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Let $\Bbbk$ be a field. Let $G$ be an affine algebraic group over $\Bbbk$. Let $X$ be a scheme over $\Bbbk$. Let $G$ act on $X$ with the action morphism $\sigma:G\times X\to X$. There are two morphisms from $G\times X$ to $X$ $$\sigma:G\times X\to X,\quad \mathrm{pr}:G\times X\to X.$$

For $\mathcal F$ a (quasi-coherent) sheaf on $X$, a linearization is an isomorhpism $$\phi:\mathrm{pr}^*\mathcal F\xrightarrow{\cong}\sigma^*\mathcal F$$ satisfying a suitable cocycle condition.

When $\mathcal F=\mathcal O_X$, there is a canonical linearization. Canonically we can identify $\mathrm{pr}^*\mathcal O_X\cong\mathcal O_{G\times X}$ and $\sigma^*\mathcal O_X\cong\mathcal O_{G\times X}$ and $\phi:\mathrm{pr}^*\mathcal O_X\to \sigma^*\mathcal O_X$ under these identification is the identify $1_{\mathcal O_{G\times X}}:\mathcal O_{G\times X}\to\mathcal O_{G\times X}$.

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When $\mathcal F=\Omega_{X/\Bbbk}$, there should also be a canonical choice of $\phi:\mathrm{pr}^*\Omega_{X/\Bbbk}\to\sigma^*\Omega_{X/\Bbbk}$. However it is unclear to me at the moment.

Can you describe what is the canonical $\phi:\mathrm{pr}^*\Omega_{X/\Bbbk}\cong\sigma^*\Omega_{X/\Bbbk}$ ?

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  • $\begingroup$ The same question on MSE got ana answer. $\endgroup$
    – Yikun Qiao
    Commented Oct 14, 2022 at 10:02

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