$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\pr{pr}$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.


1 Answer 1


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.

  • $\begingroup$ 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) $? $\endgroup$
    – user267839
    Nov 20, 2022 at 23:27
  • 1
    $\begingroup$ @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. $\endgroup$
    – Libli
    Nov 21, 2022 at 6:45
  • $\begingroup$ 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. $\endgroup$
    – user267839
    Nov 21, 2022 at 9:44
  • $\begingroup$ @JustusC : I think i may be able to do that. But I would need to write commutative diagrams, which I can not do in a comment. It will be easier if you ask a separate question. $\endgroup$
    – Libli
    Nov 21, 2022 at 13:18
  • $\begingroup$ thank you a lot. I will post it next day $\endgroup$
    – user267839
    Nov 21, 2022 at 14:09

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