Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 10

Work over the complex numbers, the usual scalar field for the finite-group character theory used below. A group representation on a nonzero vector space $V$ is a homomorphism $D:G\to GL(V)$, so $D(gh)=D(g)D(h)$ and $D(e)=I$.

A faithful representation is injective: distinct group elements have distinct representing maps. Equivalently, its kernel is just $\{e\}$.

An irreducible representation has no nonzero proper invariant subspace: if $W\subseteq V$ and $D(g)W\subseteq W$ for every $g\in G$, then $W=\{0\}$ or $W=V$. Irreducibility concerns invariant subspaces, whereas faithfulness concerns whether any group element is represented by the identity.

The homomorphism law and $g{g}^{-1}=g^{-1}g=e$ giveThus $D(g^{-1})$ is both a left and right inverse of $D(g)$, provingIf only multiplicativity is initially stated, $D(e{)}^2=D(e)$ together with invertibility of representation matrices first implies $D(e)=I$. This explains why a group representation takes values in the general linear group rather than arbitrary square matrices.

The elements $1,-1$ commute with the whole quaternion group, so they form singleton classes. Each of the three remaining pairs is a conjugacy class: for example conjugation by $j$ sends $i$ to $-i$, and likewise cyclically. No such pair can mix with a different pair, because the quotient by $\{\pm 1\}$ is abelian and its cosets $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$ are preserved by conjugation. HenceFor a finite group over $\mathbb{C}$, the number of inequivalent irreducible representations equals the number of conjugacy classes, so there are five. The sum of squares of irreducible representation dimensions is the group order, eight. With five positive integer dimensions, the only possibility isIndeed five one-dimensional contributions give five, and replacing one by a two-dimensional contribution adds exactly three; any further enlargement would exceed eight. These are the dimensions used in the character table of the quaternion group.

For a one-dimensional character, write $\chi (i)=ϵ$ and $\chi (j)=\eta$. Since $\chi (-1)=1$, the relations $i^2=j^2=-1$ require $ϵ,\eta \in \{\pm 1\}$. Also $ij=k$ gives $\chi (k)=ϵ\eta$. The four sign choices therefore supply all four linear characters.

For the remaining two-dimensional character ${\chi }_2$, its value at the identity is two. Column orthogonality between the identity and $-1$ gives $4+2{\chi }_2(-1)=0$, so ${\chi }_2(-1)=-2$. Between the identity and each of the other classes, the four linear-character values sum to zero, forcing the remaining character values to vanish. Thus the character table isHere column orthogonality is the standard finite-group character theorem ${\sum }_{\rho }{\chi }_{\rho }(e{)}^{\ast }{\chi }_{\rho }(g)=0$ for $g\ne e$. The rows also have unit norm and are mutually orthogonal in the character inner product with the displayed class-size weights. This completes the character table of the quaternion group.

Taking traces of the given matrices gives the character on the five classes,For a complex finite-group representation, the multiplicity of an irrep $\rho$ is its character inner product with ${\chi }_D$:Every linear character has value one at both $1$ and $-1$, so its multiplicity is $(4-4)/8=0$. For the two-dimensional character,ThereforeThe dimensions check: $2\times 2=4$.

The choice of scalar field matters here, even though the displayed matrices are real. Over $\mathbb{R}$, this particular four-dimensional representation is irreducible: for any nonzero real vector $v$, the vectors $v,D(i)v,D(j)v,D(k)v$ are mutually orthogonal with the same nonzero norm. This follows from the skew-symmetry and quaternion multiplication identities of the three matrices. A nonzero invariant real subspace contains all four vectors and is therefore the whole space. Its complexification is the two-copy decomposition above.