PUMaC 2017 · 加试 · 第 5 题
PUMaC 2017 — Power Round — Problem 5
题目详情
Problem 5.2.6. (15 points) Imagine a Dynkin diagram with a “branching point” – there is some vector with three lines coming off of it. Show that one of those lines must have length one (i.e. only one edge), and show that of the other two, either • if one has length 1 then the third may have any length • if one has length 2 then the other’s length may not be more than 4. Theorem 5.2.1 ( Classification of Dynkin diagrams ) . The following are the only possible irreducible Dynkin diagrams: Proof. Problems 5.2.1 and 5.2.3-7. This is an amazing result! We gave only a handful of restrictions on our definition of these diagrams – and yet we get such a restricted and asymmetrical list of options. Let’s now look into how this relates to the Lie algebra content that we spent so much time building up. PUMaC 2017 Power Round Page 21 5.3 Dynkin diagrams of Lie algebras (40 points) We are now prepared to look at root systems and bases of our simple Lie algebras! In doing so, with a little help from Jean-Pierre Serre, we will be able to see, pictorially, precisely 8 every isomorphism between the classical Lie algebras. Definition 5.3.A. The Dynkin diagram of a Lie algebra is the Dynkin diagram corre- sponding to the vector space of its roots. Definition 5.3.B. e denotes the N × N matrix having the entry 1 at row i and column i,j j , and zeros elsewhere (where the value of N will be clear from context). T For a worked example, take L = so . As L ’s elements satisfy x R + R x = 0, we 2 n 2 n 2 n [ ] M P know that x can be written in the form for any n × n matrix M and n × n T Q − M matrices P and Q both equal to their negative transpose. { } n ∑ H is the set of diagonal matrices in L , so H = c ( e − e ) | c ∈ C . We can i i,i i + n,i + n i i =1 give L the additional basis elements m = e − e ( i 6 = j ), p = e − e i,j i,j j + n,i + n i,j i,j + n j,i + n T ( i < j ), and p to go along with those of H . (It is straightforward to check that this is i,j indeed a basis of L – try writing down the matrices m and p to verify this.) We find, i,j i,j for an arbitrary element h ∈ H of the form specified in H ’s specification, [ h, m ] = ( c − c ) m i,j i j i,j [ h, p ] = ( c + c ) p i,j i j i,j T T [ h, p ] = − ( c + c ) p i j i,j i,j so our roots are particularly convenient: root eigenspace λ − λ span { m , m } i j i,j j,i λ + λ span { p , p } i j i,j j,i ∗ For H , a vector space having the basis { λ } , we see that { λ − λ , λ − λ , . . . , λ − i 1 2 2 3 n − 1 λ , λ + λ } is a base; this base will be what we use to build our Dynkin diagram. n n − 1 n Computation gives that − 2 i = j 〈 λ − λ , λ − λ 〉 = − 1 | i − j | = 1 i i +1 j j +1 − 0 otherwise { − 1 i = n − 2 〈 λ − λ , λ + λ 〉 = i i +1 n − 1 n − 0 otherwise { − 1 i = n − 2 〈 λ + λ , λ − λ 〉 = n − 1 n i i +1 − 0 otherwise Since e ( v, w ) = 〈 v, w 〉 · 〈 w, v 〉 , our Dynkin diagram is shown in Figure 6. 8 The classical Lie algebras are the four spaces defined in Section 2.2: sl , so , sp . n n 2 n PUMaC 2017 Power Round Page 22 Figure 6: In this diagram, we use the shorthands α = λ − λ and β = λ + λ . i i i +1 n n − 1 n Theorem 5.3.1 ( Serre’s theorem ) . Two Lie algebras with the same Dynkin diagram are isomorphic. Serre’s theorem is the final piece of the puzzle; now, you will put to the test what we have learned to find the bases of the other classical Lie algebras.
解析
Problem 5.3.1. (10 points each) For n ≥ 1 in all cases... (i) draw the Dynkin diagram for so 2 n +1 PUMaC 2017 Power Round Page 15 (ii) draw the Dynkin diagram for sl n (iii) draw the Dynkin diagram for sp 2 n and show your reasoning. In particular, many details of the computations were omitted above, but you will be expected to (briefly) justify the numbers you find. Solution : T (i) As L ’s elements satisfy x R + R x = 0, we know that x can be written in the form 2 n 2 n T T 0 C − B B M P where B and C each have n rows and 1 column and for any n × n T − C Q − M matrix M and n × n matrices P and Q both equal to their negative transpose. { } n ∑ H is the set of diagonal matrices in L , so H = c ( e − e ) | c ∈ C (in- i i,i i + n,i + n i i =1 dexed from zero this time). We can give L the additional basis elements m = i,j T e − e (1 ≤ i 6 = j ≤ n ), p = e − e (1 ≤ i < j ≤ n ), p , i,j j + n,i + n i,j i,j + n j,i + n i,j B = e − e (1 ≤ i ≤ n ) and C = e − e (1 ≤ i ≤ n ) to go along i i, 0 0 ,i + n i 0 ,i i + n, 0 with those of H . We find, for an arbitrary element h ∈ H of the form specified in H ’s specification, [ h, m ] = ( c − c ) m i,j i j i,j [ h, p ] = ( c + c ) p i,j i j i,j T T [ h, p ] = − ( c + c ) p i j i,j i,j [ h, B ] = c B i i i [ h, C ] = − c C i i i so our roots are: root eigenspace λ − λ span { m , m } i j i,j j,i λ + λ span { p , p } i j i,j j,i λ span { B } i i − λ span { C } i i ∗ For H , a vector space having the basis { λ } , we see that { λ − λ , λ − λ , . . . , λ − i 1 2 2 3 n − 1 λ , λ } is a base; this base will be what we use to build our Dynkin diagram. We set n n α = λ − λ and β = λ . When i < n if e = m then h = e − e − i i i +1 n α i,i +1 α i,i i + n,i + n i i e + e . e = B gets h = 2( e − e ). We compute i +1 ,i +1 i + n +1 ,i + n +1 n n,n 2 n − 2 n β β 2 e i = j α j [ h , e ] = − e | i − j | = 1 α α α j i j − 0 otherwise { − 2 e i = n − 1 α i [ h , e ] = β α i − 0 otherwise { − e i = n − 1 β [ h , e ] = α β i − 0 otherwise PUMaC 2017 Power Round Page 16 Computation gives that − 2 i = j 〈 λ − λ , λ − λ 〉 = − 1 | i − j | = 1 i i +1 j j +1 − 0 otherwise { − 2 i = n − 1 〈 λ − λ , λ 〉 = i i +1 n − 0 otherwise { − 1 i = n − 1 〈 λ , λ − λ 〉 = n i i +1 − 0 otherwise Since e ( v, w ) = 〈 v, w 〉 · 〈 w, v 〉 , our Dynkin diagram is: { } n − 1 ∑ (ii) H is the set of diagonal matrices in L , so H = c ( e − e ) | c ∈ C . We can i i,i n,n i i =1 give L the additional basis elements e (1 ≤ i 6 = j ≤ n ) to go along with those of i,j H . We find, for an arbitrary element h ∈ H of the form specified in H ’s specification, [ h, e ] = ( c − c ) e , so our roots are λ − λ with eigenspace span { e } . i,j i j i,j i j i,j ∗ For H , a vector space having the basis { λ } , we see that { λ − λ , λ − λ , . . . , λ − i 1 2 2 3 n − 1 λ } is a base; this base will be what we use to build our Dynkin diagram. We set n α = λ − λ . When i < n if e = e then h = e − e . We compute i i i +1 α i,i +1 α i,i i +1 ,e +1 i i 2 e i = j α j [ h , e ] = − e | i − j | = 1 α α α j i j − 0 otherwise Computation gives that − 2 i = j 〈 λ − λ , λ − λ 〉 = − 1 | i − j | = 1 i i +1 j j +1 − 0 otherwise Since e ( v, w ) = 〈 v, w 〉 · 〈 w, v 〉 , our Dynkin diagram is where ` = n − 1. T ˜ ˜ (iii) As L ’s elements satisfy x R + R x = 0, we know that x can be written in the form 2 n 2 n [ ] M P for any n × n matrix M and n × n matrices P and Q both equal to their T Q − M transposes. { } n ∑ H is the set of diagonal matrices in L , so H = c ( e − e ) | c ∈ C . We i i,i i + n,i + n i i =1 can give L the additional basis elements m = e − e (1 ≤ i 6 = j ≤ n ), i,j i,j j + n,i + n T T p = e + e (1 ≤ i < j ≤ n ), p , p = e (1 ≤ i ≤ n ), and p to go i,j i,j + n j,i + n i,i i,i + n i,j i,i along with those of H . We find, for an arbitrary element h ∈ H of the form specified in H ’s specification, [ h, m ] = ( c − c ) m i,j i j i,j [ h, p ] = ( c + c ) p i,j i j i,j T T [ h, p ] = − ( c + c ) p i j i,j i,j PUMaC 2017 Power Round Page 17 (where in the latter two lines i may equal j ) so our roots are: root eigenspace λ − λ span { m , m } i j i,j j,i λ + λ span { p , p } i j i,j j,i 2 λ span { p } i i,i T − 2 λ span { p } i i,i ∗ For H , a vector space having the basis { λ } , we see that { λ − λ , λ − λ , . . . , λ − i 1 2 2 3 n − 1 λ , 2 λ } is a base; this base will be what we use to build our Dynkin diagram. We set n n α = λ − λ and β = 2 λ . When i < n if e = m then h = e − e − i i i +1 n α i,i +1 α i,i i + n,i + n i i e + e . e = p gets h = e − e . We compute i +1 ,i +1 i + n +1 ,i + n +1 β n,n β n,n 2 n, 2 n 2 e i = j α j [ h , e ] = − e | i − j | = 1 α α α j i j − 0 otherwise { − 2 e i = n − 1 α i [ h , e ] = β α i − 0 otherwise { − e i = n − 1 β [ h , e ] = α β i − 0 otherwise Computation gives that − 2 i = j 〈 λ − λ , λ − λ 〉 = − 1 | i − j | = 1 i i +1 j j +1 − 0 otherwise { − 2 i = n − 1 〈 λ − λ , λ 〉 = i i +1 n − 0 otherwise { − 1 i = n − 1 〈 λ , λ − λ 〉 = n i i +1 − 0 otherwise Since e ( v, w ) = 〈 v, w 〉 · 〈 w, v 〉 , our Dynkin diagram is: