Brian Bi
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Exercise 4.3.1 Etingof confirmed that the book contains an error and the condition that $$f(xi) = \sqrt{-1}f(x)$$ should be replaced by $$f(ix) = \sqrt{-1}f(x)$$.

Let $$V$$ denote the space of functions defined in the exercise. The action of left-multiplication by $$i$$ in $$Q_8$$ splits the latter into two orbits, $$A = \{1, i, -1, -i\}$$ and $$B = \{k, -j, k, j\}$$. If $$f \in V$$, then $$f$$ is uniquely determined by the value it takes on one element from each orbit, say $$1$$ and $$k$$. Therefore $$V$$ is a vector space with dimension 2 and has a basis given by $$f_1, f_2$$ where $$f_1(1) = f_2(k) = 1$$ and $$f_1(k) = f_2(1) = 0$$.

Suppose $$f \in V, g \in Q_8$$. Then for each $$x \in Q_8$$, we have $$g \circ f(ix) = f(ixg) = \sqrt{-1} f(xg) = \sqrt{-1} g \circ f(x)$$, thus $$g \circ f \in V$$ also. It is clear that $$1 \circ f = 1$$ and $$gh \circ f = g \circ (h \circ f)$$. So $$V$$ is a valid representation of $$Q_8$$.

Direct computation shows that \begin{align*} i \circ f_1 &= if_1 \\ i \circ f_2 &= -if_2 \\ j \circ f_1 &= -if_2 \\ j \circ f_2 &= -if_1 \\ k \circ f_1 &= -f_2 \\ k \circ f_2 &= f_1 \end{align*} so in matrix form, $\rho(i) = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \qquad \rho(j) = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \qquad \rho(k) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ By inspection, there is no nontrivial common eigenspace of these matrices, so $$V$$ is irreducible. (Note that (4.3.1) involves the same set of three matrices, but assigned differently. This can be viewed as a consequence of the fact that $$Q_8$$ has an automorphism that sends $$i, j, k$$ to $$j, k, i$$ respectively.)