha+1) \ar[ul] }$$ where every arrow is an isomorphism?</p> http://mathoverflow.net/questions/63446/the-inverse-of-determinant-line-bundle/63482#63482 Answer by Fernando Muro for the inverse of determinant line bundle? Fernando Muro 2011-04-29T22:45:38Z 2011-05-15T14:33:04Z <p>A <strong>Picard groupoid</strong> 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 <strong>inverse object</strong> 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 $\varphi=\bar{\varphi}(1_x\otimes \psi)$. </p> <p>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!</p>