p I Hecke I ( 1) i Tr(T Λ n i V ) T i = 0.

Transcription

1 Literal notes 0.. Where are we? Recall that for each I fset and W I Rep(ǦI ), we have produced an object F I,W I D ind (X I ). It is the proper pushforward along paws of the moduli of Shtukas: F I,W I := (p I )! Sat(W I ) Sht I p I Hecke I X I Bun G Frob Bun G Bun G. where Sat(W I ) is regarded as an object in the constructible derived category of Sht I. Remark 0.. The geometry of Sht I equips F I,W I with equivariance structure with respect to the partial Frobenii: F {i} : Frob {i}( F I,W I ) F I,W I Cayley-Hamilton. Suppose x is a closed point of X, and V Rep(Ǧ). We would like to study F K {0},W K V XK {x} Dind (X K {x}). Hence, we consider K fset and W K Rep(Ǧ) fixed as well from now on. The equivariance data for the partial Frobenius Frob {0} now give rise to an endomorphism: F := F deg(x) {0} : F K {0},W K V F X K {x} K {0},W K V X K {x} since Frob deg(x) {0} acts as identity on {x}. The way we will interpret Cayley-Hamilton can be summarized by the slogan: F satisfies its own characteristic equation The trace operator. The usual statement of Cayley-Hamilton is the following equation for every endomorphism T acting on a vector space V of dimension n: (0.) ( ) i Tr(T Λ n i V ) T i = 0. In our setting, the role of the trace operators Tr(T Λ n i V ) will be played by the endomorphism S Λ n i V,x. Proposition 0.2 (V. Lafforgue s Proposition 7.). Suppose n = dim(v ). Then the following identity holds in the endomorphism ring of F K {0},W K V XK {x} : (0.2) ( ) i S Λ n i V,x F i = 0. Remark 0.3. (a) The endomorphism S Λ n i V,x commutes with F (hence F i ). (b) The identity (0.2) is useful because it imposes a finiteness condition on the operator F.

2 2 (c) You may wonder: the endomorphism F is only defined when we restrict to a closed point how can it be useful in practice? The answer is that ultimately we are interested in the (Hecke-finite part of the) restriction of F I,V I to the generic point η I of X I : F I,V I := ( F I,V I ) Hf η. I One may then use generality of lisse sheaves to pass from properties at a closed point to those at η I Proof of the usual Cayley-Hamilton. Assume that T is an endomorphism on the vector space V of dimension n. We want to prove (0.). The proof will explain exactly what the endomorphism appearing on the left-hand-side is. In fact, we claim: n + ( ) i Tr(T Λ n i V ) T i = C n (A n+ ). Now, what is the right-hand-side? Fix an integer m 0, and let L be an endomorphism of V (+m). The notation C m (L) denotes the following endomorphism of V : a C m (L) : V V V m (V m ) L V V m (V m ) a Do not erase. Later we will modify this diagram. T m V V m (V m ) V and the endomorphism A m denotes the anti-symmetrizer on V m : A m := sgn(σ) σ. m! σ S m Then, it is clear that the usual Cayley-Hamilton follows from the above identity, as A n+ = 0. In fact, we claim something even stronger. Let S (i+) n+ denote the subset of S n+ consisting of permutations for which the cycle containing {0} has length i +. Then we may decompose A n+ into: A = A (i+) n+, where A(i+) n+ := sgn(σ) σ, (n + )! σ S (i+) n+ and the claim is that ( ) i (0.3) n + Tr(T Λ n i V ) T i = C n (A (i+) n+ ). The proof is a straightforward computation based on the following: Observation 0.4. a (a) Suppose ρ S m is a permutation fixing {0}. Then C m (ρlρ ) = C m (L). (b) Suppose L = L 0 L, where L 0 applies to the first (i + )-factors and L to the last (m i)-factors. Then C m (L) = C m i ( L ) C i (L 0 ).

3 3 (a) (b) (c) Suppose σ 0 S i+ is the permutation (0 i). Then C i (σ 0 ) = T i. (d) Suppose A m is the anti-symmetrizer on V m. Then C m ( A m ) = Tr(T Λ m V ). a Do not erase. Later we will modify these statements. We will now perform the calculation that proves (0.3). C n (A (i+) n+ ) = (n + )! = (n + )! = n + σ S (i+) n+ ( n i (n i)! sgn(σ) C n (σ) ) i! ( ) i sgn(τ) C n (σ 0 τ) τ S n i ( ) i sgn(τ) C i (σ 0 ) C n i ( τ) τ S n i where σ 0 is the permutation (0 i). It follows that C n (A (i+) n+ ) = n + (c) ( ) i sgn(τ) F i C n i ( τ) (n i)! τ S n i (d) = ( )i n + F i C n i ( A n i ) = ( )i n + F i Tr(T Λ n i V ) One word about (c). This is a cool trick. For i =, the diagram corresponds to: C (σ 0 ) : V which is equal to: V V V V V V V σ 0 V V V T V V V T V V V V. V This latter composition is obviously equal to T. The general case follow from induction you will see it clearly from the following way of writing the composition: C i (σ 0 ) : V i V (V V ) i T ( T ) (i ) V (V V ) i i V Generalization. One would like to generalize the above tensorial proof of Cayley- Hamilton to the endomorphism F on F K {0},W K V X K {x}. (a) As a first observation, the endomorphism F takes place in the category D ind (X K {x}), and yet we want to utilize constructions such as the unit map V V V m (V m ) of Ǧ-representations. Hence we should be studying functors: Φ : Rep(Ǧ) Dind (X K {x}) aut, V F K {0},W K V with F -action. XK {x} Here, C aut denotes the category of pairs (c, F ) with c C and F Aut(c).

4 4 (b) The usual Cayley-Hamilton would correspond to the functor Φ : Rep(GL n ) Vect aut, V oblv(v ) with T -action. (c) In order to formulate trace (at least in a way applicable for us), we will need additional data that I will presently explain. Suppose C I is a family of categories parametrized by I fset. For us, C I := D ind (X K {x} I ). For each I fset, let Rep(Z I, C I ) denote the category of pairs (c, {F i } i I ) where c C I and {F i } i I is an I-family of automorphisms of c. Suppose we have a system of functors (where G is any linear algebraic group): Φ I : Rep(G I ) Rep(Z I, C I ) with functoriality along I J, i.e., a commutativity constraint: Rep(G I ) Rep(Z I, C I ) Rep(G J ) Rep(Z J, C J ) satisfying a compatibility condition for compositions: I J K. Fixing a G-representation V of dimension n, there is a well-defined notion of trace of F on U Rep(G) as an operator: S U : Φ {}() Φ {} (V U U ) = R Φ {,2,3} (V U U ) R R(F (0,,0) ) R Φ {,2,3} (V U U ) = Φ {} (V U U ) Φ {} ( ). We can state Cayley-Hamilton in this context: Theorem 0.5. The following equality holds in End(): ( ) i S Λ n i V F i = 0. The application we have in mind is for Φ I : Rep(ǦI ) Rep(Z I, D ind (X K {x} I )) given by the functor: V I F K {0},W K V I X K {x} with {F i} i I -action Proof of the general theorem. The proof of the general theorem amounts to establishing the analogues of Observations 0.4, with the following definition of C m (L) (now for L an endomorphism of the G-representation V (+m) ): C m (L) : Φ {}() Φ {} (V V m (V m ) ) Φ {}(L ) Φ {} (V V m (V m ) ) = R Φ {,2,3} (V V m (V m ) ) R(F (0,,0)) R Φ {,2,3} (V V m (V m ) ) = Φ {} (V V m (V m ) ) Φ {}( ). Among the four statements, (a) and (b) remain the same. The analogues of (c) and (d) state:

5 5 (c ) Suppose σ 0 S i+ is the permutation (0 i). Then C i (σ 0 ) = F i. (d ) Suppose A m is the anti-symmetrizer on V m. Then C m ( A m ) = S Λ m V. I will present the proofs of (d ) and (c ). The proof of (d ) is straightforward, and after seeing it, you will be convinced that the analogous of (a) and (b) can be carried out in a similar manner. The proof of (c ) is more substantive Proof of (d ). I first claim that there exists an embedding: ι : Λ m V (Λ m V ) V m (V m ) of G {,2} -representations, such that the following diagram of G {} -representations commute : triv unit V m (V m ) Λ m V (Λ m V ) ι V m (V m ) unit A m Λ m V (Λ m V ) ι V m (V m ) triv triv. (Aside: how do we write such an ι? In general, given maps between G-representations V W S V, such that T S =, the map S T : W W V V does the job.) Now, the statement (d ) follows from the commutative diagram (left-hand-side = red, righthand-side = blue): T Φ {} (V V m (V m ) ) Φ {} (V Λ m V (Λ m V ) ) ι A m Φ {} (V V m (V m ) ) Φ {} (V Λ m V (Λ m V ) ι Φ {} (V V m (V m ) ). The middle square commutes by functoriality of Φ {,2,3} because ι is induced from a morphism of G {,2} -representations Proof of (c ). We proceed by three steps: Draw the second square under the first, then modify into the following diagram.

6 The case i =. We apply the trick from before in re-organizing the composition to get rid of σ 0. The objective is to show that the red hexagon commutes. Φ {} (V V V ) R Φ {,2} (V triv) R Φ {,2} (V (V V )) F ( ) R(F (,0) ) id R(F (,0) ) R Φ {,2} (V triv) R Φ {,2} (V (V V )) Φ {} (V V V ) Φ {} ( ) Now, all squares obviously commute except the one with ( ). To get that one to work out, we need compatibility of the commutativity constraint along {} {, 2} {}: R{,2} {} R {} {,2} (V ) R {,2} {} Φ {,2} (V triv) F R {,2} {} R {} {,2} (F ) R{,2} {} R {} {,2} (V ) R {,2} {} (F (,0) ) R{,2} {} Φ {,2} (V triv) The induction step (modulo something). We would now like to perform the induction step, assuming the result for i, i.e., commutativity of: (i ) Φ {} (V (V V ) (i ) ) R Φ {,,2i } (V (V V ) (i ) ) F i R(F (,0,,,,0,0) ) (i ) Φ {} (V (V V ) (i ) ) R Φ {,,2i } (V (V V ) (i ) ). In fact, we will need the commutativity of a similar diagram for (0.4) Φ {,2} (U V ) F (0,i ) Φ {,2} (U V ) which will be established in the next step. We assume this for now. It remains to show that the following (red) diagram commutes: Φ {} (V V V ) R Φ {,2,3} (V V V ) F i R(F (0,0,i ) ) Φ {} (V V V ) R Φ {,2,3} (V V V ) ( ) R Φ {,,2i+} (V (V V ) i ) R(F (0,0,,0,,,0,0) ) R Φ {,,2i+} (V (V V ) i ) F ( ) R(F (,0,0) ) R(F (,0,,,0,0) ) Φ {} (V V V ) R Φ {,2,3} (V V V ) R Φ {,,2i+} (V (V V ) i )

7 7 where (a) the ( ) square commutes by induction hypothesis, applied to U = V V. (Along horizontal arrows, we are growing (i + )-copies of V V on the last copy of V.) (b) the ( ) square commutes by an argument as in step (applied to {2} {, 2} {2}) shows that R(F (0,0,) ) acts as F. Hence R(F (,0,i ) ) acts as F i The missing link. We now justify the missing link, i.e., the commutativity with excursion for (0.4). Indeed, for each i, the statement for follows from that for Φ {,2} (U V ) F (0,i) Φ {,2} (U V ) F i by defining C I := C {} I and Φ I : Rep(G I ) Rep(Z I, C I ) as Φ I (V I ) := Φ {} I (U V I ) which again gives a compatible system of functors. Remark 0.6. Note that even if {Φ I } I fset arose from a tensor functor between symmetric monoidal categories, { Φ I } I fset will no longer form a tensor functor.