Let $X$ be a smooth algebraic variety over $\mathbb{C}$ of dimension $n$, $\Omega_X = \Lambda^n \; \Omega^1_{X}$. In D-Modules, Perverse sheaves and Representation theory it is stated that

$$\mathcal{End}_{\mathbb{C}}(\Omega_X) \simeq \Omega_X \otimes_{\mathcal{O}_X} \mathcal{End}_{\mathbb{C}}(\mathcal{O}_X) \otimes_{\mathcal{O}_X} \Omega_{X}^{\otimes^{-1}}$$

as sheaves of rings, where the left and right $\mathcal{O}_X$ module structure on $\mathcal{End}_{\mathbb{C}}(\mathcal{O}_X)$ are given by the left and right multiplication of $\mathcal{O}_X$ regarded as a subring of $\mathcal{End}_{\mathbb{C}}(\mathcal{O}_X)$.

I have some questions about this statement:

  1. Why do we need left and right actions of $\mathcal{O}_X$? $\mathcal{O}_X$ is commutative, so left and right actions should be interchangeable. Where do we use the left and where the right?
  2. Which is the ring structure on the right-hand side? Whether the left one has a natural ring structure, and the last two elements of the tensor product have a natural ring structure, I don't understand how to give the whole right-hand side a ring structure.
  1. $End_{\Bbb C}(\mathcal{O}_X)$ is not commutative, and thus the left/right actions of $\mathcal{O}_X$ do different things. Consider the example $X=\Bbb A^1$, where $End_{\Bbb C}(\mathcal{O}_X)(X)=\Bbb C\langle x,\partial x \rangle/([\partial x,x]=1)$.

  2. Recall that we can describe the multiplication map of an $R$-algebra $S$ as a particular map $S\otimes_R S\to S$. We apply this logic to the sheaves of rings to get a map $End_{\Bbb C}(\Omega_X)\otimes_{\mathcal{O}_X} End_{\Bbb C}(\Omega_X) \cong \Omega_X\otimes_{\mathcal{O}_X}End_{\Bbb C}(\mathcal{O}_X)\otimes_{\mathcal{O}_X}\Omega_X^{-1}\otimes_{\mathcal{O}_X} \Omega_X\otimes_{\mathcal{O}_X}End_{\Bbb C}(\mathcal{O}_X)\otimes_{\mathcal{O}_X}\Omega_X^{-1} \to \Omega_X\otimes_{\mathcal{O}_X} End_{\Bbb C}(\mathcal{O}_X)\otimes_{\mathcal{O}_X}End_{\Bbb C}(\mathcal{O}_X)\otimes_{\mathcal{O}_X}\otimes \Omega_X^{-1}$ given by $\Omega_X\otimes_{\mathcal{O}_X}\Omega_X^{-1}\cong \mathcal{O}_X$ and now we can use the multiplication map on $End_\Bbb C(\mathcal{O}_X)$ to get a map to $\Omega_X\otimes_{\mathcal{O}_X} End_{\Bbb C}(\mathcal{O}_X)\otimes_{\mathcal{O}_X}\Omega_X^{-1}$.

  • $\begingroup$ Just to be clear: in the tensor product we use the left action of $\mathcal{O}_X$ in the tensor product with $\Omega_X$ and the right action in the tensor with $\Omega_X^{\otimes -1}$, right? $\endgroup$ – Federico May 3 '18 at 8:09
  • $\begingroup$ Yes, and those two actions coincide. $\endgroup$ – KReiser May 3 '18 at 13:58
  • $\begingroup$ What do you mean saying that they coincide? $\endgroup$ – Federico May 3 '18 at 21:40
  • $\begingroup$ The module $\Omega_X$ may be considered as an $\mathcal{O}_X-\mathcal{O}_X$ bimodule, and the left module structure is the same as the right module structure. $\endgroup$ – KReiser May 3 '18 at 22:17

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