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18.755 Problems 1 solutions 1. In class Friday we considered a group H = Hx,ξ,z | x ∈ R, ξ ∈ R, z = eiζ ∈ S 1 with the multiplication law Hx1 ,ξ1 ,z1 · Hx2 ,ξ2 ,z2 = Hx1 +x2 ,ξ1 +ξ2 ,z1 z2 eix1 ξ2 . (This was a group of linear transformations of an infinite-dimensional vector space of functions on R.) Can you find a collection of finite matrices satisfying this multiplication law? The answer is no, you cannot find such a collection of matrices all of which are distinct. More precisely Theorem.. Suppose V is an n-dimensional vector space over a field k, and that ρ: H → GL(V ) is a homomorphism of H into the group of invertible linear transformations of V . For every zℓ ∈ S 1 which is an ℓth root of unity, with ℓ a prime not equal to the characteristic of k, and larger than the dimension of V , the linear transformation ρ(H0,0,zℓ ) is equal to the identity. (The problem does not ask for a group, but only that the multiplication law be respected. I will omit the very easy and dull analysis of that more general case.) Proof. Replacing k by its algebraic closure, we may assume that k is algebraically closed. Notice first that for any z = eiθ , −1 −1 . H0,0,z = H1,0,1 H0,θ,1 H1,0,1 H0,θ,1 The group element H0,0,z is therefore a commutator; so its image in GL(V ) is a commutator, and therefore det(ρ(H0,0,z ) = 1. For the particular zℓ in the theorem, ρ(H0,0,zℓ ) is a linear transformation of (finite) order ℓ, assumed to be relatively prime to the characteristic. It follows that H0,0,zℓ must be diagonalizable, with eigenvalues various ℓth roots of 1 in k. We can write the eigenspace decomposition as M V = Vζ . ζ ℓ =1 Now it is also true that H0,0,zℓ belongs to the center of H, so ρ(H0,0,zℓ ) commutes with all of the linear transformations ρ(Hx,ξ,z ). These linear transformations therefore respect the eigenspace decomposition. In matrix terms, using a basis of eigenvectors of ρ(H0,0,zℓ ), all the matrices ρ(Hx,ξ,z ) are block diagonal. Each block separately defines a group homomorphism ρζ : H → GL(Vζ ) 1 2 exactly as in the theorem. We can therefore apply the commutator argument to conclude 1 = det(ρζ (H0,0,zℓ )) = ζ dim Vζ . But we chose ℓ to be a prime larger than dim V and ζ to be an ℓth root of 1. This equation therefore implies that either Vζ = 0, or ζ = 1. The conclusion is that ρ(H0,0,zℓ ) is the identity, as we wished to show. Q.E.D. This proves that any group homomorphism of H into finite matrices must have a nontrivial kernel. At least in characteristic zero, similar arguments will show that the kernel must include all the roots of unity in S 1 . The hint was more or less intentionally misleading. The matrices 1 x θ N = Nx,ξ,θ = 0 1 ξ | x, ξ, θ ∈ R 0 0 1 form a group with center R (the θ coordinate. In particular, this means that the subgroup 2πZ ⊂ R is normal. The quotient group N/2πZ is H, but the quotient of a group of matrices need not be a group of matrices. Mathematicians call the matrix group N the “Heisenberg group,” but the group that’s actually interesting in physics is the quotient H, which is not a matrix group at all. 2. Same question with R and S 1 replaced by Z/pZ (integers mod a prime p) and the group µp of complex pth roots of 1: Hx1 ,ξ1 ,z1 · Hx2 ,ξ2 ,z2 = Hx1 +x2 ,ξ1 +ξ2 ,z1 z2 e2πix1 ξ2 /p (xi ∈ Z/pZ, ξi ∈ Z/pZ, zi ∈ µp ). (To make sense of this formula, notice that if m ∈ Z/pZ, then e2πim/p is a well-defined pth root of 1. The formula defines a group of order p3 .) Can you find a group of finite matrices satisfying this multiplication law? How small can you make the matrices? Phrased without matrices, we are looking for a vector space V over a field k and a homomorphism ρ: H → GL(V ); what makes it interesting is looking for an injective homomorphism ρ. There is an easy way to do this, for any finite group H and any field k: define V to be the |H|-dimensional vector space of all k-valued functions on H, and define [ρ(h)f ](x) = f (h−1 x). Another way to say the same thing is that we are realizing H as a subgroup of the group of permutations of the set H; then realizing those permuations of |H| elements as |H| × |H| (permutation) matrices. In any case the size of the required matrices is |H|, in this case p3 . But smaller is possible. As noted in the problem, Z/pZ can be identified with the group µp of pth roots of 1 by sending m + pZ to e2πim/p . If we do this, our group law looks like Hx1 ,ξ1 ,t1 · Hx2 ,ξ2 ,t2 = Hx1 +x2 ,ξ1 +ξ2 ,t1 +t2 +x1 ξ2 (xi ∈ Z/pZ, ξi ∈ Z/pZ, ti ∈ Z/pZ). 3 This is precisely the law for multiplying the 1 Hx,ξ,t = 0 0 3 × 3 matrices x t 1 ξ 0 1 with entries in Z/pZ. So over a field with p elements, we can use 3 × 3 matrices. I believe that this is the smallest possible size in characteristic p, but I have not written down a proof. Over the complex numbers, we can use p × p matrices; or better, we can think of V as the space of functions on the group Z/pZ. Exactly as in the case of functions on R, we use three operations on functions: [Qx f ] (y) = f (x − y) [Pξ f ] (y) = e −2πiξy/p [Zz f ] (y) = zf (y) (x, y ∈ Z/pZ) f (y) (ξ, y ∈ Z/pZ) (z ∈ µp , y ∈ Z/pZ) Composing them gives Hx,ξ,z = Qx Pξ Zz , [Hx,ξ,z f ] (y) = ze−2πiξy/p f (y − x). The multiplication law for the Hs can be computed exactly as we did in class for the case of R instead of Z/pZ; it is the one in the problem. Therefore we’ve realized the group H as a linear transformations of the p-dimensional complex vector space V of functions on Z/pZ. If we use the basis of V consisting of functions that are 1 at one group element and zero at the others, then the matrices Qx look like 0 0 0 ··· 1 0 0 ··· 0 0 0 ··· 0 1 0 ··· 0 0 0 ··· 0 0 1 ··· .. .. .. .. . . . . 1 0 0 ··· 0 0 0 ··· 0 1 0 ··· 0 0 0 ··· 0 0 1 ··· 0 0 0 ··· . . .. .. .. .. . . with the 1s appearing x spaces below or to the left of the diagonal and p − x spaces above or to the right of the diagonal. (Here as almost always, I may have the signs wrong.) The matrices Pξ are diagonal, with entries 1 = e0πiξ/p , e2πiξ/p , e4πiξ/p , and so on. The matrices Zz are z Id. A general matrix Hx,ξ,z has the shape in the display above, but the entries are pth roots of unity instead of 1. This construction works in any field k admitting a nontrivial pth roots of 1; in particular, in any algebraically closed field of characteristic not p. It turns out that these are the smallest matrices by which H may be realized over such fields. In characteristic p, we saw that we can use 3 × 3 matrices I have not thought carefully about how small the matrices can be over fields of characteristic not p in which there are no nontrivial pth roots of 1. My guess would be rp, where r is the degree of the field extension generated by adjoining a pth root of 1. (Necessarily r ≤ p, and it’s easy to show that you can realize H using rp × rp matrices.)