# Determinant bundle of a tensor product

Let $X$ be a ringed space (for example, a scheme or a manifold). If $V$ is a locally free $\mathcal{O}_X$-module of rank $n$, then $\mathrm{det}(V) := \Lambda^n V$ is a locally free $\mathcal{O}_X$-module of rank $1$, called the determinant of $V$. Actually $\mathrm{det}$ is a functor. Now I wonder how to give a slick proof of the well-known (?) fact that there is a natural isomorphism $$\mathrm{det}(V \otimes W) \cong \mathrm{det}(V)^{\otimes m} \otimes \mathrm{det}(W)^{\otimes n},$$ where $V$ is locally free of rank $n$ and $W$ is locally free of rank $m$. For this I would like to construct a map globally and basis-free and then show that it is an isomorphism locally, hence an isomorphism. A typical local generator of $\mathrm{det}(V \otimes W)= \Lambda^{n \times m}(V \otimes W)$ is $$(v_{11} \otimes w_{11}) \wedge \dotsc \wedge (v_{1m} \otimes w_{1m}) \wedge \dotsc \wedge (v_{n1} \otimes w_{n1}) \wedge \dotsc \wedge (v_{nm} \otimes w_{nm}).$$ To which element of $\Lambda^n(V)^{\otimes m} \otimes \Lambda^m(W)^{\otimes n}$ should we map this?

Note that SE/571839 is a very similar question, but I would like to have an abstract proof like indicated in the last paragraph of the accepted answer. (In fact, I want to prove a similar formula in an arbitrary cocomplete symmetric monoidal $\mathbb{Q}$-linear category, where $V$ is called locally free of rank $n$ if $\Lambda^n V$ is invertible and $\Lambda^{n+1} V = 0$. Here, no local bases are available.)

Incomplete thoughts:

Thinking of the elements as written in an $n \times m$ grid, send it to the element

$\alpha = \bigotimes_{j=1}^m (v_{1j} \wedge \cdots \wedge v_{nj}) \otimes \bigotimes_{i=1}^n (w_{i1} \wedge \cdots \wedge w_{im})$.

(so, putting the $v$'s together in columns and the $w$'s in rows.)

To see that this is well-defined, it suffices to show that it is zero if two adjacent terms in the long wedge product are equal.

For instance if $v_{11} \otimes w_{11} = v_{12} \otimes w_{12}$, then either both are $0$, in which case we're done, or $v_{11} = \lambda v_{12}$, $w_{11} = \frac{1}{\lambda} w_{12}$ for some scalar $\lambda \ne 0$. Then the second equality forces $\alpha = 0$ by the $i=1$ factor on the $W$ part. This covers all the cases where we exchange two elements in the same row. (It also covers cases where we exchange elements in the same column. So maybe it is already sufficient.)

But there are still the "line break" equalities to consider, and I'm not sure how to complete the argument, sorry. (This feels very reminiscent of Fulton's proof of Sylvester's Lemma in his Young Tableaux book, with a clever recursive argument for this last case.)

Edit: Here's a thought. Rather than using "line break" inequalities, we'll go through the entries of the grid in a back-and-forth order. So, first we consider the equalities along the first row. Then, we consider the equality

$$v_{1m} \otimes w_{1m} = v_{2m} \otimes w_{2m},$$

comparing the "last entries in the first two rows". It's clear that $\alpha = 0$ in this case since it is a "column equality" (we use the $j=m$ factor -- the last one -- in the $V$ part). Then we work backwards along row 2, then forward along row 3, and so on. Thus at every step, we are either using a "row equality" or a "column equality" to conclude that $\alpha = 0$, so at the end, we conclude that the expression is alternating in all $n\cdot m$ wedges.

• Thank you. To show that the map is alternating, don't we also have to consider non-pure tensors? Usually I know tricks how to avoid this, but not here ... Also notice that I don't work over a field. Commented Dec 15, 2014 at 17:09
• Ah. Perhaps it's better to work over $\mathbb{Q}$ here (let's assume that $\mathcal{O}_X$ is a sheaf of $\mathbb{Q}$-algebras). So alternating = antisymmetric, and the latter should be easier to check since we don't have to decide when two tensors are equal. And in fact we don't have to consider non-pure tensors. Commented Dec 15, 2014 at 17:15
• I still wonder if it can be completed. The last step in the argument would go something like this: assume that the line-break equality holds, and consider the expression as a function of $(n-1)(m-1)$ arguments. Then show that it is identically zero, by showing, using the very same methods, that it is alternating in too many vectors (i.e. the condition $\bigwedge^{n+1}V = 0$ or similar -- which you say holds at your level of generality). Commented Dec 15, 2014 at 19:37
• @MartinBrandenburg I think I have a complete proof now. Commented Dec 15, 2014 at 20:55
• For the last part, seems like to swap $ij$ with $kl$ you could swap $ij\leftrightarrow kj$, then $kj\leftrightarrow kl$, and finally $ij\leftrightarrow kj$ again. Each of these are row/column swaps and hence introduce a negative sign by the first part, and since there are three swaps the overall effect is to invert the sign. Clearly the composition of the three transpositions is $ij\leftrightarrow kl$ as desired, so this shows the result is alternating in all pairs.
– BHT
Commented Aug 12, 2021 at 20:46