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I am confused as to why a regularised determinant is used in the definition of Quillen's metric on the determinant line bundle defined over the space of $\bar{\partial}$ operators on a (hermitian) vector bundle over a compact Riemann surface. I mean, why not use the metric induced by the vector bundle just by itself? Is the determinant introduced so as to define a smooth metric? For the record I am referring to Quillen's paper "On the determinants of Cauchy-Riemann operators" over a Riemann surface.

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What do you mean by "the metric induced by the vector bundle just by itself?" (Do you mean the restrictions of the $L^2$ metric?) –  Paul Jul 1 '11 at 19:11
Yes Paul, I mean the restriction of the $L^2$ metric. –  Vamsi Jul 3 '11 at 11:20
The problem with the L^2 metric is that one cannot extend it over those points (in the space of $\bar\del$ operators) where the kernel and cokernel jump. In other words, if you stratify your space of of $\bar\del$ operators according to the dimension of the kernel, then over ay stratum the kernel and cokernel forms a bundle, and you can take the $L^2$ metric (and the induced metric on the determinant).The point of Quilen's metric is to deal with what happens across strata, by taking into account the eigenvalues that become zero. The basic property of regularized determiants that make this –  Paul Jul 3 '11 at 13:29
(cont) work is that if you remove finitely many eigenvalues, the regularized determinant of what's left times the product of the ones you removed equals the regularized determinant of all (non zero) eigenvalues. So you can patch together the metric across the strata. –  Paul Jul 3 '11 at 13:31

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