Let $R$ be a commutative Noetherian ring and let $M$ be a finitely generated $R$-module.
Definition. $M$ is called totally reflexive (or $G-\dim_RM = 0$) if it satisfies the following conditions:
(i) $\mathrm{Ext}_R^i(M, R) = 0$ for all $i >0$
(ii) $\mathrm{Ext}_R^i(M^*, R) = 0$ for all $i >0$, where $M^* = \mathrm{Hom}_R(M, R)$.
(iii) The canonical map $\delta_M : M \to M^{**}$ is an isomorphism, where $M^{**} = \mathrm{Hom}_R(M^*, R)$.
Totally reflexive modules have many properties similar to projective modules.
Note. (a) If $(R, \mathfrak{m})$ is a local ring and $M$ is projective, then $M$ is free.
(b) If $(R, \mathfrak{m})$ is a Gorenstein local ring and $M$ is totally reflexive, then $M$ is maximal Cohen-Macaulay.
Question. Suppose $(R, \mathfrak{m})$ is a local ring and $M$ is totally reflexive. Is it true that $\dim M = \dim R$?