Everything here has the Zariski topology.
Let $T=(\Bbb{C}^*)^d$, and define an action of $T$ on $\Bbb{C}^n$ by $$t\cdot x=(t^{\mathbf{a}_1}x_1,\ldots, t^{\mathbf{a}_n}x_n).$$ Here $\mathbf{a}_1,\ldots,\mathbf{a}_n$ are fixed vectors in $\Bbb{Z}^d$, and for a vector $\mathbf{b}\in\Bbb{Z}^d$ and a point $t\in T$, we set $t^{\mathbf{b}}=t_1^{b_1}\cdots t_d^{b_d}$. Let $A=[\mathbf{a}_1, \ldots, \mathbf{a}_n]\in \mathbb{Z}^{d\times n}$.
Let $R=\Bbb{C}[x_1,\ldots,x_n]$ be the coordinate ring of $\Bbb{C}^n$. The action of $T$ on $\Bbb{C}^n$ induces an (a priori right) action of $T$ on $R$ via $x^{\mathbf{u}}\cdot t = t^{A\mathbf{u}}x^{\mathbf{u}}$ and then extending by $\mathbb{C}$-linearity. This then induces a $\mathbb{Z}^d$-grading of $R$: set $R_{\mathbf{b}}= \mathbb{C}\{x^{\mathbf{u}}\mid A\mathbf{u}=\mathbf{b}\}.$
One can show that the category of $T$-equivariant quasicoherent $\mathcal{O}_{\mathbb{C}^n}$-modules is equivalent to the category of $\mathbb{Z}^d$-graded $R$-modules. My question is then this:
Let $\sharp$ indicate one of unbounded, bounded, bounded below, or bounded above. Consider the following triangulated categories:
- $D_T^\sharp(\mathcal{O}_{\mathbb{C}^n})$, the equivariant $\sharp$ derived category (in the sense of Bernstein) of $T$-equivariant quasicoherent $\mathcal{O}_{\mathbb{C}^n}$-modules; and
- $D^\sharp(\operatorname{GrMod}(R))$, the $\sharp$ derived category of $\operatorname{GrMod}(R)$, where $\operatorname{GrMod}(R)$ is the (abelian) category of $\mathbb{Z}^d$-graded $R$-modules and (degree $0$) graded $R$-module homomorphisms.
Are these two triangulated categories equivalent?
