Lecture 11: Koszul Complexes

Lecture 11: Koszul Complexes 11 March 16 - Koszul complexes 11.1 More on the Hattori-Stallings Dennis trace Recall from Lemma 8.5 that the cocenter 𝑅 /[𝑅, 𝑅 ] = 𝐶 (𝑅 ) receives a universal trace map 𝜏 (𝑃 , 𝜑 ) ∈ 𝐶 (𝑅 ) where 𝑃 is a finitely generated projective and 𝜑 ∈ End(𝑃 ). In fact, if 𝑅 is Noetherian and of finite homological dimension, you can extend 𝜏 to 𝜏 (𝑀 , 𝜑 ) where 𝑀 is any finitely generated module. To do so, choose a finite projective resolution 0 → 𝑃 −𝑛 → 𝑃 −𝑛+1 → · · · → 𝑃 0 → 𝑀 → 0 (which exists because 𝑅 has finite homological dimension). Then we can • ) and this will be unique up to homotopy. Define lift 𝜑 to 𝜑˜ ∈ End(𝑃𝑀 ∑︁ (−1)𝑖 𝜏 (𝑃 −𝑖 , 𝜑˜ −𝑖 ) 𝜏 (𝑀 , 𝜑 ) = 𝑖 is a fully faithful functor to the homotopy category of complexes. Moreover, which is well-defined because 𝑀 ↦→ 𝜏 is additive on short exact sequences of modules. • 𝑃𝑀 Corollary 11.1: If 𝑅 is a finite-dimensional algebra of finite homological dimension over an algebraically closed field 𝑘 , then 𝐽 (𝑅 ) ⊂ [𝑅, 𝑅 ]. Proof. Lemma 11.2: For 𝑀 ∈ 𝑅 -Mod and 𝜑 ∈ End𝑅 (𝑀 ), we can find a 𝜑 -invariant Jordan-Holder series of 𝑀 .  𝑑𝑖 Proof. Consider 𝜑 | Soc(𝑀 ) : Soc(𝑀 ) → Soc(𝑀 ), where Soc(𝑀 ) = 𝑖 𝐿𝑖 is the socle of 𝑀 . Then 𝜑 induces an 𝑅 -linear map 𝐿𝑖𝑑𝑖 → 𝐿𝑖𝑑𝑖 i.e. an element of End𝑅 (𝐿𝑖𝑑𝑖 ) = Mat𝑑𝑖 (𝑘 ) (use Schur’s lemma) and this matrix has an eigenvector, which generates a 𝜑 -invariant irreducible submodule in 𝑀 . Then by inducting on the length of 𝑀 , we get a 𝜑 -invariant Jordan-Holder series. □   Thus, 𝜏 (𝑀 , 𝜑 ) = 𝑖 𝜏 (𝐿𝑖 , 𝜆𝑖 ) = 𝑖 𝜆𝑖 𝜏 (𝐿𝑖 , 1) where 𝜆𝑖 ∈ 𝑘 . It follows that the elements 𝜏 (𝐿𝑖 , 1) ∈ 𝐶 (𝑅 ) generate 𝐶 (𝑅 ) as a vector space over 𝑘 (use Lemma 8.5 or Example 8.6). We conclude that 𝐶 (𝑅 ) has dimension (over 𝑘 ) at ¯ It’s easy to most the number of irreducibles 𝐿𝑖 . On the other hand, let 𝑅¯ := 𝑅 /𝐽 (𝑅 ) and note that 𝐶 (𝑅 ) ↠ 𝐶 (𝑅). ¯ and 𝐽 (𝑅 ) ⊂ [𝑅, 𝑅 ]. ¯ = 𝑘 #𝐿𝑖 , so 𝐶 (𝑅 )  𝐶 (𝑅) □ see that 𝐶 (𝑅) Question : Is there a way to prove this without using the trace map? 11.2 Minimal resolutions and Koszul rings Given a module 𝑀 , how can we find its minimal resolution? For certain algebras called Koszul algebras, their minimal resolutions are called Koszul complexes. One great reference is [5, Section 2]. Let 𝐴 be a nonnegatively graded algebra over an algebraically closed field 𝑘 with 𝐴0 semisimple. We will be interested in the case 𝐴0 = 𝑘 so we can write 𝐴 = 𝑘 ⊕ 𝐴>0 .  Remark 11.3: An elementary property of minimal resolutions for graded modules is that if 𝑀 = 𝑖 ⩾0 𝑀𝑖 , then 𝑃 −𝑖 must be concentrated in degrees 𝑖 and higher, since the projective cover 𝑃 ↠ 𝑀 is an isomorphism in the bottom degree (use that 𝐴0 = 𝑘 is semisimple). We will need the following technical lemma. Lemma 11.4: Let 𝑀 be a finitely generated graded module over 𝐴. Then the following properties are equivalent: (i) 𝑀 is generated by degree 𝑖 elements, (ii) 𝑀 ⊗𝐴 𝑘 is concentrated in degree 𝑖 , (iii) Hom𝐴 (𝑀 , 𝑘 ) is concentrated in degree −𝑖 . Proof. Lemma follows from the Nakayama lemma together with the fact that Hom𝐴 (𝑀 , 𝑘 ) = Hom𝐴0 (𝑀 /𝐴 >0 𝑀 , 𝑘 ) = (𝑀 /𝐴 >0 𝑀 ) ∗ . □ Definition 11.5: We say that 𝐴 is Koszul if 𝑃 −𝑖 is generated by degree 𝑖 elements. Equivalently, Tor𝑖𝐴 (𝑘 , 𝑘 ) (where each of the 𝑘 are in degree 0) is concentrated in degree 𝑖 , which is equivalent to Ext𝑖𝐴 (𝑘 , 𝑘 ) is concentrated in degree −𝑖 (use Lemma 11.2 above). Theorem 11.6: a) Koszul rings are quadratic, i.e. 𝐴 = 𝑇 (𝑉 )/⟨𝐼 ⟩, where 𝑇 (𝑉 ) is the tensor algebra for a vector space 𝑉 and 𝐼 is a subspace of 𝑉 ⊗ 𝑉 . b) If 𝐴 is Koszul, then Ext𝐴• (𝑘 , 𝑘 ) = 𝐴! , where 𝐴! is the dual quadratic algebra 𝑇 (𝑉 ∗ )/⟨𝐼 ⊥ ⟩. Example 11.7: Let 𝐴 = 𝑇 (𝑉 ), so 𝐼 = 0. Then the dual quadratic algebra is 𝐴! = 𝑇 (𝑉 ∗ )/⟨𝑉 ∗ ⊗ 𝑉 ∗ ⟩ = 𝑘 ⊕ 𝑉 ∗ . Hence Ext𝐴 (𝑘 , 𝑘 ) is only nonzero in degrees 0 and 1. 𝑘 = 𝑇 (𝑉 )/⟨𝑉 ⟩ then has a free resolution in degrees 0 and 1. Example 11.8: Let 𝐴 = Sym(𝑉 ) = 𝑇 (𝑉 )/⟨∧2𝑉 ⟩. Then 𝐴! = 𝑇 (𝑉 ∗ )/⟨Sym2 (𝑉 ∗ )⟩ = Definition 11.9: The 𝑑 th Veronese subalgebra 𝐴 (𝑑 ) is • 𝑉 ∗. ∞ 𝑛=0 𝐴𝑛𝑑 . Let us mention the following theorem without a proof (see [3] for details). Theorem 11.10: If 𝐴 is a finitely generated commutative algebra, 𝐴 (𝑑 ) is Koszul for large 𝑑 . Remark 11.11: Using the approach of [6, Section 2] or [10] (see also Remark 12.2 below) one can easily prove (using Serre’s vanishing theorem) that for every 𝑚 ∈ Z⩾0 and large enough 𝑑 (depending on 𝑚 ) the algebra 𝐴 (𝑑 ) has the following property: 𝑃 −𝑖 is generated by degree 𝑖 elements for 𝑖 ⩽ 𝑚 . The statement of Theorem 11.10 is stronger, and the proof is more involved. 11.3 Koszul complexes Remark 11.12: Assume 𝐴 = 𝑇 (𝑉 )/⟨𝐼 ⟩ is quadratic. Then 𝑛 −2  ∑︁ 𝑛 ⊗𝑛 ⊗𝑖 ⊗𝑛 −𝑖 −2 𝑉 ⊗𝐼 ⊗𝑉 . 𝐴𝑛 = 𝑇 (𝑉 )/⟨𝐼 ⟩𝑛 = 𝑉 / 𝑖 =0 Define 𝑅𝑛 := 𝑛 −2  𝑉 ⊗𝑖 ⊗ 𝐼 ⊗ 𝑉 ⊗𝑛 −𝑖 −2 𝑖 =0 to be the intersection rather than the sum. Then 𝑅𝑛 = (𝐴𝑛! ) ∗ ,  𝑛 −2 ∑︁ ∗ ∗⊗𝑛 ∗ ⊗𝑖 ⊥ ∗ ⊗𝑛 −𝑖 −2 = 𝐴𝑛! . (𝑉 ) ⊗ 𝐼 ⊗ (𝑉 ) 𝑅𝑛 = 𝑉 / (1) 𝑖 =0 Definition 11.13: The Koszul complex, K• , is a complex of free 𝐴-modules · · · → 𝐴 ⊗𝑘 𝑅2 → 𝐴 ⊗𝑘 𝑅1 → ∞ denoted • • • 𝐴. As (graded) vector spaces, K = 𝑛 =0 K𝑛 . The differential of K𝑛 is given by: K𝑛𝑖 −𝑛 = 𝐴𝑖 ⊗ 𝑅𝑛 −𝑖 ↩→ 𝐴𝑖 ⊗ 𝑉 ⊗ 𝑅𝑛 −𝑖 −1 → 𝐴𝑖 +1 ⊗ 𝑅𝑛 −𝑖 −1 = K𝑛𝑖 +1−𝑛 where the left map is induced by the natural embedding 𝑅𝑛 −𝑖 ⊂ 𝑉 ⊗ 𝑅𝑛 −𝑖 −1 and the right map is induced by the multiplication 𝐴𝑖 ⊗ 𝑉 → 𝐴𝑖 +1 . Definition 11.14: Let 𝑉 be a vector space. A distributive lattice of subspaces of 𝑉 is a collection of subspaces satisfying • For 𝑌 in the lattice, 𝑋 ⊂ 𝑌 is also in the lattice • For 𝑋 , 𝑌 in the lattice, 𝑋 + 𝑌 is also in the lattice • For 𝑋 , 𝑌 , 𝑍 in the lattice, 𝑋 ∩ (𝑌 + 𝑍 ) = (𝑋 ∩ 𝑌 ) + (𝑋 ∩ 𝑍 ) (distributivity). Theorem 11.15 (Theorem 11.6 cont.): a) Koszul rings are quadratic, i.e. 𝐴 = 𝑇 (𝑉 )/⟨𝐼 ⟩, where 𝑇 (𝑉 ) is the tensor algebra for a vector space 𝑉 and 𝐼 is a subspace of 𝑉 ⊗ 𝑉 . b) If 𝐴 is Koszul, then Ext𝐴• (𝑘 , 𝑘 ) = 𝐴! , where 𝐴! is the dual quadratic algebra 𝑇 (𝑉 ∗ )/⟨𝐼 ⊥ ⟩. c) Say 𝐴 is a quadratic algebra. It is Koszul iff K is exact, i.e. 𝐻 𝑖 (K) = 0 for all 𝑖 ≠ 0, iff K is the minimal resolution of the left module 𝑘 . d) Say 𝐴 is a quadratic algebra. It is Koszul iff for all 𝑛 , the 𝑛 −1 vector spaces 𝑉 ⊗𝑖 ⊗𝐼 ⊗𝑉 ⊗𝑛 −𝑖 −2 , 𝑖 = 0, . . . , 𝑛 −2, generate a distributive lattice of subspaces of 𝑉 ⊗𝑛 . Lemma 11.16: A collection of vector subspaces in a vector space 𝑊 generate a distributive lattice iff there exists a basis of 𝑊 such that every subspace is spanned by a subset of the basis. Proof. Clear. □ Remark 11.17: The distributive property for the subspaces of 𝑉 ⊗𝑛 described above is what implies the exactness of K𝑛 . Moreover, the exactness of K𝑚 , 𝑚 ⩽ 𝑛 , implies the distributive property for the subspaces of 𝑉 ⊗𝑛 . −𝑙 For a collection W = (𝑊 ; 𝑊1, . . . , 𝑊𝑛 ), where  𝑊 isa vector space and 𝑊1, . . . , 𝑊𝑛 ⊂ 𝑊 are its subspaces let 𝐾 =  𝑙 −1 𝑙 −1 𝐾 −𝑙 (W) := 𝑖 =1 𝑊𝑖 / (𝑊𝑙 +1 + . . . + 𝑊𝑛 ) ∩ 𝑖 =1 𝑊𝑖 , where 𝑙 = 0, 1, . . . , 𝑛 + 1. For example, we have 𝐾 −𝑛 −1 = 𝑛  𝑖 =1 𝑊𝑖 , 𝐾 −𝑛 = 𝑛 −1  𝑖 =1 𝑊𝑖 , 𝐾 −𝑛 +1 = 𝑛 −2  𝑖 =1  𝑊𝑖 / 𝑊𝑛 ∩  𝑛   𝑊𝑖 𝑛 𝑛 ∑︁ ∑︁ , . . . , 𝐾 −1 = 𝑊 / 𝑊𝑖 , 𝐾 0 = 𝑊 / 𝑊𝑖 . 𝑖 =3 𝑖 =2 𝑖 =1 We have the natural maps 𝐾 𝑙 → 𝐾 𝑙 +1 that make 𝐾 • = 𝐾 • (W) into a complex. Lemma 11.18: If 𝑊1, . . . , 𝑊𝑛 ⊂ 𝑊 are proper subspaces and every proper subset of {𝑊1, . . . , 𝑊𝑛 } generate a distributive lattice then 𝑊1, . . . , 𝑊𝑛 do the same iff 𝐾 • (W) is exact. Proof. It is clear that if {𝑊1, . . . , 𝑊𝑛 } generate a distributive lattice then 𝐾 • (W) is exact (for example, use Lemma 11.16). Assume now that 𝐾 • (W) is exact. We prove the claim by the induction on 𝑛 . We follow [4, Section 4.5]. We will use the following notations. Given a collection 𝑈 1, . . . , 𝑈𝑛 ⊂ 𝑈 , say that a subspace 𝐵 ⊂ 𝑈 is a splitting for (𝑈 ; 𝑈 1, . . . , 𝑈𝑛 ) if there exists 𝐶 ⊂ 𝑈 such that 𝐵 ⊕ 𝐶 = 𝑈 and (𝐵 ∩ 𝑈𝑖 ) + (𝐶 ∩ 𝑈𝑖 ) = 𝑈𝑖 . We will say that (𝑈 ; 𝑈 1, . . . , 𝑈𝑛 ) is indecomposable if 𝑈 has no proper nonzero subspaces that split (𝑈 ; 𝑈 1, . . . , 𝑈𝑛 ). The following easy facts will be extremely useful. Fact (1): The subspace 𝑈 1 ∩ . . . ∩ 𝑈𝑖 or 𝑈 1 + . . . + 𝑈𝑖 is a splitting for (𝑈 ; 𝑈 1, . . . , 𝑈𝑛 ) iff it is a splitting for (𝑈 ; 𝑈𝑖 +1, . . . , 𝑈𝑛 ). Proof. Clear. Fact (2): Assume that (𝑈 1 + . . . + 𝑈𝑖 ) ∩ (𝑈𝑖 +1 ∩ . . . ∩ 𝑈 𝑗 ) = 0 and 𝑈𝑖 +1 ∩ . . . ∩ 𝑈 𝑗 is a splitting for (𝑈 ; 𝑈 1 + . . . + 𝑈𝑖 , 𝑈 𝑗 +1, . . . , 𝑈𝑛 ). Then 𝑈𝑖 +1 ∩ . . . ∩ 𝑈 𝑗 is a splitting for (𝑈 ; 𝑈 1, . . . , 𝑈𝑛 ). □ Proof. Let (𝑈𝑖 +1 ∩ . . . ∩ 𝑈 𝑗 ) ⊕ 𝐵 be a splitting for (𝑈 ; 𝑈 1 + . . . + 𝑈𝑖 , 𝑈 𝑗 +1, . . . , 𝑈𝑛 ). Our goal is to check that it also gives a splitting for (𝑈 ; 𝑈 1, . . . , 𝑈𝑛 ). From (𝑈 1 + . . . +𝑈𝑖 ) ∩ (𝑈𝑖 +1 ∩ . . . ∩𝑈 𝑗 ) = 0 we conclude that 𝑈 1 + . . . +𝑈𝑖 ⊂ 𝐵 so 𝑈 1, . . . , 𝑈𝑖 ⊂ 𝐵 . It remains to check that 𝑈𝑘 = (𝑈𝑘 ∩ (𝑈𝑖 +1 ∩ . . . ∩ 𝑈 𝑗 )) + (𝑈𝑘 ∩ 𝐵 ) for 𝑘 = 𝑖 + 1, . . . , 𝑗 . This is □ clear since 𝑈𝑖 +1 ∩ . . . ∩ 𝑈 𝑗 ⊂ 𝑈𝑘 . Fact (2’): Assume that (𝑈 1 ∩ . . . ∩ 𝑈𝑖 ) ∩ (𝑈𝑖 +1 + . . . + 𝑈 𝑗 ) = 0 and 𝑈 1 ∩ . . . ∩ 𝑈𝑖 is a splitting for (𝑈 ; 𝑈𝑖 +1 + . . . + 𝑈 𝑗 , 𝑈 𝑗 +1, . . . , 𝑈𝑛 ). Then 𝑈 1 ∩ . . . ∩ 𝑈𝑖 is a splitting for (𝑈 ; 𝑈 1, . . . , 𝑈𝑛 ). Proof. Same proof as the one of Fact 2. □ Let us now return to the proof. Without losing the generality, we can assume that W = (𝑊 ; 𝑊1, . . . , 𝑊𝑛 ) is indecomposable and all 𝑊𝑖 are nonzero (and proper). It then follows (use that by the inductive assumption, 𝑊1 ∩ 𝑊2, 𝑊3, . . . , 𝑊𝑛 ⊂ 𝑊 , 𝑊1, . . . , 𝑊𝑛 −2, 𝑊𝑛 −1 + 𝑊𝑛 ⊂ 𝑊 form distributive lattices and then apply Fact 1) that: 𝑊1 ∩ 𝑊2 = 0, 𝑊𝑛 −1 + 𝑊𝑛 = 𝑊 . (2) We can assume that 𝑛 ⩾ 4 (for 𝑛 = 3 the statement is clear, use exactness of 𝐾 • (W)). Assume that 𝑛 = 4. We have 𝑊1 ∩ 𝑊3 ∩ 𝑊4 = 0 = 𝑊2 ∩ 𝑊3 ∩ 𝑊4 (use Fact 1). We also have (𝑊1 +𝑊2 ) ∩𝑊3 ∩𝑊4 = ((𝑊1 +𝑊2 ) ∩𝑊3 ) ∩ ((𝑊1 +𝑊2 ) ∩𝑊4 ) = ((𝑊1 ∩𝑊3 ) + (𝑊2 ∩𝑊3 )) ∩ ((𝑊1 ∩𝑊4 ) + (𝑊2 ∩𝑊4 )). We claim that the intersection ((𝑊1 ∩ 𝑊3 ) + (𝑊2 ∩ 𝑊3 )) ∩ ((𝑊1 ∩ 𝑊4 ) + (𝑊2 ∩ 𝑊4 )) is zero. Indeed, if 𝑎 + 𝑏 = 𝑐 + 𝑑 for some 𝑎 ∈ 𝑊1 ∩ 𝑊3 , 𝑏 ∈ 𝑊2 ∩ 𝑊3 , 𝑐 ∈ 𝑊1 ∩ 𝑊4 , 𝑑 ∈ 𝑊2 ∩ 𝑊4 then 𝑎 − 𝑐 = 𝑑 − 𝑏 must lie in 𝑊1 ∩ 𝑊2 = 0 i.e. 𝑎 = 𝑐 ∈ 𝑊1 ∩𝑊3 ∩𝑊4 = 0, 𝑑 = 𝑏 ∈ 𝑊2 ∩𝑊3 ∩𝑊4 = 0 so 𝑎 = 𝑏 = 𝑐 = 𝑑 = 0. We conclude that (𝑊1 +𝑊2 ) ∩𝑊3 ∩𝑊4 = 0. It then follows from Fact 2 that 𝑊3 ∩𝑊4 splits (𝑊 ; 𝑊1, 𝑊2, 𝑊3, 𝑊4 ) so we must have 𝑊3 ∩𝑊4 = 0 i.e. 𝑊 = 𝑊3 ⊕ 𝑊4 . It remains to note that 𝑊 = 𝑊3 ⊕𝑊4 is splitting for (𝑊 ; 𝑊1, 𝑊2, 𝑊3, 𝑊4 ), and a contradiction finishes the argument. If 𝑛 > 4. The propperty (2) implies that (𝑊 ; 𝑊1, . . . , 𝑊𝑛 ) remains acyclic after arbitrary transpositions of 𝑊1, . . . , 𝑊𝑛 −2 (by acyclic, we mean that the corresponding complex 𝐾 • is exact, it will be equal to zero in this case). So we may assume that for certain 1 ⩽ 𝑖 ⩽ 𝑛 − 3 one has 𝐴 = 𝑊1 ∩ . . . ∩ 𝑊𝑖 ≠ 0 and each 𝑖 + 1-tuple from 𝑊1, . . . , 𝑊𝑛 −2 intersects by zero. Put 𝐵 = 𝑈𝑖 +1 + . . . + 𝑈𝑛 −2 . Then (𝑊 ; 𝐴; 𝐵 ; 𝑊𝑛 −1, 𝑊𝑛 ) satisfies the assumptions of Lemma 11.18 (acyclicity follows from the fact that 𝐴 ∩ 𝐵 = 0 and 𝑊𝑛 −1 + 𝑊𝑛 = 𝑊 ) so (from 𝑛 = 4 case) we conclude that 𝐴; 𝐵 ; 𝑊𝑛 −1, 𝑊𝑛 ⊂ 𝑊 generate a distributive lattice so 𝐴 is a splitting for (𝑊 ; 𝑊1, . . . , 𝑊𝑛 ) by Fact 2′ . Since 𝐴 ≠ 0, we get a contradiction. □ Proof (of Theorem 11.15). If Tor1 (𝑘 , 𝑘 ) is concentrated in degree 1, then 𝐴 ⩾1 is generated by degree 1 elements as an 𝐴-module (use the exact sequence 0 → 𝐴 ⩾1 → 𝐴 → 𝑘 → 0 together with Nakayama). Hence, 𝐴 is generated by degree 1 elements as a ring. Let 𝑉 = 𝐴1 and write 𝐴 = 𝑇 (𝑉 )/𝐼 . We have a map 𝐴 ⊗ 𝑉 → 𝐴. Using that Tor2 (𝑘 , 𝑘 ) is concentrated in degree 2, we see that ker(𝐴 ⊗𝑉 → 𝐴) is generated by elements in 𝐴1 ⊗𝑉 = 𝑉 ⊗𝑉 . These elements considered as elements of 𝑉 ⊗ 𝑉 ⊂ 𝑇 (𝑉 ) generated the ideal ker(𝑇 (𝑉 ) ↠ 𝐴), so 𝐴 is quadratic. Exactness of Koszul complex implies Koszul: If K𝑛 is exact for 𝑛 ⩾ 1, then K is a free resolution of 𝑘 as an 0 ∗ 𝐴-module. So now we can use it to compute Ext𝐴• (𝑘 , 𝑘 ). Since 𝑅𝑛∗ → − 𝑅𝑛−1 and 𝑅𝑛∗ = 𝐴𝑛! , Ext𝑛𝐴 (𝑘 , 𝑘 ) = 𝐴𝑛! . You also have to check that this is compatible with multiplication, but after showing that, we can deduce that 𝐴 is Koszul. To be continued next lecture. □