Suppose that $\mathcal{C}$ is a model category (with a fixed model structure). Are there any known examples where $\mathcal{C}$ is a (symmetric) monoidal model category in two different ways? I.e., can there exist tensor products $\otimes_1, \otimes_2\colon \mathcal{C}\times \mathcal{C}\to \mathcal{C}$ such that both $(\mathcal{C}, \otimes_1, A)$ and $(\mathcal{C}, \otimes_2, B)$ are symmetric monoidal model categories. Can this phenomenon even occur?

The canonical place to look for an answer to this would be Balchin's book A Handbook of Model Categories (specifically in the Die Kunstkammer) but I couldn't see anything relevant.

## Best Answer

Let $k$ be a field, and let $A$ be a finite-dimensional $k$-algebra. Suppose that $A$ admits the structure of a Hopf $k$-algebra. Then $A$ is quasi-Frobenius, so the category of $A$-modules admits a model structure where the cofibrations are the injective maps, the fibrations are the surjective maps, and the weak equivalences are the stable equivalences. This is in Hovey's Model Categories book, for example.

But now suppose $A$ admits

two differentHopf $k$-algebra structures. Each one makes $Mod(A)$ into a symmetric monoidal model category.For example, suppose $k$ has characteristic $2$, and let $A = k[x]/x^2$. If I recall correctly, you can make $A$ into a Hopf $k$-algebra by letting $x$ be primitive, or by letting the coproduct on $x$ be $x\otimes x + 1\otimes x + x\otimes 1$. These are really examples Dave Benson mentioned in a comment (although he suggested looking at the derived categories rather than the module categories themselves): the former Hopf algebra is the restricted enveloping algebra of a one-dimensional Lie $k$-algebra, while the latter Hopf algebra is isomorphic to the group algebra of the cyclic group $C_2$. The two are isomorphic as $k$-algebras but not isomorphic as Hopf $k$-algebras, so you get two different monoidal model structures on $Mod(A)$. The coproducts are both co-commutative, so these monoidal model structures are indeed symmetric monoidal.