Quillen metric definition

Question

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.

Answer 1

In a nutshell, I think knowing $|D_1-D_2|$ is very small in operator norm does not enable you to control the jump of the dimension of $\ker D_1$ to $\ker D_2$. All we have is estimates like $$ |D_2 v|<<|D_1 v|+\epsilon*|v| $$ And clearly we cannot force $|D_2 v|=0$. So something "needs to be done" to bridge the difference in $L^{2}$ metric.

The other issue that Paul did not mention is that the determinant line bundle $L_{T}$ is not defined directly having fibres $\lambda(\ker T)^{*}\otimes \lambda(\textrm{co}\ker T)$. As Quillen observed, in this case if we let $\sigma_{T}=0$ if $T$ is not invertible, and $\sigma_{T}=1$ if $T$ is invertible, then this section may not be holomorphic in $T$ on the Banach manifold of index zero operators. Quillen's construction avoids the issue by working with effectively finite dimensional subspaces $F$ of the codomain $\mathcal{H}^{1}$ by considering the set of operators $U_F$ such that $\textrm{Im}(T)+F=\mathcal{H}^{1},\forall T\in U_{F}$. Quillen then dictate that the holomorphic structure on $L$ is determined by requiring the isomorphism $$\lambda(\ker T)^{*}\otimes \lambda (\textrm{co}\ker T)\cong \lambda (T^{-1}F)^{*}\otimes \lambda(F)$$ be an isomorphism of holomorphic line bundles over $U_{F}$ for any $F$.

The issue Paul mentioned is in fact written up in a very concise manner in the paper (Quillen claimed "...it is easy to see that..."!). The first step is realizing we have an isomorphism $$ \lambda(\ker D)^{*}\otimes \lambda (\textrm{co} \ker D)\cong \lambda (F^{0}_{\alpha})\otimes \lambda (F^{1}_{\alpha}) $$ where $F^{0}_{\alpha}$ and $F^{1}_{\alpha}$ be the subspace of eigenvectors of $D^{*}D$ or $DD^{*}$ of eigenvalue less than $\alpha$. This step can be seen explicitly as $D^{*}$ maps the eigenvectors of $DD^{*}$ with eigenvalue less than $\alpha$ bijectively to the eigenvectors of $D^{*}D$ with eigenvalue less than $\alpha$. The second step is then realizing the inner product changes by $e^{-\xi'_{>\alpha}(0)}$ in this isomorphism. So since $\alpha$ is arbitrary to account for all possible "jumps" it is suffice to multiply the original inner product by the analytic torsion. However, to fill in the detail of "it is easy to see" required quite some effort and is not altogether a trivial task even if one invokes general $\Psi DO$ theory like in Ray-Singer's paper.

I learned most of the subject from C. Soule and H.Gillet via their book "Lectures on Arakelov theory", Chapter VI. Hopefully what I wrote above helps.