Let $(G,m)$ be a group scheme unipotent, and $L$ a local system of rank 1 on $G$ such that: $m^*(L) \simeq L \boxtimes L $.
Then why is $L$ an equivariant sheaf on $G$ with the action the conjugation.
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$\begingroup$ $L$ looks to me like something that deserves to be called a central extension of $G$, or perhaps a multiplicative torsor in Grothendieck's sense (I can't recall where it's defined). Think of $G$ as the trivial sheaf over itself, and then in the category of sheaves try to emulate the proof that the conjugation action of a group lifts to act on central extensions. Hope this helps. $\endgroup$– David Roberts ♦Commented May 24, 2014 at 18:16
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$\begingroup$ More naively, what you are asking for is the analogue of the statement that a 1-dimensional representation $\rho: G \to \mathbb C^\times$ of a group is a class function. The condition $m^\ast L = L \boxtimes L$ just says (roughly) that the stalk $L_{gh}$ at $gh$ is identified with $L_g \otimes L_h$. Thus we have $L_{ghg^{-1}} \simeq L_g \otimes L_h \otimes L_{g^{-1}} \simeq L_g \otimes L_{g^{-1}} \otimes L_h \simeq L_h$, which is (roughly) what it means to be $G$ equivariant. To get rid of the "roughly" you will need to express the above in terms of diagrams... $\endgroup$– Sam GunninghamCommented May 25, 2014 at 4:00
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$\begingroup$ Yes i understand that my problem is from those isomorphism goingo to this definition of equivariant sheaves: a $G$-equivariant sheaf is a pair $(F,\theta)$, where $F$ is a sheaf on $X$ and $\theta:d_1^*F \to d_0^*F$ is an isomorphism of sheaves satisfying the cocycle conditions: $d_2^*\theta \circ d_0^*\theta = d_1^*\theta$ and $s_0^*\theta = \mathrm{id}_F$. with: $s_0(x)=(1,x)$ and $d_i:X_n \to X_{n-1}$ defined by - $d_0(g_1,\dots,g_n,x) = (g_2,\dots,g_n,g_1^{-1}x)$, - $d_i(g_1,\dots,g_n,x) = (g_1,\dots,g_ig_{i+1},\dots,g_n,x)$ if $0<i<n$, - $d_n(g_1,\dots,g_n,x) = (g_1,\dots,g_{n-1},x)$. $\endgroup$– João DiasCommented May 25, 2014 at 14:48
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