Page 276 in the book Differential Topology and Quantum Field theory by C. Nash, describes a "generalization of determinant of linear map" as follows: for linear map

$O:{V} \to {W}$

its determinant is (No further description in the book)

$\det O \in {\left( {{\Lambda ^{\max }}V} \right)^*} \otimes \left( {{\Lambda ^{\max }}W} \right)$

where ${{\Lambda ^{\max }}V}$ denotes maximal exterior power of a vector space $V$.

Here's my Question: When $V$ and $W$ is of the same dimension, the construction of $\det O$ is obvioius, but I just can't figure it out the detail when they have different dimension. Can anyone point out the definition of this generalized determinant?

Thanks!

• (The construction when the dimension is the same is not obvious really, because the result will depend on bases...) – Mariano Suárez-Álvarez Jun 20 '11 at 15:56
• If the dimensions coincide (say, $\dim V = \dim W = n$) then one gets functorially a linear map $\Lambda^n V \to \Lambda^nW$ which one defines to be the determinant. As Mariano points out, you only get a number if you choose a basis and the number will depend on the basis, even though the linear map is well defined in general. – José Figueroa-O'Farrill Jun 20 '11 at 16:31
• If the dimensions do not coincide, but if you have some additional structure, such as an inner product, then there are ways to define the determinant and this is probably the context in which Nash is working. – José Figueroa-O'Farrill Jun 20 '11 at 16:39
• Thanks Mariano and José. Indeed a basis is required to obtain a number. – Lelouch Jun 22 '11 at 10:03

Let $f: \mathbb{R}^k \to \mathbb{R}^m$ be a linear map. If $k < m$, then we let $g: \mathbb{R}^m \to \mathbb{R}^m$ to be any projection to the image of $f$. The rank of $g$ is at most $k$, so $\det g$ is zero. Because $g \circ f =f$, we have: $$0 = (\det g) \cdot (\det f) = \det (g \circ f) = \det f.$$ If $k > m$, there exists a subspace of $\mathbb{R}^k$ on which the restriction of $f$ is an isomorphism to the image of $f$, and we let $h: \mathbb{R}^k \to \mathbb{R}^k$ be the composite of $f$ followed by the inverse of that restriction. We have $f \circ h = f$, and $h$ has rank at most $m$, hence determinant zero. Therefore, $\det f = 0$.
Incidentally, there is a way to get a meaningful "determinant" from a map $V \to W$, but it involves turning it into a two-term complex of vector spaces. Two-term complexes and maps of vector spaces both happen to look the same, since they are described by the same data, but they admit different natural operations (e.g., we like to compose maps but not complexes). This paper by Knudsen and Mumford describes the determinant as a symmetric monoidal functor from the category $C^\bullet is$, whose objects are bounded chain complexes of finite rank vector spaces and whose morphisms are quasi-isomorphisms, to the category $Pis$, whose objects are graded lines, and whose morphisms are homogeneous isomorphisms. In other words, if you are given a two-term complex $V \to W$, with $W$ in degree zero, determinant produces a graded line canonically isomorphic to $(\bigwedge^{max}V)^\vee \otimes (\bigwedge^{max} W)$ in degree $\dim V$. Note that the determinant in this case is not a distinguished element of that line, although a choice of bases of $V$ and $W$ seems to yield a distinguished basis element in the line.
• @Thomas: Section 3 of Deligne's paper is a sequence of analogies between Hermitian geometry and nonarchimedean geometry culminating in parallel definitions of intersection number. I think these analogies have become stronger in the last 25 years. In particular, Berkovich's $\mathbb{R}$-tree description of nonarchimedean curves appears to be an infinite negative curvature limit of Manin's hyperbolic interior of a complex curve. – S. Carnahan Jun 23 '11 at 5:51