Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a nontrivial ringed space (i.e. all stalks are nonzero). To every locally free module $M$ on $X$ of constant rank $n$ we can associated it's determinant $\det(M)$, which is a line bundle and is defined as the $n$th exterior power of $M$. We can also define $\det(M)$ if the rank is not assumed to be constant. It's surely locally constant, so $X$ is partitioned into the open subsets $(X_n)_{n \geq 0}$, where $M$ has constant rank $n$, and we can glue the $\det(M|_{X_n})$ to get the determinant $\det(M)$. There is no doubt that this is well-defined, but it seems to me a bit uncanonical.

For example look at the functoriality: If $f : M \to N$ is a homomorphism, then $\det(f) : \det(M) \to \det(N)$ is first defined on the intersections on the open subsets on which $\det(M)$ and $\det(N)$ are defined, and then glued. Then a small argument is needed to prove that this is, indeed, a functor. Isn't this ugly? Therefore:

Question Is there a characterization of the functor $\det(-)$ from locally finite free modules to line bundles which does not depend on partitions of $X$?

For example, for every $n \geq 0$, the functor $\det(-)$ on locally free modules of rank $n$ has a universal property, namely $\text{Hom}(\det(M),L)$ corresponds to alternating maps $M^n \to L$. You can write down a similar universal property when $n$ is a fixed locally constant function on $X$. But my question is if we can do it without fixing $n$.

share|improve this question
I don't think there's anyting uncanonical about this... Unless I'm confused, you're just partitioning $X$ into its connected components and "gluing" along their empty intersection. So the issue in the second paragraph never arises. –  Dave Anderson Feb 7 '11 at 17:42
Universal properties of the determinant are discussed in this MO question: <a href="mathoverflow.net/questions/7124/…;. The point is rather than glue you simply resolve your $M$ by a finite complex of locally free sheaves, and thus avoid any jumping in rank. –  David Ben-Zvi Feb 7 '11 at 18:04
Dave and David, thanks for the comments! –  Martin Brandenburg Feb 7 '11 at 19:32
Does Dmitri's suggestion work more generally? e.g. suppose I have a finite complex of vector bundles on a singular variety. (I realize this should be a comment, but I don't have enough rep!) Actually, that direct sum seems unfortunate if the goal is to always get a line bundle.... –  martin Dec 5 '11 at 19:48
add comment

2 Answers

up vote 5 down vote accepted

There is a paper by Knudsen and Mumford that gives a thorough treatment of determinants of perfect complexes. Mumford has put a copy online here.

Edit: Knudsen's 2002 paper that Theo Buehler linked in the comment below appears to resolve the question in a more canonical way than the older paper. It also has a neat letter by Grothendieck at the end.

share|improve this answer
This paper characterizes the functor $\det$ with a list of properties. Really interesting! –  Martin Brandenburg Feb 8 '11 at 10:18
Now that I look at it again, I see that the paper answers an interesting question, but one that is different from the one you asked. In particular, condition (v) in the definition of determinant functor seems to point back to the use of top exterior power in the beginning of the paper. –  S. Carnahan Feb 8 '11 at 10:54
That's right... –  Martin Brandenburg Feb 8 '11 at 11:36
Knudsen has revisited this work recently: projecteuclid.org/euclid.mmj/1028575741 –  Theo Buehler Feb 8 '11 at 12:22
add comment

How about defining det(M) as ⨁kExtkSym(M*)(O(X),Sym(M*))? Here Sym(M*) acts on O(X) by augmentation map. Ext and Sym are functorial, hence det should also be functorial.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.