# Example of a line bundle not admitting a $\operatorname{PGL}(n+1)$-linearization in Mumford's GIT

$$\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Proj{Proj}\DeclareMathOperator\Pic{Pic}$$I have a question about an example for a line bundle not admitting a $$G$$-linearization from Mumford's GIT, page 33:

We consider the action of $$\PGL(n+1)$$ on projecive space $$\mathbb{P}^n= \Proj k[X_0,\dotsc, X_n]$$. Observe that $$\PGL(n+1)$$ is given as the open affine subscheme of

$$\mathbb{P}^{n^2+2n} = \Proj k[a_{00},\dotsc, a_{0n}; a_{10}, \dotsc , a_{nn}]$$

complementary to the determinants hypersurface $$\det(a_{ij})=0$$. The action morphism $$\sigma: \PGL(n+1) \times \mathbb{P}^n \to \mathbb{P}^n$$ is determined by

$$\begin{gather*} \sigma^*(\mathcal{O}_{\mathbb{P}^n}(1)) \cong p_1^*(\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)) \otimes p_2^*(\mathcal{O}_{\mathbb{P}^{n}}(1)) \\ \sigma^*(X_i)= \sum_{j=0}^n p_1^*(a_{ij}) \otimes p_2^*(X_j) \end{gather*}$$

where $$p_1, p_2$$ are projections canonical projections.

Mumford claims that $$\mathcal{O}_{\mathbb{P}^n}(1)$$ admits no $$\PGL(n+1)$$-linearization, because the restriction of $$\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)$$ to the open subscheme $$\PGL(n+1)$$ has order $$n+1$$ in $$\Pic[\PGL(n+1)]$$, and is therefore not trivial.

My question is why the fact that $$\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)$$ restricted to the affine open $$\PGL(n+1)$$ is not trivial, implies that $$\mathcal{O}_{\mathbb{P}^n}(1)$$ admits no $$\PGL(n+1)$$-linearization?

[Indeed, $$\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)$$ has order $$n+1$$ in $$V$$ because $$\Pic(\mathbb{P}^{n^2+2n}) \to \Pic(\mathbb{P}^{n^2+2n} \backslash V(\det(a_{ij}))= \Pic(\PGL(n+1))$$ induces an isomorphism $$\Pic(\mathbb{P}^{n^2+2n} \backslash V(\det(a_{ij})) \cong \mathbb{Z}/(\deg(\det(a_{ij}))\mathbb{Z}$$.]

To turn it another way round, why if $$\mathcal{O}_{\mathbb{P}^n}(1)$$ would admit a $$\PGL(n+1)$$-linearization, then the restriction of $$\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)$$ to $$\PGL(n+1)$$ must be trivial? I conjecture that this argument can somehow reduced to an easy comparison of orders of group elements in groups $$\Pic(X)$$, $$\Pic^G(X)$$ but I do not see how it can be directly related.

Maybe it somehow helps to know that we have always a morphism of groups $$\Pic^G(X) \to \Pic(X)$$ which is not neccessarily injective.

• Isn't $p_1 : \operatorname{PGL}(n + 1) \times \mathbb P^n \to \operatorname{PGL}(n + 1)$, not $p_1 : \operatorname{PGL}(n + 1) \times \mathbb P^n \to \mathbb P^n$? Nov 10, 2022 at 23:03
• @LSpice: yes, thank you. Nov 10, 2022 at 23:09

$$PGL(n+1)$$-linearization of $$\mathcal O_{\mathbb P^n}(1)$$ could be used to produce isomorphism $$\gamma : p_2^*(\mathcal O_{\mathbb P^n}(1)) \to \sigma^*(\mathcal O_{\mathbb P^n}(1))$$, hence, taking to attention isomorphism $$\sigma^*(\mathcal{O}_{\mathbb{P}^n}(1)) \cong p_1^*(\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)) \otimes p_2^*(\mathcal{O}_{\mathbb{P}^{n}}(1))$$ which you have mentioned, we have equality of Picard classes $$([\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)],[\mathcal{O}_{\mathbb{P}^{n}}(1)])=(0,[\mathcal{O}_{\mathbb{P}^{n}}(1)])$$ in $$Pic(PGL(n+1)) \times Pic(\mathbb P^n) = Pic(PGL(n+1) \times \mathbb P^n)$$, so, $$[\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)] = 0$$ in $$Pic(PGL(n+1))$$
How to construct such $$\gamma$$? $$PGL(n+1)$$-linearization of $$\mathcal O_{\mathbb P^n}(1)$$ is a choice of morphism $$\tilde{\sigma} : PGL(n+1) \times L \to L$$ (where $$L$$ is a total space of the line bundle $$\mathcal O_{\mathbb P^n}(1)$$) such that the diagram $$\require{AMScd} \begin{CD} p_2^*(\mathcal O_{\mathbb P^n}(1)) = PGL(n+1) \times L @>\tilde{\sigma}>> L\\ @V{}VV @V{}VV \\ PGL(n+1) \times \mathbb P^n @>{\sigma}>> \mathbb P^n \end{CD}$$ commutes. Total space of the line bundle $$\sigma^*(\mathcal O_{\mathbb P^n}(1))$$ is by definition a fiber product $$(PGL(n+1) \times \mathbb P^n) \times_{\mathbb P^n} L$$, hence, by universal property of fiber products the morphism $$\gamma : p_2^*(\mathcal O_{\mathbb P^n}(1)) \to \sigma^*(\mathcal O_{\mathbb P^n}(1))$$ of $$PGL(n+1) \times \mathbb P^n$$-varieties exists. $$\gamma$$ is an isomorphism since for every $$g \in PGL(n+1)$$ induced morphism $$\gamma_g : O_{\mathbb P^n}(1) \to g^* O_{\mathbb P^n}(1)$$ is an isomorphism.