New version page

HARVARD MATH 126 - Lecture 8

Upgrade to remove ads
Upgrade to remove ads
Unformatted text preview:

Math 126 Lecture 8 Summary of results Application to the Laplacian of the cube Summary of key results Every representation is determined by its character If j is the character of a representation and c is an irreducible character then j c is the number of times that an irreducible with character c occurs in the decomposition of the representation into a direct sum of irreducibles The group G G acts on G by left and right multiplication and the corresponding representation of G G on F G decomposes as F G W1UW1 WkUWk The decomposition of F G In the decomposition F G W1UW1 WkUWk the Wi range over the irreducible representations of G Each summand in the above decomposition is irreducible under G G they are all distinct and k of conjugacy classes of G The elements of G G of the form a e act on F G by af x f a 1x This is the left regular representation and will be denoted by l On each summand the element a e acts as ri a UI So dim Hom l G F G F G S ni2 G The centralizer of the left regular representation is the right regular representation The elements of the form e b act on F G by e b f x f xb This gives the right regular representation denoted by r As l a r b f x f a1xb r b l a f x we see that r b l a l a r b for all a and b in G So r b Hom l G F G F G The r b are independent and there are G of them They form a basis of Hom l G F G F G Every element of Hom l G F G F G can be written uniquely as cb r b where the cb are constants The Laplacian of the graph of the cube 2 3 4 1 3 2 4 Let V denote the set of vertices of the cube So V 8 and in the labeling in the figure V 1 2 3 4 1 2 3 4 The Laplacian L is 1 the operator on F V where Lf x f x f y where the sum 3 is over all y which 1 are connected to x by an edge For example Lf 1 f 1 f 2 f 3 f 4 3 The eigenvalues of the Laplacian 2 3 4 1 3 2 4 We will use group theory to compute the eigenvalues of L The group S4 acts as rotational symmetries of the cube and acts transitively on V The identity has 8 fixed points The three cycles act as rotations through 120o or 240o about an axis joining diagonally opposite vertices and so have 2 fixed points For example 234 fixes 1 and 1 All other elements have no fixed points So the character j of the representation of S4 on F V is given by j e 8 j abc 2 j 0 on all other types The decomposition of F V We have computed that j e 8 j abc 2 j 0 on all other types The character table of S4 is The left hand column labels the representations as we shall explain later We see that j c 2 2 2 8 8 2 0 while j c 1 for each of the other irreducible representations We see that F V decomposes as a direct sum of four irreducible representations of dimensions 1 3 3 and 1 The multiplicities of L Being a nearest neighbor is preserved by the action of S4 on V This implies that L Hom S4 F V F V So if f F V is an eigenvector of L say Lf lf Then af is an eigenvector of L with the same eigenvalue for any a G L af a Lf a lf l af So the set of eigenvectors of L with eigenvalue l is an invariant Subspace and hence must be a sum of irreducibles Thus the eigenvalues of L must occur with multiplicities 1 3 3 and 1 For this argument all we used was that L Hom S4 F V F V But we will in fact be able to determine the eigenvalues from group theory and the form of L The Laplacian and the adjacency matrix We can write L I 1 3 A where A is the adjacency matrix of the graph Here A is the 8 by 8 matrix whose rows and columns are labeled by the vertices of the graph and an the entry in the i j position is 1 if I and j are joined by an edge and 0 otherwise In our case A 0 1 1 1 0 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 1 0 1 0 0 0 If we find the eigenvalues of A then we can derive the eigenvalues of A We shall find that the eigenvalues of A are 3 1 1 3 with multiplicities 1 3 3 1 F V as a subspace of F G We will let G S4 and H e 234 243 the isotropy group of the vertex 1 Much of what we have to say will only depend on the fact that G acts transitively on V so that we can we can identify V With G H where H is the isotropy group of a point v V In our case v is the vertex 1 To each f F V assign F F G by the rule F a f av a G Notice that if h H then F ah f ahv f av F a In other words F is invariant under the right action of H on F G In symbols we write this as F F G H r Conversely suppose that F F G H r Then we may define f w F b where bv w If we chose a different b we would get the same value since b bh for an h H We have given a G equivalence between F V and F G H r Extending the operator A to F G We have identified F V with F G H r which is a subspace of F G which is invariant under the action of the subgroup G H of G G The operator A can be considered as an operator on this subspace F G H r and commutes with the action of G H We can extend the operator A in many ways so as to get an operator B on all of F G which commutes with the action of G H on F G For example we can choose an invariant complement under G H to F G H r in F …


View Full Document
Download Lecture 8
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture 8 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture 8 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?