Lecture 24: Gelfand-Kirillov Dimension

Lecture 24: Gelfand-Kirillov Dimension 24 May 11 - Gelfand-Kirillov dimension 24.1 Growth of algebras and Gelfand-Kirillov dimension Recall that we defined the growth of a (finitely generated) algebra as follows: pick a finite-dimensional space of generators 𝑉 , which gives us a (surjective) homomorphism 𝑇 𝑉 ↠ 𝐴 and induces a filtration of 𝐴 by setting    ⊗𝑖 𝑉 𝐴 ⩽𝑛 = im . 𝑖 ⩽𝑛 Let 𝑑 (𝑛 ) := dim(𝐴 ⩽𝑛 ). Then we saw that the order of growth was independent of 𝑉 . Recall that 𝐴 has • subexponential growth if 𝑑 (𝑛 )√︁< 𝑐𝑛𝛼 for some 𝑐 for all 𝛼 > 1 • exponential growth if lim sup 𝑛 𝑑 (𝑛 ) > 1 • polynomial growth if there exists 𝑐 , 𝛿 such that 𝑑 (𝑛 ) ⩽ 𝑐 𝑛𝛿 . Definition 24.1: The Gelfand-Kirillov dimension of an algebra 𝐴 is inf {𝛿 | ∃𝑐 , 𝑑 (𝑛 ) ⩽ 𝑐 𝑛𝛿 }. That is, GKdim(𝐴) ∈ R⩾0 ∪ {∞}, and this is well-defined since another function 𝑑 ′ such that there exists 𝑎 ⩾ 1 with 𝑑 ′ (𝑛 /𝑎 ) ⩽ 𝑑 (𝑛 ) ⩽ 𝑑 ′ (𝑎𝑛 ) will lead to the same value. Example 24.2: The GK dimension of 𝑘 [𝑥 1, . . . , 𝑥𝑛 ] is 𝑛 . Remark 24.3: There are similar definitions for finitely generated groups; if they have exponential growth, then they’re called hyperbolic groups. (There is not much similarity in the methods and theorems though.) Remark 24.4: If 𝐴 ⩽𝑛+1 = 𝐴 ⩽𝑛 , then 𝐴 = 𝐴 ⩽𝑛 and 𝑑 (𝑛 ) will eventually be dim 𝐴. Hence, if dim 𝐴 < ∞, GKdim(𝐴) = 0. Otherwise, we’ll always have 𝑑 (𝑛 ) ⩾ 𝑛 + 1, so the GK dimension will be ⩾ 1. Lemma 24.5: a) If GKdim(𝐴) < 1, then 𝐴 is finite-dimensional so GKdim(𝐴) = 0. b) GKdim(𝐴 [𝑡 ]) = GKdim(𝐴) + 1 c) GKdim(𝐴 [𝑎 −1 ]) = GKdim(𝐴) if 𝑎 is central and regular. a) If 𝑑 (𝑛 + 1) = 𝑑 (𝑛 ) for some 𝑛 , then 𝐴 ⩽𝑛+1 = 𝐴 ⩽𝑛 . So 𝐴 = 𝐴 ⩽𝑛 and 𝑑 (𝑛 ) will eventually be dim 𝐴. Hence, if dim 𝐴 < ∞, GKdim(𝐴) = 0. Otherwise, we’ll always have 𝑑 (𝑛 + 1) ⩾ 𝑑 (𝑛 ) + 1, so 𝑑 (𝑛 ) ⩾ 𝑛 and the GK dimension will be ⩾ 1. b) Let 𝐵 = 𝐴 [𝑡 ] and 𝑉𝐵 = 𝑉𝐴 ⊕ 𝑘 𝑡 . Then 𝐵 ⩽𝑛 = 𝑡 𝑛 𝐴 ⩽0 ⊕ 𝑡 𝑛 −1𝐴 ⩽1 ⊕ · · · ⊕ 𝐴 ⩽𝑛 , so dim 𝐵 ⩽𝑛 ⩽ (𝑛 + 1) dim 𝐴 ⩽𝑛 . Also, dim 𝐵 ⩽𝑛 ⩾ 𝑛 dim 𝐴 ⩽𝑛 , so GKdim(𝐴 [𝑡 ]) = GKdim(𝐴) + 1. c) Again, add 𝑎 −1 to the space of generators, Then dim 𝐴 ⩽𝑛 ⩽ dim(𝐵 ⩽𝑛 ) ⩽ dim(𝐴 ⩽2𝑛 ) because 𝐵 ⩽𝑛 ↩→ 𝐴 ⩽2𝑛 via multiplication by 𝑎𝑛 . So GKdim(𝐴 [𝑎 −1 ]) = GKdim(𝐴). □ Proof. Part b) implies that GKdim(𝑘 [𝑥 1, . . . , 𝑥𝑛 ]) = 𝑛 . 24.2 Warfield’s Theorem The GK dimension of a noncommutative ring can take any value ⩾ 2. Theorem 24.6 (Warfield): For any real 𝛿 ⩾ 2, there exists an algebra with 2 generators whose GK dimension is 𝛿 . Proof. Part b) of Lemma 24.1 implies we only have to show this for 𝛿 ∈ (2, 3). We will construct a quotient of 𝑘 ⟨𝑥 , 𝑦 ⟩ by monomials. Fix a monotonically increasing sequence 𝛾𝑛 , 𝑛 = 1, . . . ,, let 𝐼 ⊂ 𝑘 ⟨𝑥 , 𝑦 ⟩ be the ideal spanned by the monomials of degree at least 3 in 𝑦 and the monomials 𝑥 𝑖 𝑦𝑥 𝑗 𝑦𝑥 𝑘 , 𝑗 < 𝛾𝑛 , 𝑛 = 𝑖 + 𝑗 + 𝑘 . The quotient 𝑘 ⟨𝑥 , 𝑦 ⟩/𝐼 is a graded algebra 𝐴; let 𝐴𝑛 be the component of degree 𝑛 . Then   𝑛 + 2 − 𝛾𝑛 dim(𝐴𝑛 ) = 1 + 𝑛 + 2 where the 1 + 𝑛 comes from monomials of degree 0, 1 in 𝑦 . If we take 𝑞 ∈ (0, 1) and set 𝛾𝑛 = 𝑛 − [𝑛𝑞 ], then GKdim(𝐴) = max(2, 2𝑞 + 1). Hence this gives you anything in (2, 3). □ Remark 24.7: This doesn’t happen for finitely presented monomial algebras. Notice that for every finitely generated algebra 𝐴, one can find a finitely generated monomial algebra 𝐴¯ with the same growth function by setting 𝐴¯ to be the associated graded for a filtration on 𝐴. But the same construction for 𝐴 finitely presented does not imply that 𝐴¯ is finitely presented. 24.3 Bergman gap theorem The proof of this theorem is presented in [8] chapter 2. Theorem 24.8 (Bergman gap): There is no finitely generated algebra whose GK dimension is strictly between 1 and 2. Proof. The theorem follows from the below proposition. To reduce to a graded algebra generated in degree 1, we can reduce to 𝐴 = 𝑘 ⟨𝑥 1, . . . , 𝑥𝑛 ⟩/𝐽 where 𝐽 is a monomial ideal. Then take the associated graded, first by total degree, then by lexicographical order. Then either dim 𝐴𝑑 ⩾ 𝑑 , which implies that the GK dimension is at least 2, or there exists 𝑑 with dim 𝐴𝑑 < 𝑑 , which (by the proposition) implies that dim 𝐴𝑛 is bounded by a constant and the GK dimension is at most 1. □ Proposition 24.9: If 𝐴 is a graded algebra generated in degree 1 and there exists 𝑑 such that dim 𝐴𝑑 < 𝑑 , then GKdim(𝐴) ⩽ 1. Proof. WLOG we can assume that all relations are monomial in degree 𝑑 . To prove this, we define “allowed words”, where a word is allowed iff all its subwords of length 𝑑 are allowed. Let 𝑆 be the set of allowed words of degree 𝑑 and suppose |𝑆 | ⩽ 𝑑 . Then the number of allowed words of degree 𝑁 is bounded. This reduces to Lemma 24.10: Assume that there at most 𝑑 allowed words of length 𝑑 . Then for 𝑛 ⩾ 2𝑑 , every allowed word of length 𝑛 has the form 𝑤 = 𝑤 1𝑤 2𝑤 3 where 𝑤 2 is 𝑝 -periodic for 𝑝 ⩽ 𝑑 , |𝑤 1 |, |𝑤 3 | ⩽ 𝑑 − 𝑝 , and |𝑤 2 | ⩾ 𝑑 + 𝑝 . (A finite word 𝑥 1 · · · 𝑥𝑛 is 𝑝 -periodic if 𝑥𝑖 +𝑝 = 𝑥𝑖 when 𝑖 , 𝑖 + 𝑝 ∈ [1, . . . , 𝑛 ].) Proof. We induct on |𝑤 |. The base case is |𝑤 | = 2𝑑 . Such a word will have 𝑑 + 1 subwords of length 𝑑 , but since there are only 𝑑 distinct allowed words, at least two of these coincide and we have the desired periodicity. Now we need the following: Lemma 24.11: If a periodic word with minimal period 𝑝 contains two equal subwords of length ⩾ 𝑝 − 1, then they are 𝑛𝑝 letters apart. Proof. Extend the word to an infinite 𝑝 -periodic word. Suppose the equal subwords are 𝑥𝑖 +1 · · · 𝑥𝑖 +𝑟 and 𝑥 𝑗 +1 · · · 𝑥 𝑗 +𝑟 with 𝑟 ⩾ 𝑝 − 1. Then the subwords 𝑥𝑖 𝑥𝑖 +1 · · · 𝑥𝑖 +𝑝 −1 and 𝑥 𝑗 𝑥 𝑗 +1 · · · 𝑥 𝑗 +𝑝 −1 are each a full period of the word 𝑥 . Since 𝑥𝑖 +𝑞 = 𝑥 𝑗 +𝑞 for all 1 ⩽ 𝑞 ⩽ 𝑟 , then 𝑥𝑖 = 𝑥 𝑗 also. So 𝑥 also has equal subwords 𝑥𝑖 · · · 𝑥𝑖 +𝑝 −1, 𝑥 𝑗 · · · 𝑥 𝑗 +𝑝 −1 . Let the word have length 𝑚 and 1 ⩽ ℓ ⩽ 𝑚 , and let 𝑡 be an integer such that ℓ + 𝑡 𝑝 = 𝑖 + 𝑠 for 0 ⩽ 𝑠 ⩽ 𝑝 − 1. Then 𝑥 ℓ +( 𝑗 −𝑖 ) = 𝑥 ℓ +( 𝑗 −𝑖 )+𝑡 𝑝 = 𝑥𝑖 +𝑠 +( 𝑗 −𝑖 ) = 𝑥 𝑗 +𝑠 = 𝑥𝑖 +𝑠 = 𝑥 ℓ +𝑡 𝑝 = 𝑥 ℓ so 𝑥 has period 𝑗 −𝑖 . Thus 𝑥 has period equal to the greatest common divisor of 𝑝 , 𝑗 −𝑖 and the minimality of 𝑝 implies that 𝑝 | 𝑗 − 𝑖 as desired. □ Now we finish the proof of the lemma. Write 𝑤 = 𝑥 1𝑤 ′ and 𝑤 ′ = 𝑤 1′ 𝑤 2′ 𝑤 3′ . If |𝑤 1′ | < 𝑑 − 𝑝 , there’s nothing to do. Otherwise, in 𝑥 1𝑥 2 · · · 𝑥𝑑 , find two coinciding length 𝑑 words. These intersect 𝑤 2 by at least 𝑝 − 1, so their intersections with 𝑤 2′ differ by a shift by 𝑛 and 𝑝 |𝑛 . One of them ends at 𝑥 2𝑑 −𝑝 (or to the left) so it contains 𝑥𝑑 −𝑝 +1 . □ Hence 𝑥𝑑 −𝑝 +1 = 𝑥𝑑 −𝑝 +1+𝑛 = 𝑥𝑑 +1 . This finishes the proof of the proposition. □ 24.4 Ufnarovskii graph Another way of working with allowed words is via the overlap graph, called the Ufnarovskii graph; the proof of the theorem can also be interpreted via the graph. Consider an oriented graph 𝑈 whose vertices are allowed length 𝑑 words and which has an edge between 𝑤 1 and 𝑤 2 iff 𝑤 2 is obtained from 𝑤 1 by removing the first letter and adding a letter at the end. Then paths of length 𝑛 − 𝑑 correspond to allowed words of degree 𝑛 ⩾ 𝑑 . The proof of the Bergman gap theorem can be restated as follows: if there are at most 𝑑 allowed words of length 𝑑 , show that 𝑈 contains at most one oriented cycle. Then any path in the graph enters the cycle at most once, traverses the cycle, then leaves the cycle; this is the factorization 𝑤 = 𝑤 1𝑤 2𝑤 3 in the lemma above (see [1, Section VI.4]). 24.5 Smoktunowicz and Berele theorems We state without proof two related results: Theorem 24.12 (A. Smoktunowicz): The Gelfand-Kirillov dimension of a graded domain cannot fall within the open interval (2, 3). Theorem 24.13 (Berele): Finitely generated PI algebras have finite GK dimension. 24.6 GK dimension of a module Definition 24.14: We can likewise define the Gelfand-Kirillov dimension of a finitely generated module over 𝐴 by defining 𝑑𝑀 (𝑛 ) = dim 𝑀⩽𝑛 where we pick generators 𝑊 ⊂ 𝑀 and 𝑀⩽𝑛 = 𝐴 ⩽𝑛 · 𝑊 , and setting GKdim(𝑀 ) = inf {𝛿 | ∃𝑐 , 𝑑𝑀 (𝑛 ) ⩽ 𝑐 𝑛𝛿 }. Again, this is not dependent on the choices of 𝑊 . Definition 24.15: We say that the GK dimension is exact for modules over an algebra 𝐴 if for 𝑀 ⊃ 𝑁 , GKdim(𝑀 ) = max(GKdim(𝑁 ), GKdim(𝑀 /𝑁 )). Example 24.16: GK dimension is exact for finitely generated modules over Noetherian PI algebras. Suppose that 𝐴 is an algebra with commutative associated graded (which also is then automatically finitely generated, hence Noetherian). Then the GK dimension is exact for (f.g. modules over) 𝐴, because Proposition 24.17: In this case, GKdim(𝑀 ) is the dimension of the support of the gr(𝐴) module gr(𝑀 ) =  𝑀⩽𝑑 /𝑀⩽𝑑 −1 . ¯ Given an In fact, there is a closer relation between the commutative and noncommutative pictures. Let gr 𝐴 = 𝐴.  ¯ increasing filtration on 𝐴 such that 𝐴 is commutative, let a good filtration on 𝑀 be a filtration such that 𝑀 = 𝑀⩽𝑑 ,  𝑀⩽𝑑 = 0, 𝐴 ⩽1 𝑀⩽𝑛 ⊂ 𝑀⩽𝑛+1 , and gr 𝑀 = 𝑀¯ is a finitely generated 𝐴¯ module. ¯ Lemma 24.18: For 𝐴, 𝑀 as above, the (set theoretic)  supp(gr 𝑀 ) ⊂ Spec(𝐴) and does not depend on the  support ¯ ¯ choice of filtration. Moreover, the class of 𝑀 in 𝐾 𝐴-mod𝑆 (the Grothendieck group) is independent of the choice ¯ ¯ of the filtration, where 𝐴-mod 𝑆 is the category of finitely generated 𝐴-modules with set-theoretic support contained in 𝑆 . ¯ Remark 24.19: The expression “set-theoretic support” refers to thinking of finitely generated 𝐴-modules as ¯ and 𝑀 is ¯ ¯ coherent sheaves on Spec(𝐴). Closed subsets 𝑆 ⊂ Spec(𝐴) correspond to radical ideals 𝐼𝑆 ⊂ 𝐴, set-theoretically supported on 𝑆 iff every element of 𝑀 is annihilated by some power of 𝐼𝑆 . Note that being scheme-theoretically supported on 𝑆 would instead mean that 𝑀 is annihilated by 𝐼𝑆 , which is stronger. ′ , find 𝑚 such that Proof (of lemma, sketch). Given two good filtrations 𝑀⩽𝑑 and 𝑀⩽𝑑 ′ 𝑀⩽𝑑 −𝑚 ⊂ 𝑀⩽𝑑 ⊂ 𝑀⩽𝑑 −𝑚+1 . Inducting on 𝑀 , we can reduce to the situation when 𝑚 = 0 and ′ ⊂ 𝑀⩽𝑑 +1 . 𝑀⩽𝑑 ⊂ 𝑀⩽𝑑 Let 𝑁 =  ′ ′ 𝑀⩽𝑑 /𝑀⩽𝑑 −1, 𝑁 =  ′ 𝑀⩽𝑑 /𝑀⩽𝑑 . Then there are short exact sequences 0 → 𝑁 ′ → 𝑀¯ → 𝑁 → 0 0 → 𝑁 → 𝑀¯ ′ → 𝑁 ′ → 0 which shows both statements. □ ¯ Remark 24.20: 𝑀¯ is naturally graded, but the class of 𝑀¯ in the Grothendieck group of graded 𝐴-modules may depend on the choice of the filtration. This is because one can equip 𝑁 , 𝑁 ′ with a grading so that the first displayed SES is one of the graded modules, but the arrows in the second one will not agree with the grading. 24.7 Poincare series Theorem 24.21 (Stephenson-Zhang): If 𝐴 is right (or left) Noetherian, it has subexponential growth. Lemma 24.22: √ a) 𝐴 has exponential growth iff 𝑎𝑛 = dim(𝐴𝑛 ) has exponential growth iff lim sup 𝑛 𝑎𝑛 > 1. b) For a sequence 𝑎 (𝑛 ) of exponential growth, there exist 0 < 𝑟 1 < 𝑟 2 < · · · such that 𝑎 (𝑟𝑘 ) < 𝑘 −1 ∑︁ 𝑎 (𝑟𝑘 − 𝑟𝑖 ) . 𝑖 =1 Proof. Having fixed 𝑎 (1), . . . , 𝑎 (𝑚 ), there are infinitely many 𝑛 such that 𝑎 (𝑛 ) ⩾ 𝛼 𝑟𝑖 𝑎 (𝑛 − 𝑟𝑖 ), 𝑖 = 1, . . . , 𝑚 . We can make the choice such that 𝛼 𝑟𝑘 > 2𝑘 . □ Proof (of theorem). Apply this to 𝑎 (𝑛 ) = dim 𝐴𝑛 . Inductively choose 𝑥𝑖 ∈ 𝐴𝑟𝑖 such that 𝑥𝑘 ∉ Theorem 24.23: Let 𝐴 = 𝑘 ⊕ Then  𝑖 ⩾1 𝐴𝑖 𝑘 −1 𝑖 =1 𝑥𝑖 𝐴𝑘 −𝑖 . be right (or left) Noetherian of right (or left) finite homological dimension. ℎ (𝑡 ) = 1 , 𝑞 (𝑡 ) ∈ Z[𝑡 ] 𝑞 (𝑡 ) where ℎ (𝑡 ) is the Hilbert series and 𝑞 (𝑡 ) is a polynomial whose roots are all roots of unity. □ Proof. We must have 𝑞 (𝑡 ) = ∑︁ (−1)𝑖 dim Tor𝑖𝐴 (𝑘 , 𝑘 )𝑡 𝑖 (i.e. the graded Euler characteristic of Tor𝐴 (𝑘 , 𝑘 )).  All the roots  𝑧𝑖 of 𝑞 (𝑡 ) must have |𝑧𝑖 | ⩾ 1; otherwise, 𝑎𝑛 𝑡 𝑛 has radius of convergence < 1 and 𝐴 has exponential growth. But 𝑧𝑖 = 1 because their product will be 1/the leading coefficient of 𝑞 (𝑡 ). So all the roots satisfy |𝑧𝑖 | = 1. Since they are roots of a polynomial in Z[𝑡 ], they are roots of unity. □ Conjecture 24.24 (Polishchuk-Positselski): The Hilbert series of a Koszul algebra is rational. Moreover, if both 𝐴 and 𝐴! have finite GK dimension, then they have the Hilbert series of a symmetric tensor exterior. Conjecture 24.25 (Anick): Assume 𝐴 is right Noetherian. If both GKdim(𝐴) and hdim(𝐴) are finite, then the Hilbert series of 𝐴 equals that of the symmetric algebra.