The product rule for derivations on finite dimensional split semi-simple Lie algebras over a field of characteristic zero

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1 The product rule for derivations on finite dimensional split semi-simple Lie algebras over a field of characteristic zero Richard L. Kramer Department of Mathematics Iowa State University Ames, Iowa 50011, USA Abstract In this article we consider maps π : R R on a non-associative ring R which satisfy the product rule πab = πab + aπb for arbitrary a, b R, calling such a map a production on R. After some general preliminaries, we restrict ourselves to the case where R is the underlying Lie ring of a finite dimensional split semi-simple Lie algebra over a field F of characteristic zero. In this case we show that if π is a production on R, then π necessarily satisfies the sum rule πa + b = πa + πb, that is, we show that the product rule implies the sum rule, making π a derivation on the underlying Lie ring of R. We further show that there exist unique derivations on the field F, one for each simple factor of R, such that appropriate product rules are satisfied for the Killing form of two elements of R, and for the scalar product of an element of F with an element of R. Key words: derivation, product rule, Lie ring, Lie algebra, non-associative ring, non-associative algebra 1 Preliminaries Definition 1 Let R be a non-associative ring. A map π : R R is called a production on R provided that πxy = πxy + xπy for all x, y R. A production which also satisfies πx + y = πx + πy for all x, y R is called a derivation on R. address: ricardo@iastate.edu Richard L. Kramer. Preprint submitted to Journal of Algebra 2 December 2006

2 Proposition 2 Let R be a non-associative ring and let π : R R be a production on R. Then π0 = 0. PROOF. We calculate π0 = π0 0 = π π0 = 0. Proposition 3 Let R be a non-associative ring with identity and let π : R R be a production on R. Then π1 = 0 and 2π 1 = 0. PROOF. To see that π1 = 0, we calculate π1 = π1 1 = π11 + 1π1. To see that 2π 1 = 0 whenever x 2 = 1, we calculate 0 = π1 = π 1 1 = π π 1 = 2π 1. Example 4 Let R be the ring Z/4 of integers modulo 4. Then it is easy to see that the productions on R are precisely the maps π : R R such that π0 = π1 = 0 and π2, π3 {0, 2}. Thus there are exactly four productions on the ring R = Z/4. Remark 5 Note that Example 4 shows that the second conclusion of Proposition 3 cannot in general be improved to read π 1 = 0, even in the case of commutative rings with identity. Lemma 6 Let S, A, B, T and C be abelian groups. Suppose that we are given a bilinear map denoted by juxtaposition A B C. Suppose that we have bilinear maps also denoted by juxtaposition S A A and S C C such that the following diagram with the obvious maps commute: S A B S C A B C sab = sab Suppose further that we also have bilinear maps also denoted by juxtaposition B T B and C T C such that the following diagram with the obvious maps commute: A B T C T A B C abt = abt Suppose that we have maps π A : A A, π B : B B, and π C : C C which jointly satisfy the equation the following production equation for any a A and b B. π C ab = π A ab + aπ B b 1 2

3 Then, given any s 1,..., s n S, a 1,..., a n A, and any b 1,..., b m B and t 1,..., t m T the following equation holds for π C, where we write a = n s i a i and b = m j=1 b j t j. m m π C ab + s i π C a i b j t j = s i π C a i b + π C ab j t j 2 j=1 j=1 Remark 7 Note that expressions such as sabt with s S, a A, b B and t T are unambiguous, because of the associativity represented by the commutative diagrams above. However, sct with c C in general is ambiguous, since it is in general possible that sct sct. Of course, if c can be written as c = a i b i with a i A and b i B, then sct = s a i b i t = sa i b i t = sai b i t = sa i b i t = sa i b i t = sa i b i t = s a i b i t = sct. In particular, sπ C abt = sπ A ab + aπ B bt is unambiguous. Remark 8 It should be emphasized that 1 is the only assumption made about the maps π A, π B and π C. In particular no assumption is made that any of the maps are additive. Since the map A B C is bilinear, it is easy to prove that π C 0 C = 0 C. We need only note that π C 0 C = π C 0 A 0 B = π A 0 A 0 B + 0 A π B 0 B = 0 C + 0 C = 0 C. However, without knowledge of the linear map A B C, this is all that can be said, and nothing analogous need hold for π A and π B. In fact, if the bilinear map A B C is the trivial map a, b 0, then 1 reduces to simply saying that π c 0 C = 0 C, so that π A and π B are totally arbitrary, as is π C, so long as it maps 0 C to 0 C. PROOF. π C ab = π A ab + aπ B b m n = π A a b j t j + s i a i π B b j=1 m = π A ab j t j + s i a i π B b j=1 m = π C ab j aπ B b j t j + s i π C a i b π A a i b j=1 m n = π C ab j s i a i π B b j t j j=1 m + s i π C a i b π A a i b j t j j=1 m m = π C ab j t j + s i π C a i b s i π A a i b j + a i π B y j t j j=1 j=1 3

4 m m = π C xy j t j + s i π C a i b s i π C a i b j t j j=1 j=1 Corollary 9 Let R be a non-associative ring and let π : R R be a production on R. Let x i R for 1 i n and y j R for 1 j m. Then we have m m πxy + πx i y j = πx i y + πxy i 3 j=1 where x = n x i and y = m j=1 y j. j=1 PROOF. Apply Lemma 6 with A = B = C = R and S = T = Z, together with the obvious bilinear maps. Let π A = π B = π C = π, s i = t j = 1, a = x and b = y. All of the hypotheses are satisfied, and 2 reduces to 3. Corollary 10 Let R be a non-associative ring and let π : R R be a production on R. Let x i, y i R for 1 i n. Suppose further that x i y j = 0 whenever i j, so that xy = n x i y i, where x = n x i and y = n y i. Then we have πxy = πx i y i. PROOF. First note that πx i y j = π0 = 0 for i j, by Proposition 2. Note also that xy i =x i y = x i y i. With this in mind, Corollary 9 now says that πxy + πx i y i = πx i y i + πx i y i from which the corollary follows. Corollary 11 Let R be a non-associative ring and let π : R R be a production on R. Let u, v R satisfy u 2 = v 2 = 0. Then we have πuv + vu = πuv + πvu. PROOF. Setting x 1 = y 1 = u and x 2 = y 2 = v in Corollary 9 with n = m = 2, we see that πuv + vu + πuv + πvu = πuv + πvu + πvu + πuv from which the corollary follows. 4

5 Corollary 12 Let R be a non-associative ring and let π : R R be a production on R. Let u, v R satisfy u 2 = v 2 = 0 and vu = uv. Then we have π uv = πuv. PROOF. By Proposition 2 and Corollary 11, we see that πuv + π uv = πuv + πvu = πuv + vu = π0 = 0. Corollary 13 Let R be a non-associative algebra over the commutative ring with identity Λ, and let π : R R be a production on the underlying ring of R. Let a, b R with c = ab. Then for any λ, µ Λ we have πλµc + λµπc = µπλc + λπµc. 4 PROOF. Apply Lemma 6 with A = B = C = R and S = T = Λ, together with the obvious bilinear maps. Let π A = π B = π C = π and n = m = 1 with s 1 = λ, t 1 = µ. All of the hypotheses are satisfied, and 2 reduces to 4. Corollary 14 Let R be a non-associative algebra over the commutative ring with identity Λ, and let π : R R be a production on the underlying ring of R. Let a 0, a 1, b 0, b 1 R satisfy a 0 b 1 = a 1 b 0 = 0 and a 0 b 0 = a 1 b 1 = c. Then for any λ, µ Λ we have and πλc + µc = πλc + πµc 5 πλµc + λµπc = µπλc + λπµc. 6 PROOF. First note that if a = λ 0 a 0 + λ 1 a 1 and b = µ 0 b 0 + µ 1 b 1, then the hypotheses imply that ab = λ 0 a 0 + λ 1 a 1 µ 0 b 0 + µ 1 b 1 = λ 0 µ 0 + λ 1 µ 1 c. Apply Lemma 6 with A = B = C = R and S = T = Λ, together with the obvious bilinear maps. Let π A = π B = π C = π and n = m = 2 with s i = λ i and t j = µ j. All of the hypotheses are satisfied, and the 2 reduces to 7. π λ 0 µ 0 + λ 1 µ 1 c + λ 0 µ 0 + λ 1 µ 1 π c = µ 0 π λ 0 c + λ 0 π µ 0 c +µ 1 π λ 1 c + λ 1 π µ 1 c 7 Letting λ 1 = µ 1 = 0, λ 0 = λ and µ 0 = µ in 7, we see immediately that 6 holds. Similarly, by letting λ 1 = µ 0 = 1, λ 0 = λ and µ 1 = µ we see that π λ + µ c + λ + µ πc = π λc + λπc + µπc + π µc. From this, 5 follows immediately. 5

6 Remark 15 Note that if R is an anti-symmetric non-associative algebra over Λ, that is, if x 2 = 0 for every x R, then any c = ab will satisfy the conclusions 5 and 6 of Corollary 14. To see this, we need only note that if we define a 0 = b 1 = a and b 0 = a 1 = b, then a 0 b 1 = a 2 = 0, a 1 b 0 = b 2 = 0, a 0 b 0 = ab = c, and a 1 b 1 = ba = ab = c. Note in particular that this applies to any Lie algebra. Let R be a non-associative ring which is direct sum of ideals R = R 1 R n. If π i : R i R i is a production on R i for each i = 1,..., n, then the map π : R R defined by πx x n = π 1 x π n x n for x i R i is easily seen to be a production on R. Theorem 16 provides a partial converse to this. We say that a non-associative ring is annihilator free if for any a R i which satisfies ax = xa = 0 for all x R i, we have a = 0. In case R is a direct sum R = R 1 R n, note that R is annihilator free if and only if each R i is annihilator free. Theorem 16 Let R be a non-associative ring which is direct sum of ideals R = R 1 R n, and let π : R R be a production on R. Suppose further that R is annihilator free. Equivalently, that each R i is annihilator free. Then there exist unique productions π i : R i R i such that πx x n = π 1 x π n x n for x i R i. PROOF. The uniqueness is trivial, since π i 0 = 0 for every i, so that π i x = πx for any x R i. For existence, we first need to show that πx R i whenever x R i. Let x R i. Write πx = a a n with a j R j. Given any j i and any y R j, we have a j y = πxy R j and also a j y = πxy = πxy xπy = π0 xπy = xπy R i. Thus, a j y = 0 for any y R j. Similarly, ya j = 0 for any y R j. Since R j is annihilator free, we must have a j = 0. Since a j = 0 for every j i, we have πx = a i R i, as desired. Now, we may define π i : R i R i, for any i = 1,..., n, by π i x = πx for any x R i. Clearly, each π i is a production on R i. It remains only to show that πx x n = π 1 x π n x n whenever x i R i for each i. Let x = x x n and y = y y n be arbitrary, with x i, y i R i. Making use of Corollary 10, we may calculate as follows: πx π 1 x 1 π n x n y = πxy πx 1 y... πx n y = πxy xπy πx 1 y x 1 πy... πx n y x n πy = πxy πx 1 y πx n y x x 1 x n πy = πxy πx 1 y 1 πx n y n = 0 6

7 Thus, πx π 1 x 1 π n x n y = 0 for any y R. Similarly, yπx π 1 x 1 π n x n = 0 for any y R. Since R is annihilator free, this gives πx x n = π 1 x π n x n, as desired. The proof is complete. Proposition 17 Let R be a non-associative ring with direct sum decomposition R = α I R α as an abelian group under addition, and let π : R R be a production on R. Suppose that π is additive on each of the summands R α, in the sense that for any α I and for any x, y R α, we have πx+y = πx+πy. Suppose further that for each α, β I there exists some γ I such that xy R γ for any x R α and y R β. Define δ : R R by δ α I = α I π, where R α for each α I and = 0 for all but finitely many α I. Then δ is a derivation on R. PROOF. We define m : I I I so that xy R mα,β whenever x R α and y R β. For any x R, we write x = α I, where R α for each α I and = 0 for all but finitely many α. Then we see that δ is a production as follows. δxy = δ y β α β = δ y β = γ = γ = α γ mα,β=γ π mα,β=γ mα,β=γ β π y β y β π y β = π y β + πy β α β = π y + x πy β α β = δxy + xδy To see that δ is a derivation on R, it remains only to show additivity of δ on R, which we see as follows. δx + y = δ + y α α α = δ + y α = α α π + y α 7

8 = π + πy α α = π + πy α α α = δ + δ y α α α = δx + δy Theorem 18 Let L be a Lie ring, and π : L L a production on L. Then we have π[[x, y], z] + π[[y, z], x] + π[[z, x], y] = 0 for every x, y, z L. PROOF. π[[x, y], z] + π[[y, z], x] + π[[z, x], y] = [π[x, y], z] + [[x, y], πz] + [π[y, z], x] + [[y, z], πx] + [π[z, x], y] + [[z, x]πy] = [[πx, y], z] + [[x, πy], z] + [[x, y], πz] + [[πy, z], x] + [[y, πz], x] + [[y, z], πx] + [[πz, x], y] + [[z, πx], y] + [[z, x], πy] = [[πx, y], z] + [[z, πx], y] + [[y, z], πx] + [[x, πy], z] + [[z, x], πy] + [[πy, z], x] + [[x, y], πz] + [[πz, x], y] + [[y, πz], x] = Finite dimensional split semi-simple Lie algebras over a field of characteristic zero In this section, we assume that L is a finite dimensional split semi-simple Lie algebra over a field F of characteristic zero, with, as its Killing form. Recall that any Lie algebra over an algebraically closed field of characteristic zero is split. Let L = H α L α 8 be a fixed Cartan decomposition for L, where = + is the set of nonzero roots of L, and + is the set of positive roots under a given ordering, 8

9 with the corresponding set of negative roots. We write h α for the coroot of α. Let 0 + be the set of simple positive roots, that is, the set of all α + such that there do not exist β, γ + with α = β + γ. Then { h α α + } is a basis for H. Note that [h, ] = αh for any h H and L α, and that [, x α ] =, x α h α for any L α and x α L α. Recall that h α, h β Q is rational for all α, β. Throughout this section we will assume that the map π : L L is a production on the underlying Lie ring of L. Lemma 19 Let L be a finite dimensional split semi-simple Lie algebra over a field F of characteristic zero, with Cartan decomposition 8. Suppose that πh = 0 for every h H and that for every α and every L α, we have π = 0. Then π is the trivial production πx = 0 for all x L. PROOF. First, we will show that π + x α = 0 9 whenever L α and x α L α. If one or both of and x α are 0, then we have nothing to prove, so assume that both are non-zero. Then, x α = 0. Thus, we may define y α L α by =, x α y α. Then, we have y α, x α = 1. Now, we use Theorem 18 to calculate as follows. 0 = π[[y α, x α ], x α ] + π[[x α, x α ], y α ] + π[[ x α, y α ], x α ] = π[ y α, x α h α, x α ] + π[, x α h α, y α ] + π[ y α, x α h α, x α ] = π y α, x α + y α, x α x α + π, x α y α + π yα, x α x α = π + x α + π + π x α = π + x α This proves 9, as desired. Next, we show that πx H whenever x = α, 10 where L α. It is enough to show that [h, πx] = 0 for any h H. Since [H, L] α L α, this is equivalent to showing that [h, πx] H for any h H. We do this by induction on the number k of non-zero terms occurring in the sum x = α. If k < 2, then πx = 0 by hypothesis. Similarly, if k = 2 and there is some α with, x α 0, then x = + x α, so that πx = 0 by 9. Either way, we have [h, πx] = [h, 0] = 0 trivially. So, we may assume that there exists α, α with, 0 with such that 9

10 α ±α. Note that any h H can be written in the form h = h + h for some h, h H satisfying α h = α h = 0. [h, πx] = [h + h, πx] = [h, πx] + [h, πx] = π[h, x] [πh, x] + π[h, x] [πh, x] = π[h, x] + π[h, x] [ = π h, ] [ + π h, ] α α = π αh + π αh α α H The last statement follows directly from the induction hypothesis, since we have α h = α h = 0 and, 0, so that both sums α αh and α αh have strictly fewer non-zero terms than the sum α, and therefore the images of both sums under π are in H. Thus, 10 is proved. Next, we show that πx H for any x L. 11 As before, it is enough to show that [h, πx] = 0 for any x L and h H, which in turn is equivalant to showing that [h, πx] H for any x L and h H. By writing x as x = h + α, we see that by 10. Thus, 11 is proved. [h, πx] = π[h, x] [πh, x] [ = π h, h + ] [0, x] α = π αh α H, Finally, we complete the proof of Lemma 19 by showing that πx = 0 for all x L. 12 Since πx H by 11, it is enough to show that [πx, y β ] = 0 for every β and y β L β. If we let h = πx, we see immediately that [πx, y β ] = [h, y β ] = βhy β L β. But [πx, y β ] = π[x, y β ] [x, πy β ] 10

11 = π[x, y β ] [x, 0] H, by 11, so that [πx, y β ] L β H = { 0 }. Thus, 12 holds, and the proof is complete. Theorem 20 Let L be a finite dimensional split semi-simple Lie algebra over a field F of characteristic zero. Suppose that π is a production on the underlying Lie ring of L. Then π is additive, that is, for all x, y L. πx + y = πx + πy 13 PROOF. First, we show that the Cartan decomposition 8 satisfies the hypotheses of Proposition 17. That is, we show that π is additive on each of the summands of 8. Given α and L α with 0, note that [h α, ] =, so that π is additive on F = F = L α, by Remark 15. Note also that if L α and x α L α with, x α 0, then, x α = 0 and [, x α ] =, x α h α, so that π is additive on F, x α h α = Fh α, by Remark 15. Recall that H = Fh α. For convenience, let us choose L α and x α L α for each α 0 + such that, x α = 1. Then [, x α ] = h α for all α 0 +. Note also that [, x β ] = 0 for any α, β 0 + with α β. We now show that π is additive on H. Let h, h H. Then we can write h = α 0 r αh + α and h = α 0 r + αh α with r α, r α F for α 0 +. The following calculation uses Corollary 10 together with the fact that π is additive on Fh α for all α 0 +. πh + h = π = π r α h α + r α + r α h α r αh α = π [ r α + r α, x α ] = = π [ r α + r α, x α ] π r α + r α h α 11

12 = π r α h α + r αh α = π rα h α + π r αh α = π r α h α + π r αh α = π [r α, x α ] + π [r α, x α ] = π [ r α, ] [ x α + π r α, x α ] = π r α h α + π r αh α = πh + πh Thus, we see that π is additive on H. Since π has already been seen to be additive on L α for each α, we see that the Cartan decomposition 8 satisfies the hypotheses for Proposition 17. Let δ : L L be the map whose existence is claimed in Proposition 17. Note that δ is a derivation and therefore a production on the underlying Lie ring of L, which agrees with π on H and on L α for all α. It is trivial to see that the set of all productions on the underlying Lie ring of L form a vector space over the field F, under the obvious pointwise definitions. In particular, the map π = π δ defined by π x = πx δx is a production on the underlying Lie ring of L. Furthermore, π h = 0 for all h H and π = 0 for all α and all L α. Thus, Lemma 19 implies that π is the trivial production, so that π = δ. Thus, π is a derivation on the underlying Lie ring L, and hence additive. Theorem 21 Let L be a finite dimensional split simple Lie algebra over a field F of characteristic zero, with Killing form,, and let π be a production on the underlying Lie ring of L. Then π is a derivation on the underlying Lie ring of L, and there exists a unique derivation δ : F F on the field of scalars F such that and for any λ F and any x, y L. πλx = δλx + λπx 14 δ x, y = πx, y + x, πy, 15 PROOF. The fact that π is a additive on L, and thus a derivation on the underlying Lie ring of L, is the content of Theorem 20. It follows that π is 12

13 actually Q-linear. In particular, π h α, h β x = h α, h β πx for all x L and all α, β, since h α, h β Q is rational. We will use this fact freely without mention. Note also that the uniqueness of δ is trivial, by either 14 or 15, so it enough to show the existence of a δ with the desired properties. Our first task is to define δ. For any α, we define a map δ α : F F by δ α λ = πλh α, h α, 16 for any λ F. We shall see shortly that δ α = δ β for all α, β. First, we show that δ α is a derivation on F for any α. The fact that δ α is additive is an immediate consequence of the fact that π is additive. To see that δ α is a production on F, choose any L α and x α L α such that, x α = 1. Then h α = [, x α ], so we may apply Corollary 13 with c = h α in 4 to see that and thus πλµh α + λµπh α = µπλh α + λπµh α, δ α λµ + λµδ α 1 = µδ α λ + λδ α µ, for any λ, µ F. Thus, to see that δ α is a production, we need only show that δ α 1 = 0, 17 for all α. To see this, once again we choose any L α and x α L α such that, x α = 1, and calculate as follows: δ α 1 = πh α, h α = πh α, [, x α ] = [πh α, ], x α = π[h α, ] [h α, π ], x α = π[h α, ], x α = π h α, h α, x α [h α, π ], x α = π, x α π, x α = 0 + π, [h α, x α ] 13

14 Thus, δ α is a derivation on F for any α, as claimed. Next, we show that π, x α h α, h = h α, h π, x α +, πx α, 18 for any α, h H, L α and x α L α. We see this as follows: π, x α h α, h = π [, x α ], h = [π, x α ], h + [, πx α ], h = π, [x α, h] + [h, ], πx α = h α, h π, x α +, πx α If h α, h = 0, we can rewrite this as π, x α h α, h h α, h = π, x α +, πx α. 19 If we set h = h α in 19, we see from 16 that δ α, x α = π, x α +, πx α. 20 Given any λ F and any α, we can find L α and x α L α such that λ =, x α. Thus, we may combine 20 with 18 to yield πλh α, h = δ α λ h α, h. 21 Next, we show that πh, + h, π = 0, 22 for any α, h H and L α. To see this, we calculate as follows: πh, + h, π = = = πh, [ ] h α, [πh, h α ], [ + h, h α, π xα [πh, h α ], xα [ ] + h, π h α, [ ] + h, πh α, ] + [h, πh α ], + [h, h α ], π = π[h, h α ], + [h, h α ], 14

15 = π0, = Using 22, we may show that + 0, πλh, = λ πh,, 23 for any α, λ F, h H and L α as follows: πλh, = λh, π = λ h, π = λ πh, From 23, we see immediately that πλh λπh, = 0 for any L α, so that πλh λπh H, 24 for any h H and λ F. Next, we show that πλh α = δ α λh α + λπh α, 25 for all α and λ F. By 24, we see that πλh α λπh α δ α λh α H. Thus, to show 25, it suffices to show that πλh α λπh α δ α λh α, h = 0 for any h H. We see this, using 21 and 17 as follows: πλh α λπh α δ α λh α, h = πλh α, h λ πh α, h δ α λ h α, h = δ α λ h α, h λδ α 1 h α, h δ α λ h α, h = 0 Next, we show that πh α + h β, [, x β ] = h α + h β, h α + h β π, x β, 26 for any α, β, L α and x β L β, by calculating as follows: πh α + h β, [, x β ] = [πh α + h β, ], x β = π[h α + h β, ] [h α + h β, π ], x β = π[h α + h β, ], x β [h α + h β, π ], x β = π h α + h β, h α, x β + π, [h α + h β, x β ] = h α + h β, h α π, x β + h α + h β, h β π, x β = h α + h β, h α + h β π, x β 15

16 From 26, we see immediately that π, x β = πh α + h β, [, x β ] h α + h β, h α + h β for α + β Since [, x β ] + [x β, ] = 0, we can use 27 to conclude that π, x β +, πx β = 0 for α + β 0, 28 for any α, β, L α and x β L β. Our next goal is to prove that πλ = δ α λ + λπ, 29 for any α, λ F and L α. We start by using 22 to show that for any h H, by calculating as follows: Next, we use 28 to show that πλ δ α λ λπ, h = 0 30 δ α λ, h + λπ, h = 0 + λ π, h = λ, πh = λ, πh = πλ, h πλ δ α λ λπ, x β = 0 for α + β 0, 31 foe any L β, by calculating as follows: δ α λ, x β + λπ, x β = 0 + λ π, x β = λ, πx β = λ, πx β = πλ, x β Finally, we use 20 to show that for any x α, by calculating as follows: πλ δ α λ λπ, x α = 0, 32 πλ, x α = δ α λ, x α λ, πx α = δ α λ, x α λ, πx α = δ α λ, x α + λδ α, x α λ, πx α = δ α λ, x α + λ δ α, x α, πx α 16

17 = δ α λ, x α + λ π, x α = δ α λ, x α + λπ, x α Since the Killing form is non-degenerate, we may use 30, 31 and 32 to conclude that 29 holds, as desired. We are now in a position to define the derivation δ on F. First, note that since h α = h α, we immediately see from 16 that for any α. Next, we show that δ α λ = δ α λ, 33 δ α+β λµ = δ α λµ + λδ β µ for α, β, α + β, 34 where λµ F, L α and x β L β. To see this, note that [, x β ] L α+β, since we are assuming that α + β. Using 29, we see that δ α+β λµ[, x β ] = π λµ[, x β ] λµπ[, x β ] = π[λ, µx β ] λµπ[, x β ] = [πλ, µx β ] + [λ, πµx β ] λµπ[, x β ] = [δ α λ + λπ, µx β ] + [λ, δ β µx β + µπx β ] λµπ[, x β ] = δ α λµ + λδ β µ [, x β ] + λµ [π, x β ] + [, πx β ] π[, x β ] = δ α λµ + λδ β µ [, x β ], which easily implies 34, by any choice of, x β 0 so that [, x β ] 0. By letting µ = 1 in 34, we see that δ α+β λ = δ α λ + λδ β 0 = δ α λ. Similarly, δ α+β µ = δ β µ. Thus, we have δ α = δ β = δ α+β for α, β, α + β. 35 Since L is assumed to be simple, from 33 and 35 we may conclude that δ α = δ β for α, β. 36 Note that if L were merely assumed to be semi-simple, then 36 would only follow in case α and β were roots of the same simple factor. We now define the derivation δ : F F on F by δ = δ α for any α. 37 By 36, δ is well defined. Also, δ is a derivation on F since δ α is. It remains to prove 14 and

18 Now, we show that πλh = δλh + λπh, 38 for any λ F and h H. To see this, we write h = µ αh α, with µ α F for all α 0 +, and use 25 and 37 as follows:. πλh = π λ µ α h α = π = = = λµ α h α π λµ α h α δλµα h α + λµ α πh α δλµα + λδµ α h α + λµ α πh α = δλ µ α h α + λ δµα h α + µ α πh α = δλh + λ πµ α h α = δλh + λπ µ α h α = δλh + λπh We are finally in a position to prove 14. Writing x = h+ α δ,with h H and L α for α, we use 29, 38 and 37 to compute as follows: Thus, 14 is proved. πλx = π λ h + α δ = π λh + λ α δ = πλh + πλ α δ δλxα + λπ = δλh + λπh + α δ = δλ h + + λ πh + π α α = δλx + λπx 18

19 Next, we show that δ λh α, µh β = πλh α, µh β + λh α, πµh β, 39 for any α, β and h α, h β H, by using 21 and computing as follows: δ λh α, µh β = δ λµ h α, h β = δλµ h α, h β = δλµ + λδµ h α, h β = δλ h α, µh β + δµ h β, λh α = πλh α, µh β + πµh β, λh α = πλh α, µh β + λh α, πµh β It is easy to see that δ h, = πh, + h, π, 40 for all α δ, h H and L α. We need only note that δ h, = δ0 = 0, and apply 22. It is equally easy to see that δ, x β = π, x β +, πx β, 41 for all α, β, L α and x β L β. If α + β 0, we need only note that δ, x β = δ0 = 0 and apply 28. If α + β = 0, we need only apply 20. Since every x L can be written as x = λ αh α + α where λ α F for all α 0 + and L α for all L α, it is easy to see that we can use 39, 40 and 41, together with the fact that both π and δ are additive, and that the Killing form is bilinear, to prove 15. This completes the proof of Theorem 21 Theorem 22 Let L be a finite dimensional split semi-simple Lie algebra over a field F of characteristic zero, with Killing form,, and let π be a production on the underlying Lie ring of L. Let n L = L i be the decomposition of L into its simple factors. Then π is a derivation on the underlying Lie ring of L, and there Then there exists a unique sequence of 19

20 derivations δ 1,..., δ n : F F on the field of scalars F such that and πλx = δ i λx i + λπx δ i x i, y i = πx, y + x, πy, for any λ F and any x, y L,where we write x = n x i and y = n y i with x i, y i L i for i = 1,..., n. PROOF. This is an easy consequence of Theorem 16 and Theorem 21. References [1] Jacobson, Nathan, Lie Algebras, Interscience Publishers, New York, [2] Kramer, Richard L. The undefinability of intersection from perpendicularity in the three-dimensional Euclidean geometry of lines, Geometriae Dedicata, , pp [3] Kramer, Richard L. Definability in the foundations of Euclidean geometry and the product rule for derivations, Model Theory in Bogotá Summer 2005, presented August 16, 2005, unpublished. [4] Schwabhäuser, Wolfram and Szczerba, Les law W. Relations on lines as primitive notions for Euclidean goemetry, Fundamenta Mathematicae, /75, pp