# Proposition 1.5 in Mumford's Geometric Invariant Theory

$$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\pr{pr}$$I have some problems to understand the proof of Proposition 1.5 from Mumford's Geometric Invariant Theory, p 34:

Corollary 1.5
Let $$G$$ be a connected linear algebraic group acting on an algebraic variety $$X$$, that is proper over $$k$$ (in the book a variety is a scheme $$X/k$$ such that $$\overline{X}= X \times \Spec(\overline{k})$$ is irreducible and reduced). Let $$\mathcal{L}$$ be an invertible sheaf on $$X$$, and let $$[\mathcal{L}]$$ the class regarded as the $$k$$-rational point of the Picard scheme $$\Pic(X/k)$$ associated to $$\mathcal{L}$$.
Then some power $$\mathcal{L}^n$$ is $$G$$-linearizable if and only if some multiple $$[\mathcal{L}]^n$$ of $$[\mathcal{L}]$$ is left fixed by induced $$G$$-action on $$\Hom_k(\Spec(k), \Pic (X/k))$$.

(at this point one should remark that in the book this left action by $$G$$ on $$\mathcal{Pic}_{X/k}(k)$$ is not explained in explicit terms. In the substantively similar question Corollary 1.6 in Mumford's Geometric Invariant Theory I made a remark how I think this action might work in detail).

[Second important remark before we dip into the proof:
Recall that more less by definition of the Picard functor is given by

$$\mathcal{Pic}_{X/k}(S) := H^0(G, R^1p_{1*} (\mathcal{O}_{S \times X}^*) = \\ \{ \mathcal{M} \text{ invertible sheaf on } X \times_k S \} / \{ \text{ inv. sheaves of the form } p^*_S(\mathcal{K}) \text{ for } \mathcal{K} \text{ invertible on } S \}.$$

The proof uses [cp Chap. 0, §5, (d)] the fact that the Picard functor is "almost" representable, that means precisely there exists a $$k$$-scheme $$\Pic(X/k)$$ representing the associated functor $$\text{Hom}( \ , \Pic(X/k))$$ which contains the Picard functor $$\mathcal{Pic}_{X/k}$$ in the sense that for any $$k$$-scheme $$S$$ there is a functorial inclusion

$$\iota_S: \mathcal{Pic}_{X/k}(S) \hookrightarrow \Hom_k(S,\Pic(X/k)).$$

In general that's a proper inclusion. The equality only holds if $$X \times_k S$$ admits a section over $$S$$.

The Proof. The only if is clear. Conversely, suppose $$[\mathcal{L}]^n$$ is left fixed by $$G$$. Then the claim is first that for some $$m$$, the two pullback sheaves $$\sigma^*(\mathcal{L}^{nm})$$ and $$p_2^*(\mathcal{L}^{nm})$$ (induced by the action and projection maps $$\sigma, p_2: G \times X \to X$$ on $$G \times X$$ are isomorphic. To see this, consider the see-saw exact sequence [Rem.: I never saw the term see-saw sequence. I think that is just the exact part of Leray–Serre spectral sequence for higher image sheaf]:

$$0 \to H^1(G, \mathcal{O}_G^*) \to H^1(G \times X, \mathcal{O}_{G \times X}^*) \to H^0(G, R^1p_{1*} (\mathcal{O}_{G \times X}^*).$$

Since $$H^1(G, \mathcal{O}_G^*)$$ is a finite group (Seminaire Chevalley, [9], 5-21), it is enough to show that the image of $$\sigma^*(\mathcal{L}^n) \otimes p_2^*(\mathcal{L}^n)^{-1}$$ in $$H^0(G, R^1p_{1*} (\mathcal{O}_{G \times X}^*)$$ is zero. But, by the functorial definition of $$\Pic (X/k)$$ (cf Chap. 0, §5, (d), page 23)

$$\mathcal{Pic}_{X/k}(G) = H^0(G, R^1p_{1*} (\mathcal{O}_{G \times X}^*) \subset \Hom_k(G, \Pic (X/k)).$$

But, as in the proof of proposition 1.4, it holds $$H^0(G \times X, \mathcal{O}_{G \times X}^*) \cong H^0(G, \mathcal{O}_G^*)$$ and the latter is just $$k* \times M$$, where $$M$$ is the set of characters, i.e., $$\Hom(G, \mathbb{G}_m)$$.
Choose an isomorphism $$\phi: \sigma^*(\mathcal{L}^{nm}) \to p_2^*(\mathcal{L}^{nm})$$, which is the identity on $$\{e\} \times X$$. [The rest of the proof verifies the cocycle condition $$p^*_{23} \phi \circ (1_G \times \sigma)^* = (m \times 1_x)^* \phi$$, that's fine .]

The question is why the assumption that the class $$[\mathcal{L}^n] \in \Hom_k(\Spec(k), \Pic (X/k))$$ is fixed by $$G$$-action, implies that the pullback sheaves $$\sigma^*(\mathcal{L}^{nm})$$ and $$p_2^*(\mathcal{L}^{nm})$$ are isomorphic, or as remarked that's equivalent to to the question why the images of classes $$[\sigma^*(\mathcal{L}^{n})]$$ and $$[p_2^*(\mathcal{L}^{n})]$$ in $$H^0(G, R^1p_{1*} (\mathcal{O}_{G \times X}^*)) \subset \Hom_k(G, \Pic (X/k))$$ are identical?

To rephrase it in other terms, the maps $$\sigma, p_2: G \times X \to X$$, which are given on geometric points by $$(g,x) \mapsto g \cdot x$$, respectively $$(g,x) \mapsto x$$, map the classes $$[\mathcal{M}] \in \mathcal{Pic}_{X/k}(k)$$ to classes in $$\mathcal{Pic}_{X/k}(G)$$ via taking $$[\mathcal{M}]$$ to the pullback $$[\sigma^*\mathcal{M}]$$, respectively $$[p_2^*\mathcal{M}]$$.
How do these operations by $$\sigma, p_2$$ look like in explicit terms as maps between $$\Hom(\Spec(k), \Pic(X/k))$$ and $$\Hom(G, \Pic(X/k))$$? Especially how to construct explicitly from the pullback of $$[\mathcal{L}]^n$$ by $$\sigma$$ and $$p_2$$ elements in $$\Hom_k(G, \Pic (X/k))$$ representing the classes of the images of $$\sigma^*(\mathcal{L}^{n})$$ and $$p_2^*(\mathcal{L}^{n})$$?

Pictorally, the action and projection morphisms $$\sigma, \pr_X$$ should induce following diagram

$$\require{AMScd} \begin{CD} \mathcal{Pic}_{X/k}(k) @>{\iota_k} >> \Hom(\Spec(k), \Pic(X/k)) \\ @VV\sigma^*, p_2^*V @VVf_{\sigma^*}, f_{\pr_X^*}V \\ \mathcal{Pic}_{X/k}(G) @>{\iota_G}>> \Hom(G, \Pic(X/k)) \end{CD}$$

and I'm interested in the explicit structure of the right vertical maps $$f_{\sigma^*}, f_{\pr_X^*}: \Hom(\Spec(k), \Pic(X/k)) \to \Hom(G, \Pic(X/k))$$ making the diagram commutative with respect $$\sigma^*, \pr_X^*$$ on the left and what they do with $$[\mathcal{L}^n] \in \Hom_k(\Spec(k), \Pic (X/k))$$.

My conjecture is that the image of $$p_2^*(\mathcal{L}^{n})$$ in $$\Hom_k(G, \Pic (X/k))$$ should represent a constant map with image be the $$k$$-point $$[\mathcal{L}^n]$$, while $$\sigma^*(\mathcal{L}^{n})$$ the orbit map of $$[\mathcal{L}^n]$$ induced by the action of $$G$$ on $$k$$-valued points of $$\Pic (X/k)$$. This would suggest that $$f_{\sigma^*}$$ and $$f_{\pr_X^*}$$ should be explicitly given by

$$[x] \mapsto f_{\sigma^*}([x]) := (g \mapsto g \cdot [x])$$

and respectively

$$[x] \mapsto f_{\pr_X^*}([x]) := (g \mapsto [x])$$

i.e. the constant map, where $$[x]: \Spec(k) \to \Pic(X/k)$$ is any geometric $$k$$-point of $$\Pic(X/k)$$ and $$g \cdot [x]:= [g^*x]$$ the induced action on Picard group via pullback. Having this, we assumed $$G$$ to fix $$[\mathcal{L}^n]$$, therefore these the images of $$[\mathcal{L}^n]$$ by these maps would coinside as elements in $$\Hom_k(G, \Pic (X/k))$$ and should give isomorphic line bundles over $$G \times X$$.

Therefore if the $$f_{\sigma^*}$$, $$f_{\pr_X^*}$$ would be given like I conjecture, this would be consistent with the tacitly used claim in the proof that $$[\sigma^*(\mathcal{L}^{n})]$$ and $$[p_2^*(\mathcal{L}^{n})]$$ are identical as elements in $$\mathcal{Pic}_{X/k}(G) \subset \Hom_k(G, \Pic (X/k))$$. But I not see how to verify that $$f_{\sigma^*}$$, $$f_{\pr_X^*}$$ have this form.

I think the fist part of the proof of 1.5 may be rephrased as follows. The hypothesis that $$[\mathcal{L}^n]$$ is fixed by the $$G$$-action exactly means that for all $$g \in G$$, there is an isomorphism: $$\sigma^* \mathcal{L}^n \big|_{\{g\} \times X} \simeq p_2^* \mathcal{L}^n|_{\{g\} \times X}.$$ Put differently, for all $$g \in G$$, there is an isomorphism:
$$\sigma^* \mathcal{L}^n \otimes \left(p_2^* \mathcal{L}^n \right)^{-1} \big|_{\{g\} \times X} \simeq \mathcal{O}_X.$$ By the Seesaw Theorem, this means that $$\sigma^* \mathcal{L}^n \otimes \left(p_2^* \mathcal{L}^n\right)^{-1}$$ is the pull-back of a line bundle on $$G$$. But a Theorem of Chevalley implies $$H^{1}(G, \mathcal{O}_{G}^*)$$ is finite, hence $$\sigma^* \mathcal{L}^n \otimes \left(p_2^* \mathcal{L}^n \right)^{-1}$$ is torsion and we are done.
• this argument I understand. but do you maybe know if it's also possible to find out the explicit form of the morphisms $G \to \text{Pic}(X/k)$ on the right side which corresponds to the classes $[ \sigma^*(\mathcal{L}^{n})]$ and $[ p_2^*(\mathcal{L}^{n})]$ in $\mathcal{Pic}_{X/k}(G)$? Nov 20, 2022 at 23:27
• @JustusC : your conjecture about the morphisms $f_{\sigma}$ and $f_{pr^*_X}$ should be correct. I think you just have to use the fact that the morphism $i_G$ sends a line bundle $L$ on $X \times G$ to the map $g \longrightarrow L|_{\{g\} \times X}$ and the fact that the diagram is commutative. Nov 21, 2022 at 6:45
• could you give a sketch (if that's not too long) or a reference where I can look up a proof of that $i_G$ is given that way? It looks of course heuristically plausible, but I nowhere found literature cointaining the formal proof which affirms this. Nov 21, 2022 at 9:44