Brian Bi

Exercise 3.6.1 Let $$\{w_1, \ldots, w_m\}$$ be a basis of $$W$$. Let $$\{x_1, \ldots, x_n\}$$ be a basis of $$V/W$$. It is easy to construct a basis of $$V$$ by combining the two bases. Explicitly, a basis of $$V$$ is given by $$\{y_1, \ldots, y_{m+n}\}$$ where $$y_1 = w_1, \ldots, y_m = w_m$$, and for each $$i = 1, 2, \ldots, n$$, $$y_{m+i}$$ is arbitrarily chosen from $$\pi^{-1}(x_i)$$ (where $$\pi$$ denotes the projection map from $$V$$ to $$V/W$$).

Let $$a \in A$$ be given. For each $$i = 1, 2, \ldots, m+n$$, write $$ay_i = \sum_{j=1}^{m+n} c_{ij} y_j \label{eqn:components}$$ Then, $$\chi_V(a) = \sum_{i=1}^{m+n} c_{ii} \label{eqn:chiV}$$ Observe that $$aw_i = \sum_{j=1}^m c_{ij} w_j$$ since $$w_i = y_i$$. There are no contributions from higher $$j$$ since $$W$$ is a subrepresentation. Thus, $$\chi_W(a) = \sum_{i=1}^m c_{ii} \label{eqn:chiW}$$ Fix $$k \in \{1, 2, \ldots, n\}$$ and apply the projection map to both sides of $$(\ref{eqn:components})$$, putting $$i = m + k$$. We obtain \begin{equation*} ax_k = a\pi(y_i) = \pi(ay_i) = \sum_{j=m+1}^{m+n} c_{ij} x_{j-m} \end{equation*} where there are no contributions from lower $$j$$ since $$\pi(y_j) = 0$$ for all $$j \leq m$$. The coefficient of $$x_k$$ on the RHS is obtained by setting $$i = m + k$$ and $$j-m = k$$ in $$c_{ij}$$, therefore: \begin{equation*} \chi_{V/W}(a) = \sum_{k=1}^n c_{m+k,m+k} = \sum_{i=m+1}^{m+n} c_{ii} \end{equation*} Together with $$(\ref{eqn:chiV})$$ and $$(\ref{eqn:chiW})$$ this implies that $$\chi_V = \chi_W + \chi_{V/W}$$.