Consider $J^{g-1}$, the variety of degree $g-1$ line bundles on a compact Riemann surface of genus $g$. Recall that $J^{g-1}$ is a torsor for the Jacobian, thus has dimension $g$. We can produce elements in $J^{g-1}$ by choosing $g-1$ points $p_1,\ldots,p_{g-1}$ and constructing the associated line bundle to the divisor $p_1+\cdots+p_{g-1}$. In this way, we obtain a $g-1$-dimensional family of elements of $J^{g-1}$, forming the so-called $\Theta$ divisor. Elements of the $\Theta$ divisor may be distinguished from other points in $J^{g-1}$ by the fact that they represent line bundles admitting sections, i.e. $h^0(L)\neq 0$.

For degree $g-1$ line bundles, the Riemann-Roch theorem gives $h^0(L)=h^1(L)$, so over $J^{g-1}$ we see that $h^0(L),h^1(L)$ are generically zero and jump along the $\Theta$ divisor. One might imagine that there are vector bundles $V,W$ over $J^{g-1}$ of the same rank, with fibres over $L\in J^{g-1}$ given by Dolbeault $0$- and $1$-forms with coefficients in $L$ respectively, together with a bundle map $\overline{\partial}:V\rightarrow W$ which is generically an isomorphism but drops rank along $\Theta$; then $\det \overline{\partial}$ would be a section of a line bundle $\det V^*\otimes\det W$ over $J^{g-1}$ which cuts out $\Theta$. This line bundle turns out to be a well-defined object called the determinant line bundle and it was introduced by Quillen.

Quillen's construction of the line bundle proceeds by replacing the family of (2-term) Dolbeault complexes parametrized by $J^{g-1}$ by a quasi-isomorphic family of finite-dimensional (2-term) complexes and taking the determinant line bundle of this. The finite-dimensional replacement for the Dolbeault complex is given by the Dolbeault operator acting on forms lying in the first few eigenspaces of the Laplacian, roughly speaking. Then as we move along $J^{g-1}$, some of the eigenvalues may hit zero and we have a jump of $h^0(L)$.

For a good list of references for the above, see determinant line bundle (nlab).

My question: What is a simple, modern construction of the determinant line bundle on $J^{g-1}$, perhaps one which uses the tools of algebraic geometry? It may well be tautological from some point of view, in which case I would probably be unsatisfied. Also, if you wish to rhapsodize on determinant line bundles, please feel free.

My motivation: Many of the classic papers by Beauville, Narasimhan, Ramanan etc. on moduli of vector bundles expend a lot of effort working with determinant line bundles and extracting information from them in quite ad-hoc and clever ways. I'd like to understand these better.

  • 3
    $\begingroup$ In chapter 17 of Abelian Varieties, Theta Functions and the Fourier Transform, Polishchuk goes through a similar construction to the one you outline above in a characteristic free way. It uses the Fourier-Mukai transform of a line bundles on degree g-1 on your curve. I wouldn't say its any simpler than the construction you outlined but it certainly uses the tools of algebraic geometry. Although I would venture to say that the construction you outlined only uses tools of at least complex algebraic geometry. $\endgroup$
    – solbap
    Commented Mar 17, 2011 at 17:07

1 Answer 1


Let's say that your (compact!) Riemann surface is $X$ (of genus at least $1$). What does it mean to say that $J^{g-1}$ is the "variety of degree $g-1$ line bundles?" One way to formalize this is to say that there is a line bundle $\mathcal{L}$ on the product $X \times J^{g-1}$ with the property the rule $p \mapsto \mathcal{L}|_{X \times \{ p \}}$ defines a bijection between the ($\mathbb{C}$-valued) points of $J^{g-1}$ and the line bundles of degree $g-1$. (A stronger statement is that $J^{g-1}$ represents the Picard functor, but I'm going to try to sweep this under the run.)

Let $\pi \colon X \times J^{g-1} \to J^{g-1}$ denote the projection maps. Recall that the formation of the direct image $\pi_{*}(\mathcal{L})$ does not always commute with passing to a fiber. However, the theory of cohomology and base change describes how the fiber-wise cohomology of $\mathcal{L}$ varies. The main theorem states that there is a 2-term complex of vector bundles

$d \colon \mathcal{E}_0 \to \mathcal{E}_1$

that computes the cohomology of $\mathcal{L}$ universally. That is, for all morphisms $T \to J^{g-1}$, we have

$\operatorname{ker}(d_{T}) = (\pi_{T})_{*}(\mathcal{L}_{X \times T})$


$\operatorname{cok}(d_{T}) = (R^{1}\pi_{T})_{*}(\mathcal{L}_{X \times T}).$

(The most important case is where $T = \operatorname{Spec}(\mathbb{C})$ and $T \to J^{g-1}$ is the inclusion of a point.)

The complex $\mathcal{E}_{\cdot}$ is not unique, but any other complex of vector bundles with this property must be quasi-isomorphic.

In the literature, many authors construct a complex $\mathcal{E}_{\cdot}$ using an explicit procedure, but the existence is a very general theorem. I learned about this topic from Illusie's article "Grothendieck's existence theorem in formal geometry," which states the base change theorems in great generality.

In your question, you described how to convert the complex into a line bundle (take the difference of top exterior powers). But now we must check that two quasi-isomorphic complexes have isomorphic determinant line bundles. This can be checked by hand, but this statement has been proven in great generality by Mumford and Knudsen (see MR1914072 or MR0437541). The determinant of the complex $\mathcal{E}_{\cdot}$ is exactly the line bundle you are asking about.

One subtle issue is that the universal line bundle $\mathcal{L}$ is NOT uniquely determined. If $\mathcal{M}$ is any line bundle on $J^{g-1}$, then $\mathcal{L} \otimes \pi^{-1}(\mathcal{M})$ also parameterizes the degree $g-1$ line bundles in the sense described above (and in a stronger sense that I am sweeping under the rug). However, one can show that $\mathcal{L}$ and $\mathcal{L} \otimes p^{*}(\mathcal{M})$ have isomorphic determinants of cohomology by using the fact that a degree $g-1$ line bundle has Euler characteristic equal to zero.

Added: Here is one method of constructing the complex. Fix a large collection of points $\{x_1, \dots, x_n\} \subset X$ (at least $g+1$ points will do), and let $\Sigma \subset X \times J^{g-1}$ denote the subset consisting of pairs $(x_i,p) \in X \times J^{g-1}$. The universal line bundle $\mathcal{L}$ fits into a short exact sequence:

$$ 0 \to \mathcal{L}(-\Sigma) \to \mathcal{L} \to \mathcal{L}|_{\Sigma} \to 0 $$

Here $\mathcal{L}(-\Sigma)$ is the subsheaf of local sections vanishing along $\Sigma$ and $\mathcal{L}|_{\Sigma}$ is the pullback of $\mathcal{L}$ to $\Sigma$.

Associated to the short exact sequence on $X \times J^{g-1}$ is a long exact sequence relating the higher direct images under $\pi$. The first connecting map is a homomorphism

$d : \pi_{*}(\mathcal{L}|_{\Sigma}) \to R^{1}\pi(\mathcal{L}(-\Sigma))$

This is a complex $K_{\cdot}$ of vector bundles that compute the cohomology of $\mathcal{L}$ universally in the sense described earlier. Thus, its determinant is the desired line bundle.

Why is this true? There are two statements to check: that the direct images are actually vector bundles and that the complex actually computes the cohomology of $\mathcal{L}$. Both facts are consequences of Grauert's theorem. Indeed, the hypothesis to Grauert's theorem is that the dimension of the cohomology of a fiber $\pi^{-1}(p)$ is constant as a function of $p \in J^{g-1}$. Using Riemann-Roch (and the fact that we chose a large number of points), this is easily checked. The theorem allows us to conclude that both $\pi_{*}(\mathcal{L}|_{\Sigma})$ and $R^1 \pi(\mathcal{L}(-\Sigma))$ are vector bundles whose formation commutes with base change by a morphism $T \to J^{g-1}$.

An inspection of the relevant long exact sequence shows that the cohomology of the complex is $\pi_{*}(\mathcal{L})$ and $R^1\pi(\mathcal{L})$ (i.e. the complex compute the cohomology of $\mathcal{L}$). We would like to assert that this property persists if we base-change by a morphism $T \to J^{g-1}$ and work with the complex $K_{\cdot} \otimes \mathcal{O}_{T}$.

But we already observed that the formation of the direct images appearing in the complex commutes with base-change, and so the base-changed complex fits into a natural long exact sequence. Again, an inspection of this sequence shows that base-changed complex compute the cohomology of $\mathcal{L}|_{X \times T}$, and so the original complex computes the cohomology universally.

  • $\begingroup$ Thanks for the reference to Illusie. I won't accept this answer since it does not address the main point of the question, i.e. construction of the bundle. $\endgroup$ Commented Mar 17, 2011 at 13:57
  • $\begingroup$ Thanks for the response. Could you be more precise about what you would like? Would you like an explanation of how to construct the complex d : \mathcal{E}_0 \to \mathcal{E}_1 or an explanation of how to go from the complex to the line bundle or am I misunderstanding your question? $\endgroup$
    – jlk
    Commented Mar 17, 2011 at 19:51
  • $\begingroup$ The part of your answer which interests me is "In the literature, many authors construct a complex E using an explicit procedure". I would like to see such an explicit procedure, besides the one which I indicated I am aware of, from Quillen. $\endgroup$ Commented Mar 19, 2011 at 4:54
  • $\begingroup$ @Marco Gualtieri: I added a construction of the complex. $\endgroup$
    – jlk
    Commented Mar 19, 2011 at 20:11

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