Brian Bi
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$\DeclareMathOperator{\Ind}{Ind} \DeclareMathOperator{\Res}{Res}$

Problem 5.10.2(a) We will use the universal property for the tensor product of bimodules, which is as follows:

Suppose $$M$$ and $$N$$ are respectively $$(A, B)$$- and $$(B, C)$$-bimodules. If $$P$$ is an $$(A, C)$$-bimodule, then a map $$f : M \times N \to P$$ is said to be balanced if all of the following hold:
• $$f(m_1 + m_2, n) = f(m_1, n) + f(m_2, n)$$ for all $$m_1, m_2 \in M$$ and $$n \in N$$;
• $$f(m, n_1 + n_2) = f(m, n_1) + f(m, n_2)$$ for all $$m \in M$$ and $$n_1, n_2 \in N$$;
• $$f(am, n) = af(m, n)$$ for all $$m \in M, n \in N, a \in A$$;
• $$f(mb, n) = f(m, bn)$$ for all $$m \in M, n \in N, b \in B$$; and
• $$f(m, nc) = f(m, n)c$$ for all $$m \in M, n \in N, c \in C$$
The tensor product is defined as an $$(A, C)$$-bimodule $$M \otimes_B N$$ together with a balanced map $$\otimes : M \times N \to M \otimes_B N$$ such that for every $$(A, C)$$-bimodule $$P$$ and every balanced map $$f : M \times N \to P$$, there is a unique $$(A,C)$$-bimodule homomorphism $$\tilde{f} : M \otimes_B N \to P$$ such that $$f = \tilde{f} \circ \otimes$$.

The above universal property is easily proven to hold for the tensor product as defined in Problem 2.11.6, and a simple argument similar to the one for modules establishes that it uniquely characterizes the tensor product up to isomorphism.

Now consider the $$(k[H], k)$$-bimodule (that is, left $$k[H]$$-module) $$\Res^G_H V$$ together with the binary operation $$\star : k[G] \times V \to \Res^G_H V$$ given by $$a \star v = av$$. It's easy to see that $$\star$$ is balanced. Let $$f : k[G] \times V \to P$$ be a balanced map and $$P$$ a left $$k[H]$$-module. We claim that there is precisely one left $$k[H]$$-module homomorphism $$\tilde{f} : \Res^G_H V \to P$$ such that $$f = \tilde{f} \circ \star$$. In particular, we must have $$f(1, v) = \tilde{f}(1 \star v)$$ for all $$v \in V$$, that is, $$\tilde{f}(v) = f(1, v)$$. With this definition of $$\tilde{f}$$, we obtain $$(\tilde{f} \circ \star)(a, v) = \tilde{f}(av) = f(1, av) = f(a, v)$$, as required, and it is easy to check that $$\tilde{f}$$ is an isomorphism of left $$k[H]$$-modules. Since $$\Res^G_H V$$ satisfies the universal property, it is isomorphic to $$k[G] \otimes_{k[G]} V$$ as a left $$k[H]$$-module.

Problem 5.10.2(b) It is fairly clear why Remark 5.8.2 is true, so we won't bother to prove it here. By Remark 5.8.2, and functoriality of $$\Hom(V, -)$$ (i.e., it preserves isomorphisms), we have $\Hom_G(V, \Ind^G_H W) \cong \Hom_{k[G]}(V, \Hom_{k[H]}(k[G], W))$ as $$k$$-vector spaces. By the tensor-hom adjunction (Problem 2.11.6(b)), using the fact that $$V$$ is a $$(k[G], k)$$-bimodule, $$k[G]$$ is a $$(k[H], k[G])$$-bimodule, and $$W$$ is a $$(k[H], k)$$-bimodule, we have $\Hom_{k[G]}(V, \Hom_{k[H]}(k[G], W)) \cong \Hom_{k[H]}(k[G] \otimes_{k[G]} V, W)$ as $$(k, k)$$-bimodules, that is, as $$k$$-vector spaces. By part (a) of this problem and functoriality of $$\Hom(-, W)$$, we have $\Hom_{k[H]}(k[G] \otimes_{k[G]} V, W) \cong \Hom_H(\Res^G_H V, W)$ as $$k$$-vector spaces. Putting it all together, we conclude $$\Hom_G(V, \Ind^G_H W) \cong \Hom_H(\Res^G_H V, W)$$ as $$k$$-vector spaces.

Problem 5.10.2(c) We'll show that $$\Hom_{k[G]}(k[G], V) \cong k[G] \otimes_{k[G]} V$$ as left $$k[G]$$-modules using the universal property of the tensor product. The desired result will immediately follow.

Define $$\star : k[G] \times V \to \Hom_{k[G]}(k[G], V)$$ by $$(a \star v)(b) = bav$$. Then $$\star$$ is a balanced product of the $$(k[G], k[G])$$-bimodule $$k[G]$$ and the left $$k[G]$$-module $$V$$.

Let $$f : k[G] \times V \to P$$ be a balanced map where $$P$$ is a left $$k[G]$$-module. We claim that there is exactly one left $$k[G]$$-module homomorphism $$\tilde{f} : \Hom_{k[G]}(k[G], V) \to P$$ such that $$f = \tilde{f} \circ \star$$. In particular, every $$\varphi : \Hom_{k[G]}(k[G], V)$$ takes the form $$\varphi(a) = \varphi(a \cdot 1) = a\varphi(1)$$, and there is precisely one such map for each possible value of $$\varphi(1)$$; call this map $$\varphi_v$$, so that $$\varphi_v(a) = av$$. Observe that for each $$v \in V$$, we have $$f(1, v) = \tilde{f}(1 \star v) = \tilde{f}(\varphi_v)$$. Thus, we must have $$\tilde{f}(\varphi) = f(1, \varphi(1))$$ for each $$\varphi \in \Hom_{k[G]}(k[G], V)$$.

Indeed, for each $$a \in k[G]$$, $$\tilde{f}(a \star v) = \tilde{f}(b \mapsto bav) = \tilde{f}(\varphi_{av}) = f(1, av) = f(a, v)$$, so $$\tilde{f} \circ \star = f$$ as required.

Problem 5.10.2(d) We'll show that $$\Ind^G_H W$$ satisfies the universal property of the tensor product $$k[G] \otimes_{k[H]} W$$ (where $$k[G]$$ is a $$(k[G], k[H])$$-bimodule and $$W$$ is a left $$k[H]$$-module).

We must define a balanced map $$\star : k[G] \times W \to \Ind^G_H W$$. Intuitively, we want something similar to $$a \star w = (g \mapsto gaw)$$, which doesn't make sense since we haven't defined an action of $$k[G]$$ on $$W$$ yet. However, we can construct one by treating $$k[G]$$ as a direct sum of $$|G|/|H|$$ copies of $$k[H]$$ and having only one copy (i.e., the one naturally embedded in $$k[G]$$) act on $$W$$. That is, \begin{equation} \label{eqn:star1} (a \star w)(g) = \pi(ga)w \end{equation} where $$\pi : k[G] \to k[H]$$ is the $$k$$-linear map such that $$\pi(g)$$ is $$g$$ if $$g \in H$$, and 0 otherwise. Observe that for all $$h \in H, a \in k[G]$$, we have $$\pi(ha) = h\pi(a)$$.

Let $$g \in G, h \in H$$; then $$(a \star w)(hg) = \pi(hga)w = h\pi(ga)w = h(a \star w)(g)$$. This shows that $$a \star w \in \Ind^G_H W$$.

Linearity of $$\pi$$ implies that $$\star$$ satisfies the first two properties of a balanced product, as stated in 5.10.2(a). If $$a_1, a_2 \in k[G], w \in W, g \in G$$, then $$((a_1 a_2)\star w)(g) = \pi(ga_1 a_2)w = (a_2 \star w)(ga_1) = (a_1 (a_2 \star w))(g)$$. If $$a \in k[G], b \in k[H], w \in W, g \in G$$, then $$((ab)\star w)(g) = \pi(gab)w = (\pi(ga)b)w = \pi(ga)(bw) = (a \star (bw))(g)$$. Finally, the last property of balanced products in this case is just $$k$$-linearity, which obviously holds. Therefore $$\star$$ is a balanced product of a $$(k[G], k[H])$$-bimodule and a left $$k[H]$$-module.

In general, $$\im \star$$ is not the entire set $$\Ind^G_H W$$. However, observe that if $$g \in G, w \in W$$, then $$(g^{-1} \star w)(g') = \pi(g'g^{-1})w$$, therefore \begin{equation*} (g^{-1} \star w)(g') = \begin{cases} g'g^{-1} w & \text{if $g, g'$ belong to the same right $H$-coset} \\ 0 & \text{otherwise} \end{cases} \end{equation*} and in particular $$(g^{-1} \star w)(g) = w$$. Thus, if $$S$$ is a right $$H$$-transversal of $$G$$, i.e., $$S = \{g_1, g_2, \ldots, g_{|G|/|H|}\}$$ with one element chosen from each $$H$$-coset of $$G$$, then for each $$v \in \Ind^G_H W$$, we can write \begin{equation*} v = \sum_{g \in S} g^{-1} \star v(g) \end{equation*} that is, elements of the form $$g^{-1} \star w$$, where $$g \in S$$ and $$w \in W$$, span $$\Ind^G_H W$$.

Now let $$f : k[G] \times W \to P$$ be a balanced product of a $$(k[G], k[H])$$-bimodule and a left $$k[H]$$-module. We claim that there is exactly one left $$k[G]$$-module homomorphism $$\tilde{f} : \Ind^G_H W \to P$$ such that $$f = \tilde{f} \circ \star$$. In particular, for all $$g \in G, w \in W$$, we have $$f(g, w) = \tilde{f}(g \star w)$$, which fixes the value of $$\tilde{f}$$ at each element of $$\Ind^G_H W$$ of the form $$g \star w$$; we have already established that such elements span $$\Ind^G_H W$$, so there can be at most one possible $$\tilde{f}$$. Indeed, fix some right transversal $$S$$ as above and set \begin{equation} \label{eqn:tildef1} \tilde{f}(v) = \sum_{g \in S} f(g^{-1}, v(g)) \end{equation} Since $$f$$ is a balanced product, \eqref{eqn:tildef1} doesn't depend on the choice of transversal. Observe that $$\tilde{f}$$ so defined is $$k$$-linear and for each $$g' \in G$$, we have \begin{align*} \tilde{f}(g'v) &= \sum_{g \in S} f(g^{-1}, (g'v)(g)) \\ &= \sum_{g \in S} f(g^{-1}, v(gg')) \\ &= \sum_{g \in S} f(g'(gg')^{-1}, v(gg')) \\ &= g' \sum_{g \in S} f((gg')^{-1}, v(gg')) \\ &= g' \tilde{f}(v) \end{align*} where we have used the fact that $$\{gg' \mid g \in S\}$$ is also a right $$H$$-transversal of $$G$$. It is thereby established that $$\tilde{f}$$ is a left $$k[G]$$-module homomorphism. We also have that for each $$g \in S, w \in W$$, \begin{align*} \tilde{f}(g^{-1} \star w) &= \sum_{g' \in S} f(g'^{-1}, (g^{-1} \star w)(g')) \\ &= f(g^{-1}, w) \end{align*} where we have used the fact that for $$g, g' \in S$$, $$(g^{-1} \star w)(g')$$ is nonzero only when $$g = g'$$. We see that $$\tilde{f} \circ \star$$ agrees with $$f$$ on $$S^{-1} \times W$$. As we previously discussed, elements of the form $$g^{-1} \star w$$ span $$\Ind^G_H W$$, so by linearity, $$f = \tilde{f} \circ \star$$ as required.

(Note: We have used the fact that \eqref{eqn:tildef1} is independent of the choice of transversal $$S$$ in the proof of the universal property. However, the definition of $$\star$$, which establishes the isomorphism between $$\Ind^G_H W$$ and $$k[G] \otimes_{k[H]} W$$, was given as \eqref{eqn:star1} and doesn't make any reference to $$S$$. Therefore, said isomorphism is natural in the informal sense.)

Problem 5.10.2(e) Using (c) and (d) together with the tensor-hom adjunction, we find: \begin{equation*} \Hom_H(W, \Res^G_H V) \cong \Hom_H(W, \Hom_{k[G]}(k[G], V)) \cong \Hom_{k[G]}(k[G] \otimes_{k[H]} W, V) \cong \Hom_G(\Ind^G_H W, V) \end{equation*} This isomorphism of $$k$$-vector spaces is natural because each of the three isomorphisms in the above chain is natural.

Problem 5.10.2(f) We follow the Hint. By part (d), $$(\Ind^G_H V)^* \cong (k[G] \otimes_{k[H]} V)^\star$$. By Remark 5.8.2, $$\Ind^G_H(V^*) \cong \Hom_H(k[G], V^*)$$. Define a $$k$$-bilinear functional $$\langle \cdot, \cdot \rangle : \Hom_H(k[G], V^*) \times (k[G] \otimes_{k[H]} V) \to k$$ by setting \begin{equation} \label{eqn:form1} \langle f, x \otimes_{k[H]} v \rangle = f(Sx)(v) \end{equation} for all $$f \in \Hom_H(k[G], V^*), x \in k[G], v \in V$$ and extending by linearity. (Note that the text incorrectly stated the intended domain of the bilinear form, but it's clear enough from the hint what the correct domain is.) It's not obvious that this is well-defined, so we now show that this is the case. Fix $$f \in \Hom_H(k[G], V^*)$$; then define $$B_f : k[G] \times V \to k$$ by $$B_f(x, v) = f(Sx)(v)$$. It's obvious that $$B_f$$ is $$k$$-bilinear; furthermore, for each $$h \in H$$, we have $$B_f(xh, v) = f(S(xh))(v) = f(h^{-1}Sx)(v) = (h^{-1}f(Sx))(v) = f(Sx)(hv) = B_f(x, hv)$$. Thus, $$B_f$$ is a balanced map from a right $$k[H]$$-module and a left $$k[H]$$-module to $$k$$. By the universal property, there is one unique $$k$$-linear map $$\tilde{B}_f : k[G] \otimes_{k[H]} V \to k$$ such that $$B_f = \tilde{B}_f \circ \otimes_{k[H]}$$. Finally, define $$\langle f, \cdot \rangle = \tilde{B}_f$$. It can be easily seen that this agrees with \eqref{eqn:form1} and the mapping $$f \mapsto \tilde{B}_f$$ is linear.

Suppose $$f \in \Hom_H(k[G], V^*)$$ satisfies $$\tilde{B}_f = 0$$. Then $$f(Sx)(v) = 0$$ for all $$x \in k[G], v \in V$$. This implies $$f(Sx) = 0$$ for all $$x \in k[G]$$, so $$f(g^{-1}) = 0$$ for all $$g \in G$$. By linearity, $$f$$ vanishes identically. This establishes that the mapping $$\varphi : \Hom_H(k[G], V^*) \to (k[G] \otimes_{k[H]} V)^*$$ given by $$\varphi(f) = \langle f, \cdot \rangle$$ is an injection from $$\Hom_H(k[G], V^*)$$ to $$(k[G] \otimes_{k[H]} V)^*$$; that is, from $$\Ind^G_H(V^*)$$ to $$(\Ind^G_H V)^*$$. Since $$\dim \Ind^G_H(V^*) = |G| \dim(V^*)/|H| = |G| \dim(V)/|H| = \dim(\Ind^G_H V) = \dim((\Ind^G_H V)^*)$$, $$\varphi$$ is in fact an isomorphism.

Furthermore, we always have $$\langle gf, g(x \otimes_{k[H]}v) \rangle = \langle gf, gx \otimes_{k[H]}v \rangle = (gf)(S(gx))(v) = f((Sx)g^{-1}g)(v) = f(Sx)(v) = \langle f, x \otimes_{k[H]} v\rangle$$, for each $$g \in G$$. By linearity in the second argument, we can conclude that $$\langle \cdot, \cdot \rangle$$ is $$G$$-invariant. Thus, $$\varphi(gf) = (t \mapsto \langle gf, t\rangle) = (t \mapsto \langle f, g^{-1}t\rangle) = g(t \mapsto \langle f, t\rangle) = g\varphi(f)$$, so, in fact, $$\Ind^G_H(V^*) \cong (\Ind^G_H V)^*$$ as left $$k[G]$$-modules.