the one above: g σ(g) g σ(g) id 1 0 0 1! x 0 1 1 0! y i 0 0 −i! yx 0 i −i 0! y2 −1 0 0 −1! y2x 0 −1 −1 0 ! y3 −i 0 0 i! y3x 0 −i i 0 ! . The trace is the sum of the diagonal entries of a matrix, and is denoted by Tr. The trace of a matrix has the property that Tr(AB) = Tr(BA), which implies that Tr(B−1AB) = Tr(ABB−1) = Tr(A), and so the trace is constant on each conjugacy class. Hence, we can tell that two representations are conjugate if they have the same trace for all elements. If ρ andσ are homomorphisms fromGinto GL(n, C), they are conjugate if there exists a g inGL(n, C) such that g−1ρ(h)g =σ(h) for all g ∈G. If we represent a group by a homomorphism, we can define the character of (ρ, V), to be a function χ(ρ,V) : G→Csuch that g 7→Tr(ρ(g)). Note that this allows us to find the characters of these elements of D4. By computing the trace of these matrices, we find that id has a character of 2, y2 has a character of −2, and all of the other elements have a character of 0. Channels • 2022 • Volume 6 • Number 2 Page 31
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