Why is the symmetric monoidal structure on invertible modules strict?

Question

Let $N$ be an object in a symmetric monoidal category. Then the braid map $N\otimes N\to N\otimes N$ is almost never the identity, and this is the obstruction to making a symmetric monoidal category into a "strict" symmetric monoidal category, in which the functor $\otimes$ is commutative on the nose. For example, when $\otimes$ is just the categorical product in the category of sets (or any other concrete category), this is the map $(x,y)\mapsto(y,x)$ which is almost never the identity.

However, one case in which the braid map is the identity is in the category of invertible modules over a commutative ring $R$, as a full subcategory of all modules. Indeed, the braid map $R\otimes R\to R\otimes R$ is the identity, and any invertible module $I$ is locally isomorphic to $R$, so the braid map $I\otimes I\to I\otimes I$ is locally equal to the identity and hence equal to the identity.

What I'd like to have is a more conceptual explanation for why the braid map is the identity for invertible modules, which does not use the fact that they are locally free (indeed, I'm interested in this because I want to use this to prove invertible modules are locally free in a more general setting). Unfortunately, the proof cannot just be abstract nonsense involving invertibility--for example, if we work with graded modules over a commutative ring instead of ordinary modules and use the usual sign conventions, then the braid map will be -1 rather than 1 on invertible modules concentrated in odd degrees. Does anyone know of a better explanation, or know a reason I shouldn't expect there to be one?

EDIT: Inspired by Charles's answer, here's a closely related question. I'm really interested in invertible objects in the derived category, and in the derived category dualizable objects can be represented by finite chain complexes of finite-rank projective modules. Over a local ring, then, all dualizable objects have an Euler characteristic which is an integer (as opposed to an arbitrary element of $R$). Since as Charles noted, the braid map of an invertible object can be identified with its Euler characteristic as a dualizable object, this implies that the braid map of any invertible object in the derived category of a local ring is $\pm 1$ (and so if you're willing to suspend your objects if necessary, you can assume it is 1).

Thus I would be satisfied with a conceptual answer to the following question: why is the Euler characteristic of a dualizable object in the derived category of a local ring always an integer? (It may be more natural to not assume that the ring is local, in which case you should replace "integer" with "integral linear combination of idempotents".)

Answer 1

Here's how I look at this. Let $(\mathcal C,\otimes,I, \sigma)$ be a symmetric monoidal category. Recall that the dimension of a dualizable object $C \in \mathcal C$ is the trace of the identity map on $C$; it lives in $End_{\mathcal C}(I)$. EDIT: This is what Charles referred to as the Euler characteristic.

Observations:

1. Dimension is multiplicative: $\dim(C\otimes D) = \dim(C) \dim(D)$ for $C,D \in \mathcal C$ dualizable.

2. $\dim(C) = \dim(C^\vee)$.

3. If $L\in \mathcal C$ is $\otimes$-invertible, then $\dim(L)$ is invertible.

Corollary: Let $L \in \mathcal C$ be $\otimes$-invertible. Then

1. The symmetry on $L$ is given by $\sigma_{L,L} = \frac 1 {\dim L} 1_{L \otimes L}$.

2. $(\dim L)^2 = 1$. In particular, if $End_{\mathcal C}(I)$ is a domain, then $\dim L = \pm 1$.

Proof:

1. It suffices to verify this equation after tensoring with $L^{-1}$ on both sides and then capping off with some units and counits, where it becomes the true equation $\dim L = \frac 1 {\dim L} (\dim L)^2$.

2. This follows from the observations above since $L \otimes L^\vee = I$ and $\dim I = 1$.

So the fact that the braidings are trivial on $Pic(R)$ reduces to the fact that the dimension of an invertible module is $1$ rather than $-1$.

Answer 2

This is an interesting question. I.e., I have no idea how to answer it ...

Here's a little bit of context to put this in. So $C$ is a symmetric monoidal category, with unit object $R$. Let $\mathrm{Pic}(C)$ be the collection of isomorphism classes of invertible objects in $C$; it's an abelian group using $\otimes$.

There's a group homomorphism $$\eta : \mathrm{Pic}(C) \to \text{($2$-torsion subgroup of $\mathrm{Aut}(R)$).}$$ This sends an invertible object $I$ to its Euler characteristic..

Here's a different construction of $\eta$ which I find easier to understand. If $I$ is an invertible object, there is a canonical isomorphism $\mathrm{Aut}(I)\approx \mathrm{Aut}(R)$; to construct it, choose an isomorphism $I\otimes J\approx R$, so that an automorphism $f:I\to I$ gets sent to an automorphism $R\approx I\otimes J \xrightarrow{f\otimes 1} I\otimes J\approx R$. Now $\eta (I)$ is just the image of the braid map $(\tau: I\otimes I \to I\otimes I)\in \mathrm{Aut}(I\otimes I)\approx \mathrm{Aut}(R)$. This definition makes it clear that $(\eta (I))^2=1$.

So you're asking why $\eta$ has trivial image when $C$ is a category of modules over a commutative ring. As you point out, in graded modules the image is $\{\pm1\}$. I got nothing here ...