Homework 1 Lie Algebras

Transcription

1 Homework 1 Le Algebras Joshua Ruter February 9, 018 Proposton 0.1 Problem 1.7a). Let A be a K-algebra, wth char K. Then A s alternatve f and only f the folowng two lnear) denttes hold for all a, b, y A. ab)y + ba)y = aby) + bay) yab) + yba) = ya)b + yb)a Proof. Frst suppose that A s alternatve. Let a, b, y A. Defne x = a + b. Then the alternatve dentty x )y = xxy) mples a + b) y = a + b)a + b)y) = a + ab + ba + b )y = a + b)ay + by) = a )y + ab)y + ba)y + b )y = aay) + aby) + bay) + bby) and the dentty yx ) = yx)x mples = ab)y + ba)y = aby) + bay) ya + b) ) = ya + b))a + b) = ya + ab + ba + b ) = ya + yb)a + b) = ya ) + yab) + yba) + yb ) = ya)a + ya)b + yb)a + yb)b = yab) + yba) = ya)b + yb)a so the two desred lnear denttes hold n A. Note that we dd not yet nvoke char K, so these denttes are true regardless of characterstc n an alternatve algebra.) Now suppose that A s an algebra n whch these two lnear denttes hold. Settng a = b = x, we get xx)y + xx)y = xxy) + xxy) = xx)y = xxy) yxx) + yxx) = yx)x + yx)x = yxx) = yx)x Provded char K, these mply the alternatve denttes. xx)y = xxy) = xx)y = xxy) yxx) = yx)x = yxx) = yx)x Thus n char K, these two lnear denttes are equvalent to the standard alternatve denttes. 1

2 Proposton 0. Problem 1.7b). Let A be an alternatve K-algebra and E an extenson feld of K. Then E K A s an alternatve E-algebra. Proof. We just need to check that the alternatve denttes hold on a bass of E K A. We use the usual bass, {x a : x E, a A}. Let x a, y b E K A. We need to show x a) ) y b)? = x a) x a)y b)) y b) x a) )? = y b)x a)) x a) Frst, recall that multplcaton n E K A s defned by x a)y b) = xy) ab) where xy) s multplcaton n E and ab) s multplcaton n A. We start wth the the top dentty. x a) ) y b) = xx aa)y b) = xx)y aa)b = xxy) aab) = x a)xy ab) = x a) x a)y b)) Note that aa)b = aab) s one of the alternatve denttes n A, and xx)y = xxy) s the assocatvty of the product n E, snce E s a feld. The second dentty follows n a smlar manner. y b) x a) ) = y b)xx aa) = yxx) baa) = yx)x ba)a = yx ba)x a) = y b)x a)) x a) Lemma 0.3 for Problem 1.8a). Let A be an assocatve K-algebra wth char K. Let A +, ) be the assocated Jordan algebra, wth multplcaton defned by x y = 1 xy + yx) Two adjacent letters such as xy wth no symbol refers to multplcaton n A.) Let ad : A + End + K A) be the adjont map ad ax) = a x. If ad s a representaton of Jordan algebras, then aax xa) = ax xa)a for all a, x A.

3 Proof. Recall that the Jordan multplcaton nduced on End + K A) s φ ψ = 1 φψ + ψφ) where the products φψ and ψφ are functon composton. For a A +, we consder how ad a a and ad a ad a act on x A +. ad a a x) = a a) x ) 1 = a + a ) x = a x char K ) = 1 ) a x + xa Snce A s assocatve, we don t need to bother wth parentheses n wrtng a x. ad a ad a )x) = 1 ad a ad a + ad a ad a )x) = 1 ad a ad a x + ad a ad a x) = ad a ad a x)) char K ) = ad a a x) = a a x) ) 1 = a ax + xa) = 1 aax + xa) + ax + xa)a) 4 = 1 aax) + axa) + axa) + xa)a) 4 = 1 4 a x + xa + axa) Suppose ad s a representaton of Jordan algebras. Then t must preserve, that s, ad a b = ad a ad b for a, b A +. In partcular, ad a a = ad a ad a for all a A, so the mage of x under these maps s equal, so we get the followng relaton. 1 a x + xa ) = 1 4 a x + xa + axa) = a x + xa = a x + xa + axa = a x + xa = axa = axa + axa = a x axa = axa xa = aax xa) = ax xa)a Thus ad beng a homomorphsm of Jordan algebras mples ths relaton. 3

4 Proposton 0.4 Problem 1.8a). Let A be an assocatve K-algebra wth char K. Let A +, ) be the assocated Jordan algebra, and let ad : A + End + K A) be the adjont map ad ax) = a x. Ths s not always a representaton. Proof. By the prevous lemma, f ad s a representaton of Jordan algebras, then aax xa) = ax xa)a That s, a commutes wth [a, x] for every a, x A. Ths s not true n every assocatve algebra. Here s a concrete example n gl R). ) ) a = x = a[a, x] [a, x]a As the reader can probably tell, these matrces were chosen of the top of my head, so t s clear ths equaton nearly always fals for randomly chosen matrces. Hence, ad s not always a representaton of Jordan algebras. Proposton 0.5 Problem 1.8b). Let A be an assocatve K-algebra where char K. For a A, defne maps L a : A A U a : A A x a x = 1 ax + xa) x axa Let V be a vector subspace of A. If V s nvarant under all L a, then t s also nvarant under all U a. If A has a unt 1 A, then the converse holds as well: f V s nvarant under all U a, then t s nvarant under L a Proof. Let V be a subspace nvarant under all L a for a A. That s, v V = L a v) = a v = 1 av + va) V Let v V. We need to show that U a v) = ava V. From the computaton n Lemma 0.3, L a a v) = 1 a v + va ) L a L a )v) = 1 4 a v + va + ava) By hypothess, L a a v) and L a L a )v) le n V, so the dfference ) L a L a )v) L a a v) = 1 a v + va + ava) 1 a v va ) = ava must le n V. That s, U a v) V, so V s nvarant under U a. Now for the other drecton, suppose that V s nvarant under all U a. We now need to assume A has a unt 1 A, whch we just denote by 1. Let a A, v V. Then U a+1 v) = a + 1)va + 1) = ava + v + av + va = U a v) + v + av + va 4

5 We know that U a v) and v le n V, so the dfference av + va) must le n V as well. Thus so V s nvarant under L a. L a v) = 1 av + va) V Proposton 0.6 Problem.5). Let K be a feld. There are exactly two Le K-algebras of dmenson two, up to somorphsm. There s the abelan Le algebra, whch we denote L ab, and there s the non-abelan Le algebra L x = span K {x, y} wth the bracket defned by [x, y] = x. Proof. Frst, note that these two are n fact Le algebras, and that all Le brackets n them are defned by extendng lnearly and usng the basc Le algebra relatons. For example, n L x, [y, x] = [x, y] = x It s suffcent to prove that any non-abelan Le algebra of dmenson two over K s somorphc to L x. Let L be an arbtrary non-abelan two-dmensonal K-Le algebra. We can choose a bass for L, {a, b}. If [a, b] = 0, then all brackets are zero and L s abelan, so we must have [a, b] 0. Defne x = [a, b], and extend t to a bass {x, y } of L. Then we can wrte x, y n terms of the bass a, b, that s, Then x = λa + µb y = γa + ηb λ, µ, γ, η K [x, y ] = [λa + µb, γa + ηb] = λη µγ)[a, b] = λη µγ)x span{x } If λη µγ = 0, then [x, y ] = 0 and L would be abelan, so t must be nonzero. So we can defne ) y 1 = y λη µγ Then by lnearty of the bracket, [x, y ] = 1 λη µγ ) [x, y ] = Thus we have an somorphsm of Le algebras, L L x x x, y y ) λη µγ [a, b] = [a, b] = x λη µγ Proposton 0.7 Problem.6). Let K be an algebracally closed feld of characterstc. Then sl K) = so 3 K). Proof. We have the usual bass {e, f, h} where e = e 1, f = e 1, h = e 11 e for sl K), wth the followng bracket relatons: [e, f] = h [h, e] = e [h, f] = f 5

6 Snce char K, the matrces so 3 K) are skew symmetrc, so they have the obvous bass {a, b, c} where a = e 1 e 1, b = e 13 e 31, c = e 3 e 3. The bracket relatons for a, b, c are Defne [a, b] = c [b, c] = a [c, a] = b H = a E = b + c F = b c Recall that K s algebracally closed, so x + 1 = 0 has two roots. We can choose one of these and call t. That s, K contans an embedded copy of Q).) It s clear that {E, F, H} forms a bass for so 3 K). Then one can check that the brackets of H, E, F are [E, F ] = H [H, E] = E [H, F ] = F Thus we may defne a map φ : so 3 K) sl K) by E e F f H h and extendng lnearly. Ths maps a bass to a bass, so t s a vector space somorphsm. It preserves the brackets on the bass, so t s a Le algebra homomorphsm. Thus φ s an somorphsm of Le algebras. Note: The choce of H, E, F as the correspondng bass for so 3 K) n the prevous theorem was not easy to fnd. When a proof such as ths s presented, the reader may feel ntmdated by the fact that the wrter draws such a choce out of thn ar, as f conjured by magc. However, t took over an hour of workng out some brute force calculatons and many faled guesses to determne ths bass for so 3 K). I knew that I needed an element H so 3 K) that acted lke h n sl K), that s, I needed an H wth egenvectors E, F wth egenvalues ±. So I calculated the matrx of ad H for arbtrary H = H 1 a + H b + H 3 b so 3 K), and used a computer algebra system to compute the characterstc polynomal of ths matrx. I then choose H 1, H, H 3 so that the characterstc polynomal would have roots 0,,, and then worked out the correspondng egenvectors, whch I then called E and F. Lemma 0.8 Problem.7, char case). Let K be a feld wth char K 0. Let {e, f, h} be the usual bass for sl K) where e = e 1, f = e 1, h = e 11 e. Let H = Kh = spanh). The only subalgebras of sl K) contanng H are H tself, all of sl K), and V e = spanh, e) and V f = spanh, f). Proof. Recall that [e, f] = h [h, e] = e [h, f] = f Let V be a subalgebra of sl K) contanng H. If dm V = 1, then V = H, and f dm V = 3 then V = sl K). Suppose dm V =. Then V = spanh, x) where x s some matrx. We can wrte x n terms of the bass {e, f, h} as x = x 1 e + x f + x 3 h. We can assume x 3 = 0, snce spanh, x) = spanh, x 1 e + x f) Then [h, x] = [h, x 1 e + x f] = x 1 [h, e] + x [h, f] = x 1 e x f 6

7 Snce V s a subalgebra, [h, x] V. Snce [h, x] doesn t depend on h, t must be n the span of x. That s, [h, x] = λx for some λ K. Thus x 1 e x f = λx 1 e + x f) = λx 1 e + λx f Snce {e, f, h} s a bass of sl K), the coeffcents are unque, so ths mples x 1 = λx 1 and x = λx. We can t have x 1 = x = 0, snce then V = spanh). Now the fact that char K comes nto play. If x 1 0, then λ = and x = 0. If x 0, then λ = and x 1 = 0. Thus the only possble values for x are x = e and x = f. Thus the only two dmensonal subalgebras of sl K) contanng H are V e and V f. Lemma 0.9 Problem.7, char = case.). Let K be a feld wth char K =. Let {e, f, h} be the usual bass for sl K) where e = e 1, f = e 1, h = e 11 e. Let H = Kh = spanh). Then any vector subspace of sl K) contanng H s a subalgebra. Proof. Let V be a vector subspace of sl K) contanng H. If dm V = 1, then V = H, or f dm V = 3 then V = sl K); both of these are subalgebras. Suppose dm V =. Then extend {h} to a bass of V, V = spanh, x), where x = x 1 e + x f + x 3 h. Usng the computaton n the prevous lemma, [h, x] = x 1 e x f Snce char K =, ths s actually zero. That s, [h, x] V = spanh, x), so V s a subalgebra. Proposton 0.10 Problem.7). Let K be a feld. Let {e, f, h} be the usual bass for sl K), and let H = spanh). We classfy the subalgebras of sl K) contanng H. 1. If char K, then the only such subalgebras are H, sl K), V e = spanh, e), and V f = spanh, f).. If char K =, then any vector subspace V contanng H s a subalgebra. Incdentally, f V sl K) then V s abelan.) Proof. Use prevous two lemmas. Proposton 0.11 Problem.8). Let A = K[x 1,..., x n ]. The dervaton Le algebra Der K A) s { } Der K K[x 1,..., x n ]) = p x p A Proof. As a K-algebra, A s generated by K and x 1,..., x n. Let D Der K A). Then D s determned by D1), Dx ), snce D s K-lnear. We know that D1) = 0, because 1Dx) = D1x) = D1)x + 1Dx) = D1)x = 0, x A = D1) = 0 7

8 because A s an ntegral doman. Defne p = Dx ). Then we clam that Dx k ) = p kx k 1. Ths s true for k = 1 by defnton of p. Then by nducton on k, D x k ) ) = D x x k 1 = D x ) x k 1 = p x k 1 = p x k 1 + x D ) x k 1 + x p k 1)x k p k 1)x k 1 = p 1 + k 1)x k 1 = p kx k 1 Ths completes the nductve step. Thus D acts as p kernel of, we can wrte D as x n D = p x =1 x on x. Snce x j for j s n the Defnton 0.1. The kth superdagonal of a matrx a = a j ) s the kth dagonal above the man dagonal, also known as the entres a,+k. For example, the 0th superdagonal s the man dagonal, and the 1st superdagonal s the dagonal just above the man dagonal. Defnton 0.. The group UT n K) s the group of upper trangular matrces wth ones along the man dagonal. Defnton 0.3. The set UT m n K) s the subset of UT m n K) wth zeros on the 1st, nd,..., mth superdagonals. For example, UT 0 nk) = UT n K), UT 1 nk) has zeroes along the 1st superdagonal, and UT nk) has zeroes along the 1st and nd superdagonals. UTn n K) has zeroes everwhere except the man dagonal and the top rght corner entry, and UTn n 1 K) s just the dentty. More precsely, a = a j ) UT m n K) = a,+1 =... = a,+m = 0 Lemma 0.1 for Problem.9a). Let a = a j ) UT n K). Let c = a 1 = c j ). Then a,+1 + c,+1 = 0 for = 1,..., n 1. That s, the 1st superdagonal of a 1 s the negatve of the 1st superdagonal of a. Another way to say ths s that a + a 1 UT 1 nk). Proof. By defnton, aa 1 = I. We also know the j-th entry of ac by the formula ac) j = k a k c kj The j-th entry of I s δ j. Consder the, + 1)-th entry of aa 1 = I. Snce t s a non-dagonal entry of I, t s zero. It s also equal to ac),+1 = k a k c k,+1 = a 1 c 1, a c,+1 + a,+1 c +1, a n c n,+1 8

9 Snce a, a 1 G, we know that a = c = 1, and a j = c j = 0 for > j. So the only nonzero terms are a c,+1 and a,+1 c +1,+1. Thus 0 = a c,+1 + a,+1 c +1,+1 = c,+1 + a,+1 Lemma 0.13 for Problem.9a). UT m n K) s a subgroup of UT n K). Proof. The case m = 0 s just the statement that UT n K) s a subgroup of tself, so assume m 1. We need to show that UT m n K) s closed under nverson and multplcaton. We ll do nverson frst. We want to show that UT m n K) s closed under nverson. Frst, for the base case m = 1, let a UT 1 nk). By the prevous lemma, the entres on the 1st superdagonal of a 1 are the negatve of entres on the 1st superdagonal of a, whch are zero, so a 1 UT 1 nk). Now we nduct on m. We assume as nductve hypothess that UT 1 nk), UT nk),..., UT k nk) are closed under nverson. Let a = a j ) UT k+1 n K), and let a 1 = c = c j ). We need to show that a 1 = c UT k+1 n K); that s, we need to show that c,+k+1 = 0. Consder the, + k + 1)-th entry of aa 1 = I. Snce t s not on the man dagonal, t s zero. On the other hand, t s also equal to 0 = aa 1 ),+k+1 = t a t c t,+k+1 = a 1 c 1,+k a c,+k+1 + a,+1 c +1,+k a n c n,+k+1 Usng the fact that a j = c j = 0 for > j, and a = c = 1, all terms before a c,+k+1 are zero, snce the a term s zero, and all terms after a,+k+1 c +k+1,+k+1 are zero snce the c term s zero. We also know that a = c = 1. Thus 0 = a c,+k+1 + a,+1 c +1,+k a,+k+1 c +k+1,+k+1 = c,+k+1 + a,+1 c +1,+k a,+k c +k,+k+1 + a,+k+1 Snce a UT k+1 n K), we know a,+1,..., a,+k, a,+k+1 = 0. By nductve hypothess, a 1 UT k nk), so c,+1,..., c,+k = 0. After zerong out all the terms contang these, we get 0 = c,+k+1 Thus c = a 1 UT k+1 n. Now we need to show that UT m n K) s closed under multplcaton. Let a = a j ), b = b j ) UT m n K). We need to show that ab UT m n K), so we need to show that ab),+k = 0 for k = 1,... m. ab),+k = t a t b t,+k = a 1 b 1,+k a b,+k a n b n,+k Snce a UT n K), we know that the terms before a b,+k are zero, snce the a part s zero. Smlarly, snce b UT n K), the terms after b +k,+k are also zero. Also a = b +k,+k = 1. ab),+k = b,+k + a,+1 b +1,+k a,+k Snce a UT m n K), the terms a,+1,... a,+k are zero. Snce b UT m n K), the term b,+k s also zero, so ab),+k = 0. Thus ab UT m n K). 9

10 Lemma 0.14 for Problem.9a). Let G = UT n K). Then [G, G] UT 1 nk). Proof. Frst we show ths holds for the generators of [G, G], that s, commutators aba 1 b 1 where a, b G. Let a = a j ), b = b j ), a 1 = c = c j ), b 1 = d = d j ). Then ab) j = k a k b kj a 1 b 1) j = m c m d mj So the j)th entry of aba 1 b 1 s aba 1 b 1) j = k = k ab) k a 1 b 1 ) kj ) a m b mk m t c kt d tj ) = k,m,t a m b mk c kt d tj Recall that because a G, a = 1 and a j = 0 for > j. Our clam s that aba 1 b 1 ),+1 = 0. aba 1 b 1),+1 = k,m,t a m b mk c kt d tj If a gven term a m b mk c kt d tj s non-zero, then m k t j, snce negatng any sngle one of these nequaltes would force one of the multpled entres to be zero. In partcular, f j = + 1, then m k t + 1. So there are only four nonzero terms: That s, = m = k = t = m = k and t = + 1 = m and k = t = + 1 m = k = t = + 1 aba 1 b 1),+1 = a b c d,+1 + a b c,+1 d +1,+1 + a b,+1 c +1,+1 d +1,+1 + a,+1 b +1,+1 c +1,+1 d +1,+1 = d,+1 + c,+1 + b,+1 + a,+1 By the prevous lemma, a,+1 + c,+1 = 0 and b,+1 + d,+1 = 0, so ths s zero. Thus all the commutators aba 1 b 1 le n UT 1 nk), so [G, G] UT 1 nk). Defnton 0.4. Let G be a group. The lower central seres of G s the sequence L k G) defned recursvely by L 1 G) = G and L k+1 G) = [G, L k G)]. 10

11 Lemma 0.15 for Problem.9a). Let G = UT n K). Let L k G) be the lower central seres of G. Then L k+1 G) UT k nk). Proof. The case k = 0 s just the statement that L 1 G) = G UT 0 nk) = UT n K). The case k = 1 s the content of the prevous lemma, whch establshed that L G) = [G, G] UT 1 nk). Now we nduct on k. Assume that L k G) UT k 1 n K). We need to show that L k+1 G) UT k nk). It s suffcent to show that the generators of L k+1 G) le n UT k nk), snce UT k nk) s a subgroup. A generator of L k+1 G) s of the form aba 1 b 1 where a = a j ) G and b = b j ) L k G). We want to show that aba 1 b 1 UT k nk), that s, aba 1 b 1 ),+m = 0 for m = 1,..., k. Recallng a calculaton from a prevous lemma, aba 1 b 1 ) j = r,s,t a s b sr c rt d tj Now substtute j = + m. aba 1 b 1 ),+m = r,s,t a s b sr c rt d t,+m Snce a, b, a 1, b 1 G = UT n K), all nonzero terms satsfy s r t + m. By nductve hypothess, along wth the fact that UTn k 1 K) s closed under nverson, b, b 1 K). Ths mples that UT k 1 n b s,s+1 =... = b s,s+k 1 = d t,t+1 =... = d t,t+k 1 = 0 Ths mposes further restrctons on nonzero terms of aba 1 b 1 ),+m, namely s r k and t + m k. Combnng all of our nequaltes, s r k t k + m k = 0 m k = k m for nonzero terms. But we only care about m = 1,..., k, so n all of these cases k m s false. That s to say, there are no nonzero terms, so aba 1 b 1 ),+m = 0 for m = 1,..., k. Thus aba 1 b 1 UT k n, so all generators of L k+1 G) are contaned n UT k nk), so L k+1 G) UT k nk). Ths completes the nducton. Proposton 0.16 Problem.9a). The group G = UT n K) s nlpotent. Proof. By the prevous lemma, L k+1 G) UT k nk). Clearly, UT n nk) s just the dentty, so L n+1 G) s trval. Hence G s nlpotent. Defnton 0.5. Let X n K) be the set of matrces n M n K) that are upper trangular, have ones along the dagonal, may have anythng along the frst row or last column, and are zero elsewhere. Lemma 0.17 for Problem.10a). Let A = a j ) X n K) and C = c j ) = A 1. Then for j and, j) 1, n), c j = a j. In words, all the entres of A 1 except the dagonal and the upper rght corner are just the negatve of the correspondng entry n A. 11

12 Proof. Snce A and C vansh on the off dagonals except for the frst row and column, we actually only need to show that c 1j = a 1j and c n = a n when j 1, n and 1, n. We know that AC = I, so AC) j = a k c kj = δ j k Assume j 1, n and 1, n. Then c kj 0 mples k = j or k = 1, so 0 = AC) 1j = k a k c kj = a 11 c 1j + a 1j c jj = c 1j + a 1j Thus c 1j = a 1j. And smlarly, 0 = AC) n = k a k c kn = a c n + a n c nn = c n + a n so c n = a n. Proposton 0.18 part of Problem.10a). Let X = X n K). For n 3, [X, X] = UT n n K) = {I + λe 1n λ K} where e 1n s the standard bass matrx 1 n the upper rght, zeroes elsewhere). Proof. Let A = a j ), B = b j ), C = A 1 = c j ), D = B 1 = d j ) X n K). ABA 1 B 1 ) j = AB) k CD) kj = ) ) a m b mk c kt d tj = a m b mk c kt d tj k k m t k,m,t Snce A, B, C, D are upper trangular, we know that the only nonzero terms occur when m k t j. Let E = ABA 1 B 1. We already know what E looks lke except n the frst row and last column. We consder two cases: 1. = 1, j n, < j. j = n, 1, < j In case 1), E j = E 1j = m,k,t a 1m b mk c kt d tj Snce j n, we know that d tj s nonzero only when t = 1 or t = j. E 1j = m,k a 1m b mk c k1 d 1j + a 1m b mk c kj d jj ) = a 11 b 11 c 11 d 1j + m,k a 1m b mk c kj = d 1j + m,k a 1m b mk c kj 1

13 Agan, we know that j n, so c kj s nonzero only when k = 1 for k = j, so we perform the same procedure to get E 1j = d 1j + c 1j + a 1m b mj m And we apply the same trck a 3rd tme, so b mj 0 only when m = 1 or m = j. E 1j = d 1j + c 1j + b 1j + a 1j By a symmetrc argument or just by transpose symmetry) n case ) we get E n = a n + b n + c n + d n By the prevous lemma Lemma 0.17), both of these expressons are zero. By consderng cases 1) and ), we have shown that a commutator n X s contaned n {I + λe 1n λ K} We know that ths set s a group, snce t s UT n n K), so t s closed under multplcaton, so the subgroup generated by the commutators wll also le nsde ths. If we can just show that [X, X] s not the trval subgroup, we wll be done. In the cases n = 1 and n =, the commutator subgroup s actually trval. For n = 1, X s trval, so [X, X] must also be trval. For n =, X s somorphc to K, va ) 1 λ λ 0 1 and so X s actually abelan, so [X, X] = 0. In the case n = 3, we have the concrete example A = B = where 1 0 ABA 1 B 1 = so [X, X] s nontrval. Smlar examples can be found for n > 3. Proposton 0.19 part of Problem.10a). Let X = X n K). For n 3, X has nlpotency class. Proof. We showed above that [X, X] = UTn n K). We also showed n Lemma 0.15 that [UT n K), UT n n K)] UT n 1 K) We know that UT n 1 n K) s the trval subgroup, so we can wrte the derved seres of X n the followng three equvalent ways. Thus X has nlpotency class. L 1 X) L X) L 3 X) X [X, X] [X, [X, X]] X UT n n K) {I} 13 n

14 Proposton 0.0 part of Problem.10a). Let X = X n K). [X, X]. Then the center of X s Proof. Let A = a j ) be n the center of X. Then A commutes wth the matrces B m = I+e 1m and C l = I + e ln for all m 1, l n. So AI + e 1m ) = I + e 1m )A = Ae 1m = e 1m A and smlarly for e ln. We compute the jth entry of Ae lm and e lm A. Observe that we can wrte the j-th entry of e lm as δ l δ jm. Ae lm ) j = k a k e lm ) kj = k a k δ lk δ mj = a l δ mj e lm A) j = k e lm ) k a kj = k δ l δ mk a kj = a mj δ l We know that A commutes wth e lm when l = 1 or m = n. When l = 1 we get one equaton, and when m = n we get another equaton. a 1 δ mj = a mj δ 1 a l δ nj = a nj δ l These hold for all, j. In partcular, takng = 1 n the left equaton, we get a 11 δ mj = a mj δ 11 = a mj = δ mj for all m 1. Takng j = n n the other equaton, we get a l δ nn = a nn δ l = a l = δ l for all l n. Puttng these together, A = a j ) looks exactly lke the dentty matrx except for possbly m = 1, l = n, that s, the 1n-th entry. Thus the center of X s contaned n [X, X]. We can also check that [X, X] s contaned n the center. Let B X, and C [X, X]. Then we can wrte C as C = I + λe 1n so BC = BI + λe 1n ) = B + λbe 1n CB = I + λe 1n )B = B + λe 1n B so to show that C s n the center, t s suffcent to check that e 1n commutes wth B. We can also wrte B as a sum B = I + b j e j,j where the only nonzero b j are b 1j for j 1 and b n for n. So to show that B commutes wth e 1n, t s suffcent to check that e 1j, e n commute wth e 1n for j 1, n. Ths we can check. In general, we have e j e lm = δ jl e m Specfcally when l = 1 and m = n, we get j 1 = e 1n e 1j = 0 = e 1j e 1n n = e 1n e n = 0 = e n e 1n Thus [X, X] s contaned n the center. Snce we have contanment both ways, t s equal to the center. 14