MODULARITY LIFTING THEOREMS RONG ZHOU 1. Introduction These are live-texed lecture notes from the MIT graduate class on Galois represen- tations given by Sug Woo Shin in Spring of 2014. The class expanded on the notes of Toby Gee's course from the Arizona Winter School in 2014, which it closely followed. The problem sets from the class contained some proofs of results which were stated in class and some of these were later added to the notes. Any mistakes in these notes are due to me and not the lecturer. 2. Lecture 1 Let F be a field and fix an algebraic closure F of F . We let Gal(F =F ) be the absolute Galois group of F . Definition 2.1. A Galois representation is a continuous group homomoprhism Gal(F =F ) ! GLn(R) where R is a topological ring. Most of the time we will take R to be Ql where l is a prime, such a representation will be a called an l-adic Galois representation. The first result we need about l-adic representations is that up to conjugation, the image lands in GLn(OL) where L is a finite extension of Qp. This follows from the next two lemma which actually hold in a more general context: Lemma 2.2. Let Γ be a compact topological group (in particular a Galois group is such) and let ρ :Γ ! GLn(Ql) be a continuous homomorphism. Then there exists a finite extension L=Ql such that ρ(Γ) ⊂ GLn(L). Proof. ρ(Γ) is compact Hausdorff, hence by the Baire Category theorem, the intersec- tion of a countable set of open dense subsets of ρ(Γ) is dense in ρ(Γ). Now since [ ρ(Γ) = (ρ(Γ) \ GLn(L)) L=Ql finite 1 2 RONG ZHOU and each set ρ(Γ) \ (GLn(L) is closed in ρ(Γ), it follows from the above, that some ρ(Γ) \ GLn(L) contains an open subset, hence ρ(Γ) \ GLn(L) itself is open. As Γ was profinite, we have that [ρ(Γ) : ρ(Γ) \ GLn(L)] is finite. If we choose coset representatives α1; :::; αr for ρ(Γ), there exists Li=Ql a finite ex- tension such that ρ(αi) 2 GLn(Li). Taking L to be the compositum of the Li we obtain the result. Lemma 2.3. Let Γ be a compact topological group, L=Ql a finite extension with ring of integers OL and ρ :Γ ! GLn(L) a continuous representation. Then there exists −1 g 2 GLn(L) such that gρg has image in GLn(OL) n Proof. It is enough to show that there exists a Γ invariant OL lattice Λ ⊂ L . Indeed n letting g be the change of basis matrix taking Λ to the standard basis of OL we get the result. n Define Λ0 := OL to be the standard lattice. Observe that GLOL (Λ0) ⊂ GLn(L) is −1 open, so that ρ (GLOL (Λ0)) ⊂ Γ has finite index. Let α1; :::; αr be a finite set of coset representatives in Γ. We then define r X Λ := ρ(Γ)Λ0 = αiΛ0 i=1 which is clearly a ρ(Γ) invariant lattice. Definition 2.4. Let R be a topological ring and Γ a topological group. Two continuous representations ρ1; ρ2 :Γ ! GLn(R) are said to be isomorphic if there exists and isomorphism φ : Rn ! Rn such that the following diagram commutes: ρ (γ) Rn 1 > Rn φ φ _ _ ρ (γ) Rn 2 > Rn for all γ 2 Γ As a consequence of the Lemma, a representation ρ :Γ ! GLn(Ql) with Γ compact is isomorphic to one with image in GLn(OL) Our next result will be the Brauer Nestbitt theorem, this gives a condition for when the semi simplification of two representations are isomorphic and which we will need for the study of the mod p reduction of a representation. We first need to define the notion of semisimple representation and semisimplification. As before Γ is a compact topological group. ∼ Definition 2.5. Let k be a field, and ρ :Γ ! GLn(K) = GLn(V ). Then ρ is semi simple if ∼ r ρ = ⊕i=1ρi MODULARITY LIFTING THEOREMS 3 for some irreducible ρi :Γ ! GLni (R). Definition 2.6. Let ρ :Γ ! GLn(V ) and choose 0 ⊂ V1 ⊂ ::: ⊂ Vr = V where the Vi are Γ invariant subgroups with Vi=Vi−1 irreducible. The semi simplification ss ss r ρ is the representation induced by ρ on the vector space V = ⊕i=1Vi=Vi+1 It can be shown that this is well defined and does not depend on the choice of the Vi's. Example 2.7. Consider ρ :Γ ! GL2(k) given by a(γ) b(γ) γ 7! 0 d(γ) Then ρss = a ⊕ d Theorem 2.8 (Brauer, Nesbitt). Let k be a field and Γ a (topological) group, n1; n2; ≥ 1. For i = 1; 2 and ρi :Γ ! GLni (k), assume either: i) 8γ 2 Γ we have det(1 − ρ1(γ)T ) = det(1 − ρ1(γ)T ) or ii) char k = 0(>> 0), and 8γ 2 Γ we have trρ1(γ) = trρ2(γ) ss ss Then ρ1 = ρ2 We will need the following lemma: Lemma 2.9. Let R be an associative k algebras (not necessarily commutative eg. R = k[Γ]) and let M1; :::; Mr be non-isomorphic R-simple modules which are finite dimensional over k. Then 9e1; :::; er 2 R such that eimi = mi 8i; 8mi 2 Mi eimj = 0 8j 6= i; 8mj 2 Mj r Proof. (Sketch) Upon replacing R by its image in Endk ⊕i=1 Mi wlog. we can assume, k is a finite dimensional semisimple k-algebra. ∼ Qs The Artin Wedderburn theorem then tells us that R = i=1 Mni (Di) where Di is a division algebra over k. One then deduces that r = s and R acts as Mni (Di) on Mi for th i = 1; :::; r (after reordering). Defining ei to be 1 on the i component and 0 elsewhere we obtain the result. Proof of Brauer Nesbitt: Let ρi :Γ ! GLn(V ); j = 1; 2. Since the conditions in the theorem don't change upon taking semisimplifications, we may assume the ρi are semisimple. 4 RONG ZHOU 1 2 Let M1; :::; Mr be the distinct irreducible subrepresentations of V1 ⊕ V2. Let mi ; mi 1 2 be the multiplicities of Mi in V1 and V2 respectively. It suffices to prove that mi = mi for i = 1; :::; r. Case i) Take ei as in Lemma 1.9, then the characteristic polynomial of ei on Mi is (t − 1)dim Mi and tdim Mj when j 6= i. From the equality of characteristic polynomials of 1 2 ρ1(ei) and ρ2(ei) one finds the dimensions mi and mi match. Case ii)For j = 1; 2, there exists a unique continuous map θj which makes the following diagram commute: ρ tr Γ >j GL (V ) > k n j > _ θj k[Γ] It can be shown that θ1(α) = θ2(α)8α 2 k[Γ]. Plugging in ei, we obtain θ1(ei) = 1 2 mi dim Mi and θ2(ei) = mi dim Mi. 3. Lecture 2 The Brauer Nesbitt Theorem allows us to define the reduction mod l of l-adic rep- resentations. As before L is a finite extension of Ql with ring of integers OL Definition 3.1. Let Γ be a compact group and ρ :Γ ! GLn(L) be a representation. 0 0 0 Choose a conjugate ρ of ρ such that ρ has image in GLn(OL) and let ρ be composition with GLn(OL) ! GLn(kL). The reduction of ρ mod l is then defined to be ρ := (ρ0)ss Lemma 3.2. ρ is well defined (up to isomorphism), i.e. it does not depends on the choice of ρ0 0 Proof. Let γ 2 Γ. Observe that char(ρ(γ)) = char(ρ (γ)) 2 kL[T ], the Lemma then follows by Theorem 1.8 We now recall some basic properties of number fields and local fields. In this course a number field will be a finite extension of Q and local field will be finite extension of Qp or R so we exclude the function field case. Definition 3.3. Let F be a number field. A place of F is an equivalence class of valuations on F (where two valuations are considered equivalent if they induce the same topology on F ). A place is finite if the valuation is non-archimedean, otherwise it is an infinite place. A finite place is p-adic if jpj < 1. MODULARITY LIFTING THEOREMS 5 We state some basic facts about places: 1)There are 1-1 correspondences between the following sets i) fprime ideals of OF dividing pg ii) fp-adic places of F g iii) Gal(Qp=Qp)nHomQ(F; Qp) The first bijection is given by sending a p-adic place of F to the set fa 2 OF : jaj < 1g. The bijection between ii) and iii) is given by sending a φ 2 Gal(Qp=Qp)nHomQ(F; Qp) to the place j:jp ◦ φ. 2) The infinite places of F are in one to one correspondence with Gal(C=R)nHomQ(F; C), and these are a disjoint union of of real places and complex places, where the real places are the 1 element orbits of Gal(C=R)nHomQ(F; C) and the complex places are the two elements orbits. Now for a place v of F , let Fv to be the completion of F with respect to the topology induced by v.