Brian Bi
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Problem 5.8.4 Let's start by recalling that the elements of \(\Ind^G_K V\) are functions \(G \to V\) and the elements of \(\Ind^G_H \Ind^H_K V\) are functions \(G \to (H \to V)\). These functions must satisfy certain constraints, but writing out the types helps us to identify the required isomorphism. It must have the type \(\varphi : (G \to V) \to (G \to (H \to V))\). This suggests that we might try: \begin{equation*} \varphi(f) := g, h \mapsto f(hg) \end{equation*}

Let's now check that \(\varphi\) actually maps from \(\Ind^G_K V\) to \(\Ind^G_H \Ind^H_K V\). Let \(f \in \Ind^G_K V\), so that for all \(k \in K, g \in G\), we have \(f(kg) = kf(g)\). For each \(h, h' \in H, g \in G\), we have: \begin{equation*} \varphi(f)(hg)(h') = f(h'hg) \end{equation*} and \begin{align*} (h(\varphi(f)(g)))(h') &= \varphi(f)(g)(h'h) \\ &= f(h'hg) \end{align*} Therefore, for all \(g \in G, h \in H\), we have \(\varphi(f)(hg) = h(\varphi(f)(g))\), which is one of the requirements to have \(\varphi(f) \in \Ind^G_H \Ind^H_K V\). The other requirement is that each of the maps \(\varphi(f)(g)\) actually lies in \(\Ind^H_K V\). For this, note that for each \(k \in K\), we have \((\varphi(f)(g))(kh) = f(khg) = kf(hg) = k(\varphi(f)(g)(h))\), as required.

Next, we show that \(\varphi\) is a linear map. Let \(F\) denote the base field. If \(\alpha \in F\), then for each \(f \in \Ind^G_K V\), we have \(\varphi(\alpha f) = (g, h \mapsto (\alpha f)(hg)) = (g, h \mapsto \alpha f(hg)) = \alpha(g, h \mapsto f(hg)) = \alpha \varphi(f)\), and for each \(f_1, f_2 \in \Ind^G_K V\), we have \(\varphi(f_1 + f_2) = (g, h \mapsto (f_1 + f_2)(hg)) = (g, h \mapsto f_1(hg) + f_2(hg)) = (g, h \mapsto f_1(hg)) + (g, h \mapsto f_2(hg)) = \varphi(f_1) + \varphi(f_2)\), as required.

Next, we show that \(\varphi\) is injective. Suppose \(\varphi(f) = 0\) for some \(f \in \Ind^G_K V\). Then for all \(g \in G, h \in H\), we have \(\varphi(f)(g)(h) = 0\). This implies that \(f(hg) = 0\) for all \(g, h\). But the set of elements of the form \(hg\) is all of \(G\), so \(f\) vanishes identically.

For \(V\) finite-dimensional, we could immediately conclude that \(\varphi\) is surjective because its domain and codomain have the same dimension over \(F\). There doesn't seem to be anything in the text implying that \(V\) is assumed finite-dimensional, so we will show surjectivity directly. Suppose \(j \in \Ind^G_H \Ind^H_K V\). Then let \(f : G \to V\) be defined by \(f(g) = j(g)(1)\). We then see that \begin{align*} \varphi(f)(g)(h) &= f(hg) \\ &= j(hg)(1) \\ &= h(j(g))(1) \\ &= j(g)(1\cdot h) \\ &= j(g)(h) \end{align*} therefore \(\varphi(f) = j\). Also, for each \(k \in K\), \begin{align*} f(kg) &= j(kg)(1) \\ &= k(j(g))(1) \\ &= j(g)(1\cdot k) \\ &= j(g)(k\cdot 1) \\ &= k(j(g)(1)) \\ &= kf(g) \end{align*} therefore \(f \in \Ind^G_K V\) as required.

Having established that \(\varphi\) is a linear isomorphism, we have one final task, namely to show that \(\varphi\) is an isomorphism of \(G\)-modules. So let \(f \in \Ind^G_K V\) and let \(g, g' \in G, h \in H\). Then \begin{align*} (\varphi(g'f))(g)(h) &= (g'f)(hg) \\ &= f(hgg') \\ &= \varphi(f)(gg')(h) \\ &= (g'\varphi(f))(g)(h) \end{align*} so \(\varphi(g'f) = g'\varphi(f)\) as required.

Exercise 5.8.5 The required isomorphism \(\varphi : \Ind^G_K \mathbb{C}_\chi \to \mathbb{C}[G]e_\chi\) is given by: \begin{equation*} \varphi(f) := \sum_{g \in G} f(g^{-1}) g \end{equation*} Note that: \begin{align*} \varphi(f)e_\chi &= \frac{1}{|K|}\sum_{g \in G} \sum_{k \in K} \chi(k)^{-1} f(g^{-1}) gk \\ &= \frac{1}{|K|} \sum_{g \in G} \sum_{h \in gK} \chi(g^{-1}h)^{-1} f(g^{-1})h \\ &= \frac{1}{|K|} \sum_{h \in G} \sum_{g : h \in gK} \chi(h^{-1}g) f(g^{-1})h \\ &= \frac{1}{|K|} \sum_{h \in G} \left[\sum_{g \in hK} \chi(h^{-1}g) f(g^{-1})\right] h \\ &= \frac{1}{|K|} \sum_{h \in G} \left[\sum_{k \in K} \chi(h^{-1}hk) f((hk)^{-1})\right]h \\ &= \frac{1}{|K|} \sum_{h \in G} \left[\sum_{k \in K} \chi(k) f(k^{-1}h^{-1})\right] h \\ &= \frac{1}{|K|} \sum_{h \in G} \left[\sum_{k \in K} \chi(k) k^{-1} f(h^{-1})\right] h \\ &= \frac{1}{|K|} \sum_{h \in G} \sum_{k \in K} f(h^{-1})h \\ &= \frac{1}{|K|} \sum_{h \in G} |K| f(h^{-1})h \\ &= \sum_{h \in G} f(h^{-1})h \\ &= \varphi(f) \end{align*} therefore \(\varphi(f) \in \mathbb{C}[G]e_\chi\), as required.

It is obvious that \(\varphi\) is a linear map. If \(\varphi(f) = 0\), then it is easy to see that \(f = 0\), so \(\varphi\) is injective.

Notice that for all \(g \in G, k \in K\), we have \begin{align*} gk e_\chi &= \frac{1}{|K|} gk \sum_{k' \in K} \chi(k')^{-1} k' \\ &= \frac{1}{|K|} g \sum_{k' \in K} \chi(k')^{-1} kk' \\ &= \frac{\chi(k)}{|K|} g \sum_{k' \in K} \chi(k)^{-1} \chi(k')^{-1} kk' \\ &= \frac{\chi(k)}{|K|} g \sum_{k' \in K} \chi(kk')^{-1} kk' \\ &= \frac{\chi(k)}{|K|} g \sum_{k'' \in K} \chi(k'')^{-1} k'' \\ &= \chi(k) g e_\chi \end{align*} therefore \(\mathbb{C}[G]e_\chi\) is \(|G|/|K|\)-dimensional with a basis given by \(g_1 e_\chi, \ldots, g_{|G|/|K|} e_\chi\) where the \(g_i\)'s belong to different left \(K\)-cosets of \(G\). Since \(\mathbb{C}[G]e_\chi\) has the same dimension (as a vector space) as \(\Ind^G_K \mathbb{C}_\chi\), the map \(\varphi\) is surjective.

Finally, we must show that \(\varphi: \Ind^G_K \mathbb{C}_\chi \to \mathbb{C}[G]e_\chi\) is an isomorphism of \(G\)-modules. Let \(f \in \Ind^G_K \mathbb{C}_\chi, g \in G\). Then \begin{align*} \varphi(gf) &= \sum_{h \in G} (gf)(h^{-1})h \\ &= \sum_{h \in G} f(h^{-1}g)h \\ &= g\sum_{h \in G} f(h^{-1}g)g^{-1}h \\ &= g\sum_{h' \in G} f(h'^{-1})h' \\ &= g\varphi(f) \end{align*} as required.