Math 502: Abstract Algebra

Homework 12

Lawrence Tyler Rush

<me@tylerlogic.com> In Collaboration With

Keaton Naff

January 5, 2014

http://coursework.tylerlogic.com/courses/upenn/math502/homework12

1

Let G be a finite group and ℂ(G) be the set of all ℂ-valued functions on G. Define an inner-product on the ℂ-vector space by

(f |f) := -1--∑ f (x)⋅f-(x) 1 2 #G x∈G 1 2

(a)

For each g ∈ G, let U_g ∈ End_ℂ(C(G)) be defined by f

f(xg). With this definition, we see that

--1-∑ --------- (Ug(f1)|Ug(f2)) = #G Ug (f1)(x)⋅Ug(f2)(x) x∑∈G ------ = --1- f1(xg)⋅f2(xg) #G x∈G

but because right translation in G is a bijection, we can alter the index on G to be

∑ ----- (Ug(f1)|Ug(f2)) = -1-- f1(y)⋅f2(y) #G y∈G = (f1|f2)

and hence U_g is a unitary operator for each g ∈ G.

(b)

For each g ∈ G, let U_g ∈ End_ℂ(C(G)) be defined by f

f(xg). With this definition, we see that

--1-∑ (Ug (f1)|f2) = #G Ug (f1)(x)⋅f2(x) x∑∈G = --1- f1(xg)⋅f2(x) #G x∈G

however in the above equation, we can substitute y for xg to get

(Ug(f1)|f2) =--1-∑ f1(y)⋅f2(yg-1) #G y∈G

which indicates that U_g^-1 will be the hermitian conjugate of U_g.

(c)

2

Let V be a finite dimensional vector space over a field F which is either ℂ or ℝ with S a linear operator on V .

(a) Prove S being skew-hermitian implies exp(S) is unitary

Assume that S is skew-hermitian. Let λ ∈ F be an eigenvalue of S with corresponding eigenvector v. Then we have that

∑∞ Sk ∞∑ λk λ exp(S)(v) = -k! (v) = k!(v) = e (v) k=0 k=0

with the rightmost equality coming from the Taylor series expansion of the exponential function. So we see that e^λ is an eigenvalue for exp(S). Now because S was assumed skew-hermitian, then λ ∈ √--1 ℝ, implying that λ = ai for some a ∈ ℝ. So e^ai is an eigenvalue of exp(S), but since it has magnitude of 1, then exp(S) must be unitary.

(b)

Counter-example:
Let

( i ) ( - 1 ) A = and B = 1

then both A and B are period matrices since

( ) ( ) ( ) ( ) 2 - 1 3 - i 4 1 5 i A = A = A = A =

and

( ) ( ) ( ) ( ) B2 = - 1 B3 = 1 B4 = 1 B5 = - 1 - 1 - 1 1 1

Therefore since ∑ _k=0^∞ 1-
k! converges (to e, although it’s immaterial), then exp(A) = I + a₁A¹ + a₂A² + a₃A³ + a₄A⁴ and exp(B) = b₁B¹ + b₂B² + b₃B³ + b₄B⁴ where each a_i,b_i ∈ F. In other words, exp(A) ⋅ exp(B) is a finite linear combination of 16 matrices.

Now I’d like to show that exp(A + B) is not a finite linear combination of matrices, but I am unsure how to do so.

(c) Extra Credit

3 Extra Credit

4

(a)

If we let T ∈ End_ℂ(ℂ[ℤ∕nℤ] be multiplication by b₀[0] + b₁[1] +

+ b_n-1[n- 1], then we have that

T([0]) = b [0]+ b [1]+ ⋅⋅⋅+ b [n - 1] 0 1 n-1 T([1]) = bn-1[0]+ b0[1]+b1[2]+⋅⋅⋅+ bn-2[n - 1] ... T([n - 1]) = b1[0]+ b2[1]+ ⋅⋅⋅+ b0[n - 1]

which implies that the matrix representation of T in the basis {[0],[1],…,[n - 1]} of ℂ[ℤ∕nℤ] is A_n.

(b)

(c) Extra Credit

(d) Extra Credit