# On tilting and cotilting modules

Let A be an Artin algebra and assume all modules are basic, then a classical result says that tilting modules $$T$$ are in bijection with complete cotorsion pairs $$(T^{\perp}, \check{ add(T)})$$ (with some additional assumptions on the cotorsion pairs) , where $$T^{\perp}= \{ X | Ext_A^i(T,X)=0$$ for all $$i >0 \}$$ and $$\check{ add(T)}$$ denotes the full subcategory of modules $$Y$$ with a finite $$add(T)$$ coresolution. Dually we have the same for cotilting modules $$U$$ with the dual properties and cotilting modules are in bijection with complete cotorsion pairs $$( ^{\perp} U, \hat{add(U)})$$ (with some additional assumptions on the cotorsion pairs).

Assume we have a tilting module $$T$$ and a cotilting module $$U$$ such that the following two conditions are satisfied:

1. $$T^{\perp} \cap \hat{add(U)} = \hat{add(U)}$$

2. $$^{\perp}U \cap \check{add(T)} = \check{add(T)}$$.

Is it then true that $$A$$ is Gorenstein if and only if $$U=T$$? (or maybe this is true up to adding another condition?)

Note that being Gorenstein is equivalent to all tilting modules being cotilting and also that every module of finite projective dimension has finite injective dimension and vice versa.

So the direction $$U=T$$ implies Gorenstein is easy.

For the other direction we can use that $$A$$ being Gorenstein gives us that $$T$$ is some cotilting module and 1. also gives us that $$Ext^i(T,U)=0$$ for all $$i>0$$. Now it would be enough to show that also $$Ext^i(U,T)=0$$ for all $$i>0$$ to conclude that $$T=U$$. But I do not see a good approach to prove that.

We have $$^\perp(T^\perp) = \check{\operatorname{add}}\, T$$, and $$(\check{\operatorname{add}}\, T)^\perp = T^\perp$$, and $$(^\perp U)^\perp = \hat{\operatorname{add}}\, U$$, and $$^\perp(\hat{\operatorname{add}}\, U) = {^\perp U}$$. Condition 1. and 2. are equivalent to
1. $$\hat{\operatorname{add}}\, U \subseteq T^\perp$$
2. $$\check{\operatorname{add}}\, T \subseteq {^\perp U}$$.
Taking $$^\perp$$ on the correct sides and using the above identities, it follows that 1. and 2. are equivalent. So it is enough to have one of the conditions, say 2. Take your favorite Gorenstein algebra which is not selfinjective. Then choose $$U = D(A)$$ and $$T = A$$, which implies that $$\check{\operatorname{add}}\, T = \mathcal{P}(A)$$ and $$^\perp U = \operatorname{mod}\, A$$, that is, 2. is satisfied. But $$T \neq U$$. So additional conditions have to be added in order for this to be true.