Vanishing of $\operatorname{Ext}_R(\operatorname{Tr} M,N)$ and freeness criteria $\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\Tr{Tr}\DeclareMathOperator\coker{coker}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Tor{Tor}$I am investigating the interplay between freeness criteria and Ext vanishing. A nice example is a vast literature around the Auslander-Reiten conjecture (ARC) (in the local case):
(ARC) For a Noetherian local ring $R$ and a finitely generated $R$-module $M$ such that $\Ext^i_R(M,M\oplus R)=0$ for all $i>0$, then $M$ is free.
Let $R$ be a Noetherian local ring and $M$ be finitely generated $R$-modules. The Auslander transpose of $M$ is defined as $$\Tr M:=\coker(F^*\rightarrow G^*)$$ where $G\rightarrow F\rightarrow M\rightarrow0$ is a presentation of $M$ and $\_^*=\Hom_R(\_,R)$. An interplay between freeness and Tor vanishing is the Yoshino freeness criterion: if $\Tor_1^R(Tr M,M)=0$, then $M$ is free.
I wonder if there is also an interplay between freeness criteria and vanishing of $\Ext_R^i(\Tr M,N)$ for some $R$-module $N$. Or even any sufficient condition for having $\Ext^i_R(\Tr M,N)=0$.
 A: I am not sure if this is the kind of thing you are interested in, but let me at least state the easiest to prove criteria that I know for free-ness of $M$ in terms of vanishing of certain $\text{Ext}_R^i(\text{Tr}M, -)$ .
Proposition: If $M$ is a finitely generated module over a Noetherian local ring $R$ such that $M^*\ne 0$, and $\text{Ext}_R^{1,2}(\text{Tr} M, M^*)=0$, then $M$ is free. (See https://arxiv.org/abs/1805.04568 Lemma 4.12)
Proof: By definition, one have exact sequences $0\to M^*\to F\to K \to 0$ and $0\to K \to G\to \text{ Tr} M\to 0$ for some modules $K,F,G$, where $F$ and $G$ are free. Now $\text{Ext}_R^{1}(K, M^*)\cong \text{Ext}_R^{2}(\text{Tr} M, M^*)=0$, hence $0\to M^*\to F\to K \to 0$ splits, so $M^*$ is a free module. Since $M^*$ is non-zero, and now we know it is free, so $\text{Ext}_R^{1,2}(\text{Tr} M, M^*)=0$ yields $\text{Ext}_R^{1,2}(\text{Tr} M, R)=0$, hence $M$ is reflexive, so $M\cong M^{**}$ is free.
If you have other specific kind of situations in mind, I probably will be able to help you better ...
