What is an example of a rigid object $A$ in an abelian monoidal category $\mathcal{M}$ that is not projective as an object in $\mathcal{M}$? (Since $\mathcal{M}$ is abelian projective just means that all exact sequences of the form $ 0 \to R \to Q \to P \to 0$ split.) What are sufficient conditions for such a rigid object to be projective?

## 1 Answer

If by "rigid" you mean the same thing as dualizable, an example is given by $\mathcal M = Mod_{R[G]}$ for some commutative ring $R$ and group $G$, with monoidal structure given by $\otimes_R$ and the diagonal action.

Then the object $R$ with a trivial $G$-action is the unit, hence it's rigid, but it's rarely projective: $\hom(R,-)$ is isomorphic to $(-)^G$, so it's projective if and only if the latter is exact, if and only if $H^*(G,-)$ vanishes on $R$-modules - counterexamples are given by $R= \mathbb F_p, G= $ any finite group with order divisible by $p$.

More generally, note the following : if $P$ is dualizable, with dual $P^\vee$ then $\hom(P,-)\cong \hom(\mathbf 1, P^\vee \otimes -)$, where $\mathbf 1$ is the unit. So there are two ways exactness can fail: $\hom(\mathbf 1,-)$ might not be exact (in which case, the unit is an example anyway), or $P^\vee\otimes -$ might not be exact (these are not necessary/sufficient conditions, because $\hom(\mathbf 1,-)$ might not be conservative either).

Note that in fact, the latter cannot happen if $\mathcal M$ is symmetric monoidal : indeed $P^\vee\otimes -$ is then both a left adjoint (hence right exact) and a right adjoint (hence left exact). I don't know what happens if $\mathcal M$ is just monoidal, because then $P^\vee$ might itself not be dualizable (say I fixed a side and am only referring to "left dualizable")

So in deciding if $P$ is projective, you really only have to check whether the unit $\mathbf 1$ is projective.

Note that it might still happen that the unit is not projective, yet $P$ is. Consider for instance the following : take a counterexample $\mathcal M$ as above, and an ordinary category of modules $\mathcal M'$ over a commutative ring. Then the unit of $\mathcal M\times\mathcal M'$ is not projective (it has the unit of $\mathcal M$ as a summand, and that one is not projective), yet any projective module in $\mathcal M'$ is projective in the product.