Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 10

A finite-dimensional representation of a group $G$ on a vector space $V$ is a group homomorphism $D:G\to \mathrm{GL}(V)$. Thus each $D(X)$ is invertible andHere the group identity and the identity linear map are different objects, represented by corresponding symbols. Applying the multiplication property to $X{X}^{-1}=X^{-1}X=I$ givesBy the defining property of a matrix inverse,If the definition only states the multiplication law together with invertibility, $D(I{)}^2=D(I)$ already forces $D(I)=I_V$ after multiplication by its inverse.

Use the identity relation as printed in the PDF, $XYZ=I$. Since $Z^2=I$, multiplying on the right by $Z$ gives $XY=Z$. Alsobecause $X,Y,Z$ are involutions. Thus $X$ and $Y$ commute, and so does their product $Z$ with either of them. Therefore the group is Abelian, and is the Klein four-group.

Each conjugacy class is a singleton. Over the complex numbers every irreducible representation of a finite abelian group is one-dimensional. Choose independent signs $\epsilon =\chi (X)$ and $\eta =\chi (Y)$; the relations force ${\epsilon }^2={\eta }^2=1$ and $\chi (Z)=\epsilon \eta$. Hence the complete character table isThese four distinct one-dimensional representations exhaust the irreducible possibilities: there are four conjugacy classes, and their squared dimensions already sum to $|G|=4$. The rows also satisfy character orthogonality. The local transcription damages the final symbol in the defining relation; the PDF makes the identity $I$ unambiguous.

Put $Y_p=D(Y)$. Direct matrix multiplication givesfor every real $p$. Since $D(X)=-I_2$ commutes with all matrices and $D(Z)=-Y_p$, the remaining relations also hold:All matrices are invertible, and the relations reproduce the multiplication table of the Klein four-group. ThereforeA representation need not be an orthogonal representation; imposing orthogonality on the displayed matrices would be an extra condition.

The characters are their traces, independent of $p$:In terms of part (b), $\chi ={\chi }_{-+}+{\chi }_{--}$, so the representation is reducible. For an explicit change of basis, takeThen $S^{-1}D(Y)S=\mathrm{diag}(-1,1)$ and $S^{-1}D(Z)S=\mathrm{diag}(1,-1)$, while $D(X)$ remains $-I_2$. These are the equivalent sheared representations of the Klein four-group; the parameter changes the basis, not the two irreducible constituents.