Let $n = \dim U$ and $m = \dim W$. Let $V \subseteq U \otimes W$ be a nonzero subrepresentation. Since $V$ is nonzero, it must contain some nonzero vector $u \otimes w$. Since $U$ is an irrep of $G$, the smallest $G$-invariant subspace containing $\C u$ is all of $U$. So we can find $g_1, \ldots, g_n \in G$ such that $g_1u, \ldots, g_nu$ is a basis of $U$. Similarly, we can find $h_1, \ldots, h_m \in H$ such that $h_1w, \ldots, h_mw$ is a basis of $w$. Since $V$ is a subrepresentation, we know that $(g,h) (u \otimes w) \in V$ for any $(g,h) \in G \times H$. Thus, $(g_iu) \otimes (h_jw) \in V$ for all $i,j$. This means that $V$ contains a basis for $U \otimes W$, so $V$ is all of $U \otimes W$. Thus, the only nonzero $(G \times H)$-subrepresentation of $U \otimes W$ is the entire space $U \otimes W$. So $U \otimes W$ is irreducible.

Let $T \in \End(V)$. Then $M_V(T)(g) = \tr(\phi(g) \circ T)$. Let $(h_1, h_2) \in G \times G$. \[\begin{aligned} M_V((h_1,h_2)\cdot T)g &= \tr(\phi(g) \circ ((h_1,h_2) \cdot T))\\ &= \tr(\phi(g) \circ \phi(h_2) \circ T \circ \phi(h_1)^{-1})\\ &= \tr(\phi(h_1^{-1}gh_2) \circ T)\\ &= M_V(T)(h_1^{-1}gh_2)\\ &= ((h_1,h_2)\cdot (M_V(T)))(g) \end{aligned}\] So $M_V$ is $(G \times G)$-linear.

Clearly if $V$ is irreducible, then $V^*$ is also an irreducible $G$-representation. By our lemma, this means that $V \otimes V^* \cong \End(V)$ is an irreducible $(G \times G)$-representation. Since $M_V$ is $(G \times G)$-linear, $\ker M_V$ must be a subrepresentation. Thus, $M_V$ is either zero or injective. $M_V$ is clearly nonzero since it sends the identity matrix to $\chi_V$, which is nonzero.

1. $M_{V^*}(T^*)(g) = \tr(\phi(g^{-1})^T \circ T^T) = \tr(\phi(g^{-1}) \circ T) = M_V(T)^*(g)$
2. $M_{\overline V}(\overline T)(g) = \tr(\overline{\phi(g)} \circ \overline T) = \overline{M_V(T)(g)}$
3. The computation here looks a bit longer, but it's pretty straightforward \[\begin{aligned} M_{V \oplus W}(T \oplus S)(g) &= \tr((\phi(g) \oplus \psi(g)) \circ (T \oplus S))\\ & = \tr((\phi(g) \circ T) \oplus (\psi(g) \circ S))\\ &= \tr(\phi(g) \circ T) + \tr (\psi(g) \circ S)\\ &= M_V(T)(g) + M_W(S)(g) \end{aligned}\]
4. This one also looks a bit long, but it's also straightforward \[\begin{aligned} M_{V \otimes W}(T \otimes S)(g) &= \tr((\phi(g) \otimes \psi(g)) \circ (T \otimes S))\\ & = \tr((\phi(g) \circ T) \otimes (\psi(g) \circ S))\\ &= \tr(\phi(g) \circ T) \cdot \tr (\psi(g) \circ S)\\ &= M_V(T)(g) \cdot M_W(S)(g) \end{aligned}\]
5. This follows from the first and fourth identities, using the fact that $\Hom(V,W) \cong W \otimes V^*$.
6. This computation is the trickiest, but it's still not too bad. It's pretty much all tricks we've used before \[\begin{aligned} M_{V^G}(Av_G \circ T)(g) &= \tr_{V^G}(\phi(g) \circ Av_G \circ T)\\ &= \tr_{V^G}(T \circ \phi(g) \circ Av_G) \end{aligned}\] Since $T \circ \phi(g) \circ Av_G$ acts as $T \circ \phi(g) \circ Av_G$ on $V^G$ and acts as $0$ on the orthogonal complement of $V^G$, we can trace over $V$ instead of $V^G$. So we see that \[\begin{aligned} M_{V^G}(Av_G \circ T)(g) &= \tr_V(T \circ \phi(g) \circ Av_G)\\ &= \tr_V\left( T \phi(g) \int_G \phi(h)dh\right)\\ &= \int_G \tr_V(T \circ \phi(gh))dh \end{aligned}\] Since our measure is left-invariant, this is just equal to \[\begin{aligned} M_{V^G}(Av_G \circ T)(g) &= \int_G \tr_V(T \circ \phi(g))dg\\ &= av(\tr_V(T \circ \phi(g)))\\ &= av(M_V(T)) \end{aligned}\]

Just like the proof for characters, this proof is pretty simple given the constructions we defined above. \[\begin{aligned} \inrp{M_F(S)}{M_E(T)} &= av(M_F(S) \cdot \overline{M_E(T)})\\ &= av(M_F(S) \cdot M_{\overline E}(\overline T))\\ &= av(M_{F \otimes \overline E})(S \otimes \overline T)\\ &= M_{(F \otimes \overline E)^G}(Av_G \circ (S \otimes \overline T)) \end{aligned}\] Recall that $\overline E$ is isomorphic to $E^*$. Thus, $F \otimes \overline E \cong F \otimes E^* \cong \Hom(E,F)$. This isomorphism shows us that $(F \otimes \overline E)^G = \Hom(E,F)^G = \Hom_G(E,F)$, which is $0$ by Schur's lemma. Thus, the inner product must be $0$.

1. Recall that the $(G \times G)$ action on $\End(E)$ is given by $(g,h) \cdot T = \phi(g) \circ T \circ \phi(h)^{-1}$. Thus, \[\begin{aligned} \inrp{(g,h)T}{(g,h)S} &= \tr_E(\phi(g) \circ T \circ \phi(h)^{-1} \circ (\phi(g) \circ S \circ \phi(h)^{-1})^\dagger)\\ &= \tr_E(\phi(g) \circ T \circ \phi(h)^{-1} \circ \phi(h^{-1})^\dagger \circ S^\dagger \circ \phi(g)^\dagger)\\ &= \tr_E(\phi(g) \circ T \circ \phi(h)^{-1} \circ \phi(h) \circ S^\dagger \circ \phi(g^{-1}))\\ &= \tr_E(\phi(g) \circ T \circ S^\dagger \circ \phi(g^{-1}))\\ &= \tr_E(T \circ S^\dagger)\\ &= \inrp T S \end{aligned}\]
2. We saw earlier that $\End(E)$ is an irreducible $(G \times G)$-representation. Since $M_E$ is $(G\times G)$-linear, this means it must be injective.

We will show that $4 \implies 3 \implies 2 \implies 4$. The proof that $4 \implies 3 \implies 1 \implies 4$ is essentially the same.
• $4 \implies 3:$ Since $M_V$ is a $(G \times G)$-linear map and has a finite-dimensional domain, every function in its image must be $(G \times G)$-finite
• $3 \implies 2:$ The right-$G$-action is the action of $(1 \times G) \subseteq (G \times G)$, so $(G \times G)$-finiteness implies right-$G$-finiteness.
• $2 \implies 4:$ Let $f$ be right-$G$-finite. Then $f$ is contained in a finite-dimensional representation $V \subsetneq C(G)$ of $G$ (with the right-$G$-action). Let $\alpha \in V^*$ be the linear functional $\alpha(h) := h(e)$. Now, we see that \[M_V(f \otimes \alpha)(g) = \tr_V(\phi(g) \circ f \otimes \alpha) = \alpha(g \cdot f) = (g \cdot f)(e) = f(g)\] Thus, $f$ is in the image of the matrix coefficient map $M_V$

$1 \implies 2$: Since $C(G)$ is dense in $L^2(G)$, if $C(G)^{fin}$ is dense in $C(G)$, it must also be dense in $L^2(G)$.

$2 \implies 3$: Since $C(G)^{fin} \subseteq L^2(G)^{fin}$, then $C(G)^{fin}$ being dense implies that $L^2(G)^{fin}$ is dense as well.

$3 \implies 4$: Consider $e \neq g \in G$. We will start by finding a function in $L^2(G)^{fin}$ on which $g$ acts nontrivially. First, note that there must be some function $f \in C(G)$ such that $g \cdot f \neq f$. Now, suppose for contradictin that $g$ acts trivially on all of $L^2(G)^{fin}$. Since the action is continuous and $L^2(G)^{fin}$ is dense, $g$ must act trivially on all of $L^2(G)$, which is a contradiction. Thus, there is some $\tilde f \in L^2(G)^{fin}$ upon which $g$ acts nontrivially. Then $G \tilde f$ is contained in a finite-dimensional subspace of $C(G)$ by definition of $L^2(G)^{fin}$. So $\tilde f$ is a vector in a finite-dimensional representation of $G$ upon which $g$ acts nontrivially.

$4 \implies 5$: Let $g \neq h \in G$. By (4), there is some irreducible representation $(V, \phi)$ such that $\phi(gh^{-1}) \neq \Id$. Consider $M_V(\phi(h^{-1}))$. \[\begin{aligned} M_V(\phi(h^{-1}))(g) &= \tr_V(\phi(g) \circ \phi(h^{-1}))\\ &= \chi_V(gh^{-1})\\ M_V(\phi(h^{-1}))(h) &= \tr_V(\phi(h) \circ \phi(h^{-1}))\\ &= \chi_V(\Id) \end{aligned}\] Since we can pick a metric on $V$ with respect to which $\phi(gh^{-1})$ is unitary, it must have complex eigenvalues, and those eigenvalues must be roots of unity. Thus, the only way for $\tr_V(\phi(gh^{-1}))$ to equal $\tr_V(\Id)$ is if all of the eigenvalues are 1, which means that $\phi(gh^{-1}) = \Id$. But this is impossible. Thus, $\chi_V(gh^{-1}) \neq \chi_V(\Id)$. So $M_V(\phi(h^{-1}))$ separates $g$ and $h$. And by definition, $M_V(\phi(h^{-1}))$ is in $C(G)^{fin}$.

$5 \implies 1$: We observed earlier that $C(G)^{fin}$ is a subalgebra of $C(G)$. Clearly $C(G)^{fin}$ contains a nonzero constant function. Thus, the Stone Weierstrass theorem tells us that $C(G)^{fin}$ is dense in $C(G)$ if and only if it separates points.

The operator norm of $T_K$ is \[\begin{aligned} \|T_K\|_{op} &:= \sup_{f \in L^2(X), \|f\|_{L^2(X)}=1}\|T_K(f)\|_{L^2(X)}\\ &= \sup_{\|f\|_{L^2(X)}=1} \left\|x \mapsto \int_X K(x,y) f(y)\;dy\right\|_{L^2(X)}\\ &= \sup_{\|f\|_{L^2(X)}=1} \left(\int_{X}\left(\int_X K(x,y) f(y) \;dy\right)^2 \;dx \right)^{1/2} \end{aligned}\] By Holder's inequality, \[\int_X K(x,y) f(y)\;dy \leq \left(\int_X K(x,y)^2\;dy\right)^{1/2} \|f\|_{L^2(X)}\] Therefore, \[\|T_K\|_{op} \leq \left(\int_X \int_X K(x,y)^2\;dydx\right)^2 = \|K\|_{L^2(X \times X)}\] Note that since $T_K$ is a bounded linear operator, it is continuous.

Next, we will show that $T_K$ is compact. Recall that the limit of a sequence of finite-rank operators between Hilbert spaces is a compact operator. So it is sufficient to show that $T_K$ is the limit of a sequence of finite-rank operators.

Consider the set of functions \[\left\{x,y \mapsto \sum_i f_i(x) g_i(y)\;|\; f_i, g_i \in C(X)\right\} \subseteq C(X \times Y)\] This is clearly a subalgebra of $C(X \times Y)$ containing a nonzero constant function, so the Stone-Weierstrass theorem tells us that these functions are dense in $C(X \times Y)$. Since $C(X \times Y)$ is dense in $L^2(X \times Y)$, these functions are also dense in $L^2(X \times Y)$. Thus, any $T_K$ is a limit of operators $T_{K_i}$ for $K_i$ in this subalgebra. Now, we just have to show that $T_{K_i}$ is finite-rank. Since a linear combination of operators of finite rank must also have finite rank, it is sufficient to show that $T_{f_1f_2}$ has finite rank (in fact, it has rank 1). \[T_{f_1f_2}(f)(x) = \int_X f_1(x)f_2(y)f(y)\;dy \propto f_1(x)\]

Finally, we note that if $K(x,y) = \overline {K(y,x)}$, then $T_K$ is self-adjoint. \[\begin{aligned} \inrp{T_K(f)}{g} &= \int_X T_Kf(x) \overline{g(x)}\;dx\\ &= \int_X \int_X K(x,y) f(y)\;dy \;\overline {g(x)}\;dx\\ &= \int_X f(y) \int_X K(x,y) \overline {g(x)}\;dx\;dy\\ &= \int_X f(y) \overline{\int_X K(y,x) g(x)\;dx}\;dy\\ &= \int_X f(y) \overline {T_Kg(y)}\;dy\\ &= \inrp f {T_K(g)} \end{aligned}\]

Constructing such a function $u_U$ is fairly straightforward. Pick any nonzero continuous real-valued function $w \in C(G)$ with support inside $U$. Consider $\tilde w =(w + w^*)^2$. This is clearly continuous, nonnegative, and $\tilde w = \tilde w^*$. Since it is continuous, $\int_G \tilde w\;d\mu$ is finite. So we can set \[u_U = \frac 1 {\int_G \tilde w\;d\mu} \tilde w\]

Now, we'll show that $u_U$'s convolution with $f$ satisfies the desired bound. \[ f*u_U(x) - f(x) = \int_G u_U(y^{-1}x) f(x)\;dy - f(x) \] Since $u_U = u_U^*$, we can take the inverse of $u_U$'s argument. \[ f*u_U(x) - f(x) = \int_G u_U(x^{-1}y) f(x)\;dy - f(x) \] Since our measure is left-invariant, we can do a change of variables $t = x^{-1}y$. \[ f*u_U(x) - f(x) = \int_G u_U(t) f(xt)\;dt - f(x) \] Since $u_U$ integrates, to 1, $f(x) = \int_G u_U(t) f(x)\;dt$. Thus, \[\begin{aligned} f*u_U(x) - f(x) &= \int_G u_U(t) f(xt) - u_U(t)f(x)\;dt\\ &= \int_G u_U(t)\left[f(xt) - f(x)\right]\;dt \end{aligned}\] Since $u_U$ is zero outside of $U$, we can restrict this integral to $U$ without changing its value. But then $|f(xt) - f(x)| \leq \epsilon$. So \[\begin{aligned} |f*u_U(x) - f(x)| &\leq \int_U u_U(t) \Big|f(xt) - f(x)\Big|\;dt \leq \epsilon \end{aligned}\] Therefore, $\|f*u_U - f\|_\infty \leq \epsilon$.

Earlier, we proved that these statements are all equivalent. So we only have to prove one of them. We will prove 3. Let $f \in L^2(G)$. We want to show that we can approximate $f$ by elements of $L^2(G)^{fin}$. Since $CG)$ is dense in $L^2(G)$, it is enough to show that we can approximate continuous functions. So let $f$ be continuous.

We saw that we can approximate $f$ with convolutions $f * u$. Since $u$ is real-valued and $u = u^*$, we also have $\overline u = u^*$. So convolution with $u$ is a compact, self-adjoint operator. Thus, $f*u$ is in the image of a compact, self-adjoint operator, so we can approximate it by sums of elements in the nonzero eigenspaces of $\cdot *u$. Since convolution is left-$G$-linear, the eigenspaces are $G$-finite. Thus, we have shown that we can approximate any $f \in L^2(G)$ by $G$-finite $L^2(G)$ functions, so $L^2(G)^{fin}$ is dense in $L^2(G)$.

Our orthogonality results show us that this map is injective. Furthermore, we note that the image of this map is precisely $C(G)^{fin}$. Suppose $f \in C(G)^{fin}$. Then $f \in \im M_V$ for some finite-dimensional representation $V$. Since every finite-dimensional representation can be written as a direct sum of irreducible representations, $\im M_V$ is contained in $\bigoplus_i \im M_{E_i}$. Therefore, the image of $\bigoplus_i\End(E_i)$ in $L^2(G)$ is $C(G)^{fin}$, which is dense. This implies the Peter-Weyl decomposition.

First, we will show existence. Let $n$ denote the dimension of $G$. Recall that as long as $G$ is oriented, an $n$-form on $G$ induces a measure. Furthermore, we recall that if we can find a nonvanishing $n$-form on $G$, then $G$ must be orientable. So it is sufficient to find a left-invariant nonvanishing $n$-form on $G$. Let $\Lambda^nT_eG$ denote the space of $k$-covectors on the tangent space to the identity of $G$. Pick any nonzero $\omega_e \in \Lambda^nT_eG$. Now, we can extend $\omega_e$ to a differential form on $G$. Let $L_g$ denote the automorphism of $g$ given by left-multiplication by $g$. This is continuous. $L_{g^{-1}}$ sends $g$ to $e$, so we can pull $\omega_e$ back along this map to define $\omega_g = (L_{g^{-1}})^* \in \Lambda^n T_gG$. This defines a differential $n$-form $\omega \in \Omega^n(G)$ on all of $G$. $\omega$ is left-invariant by construction. \[((L_h)^* \omega)_g = (L_h)^* \omega_{hg} = (L_h)^* (L_{(hg)^{-1}})^* \omega_e = (L_{(hg^{-1})h})^* \omega_e =(L_{g^{-1}})^* \omega_e = \omega_g\] Clearly this differential form is nonvanishing. And by negating $\omega$ if necessary, we see that $\omega$ is positive with respect to $G$'s orientation, so it defines a left-invariant measure on $G$.

First, suppose that $A \in \Hom(V,W)^G$. Then $g \cdot A = A$, so in particular we have $(g \cdot A)v = Av$ for any $v \in V$. Using the formula for $g \cdot A$, we see that $g \cdot A(g^{-1} v) = Av$ for all $g \in G, v \in V$. Multiplying both sides by $g^{-1}$, we find that $A(g^{-1}v) = g^{-1} Av$. Since this is true for all $g \in G$, we conclude that $A$ is $G$-linear. So $A \in \Hom_G(V,W)$.

Conversely, suppose that $A \in \Hom_G(V,W)$. Then $A(gv) = g(Av)$ for all $g \in G, v \in V$. So $g^{-1}A(gv) = Av$. Letting $h = g^{-1}$, we see that $g A (h^{-1}v) = Av$ for all $h \in G, v \in V$. So $A$ is in the subspace of invariants $\Hom(V,W)^G$.

1. Since $A$ is $G$-linear, we know that $\ker A, \im A$ are subrepresentations. Since $V,W$ are irreps, this implies that $\ker A$ is either $0$ or all of $V$, and $\im A$ is either $0$ or all of $W$. Thus, the only way for $A$ to be nonzero is if $\ker A = 0$ and $\im A = W$. This means that if $A$ is nonzero, it must be an isomorphism.
2. Since $A$ is a complex matrix, it has an eigenvalue $\lambda$. Clearly $\lambda \Id$ is a $G$-linear endomorphism of $V$. Thus, $A - \lambda \Id \in \Hom_G(V,W)$. But $A-\lambda \Id$ cannot be an isomorphism. So it must be $0$. Thus, $A = \lambda \Id$.

First, we will construct one such projection. Explicitly, we define \[Av_G(v) := \int_G g \cdot v \;dg\] This operation averages over the group action, which is why we named the projection $Av_G$. To show that $Av_G$ is a projection, we have to show that it restricts to the identity on its image. First, we note that the image of $Av_G$ is simply $V^G$. We see that $\im Av_G$ is contained in $V^G$. Let $v$ be any vector in $V$. Then for any $h \in G$, we have \[h \cdot Av_G(v) = h \cdot \int_G g \cdot v \;dg = \int_G (hg) \cdot v\;dg = \int_G (hg)\cdot v \;d(hg)\] the last equality follows from the left-invariance of our measure. So we see that $h \cdot Av_G(v) = Av_G(v)$, which implies that $\im Av_G \subseteq V^G$.

Furthermore, any vector of $V^G$ is itself fixed by $Av_G$. If $v \in V^G$, then $Av_G(v) = \int_G g \cdot v\;dg = \int_G v\;dg = v$. So in particular, $v \in \im Av_G$. Thus, we see that $\im Av_G = V^G$, and $Av_G$ acts as the identity on its image. So it is a projection.

Now, we will show uniqueness. First, note that $Av_G$ commutes with any other $T \in \End_G(V)$. \[Av_G \circ T(v) = \int_G g \cdot (Tv)\;dg = \int_G T (g\cdot v) \;dg = T \int_G g \cdot v\;dg = T \circ Av_G (v)\] Suppose that $P$ is another projection onto $V^G$. In particular, it is an element of $\End_G(V)$, so it commutes with $Av_G$. Thus, \[P = Av_G \circ P = P \circ Av_G = Av_G\]

Let $\inrp \cdot \cdot$ be any hermitian inner product on $V$. We can view $\inrp \cdot \cdot$ as an element of $\Hom(\overline V \otimes V, \C) \cong \Hom(\overline V, V^*)$. We can think of the $G$-invariant inner products as elements of $\Hom(\overline V, V^*)^G$. So $Av_G \inrp \cdot \cdot$ gives us a $G$-invariant inner product.

Explicitly, this just means that we can define a $G$-invariant inner product $\inrp \cdot \cdot _G$ by the formula \[\inrp v w_G := \int_G \inrp {gv}{gw}\;dg\]

Let $\inrp \cdot \cdot$ be a $G$-invariant inner product on $V$. Let $U = W^\perp$.

We note that $U$ is a subrepresentation of $V$. Let $u \in U$. By definition, $\inrp u w = 0$ for all $w \in W$. Since the inner product is $G$-invariant, $\inrp {gu} {w} = \inrp u {g^{-1}w}$. Since $W$ is a subreresentation, $g^{-1}w \in W$, so $\inrp u {g^{-1}w} = 0$. Thus, $\inrp {gu} w = 0$ for all $w \in W$, so we conclude that $gu \in U$.

Therefore, $V = W \oplus U$.

Just reply Maschke's theorem repeatedly. Since our vector space is finite-dimensional, this process must terminate.

This follows from the above corollary and Shur's lemma.

Let $A:V \to W$ be a ($G$-linear) isomorphism. Then $\psi(g)Av = A \phi(g) v$. So $\phi(g) = A^{-1}\psi(g) A$. By the cyclic property of the trace, $\tr(A^{-1} \psi(g) A) = \tr(\psi(g))$. Thus, \[\chi_V(g) = \tr(\phi(g)) = tr(A^{-1}\psi(g)A) = \tr(\psi(g)) = \chi_W(g)\]

1. Since $g$ acts on $V^*$ by $\phi(g^{-1})^T$, and transposing does not change the trace, we see that $\chi_{V^*}(g) = \chi_V(g^{-1})$.
2. Since scalar multiplication on $\overline V$ is conjugated, we have to take the complex conjugate of the entries in the matrix $\phi(g)$ to get the matrix which acts on $\overline V$. Thus, $\chi_{\overline V} = \overline{\chi_V}$.
3. $\chi_{V \oplus W}(g) = \tr (\phi(g) \oplus \psi(g)) = \tr(\phi(g)) + \tr(\psi(g)) = \chi_V(g) + \chi_W(g)$.
4. $\chi_{V \otimes W}(g) = \tr (\phi(g) \otimes \psi(g)) = \tr(\phi_V(g)) \cdot \tr(\psi_W(g)) = \chi_V(g)\chi_W(g)$.
5. $\chi_{\Hom(V,W)} = \chi_{W \otimes V^*} = \chi_V^* \cdot \chi_W$.
6. This one is more complicated. We need to compute $\chi_{V^G}$. To do so, we use a trick involving the averaging projection.

   Note that the averaging projection $Av_G:V \to V^G$ acts as the identity on $V^G$ and acts as $0$ on the orthogonal complement to $V^G$. Thus, $\phi(g) \circ Av_G$ acts as $\phi(g)$ on $V^G$ and acts as $0$ on the orthogonal complement to $V^G$. So $\tr_{V^G} \phi(g) = \tr_V (\phi(g) \circ Av_G)$. (Here $\tr_{V^G}$ denotes the trace over $V^G$ and $\tr_V$ denotes the trace over $V$)

   Therefore, \[\chi_{V^G}(g) = \tr_{V^G} \phi(g) = \tr_V (\phi(g) \circ Av_G) = \tr_V \left(\phi(g)\int_G \phi(h)dh\right) = \tr_V\int_G \phi(gh)dh\] Since our measure is left-invariant, this is just \[\chi_{V^G}(g) = \tr_V \int_G \phi(g)dg = \int_G \tr_V \phi(g)dg = \int_G \chi_V(g)dg = av(\chi_V)\]

Recall that the quaternions are defined by \[\H = \{a + jb\;|\; a,b \in \C\}\] where $jb = \overline b j$. Then $S^3 = \{a + jb \in \H\;|\; |a|^2 + |b|^2 = 1\}$. We have a natural action of $S^3$ on $\H$ by left-multiplication. This gives us a two-dimensional complex representation of $S^3$. Writing it out explicitly, we see that \[\begin{aligned} (a+jb) : 1 &\mapsto a+jb\\ (a+jb) : j &\mapsto - \overline b + j \overline a\\ \end{aligned}\] Thus, our representation is given by \[ (a+jb) \mapsto \begin{pmatrix} a & -\overline b\\b & \overline a \end{pmatrix} \] This matrix is unitary, and has determinant $|a|^2 + |b|^2 = 1$. So this is clearly a continuous bijection from $S^3$ to $SU(2)$. You can check that this bijection is a group isomorphism.

1. Using our identification of $S^3$ with $SU(2)$, we can think of points on the sphere as special unitary matrices. Unitary matrices are unitarily-diagonalizable. Clearly we can rescale these matrices so that the matrices we diagonalize with are in $SU(2)$. Finally, note that a diagonal matrix in $SU(2)$ must have the form \[\begin{pmatrix} a & 0 \\ 0 & \overline a\end{pmatrix}\] for $a \in \C$, $|a|^2 = 1$. Thus, diagonal matrices in $SU(2)$ correspond to points $e^{i\theta}$ on the sphere.

2. Since the quaternion norm is multiplicative, and elements of $S^3$ have norm 1, we see that $|g e^{i\theta}g^{-1}| = |g||e^{i\theta}||g|^{-1} = 1$. Furthermore, \[\begin{aligned} ge^{i\theta}g^{-1} &= g (\cos \theta + i \sin \theta)g^{-1}\\ &= \cos \theta + \sin \theta \; g i g^{-1} \end{aligned}\] Now, we will consider the map $\pi:g \mapsto g i g^{-1}$. Note that for unit quaternions, $g^{-1} = \overline g$, the conjugate of $g$. So we can also write this map $\pi:g \mapsto g i \overline g$. Note that \[\overline{g i \overline g} = g \overline i \overline g = -g i \overline g\] So $gi\overline g$ is purely imaginary. Furthermore, since $g, i$ and $\overline g$ are all unit quaternions, so is their product. Thus, we can think of $\pi$ as a map $\pi : S^3 \to S^2$, where we view $S^3$ as the unit quaternions and $S^2$ as the unit imaginary quaternions.

   Furthermore, $\pi$ is surjective. If we represent vectors in $\R^3$ as imaginary quaternions, then $v \mapsto g v \overline g$ is a representation of $SU(2)$ on $\R^3$ which acts by rotations. Since we can write all rotations in this form, and the rotation group $SO(3)$ acts transitively on the two-sphere, we see that $\{gi \bar g\}_{g \in SU(2)}$ covers all of $S^2$. So $\pi$ is surjective.

   So since $g e^{i\theta} g^{-1} = \cos \theta + \sin \theta \pi(g)$, we see that $\{g e^{i\theta} g^{-1}\}$ is a sphere with radius $\sin \theta$. Now, we just have to check that the intersection of this sphere with $\C$ is $e^{\pm i \theta}$. Note that $\im \pi$ is the imaginary unit quaterions, and the only imaginary unit quaternions that lie in $\C$ are $\pm i$. Thus, the intersection of $\{g e^{i\theta} g^{-1}\}$ with $\C$ is $\cos \theta \pm i \sin \theta$.

To prove this, we will use characters. For convenience, let us write $\chi_{V_k}$ as $\chi_k$. Recall that the image of $e^{i\theta} \in S^3$ in $SU(2)$ is the matrix \[\begin{pmatrix} e^{i\theta}& 0 \\ 0 & e^{-i\theta}\end{pmatrix}\] Note that the eigenspaces of this operator on $V_k$ are $\{\C z_1^\ell z_2^{k-\ell}\}_\ell$ with eigenvalues $\{e^{(2 \ell - k)i \theta}\}_\ell$. Therefore, \[\begin{aligned} \chi_k(e^{i\theta}) &= \sum_{\ell=0}^k e^{(2 \ell - k) i \theta}\\ &= \frac{e^{(k+1)i\theta} - e^{-(k+1)i\theta}}{e^{i\theta} - e^{-i\theta}}\\ &= \frac{\sin[(k+1)\theta]}{\sin\theta} \end{aligned}\] Note that all of these characters are different. This shows that all of the representations $V_k$ are distinct. Now, we will show that the characters are orthonormal.

Recall that the inner product on characters is given by \[\inrp{\chi_k}{\chi_\ell} = \int_{S^3} \chi_k(g) \overline{\chi_\ell(g)}\;dg\] Since the volume of $S^3$ is $2\pi^2$, we can write $dg = \frac 1 {2\pi^2}d\sigma$ where $d\sigma$ is the standard volume element on $S^3$. So we want to compute \[\inrp{\chi_k}{\chi_\ell} = \frac 1 {2\pi^2} \int_{S^3} \chi_k(g) \overline{\chi_\ell(g)}\;d\sigma\] Recall that characters are constant on conjugacy classes. Since every element of $SU(2)$ is conjugate to exactly two unit complex numbers, we have \[\inrp{\chi_k}{\chi_\ell} = \frac 1 {2\pi^2} \int_0^\pi \chi_k(e^{i\theta}) \overline{\chi_\ell(e^{i\theta})}\vol(\text{orbit})\;d\theta\] Above, we showed that these orbits are spheres with radius $\sin \theta$. Therefore, the volume of an orbit is $4 \pi \sin^2 \theta$. Substituting this and our expressions for the characters, we see that our inner product is \[\inrp{\chi_k}{\chi_\ell} = \frac 2 {\pi} \int_0^\pi \sin[(k+1)\theta)\sin[(\ell+1)\theta]\;d\theta\] Because sines with different frequencies are orthogonal, we conclude that \[\inrp{\chi_k}{\chi_\ell} = \delta_{k\ell}\] So our characters are orthonormal.

Finally, we will show that these are all of the irreducible representations. Suppose that $W$ was another irreducible representation. Then \[0 = \inrp{\chi_W}{\chi_k} = \int_G \chi_W(g) \overline{\chi_k(g)}\;dg\] Using the same computational tricks, we see that \[0 = \frac 2 \pi \int_0^\pi \chi_W(e^{i\theta}) \sin[(k+1)\theta]\sin\theta\;d\theta\] Since sinces form an orthonormal basis for the set of square-integrable functions on the circle, we see that $\chi_W(e^{i\theta}) = 0$, which is impossible. Thus, every irreducible representation must be isomorphic to some $V_k$.

First, we will consider the special case where the boundary maps in $C$ are all zero. This means that $H_i(C) = C_i$. Since the boundary maps are zero, the boundary map in the tensor product complex simplifies to $\partial(c \otimes c') = (-1)^{|c|} c \otimes \partial c'$. Thus, the complex $C \otimes_R$ is simply the direct sum of the complexes $C_i \otimes_R C'$, and each of these complexes is a direct sum of copies of $C'$ because $C_i$ is free. So \[H_n(C_i \otimes_R C') \cong C_i \otimes_R H_{n-i}(C') = H_i(C) \otimes_R H_{n-i}(C')\] Taking a direct sum over $i$ gives us an isomorphism \[H_n(C \otimes_R C') \cong \bigoplus_i H_i(C) \otimes_R H_{n-i}(C')\] and this is precisely what we needed to prove. The $\Tor$ terms in the theorem statement are 0 since in this case $H_i(C) = C_i$ is free.

Now, we will consider the general case. Let $Z_i, B_i \subseteq C_i$ denote the cycles and boundaries in $C_i$ respectively (i.e. the kernel and image of the boundary maps). We can construct chain complexes $Z$ and $B$ with trivial boundary maps, and these yield a short exact sequence of chain complexes $0 \to Z \to C \to B \to 0$. This is composed of the short exact sequences $0 \to Z_i \to C_i \xrightarrow{\partial} B_{i-1} \to 0$. Note that these short exact sequences split since $B_{i-1}$ is free (since it is a submodule of the free module $C_{i-1}$). Because these short exact sequences split, tensoring with $C'$ gives us another short exact sequence, so tensoring our short exact sequence of chain complexes gives another short exact sequence of chain complexes. This gives us a long exact sequence of homology groups

The homology group $H_{n-1}(B \otimes_R C')$ is shifted down a degree from what one might expect because $B_{i-1}$ appears in the short exact sequence above instead of $B_i$. It turns out that the boundary map $H_{n-1}(B \otimes_R C') \to H_{n-1}(Z \otimes_R C')$ is induced by the inclusion $B_i \subseteq Z_i\;\forall i$.

Since $B$ and $Z$ are chain complexes whose boundary maps are all 0, we can apply the computation we did above to turn $H_n(Z \otimes_R C')$ into $\bigoplus_i Z_i \otimes_R H_{n-i}(C')$ and the same for $B$. This gives us a long exact sequence \[\cdots \xrightarrow{i_n} \bigoplus_i Z_i \otimes_R H_{n-i}(C') \to H_n(C \otimes_R C') \to \bigoplus_i B_i \otimes_R H_{n-i-1}(C') \xrightarrow{i_{n-1}} \cdots\] We can split this long exact sequence up into many short exact sequences. In particular, we find short exact sequences \[0 \to \Coker i_n \to H_n(C \otimes_R C') \to \Ker i_{n-1} \to 0\] By definition, $\Coker i_n = \left(\bigoplus_j Z_j \otimes_R H_{n-j}(C')\right)/\Im i_n$. We can view $i_n$ as the map which applies $i:B \inj Z$ on the first tensor factor, and the identity on the second tensor factor. Since taking a tensor product with a fixed module is a right exact functor, we conclude that \[\left(\bigoplus_j Z_j \otimes_R H_{n-j}(C')\right)/\Im \bigoplus_j (i \otimes_R 1) \cong \bigoplus_j \left(Z_j / \Im i\right) \otimes_R H_{n-j}(C') \cong \bigoplus_j H_j(C) \otimes_R H_{n-j}(C')\] Now, we only have to show that $\Ker i_{n-1}$ is $\bigoplus_i \Tor_R(H_i(C), H_{n-i}(C'))$.

$\Tor$ is the left derived functor corresponding to taking tensor products. So taking the short exact sequence $0 \to B_i \to Z_i \to H_i(C) \to 0$ and tensoring with $H_{n-i}(C')$ yields a long exact sequence

Naturality essentially follows because all of the exact sequences and operations on them that we considered are natural.

We will only show the splitting in the case that $C$ and $C'$ are both free, although it is true that the sequence splits in the general case as well. We will show that the short exact sequence is split by constructing a homomorphism $H_n(C \otimes_R C') \to \bigoplus_i(H_i(C) \otimes_R H_{n-i}(C'))$. We observed earlier that the sequence $0 \to Z_i \to C_i \to B_{i-1} \to 0$ splits. Thus, we have a splitting map $s:C_i \to Z_i$. Using this splitting map, we can lift the quotient $Z_i \to H_i(C)$ to a map $C_i \to H_i(C)$. Similarly, using the assumption that $C'$ is free, we can construct maps $C_j' \to H_j(C')$. Now, we can construct chain complexes $H(C), H(C')$ whose modules are $H_i(C)$ and $H_j(C')$ respectively, and whose boundary maps are all trivial. Then the lifts of the quotient maps that we constructed give us chain maps $C \to H(C), C' \to H(C')$. Taking the tensor product of these chain maps gives us a chain map $C \otimes_R C' \to H(C) \otimes_R H(C')$. This induces a map on homology $H(C \otimes_R C') \to H(H(C) \otimes_R H(C'))$. Since $H(C), H(C')$ have trivial boundary maps, their tensor product does as well, so $H(H(C) \otimes_R H(C'))$ is simply $H(C) \otimes_R H(C')$. Thus, we have constructed a map $H(C \otimes_R C') \to H(C) \otimes_R H(C')$. And this map splits the short exact sequence which we constructed above.

First, we prove the result for the cube $I^n$. We give $I$ the cell structure with two vertices and one edge. We will denote the 0-cells of the $i$th copy of $I$ $0_i$ and $1_i$, and we will denote the 1-cell $e_i$. The boundary map is given by $de_i = 1_i - 0_i$. The $n$-cell in $I^n$ is $e_1 \times \cdots \times e_n$, and its boundary is \[d(e_1 \times \cdots \times e_n) = \sum_i (-1)^{i+1} e_1 \times \cdots \times de_i \times \cdots \times e_n\]

Now, write $I^n = I^i \times I^j$ with $i+j= n$. Let $e^i = e_1 \times \cdots \times e_i$ and $e^j = e_{i+1} \times \cdots \times e_n$. Then, our formula tells us that \[d(e^i \times e^j) = de^i \times e^j + (-1)^i e^i \times de^j\] as desired.

To extend this result to the general case, we will use a lemma about the naturality of the cross product.

Let us write $f_*(e^i_\alpha) = \sum_\gamma m_{\alpha \gamma}e^i_\gamma$ and $g_*(e^j_\beta) = \sum_\delta n_{\beta\delta}e^j_\delta$. Then we want to show that $(f \times g)_*(e^i_\alpha \times e^j_\beta) = \sum_{\gamma,\delta} m_{\alpha \gamma} n_{\beta\delta} (e^i_\gamma \times e^j_\delta)$. By the definition of cellular induced maps, the coefficient $m_{\alpha \gamma}$ is the degree of the composition $f_{\alpha \gamma} : S^i \to X^i /X^{i-1} \to Z^i /Z^{i-1} \to S^i$ where the first and last maps are induced by the characteristic maps for the cells $e^i_\alpha$ and $e^i_\gamma$ and the middle map is induced by the cellular map $f$. With the right choice of basepoints in the middle spaces, $f_{\alpha \gamma}$ is basepoint-preserving. The coefficients $n_{\beta\delta}$ are obtained similarly from the composition $g_{\beta\delta} : S^j \to Y^j/Y^{j-1} \to W^j/W^{j-1}\to S^j$.

The coefficients of $e_\gamma^i \times e_\delta^j$ in $(f \times g)_*(e^i_\alpha \times e^j_\beta)$ is given by the degree of the map $(f \times g)_{\alpha\beta, \gamma\delta}:S^{i + j} \to S^{i + j}$. We can obtain $(f\times g)_{\alpha\beta, \gamma\delta}$ by taking the product map $f_{\alpha\gamma} \times g_{\beta\delta}:S^i \times S^j \to S^i \times S^j$ and collapsing the $(i+j-1)$-skeleton of $S^i \times S^j$ to a point.

This means that $(f \times g)_{\alpha\beta, \gamma\delta}$ is the smash product map $f_{\alpha\beta} \wedge g_{\beta\delta}$. So the lemma we want to show boils down to proving that $\deg (f \wedge g) = \deg (f) \deg (g)$ for $f,g$ maps from spheres to themselves.

We can write $f \wedge g$ as $f \wedge \id \circ \id \wedge g$. So we only have to show that $\deg(f \wedge \id) = \deg f$ and $\deg(\id \wedge g) = \deg g$. We will do so by considering the relationship between smash products with spheres and suspension. First, we will consider circles. The smash product $X \times S^1$ can be written $X \times I / (X \times \partial I \cup \{x_0\} \times I)$. This is simply the reduced suspension $\Sigma X$ (recall that $\Sigma X$ is the quotient of the suspension $SX$ which collapses the line $\{x_0\} \times I$ to a point). If $X$ is a CW complex with 0-cell $x_0$, then this quotient $SX \to \Sigma X \cong X \wedge S^1$ just collapses a contractible subspace to a point, so it induces an isomorphism on homology. Now, let $X = S^i$. We have the following commutative diagram

Applying the homology functor to the diagram, we conclude that $Sf$ and $f \wedge \id$ have the same degree. By Proposition 2.33, $Sf$ has the same degree as $f$. So $\deg (f \wedge \id) = \deg f$ (where $\id$ is the identity map $S^1 \to S^1$). Since $S^j$ is the smash product of $j$ copies of $S^1$, we can show by induction that the formula holds when $\id$ is the identity map on $S^j$. The same argument shows that $\deg(\id \wedge g) = \deg g$.

Using this lemma, we can finish proving the proposition. Let $\Phi:I^i \to X^i$ and $\Psi:I^j \to Y^j$ be the characteristic maps of cells $e^i_\alpha \subset X$ and $e^j_\beta \subset Y$ respectively. The restriction of $\Phi$ to $\partial I^i$ is the attaching map for cell $e^i_\alpha$. By the cellular approximation theorem, we can homotope this map to a cellular map. Applying this homotopy does not affect the cellular boundary $de^i_\alpha$ because $de^i_\alpha$ is determined by an induced map on homology groups. So we can assume that $\Phi$ is cellular. By the same argument we can assume that $\Psi$ is cellular. This implies that $\Phi \times \Psi$ is cellular. A cellular map induces a chain map on the cellular chain complexes $C_*(X \times Y)$ and $C_*(Z \times W)$.

Let $e^i$ denote the $i$-cell in $I^i$ and $e^j$ denote the $j$-cell in $I^j$. We have $\Phi_*(e^i) = e^i_\alpha$, $\Psi_*(e^j) = e^j_\beta$, and $(\Phi \times \Psi)_*(e^i \times e^j) = e^i_\alpha \times e^j_\beta$. Therefore, \[d(e^i_\alpha \times e^j_\beta) = d((\Phi \times \Psi)_*(e^i \times e^j))\] Since $(\Phi \times \Psi)_*$ is a chain map, we know that \[d((\Phi \times \Psi)_*(e^i \times e^j)) = (\Phi \times \Psi)_* d(e^i \times e^j)\] We already showed the product rule on the cube $I^{i + j}$ So we know that \[(\Phi \times \Psi)_*d(e^i \times e^j) = (\Phi \times \Psi)_*(de^i \times e^j + (-1)^i e^i \times de^j)\] By our lemma, we can distribute $(\Phi \times \Psi)_*$ over these cross products, so \[(\Phi \times \Psi)_*(de^i \times e^j + (-1)^i e^i \times de^j) = \Phi_*(de^i) \times \Psi_* e^j + (-1)^i \Phi_* e^i + \Psi_*(de^j)\] Using the fact that $\Phi_*$ and $\Psi_*$ are chain maps, we find that \[\Phi_*(de^i) \times \Psi_* e^j + (-1)^i \Phi_* e^i + \Psi_*(de^j) = d\Phi_*e^i \times \Psi_*e^j + (-1)^i \Phi_*e^i \times d\Psi_*e^j\] And by definition, this is just \[de^i_\alpha \times e^j_\beta + (-1)^i e^i_\alpha \times de^j_\beta\] This is precisely what we set out to prove.

Products of CW complexes are problematic because the compactly generated CW topology is not necessarily the same as the product topology. However, both topologies have the same compact sets, so they both have the same singular simplices, which means that they have isomorphic homology groups.

Let $C = C_\bullet(X;R)$ and $C' = C_\bullet(Y;R)$ be the cellular chain complexes with coefficients in $R$. We noted above that $C \otimes_R C' = C_\bullet(X \times Y;R)$. Then the theorem follows from the algebraic Künneth formula. The naturality follows from the naturality guaranteed by the algebraic Künneth formula, combined with the fact that we can homotopy arbitrary maps to cellular maps.