Let $x_0,\ldots,x_n$ be symbols with relations $x_i=x_i^*$, $x_i x_j = x_j x_i$ and $\sum_i x_i^2 = 1$. Then I want to show that the universal $C^*$-Algebra $A$ of these relations exists and that $A\cong C(S^n)$.
My proof is almost complete, but i need help with the last step:
Notation. $\Delta B$ is the Gelfand-Space (Spectrum) for any $C^*$-Algebra $B$, ie. the set of nonzero algebra homomorphisms $\chi:B\to\mathbb{C}$. Note that for any $b\in B$, $\{\chi(b)\,:\,\chi\in\Delta B\}$ equals the spectrum of $b$.
Existence of $A$. $A$ exists, because the relations can be rewritten as $\|p(x_0,\ldots,x_n)\|=0$ for polynomials $p$ and because in any $C^*$-Algebra $B$ with elements $x_0,\ldots,x_n$ that satisfy the relations, the norms $\|x_i\|$ are bounded independently of the choice of $B$. To see that consider $x_i^2=x_i^* x_i\ge 0$, hence $x_k^2 \le \sum_i x_i^2=1$ and therefore $\|x_k\|\le 1$.
$A$ isomorphic to $C(S^n)$. Let $\pi_i:S^n\to\mathbb{R}\subset\mathbb{C}$ be the $i$-th projection. The $\pi_0,\ldots,\pi_n$ satisfy the same relations as the $x_i$. By universality of $A$ there exists a $*$-homomorphism $\varphi:A\to C(S^n)$ with $\varphi(x_i)=\pi_i$. From Stone-Weierstrass it follows that $C^*(\pi_0,\ldots,\pi_n)=C(S^n)$. So $\varphi$ is surjective.
$\varphi$ is injective. We show that $\hat\varphi:\Delta C(S^n)\to\Delta A$ defined by $\hat\varphi(\chi)=\chi\circ\varphi$ is surjective. Because then injectivity of $\varphi$ follows from
$$ \|x\| = \sup_{\chi\in\Delta A} |\chi(x)| = \sup_{\chi\in\Delta C(S^n)} |\chi\circ\varphi(x)| = \|\varphi(x)\|$$
So my question would be answered if i can show that $\hat\varphi$ is surjective!
This is where I'm stuck. My thoghts so far: I tried to take any $\tilde\chi\in\Delta A$ and find a $\chi\in\Delta C(S^n)$ with $\tilde\chi=\chi\circ\varphi$. Clearly we have to set $\chi(\pi_i):=\tilde\chi(x_i)$ and extend this to a $*$-homomorphism. $\chi$ is uniquely defined by this, if such a $\chi$ exists, because $C(S^n)=C^*(\pi_0,\ldots,\pi_n)$. But why is $\chi$ well defined?
I think that on the $*$-Algebra $C_0$ generated by the $\pi_0,\ldots,\pi_n$, $\chi$ is a well defined $*$-homomorphism. That is because every element $x\in C_0$ can be uniquely written as $x=p(\pi_0,\ldots,\pi_n)$ where $p$ is a commutative polynomial (I haven't proved that, but I think it's true), hence we can define $\chi:C_0\to\mathbb{C}$ by $$\chi(x) \equiv \chi(p(\pi_0,\ldots,\pi_n)) := p(x_0,\ldots,x_n)$$ This is well defined, because there is only one commutative polynomial $p$ with $x=p(\pi_0,\ldots,\pi_n)$. By this definition, $\chi$ is a $*$-homomorphism on $C_0$. Now I want to extend $\chi$ to $C(S^n)$. Because $C_0$ is dense in $C(S^n)$, this would follow from continuity of $\chi$, but i wasn't able to show that yet.
Edit: Alternative proof
Based on Martin Argerami's answer I'm adding this proof for future readers.
We show that $C(S^n)$ is the the universal algebra by showing that
- The projections $\pi_0,\ldots,\pi_n\in C(S^n)$ satisfy the relations
- $C(S^n)=C^*(\pi_0,\ldots,\pi_n)$
- universal property: For any $C^*$-algebra $B$ with elements $b_0,\ldots,b_n$ that satisfy the relations, there exists a $*$-homomorphism $\varphi:C(S^n)\to B$ with $\varphi(\pi_i)=b_i$ for all $i$.
Point 1. is clear, Point 2. follows from Stone Weierstrass. Let $B$ be as in point 3. Define a map $\varphi:C(S^n)\to B$ by first defining it on the dense $*$-Algebra $C_0\subset C(S^n)$ generated by $\pi_0,\ldots,\pi_n$ by
$$ \varphi(p(\pi_0,\ldots,\pi_n)) := p(b_0,\ldots,b_n) $$
for all commutative polynomials in $n+1$ variables. This is well defined, because for all $z\in C_0$ there is exactly one commutative polynomial $p$ with $z=p(\pi_0,\ldots,\pi_n)$. It's easy to see that $\varphi$ is a $*$-Homomorphism. To extend it to $C(S^n)$, we have to show that $\varphi:C_0\to B$ is continuous. Let $B_0:=C^*(b_0,\ldots,b_n)\subset B$, then $B_0$ is commutative (because the $b_i$ commute and are self adjoint). So for all $b\in B_0$ we have $\|b\|_B=\|b\|_{B_0}=\sup_{\chi\in\Delta B_0} |\chi(b)|$ (this is the norm of the Gelfand transform $\hat b$ of $b\in B_0$). So continuity follows from
$$ \|\varphi(p(\pi_0,\ldots,\pi_n))\| = \|p(b_0,\ldots,b_n)\| = \sup_{\chi\in\Delta B_0} |\chi(p(b_0,\ldots,b_n))| = \sup_{\chi\in\Delta B_0} |p(\chi(b_0),\ldots,\chi(b_n))| \le \sup_{t=(t_0,\ldots,t_n)\in S^n} |p(t_0,\ldots,t_n)| = \sup_{t\in S^n} |p(\pi_0(t),\ldots\pi_n(t))| = \|p(\pi_0,\ldots,\pi_n)\|_{C(S^n)} $$
where we used that $(\chi(b_0),\ldots,\chi(b_n))\in S^n$ because $\sum \chi(b_i)^2 = \chi(\sum b_i^2)=\chi(1)=1$.
Edit: Another proof using multivariable functional calculus
Thanks to Martin Argerami for introducing me to multivariable functional calculus, which motivated me to do yet another proof.
Multivariable functional calculus. For $n$ elements $x_1,\ldots,x_n$ of a unital $C^*$-algebra, their joint spectrum is defined as
$$ \sigma(x_1,\ldots,x_n) := \{ (\chi(x_1),\ldots,\chi(x_n))\in\mathbb{C}^n \,|\, \chi\in\Delta C^*(x_1,\ldots,x_n,1) \} $$
If the elements $x_1,\ldots,x_n,x_1^*,\ldots,x_n^*$ commute, there is a unique (isomorphic) $*$-isomorphism
\begin{align*} C(\sigma(x_1,\ldots,x_n)) &\to C^*(x_1,\ldots,x_n,1) \\ f &\mapsto f(x_1,\ldots,x_n) \end{align*}
with $1(x_1,\ldots,x_n)=1_A$ and $p_k(x_1,\ldots,x_n)=x_k$, where $p_k$ is the projection $p_k(t_1,\ldots,t_n)=t_k$. As in the single variable functional calculus, we have
$$ \sigma(f(x_1,\ldots,x_n)) = f(\sigma(x_1,\ldots,x_n)) \in \mathbb{C} $$
Further for all continuous $f:\sigma(x_1,\ldots,x_n)\to\mathbb{C}^n$, write $f=(f_1,\ldots,f_n)$ with $f_i\in C(\sigma(x_1,\ldots,x_n))$, we also have
$$ \sigma(f(x_1,\ldots,x_n)) = f(\sigma(x_1,\ldots,x_n)) \in \mathbb{C}^n $$
Proof that $C(S^n)$ is the universal algebra. Let $A$ be the universal $C^*$-algebra of $x_1,\ldots,x_n$ (which exists, as noted above). Because the $x_1,\ldots,x_n$ commute and are selfadjoint, multivariable functional calculus induces an isomorphism
$$ C(\sigma(x_1,\ldots,x_n)) \cong C^*(x_1,\ldots,x_n,1)=A $$
It remains to show that $\sigma(x_1,\ldots,x_n)=S^n$. For all $\chi\in\Delta C^*(x_1,\ldots,x_n,1)=A$, we have $\chi(x_i)\in\mathbb{R}$ because $x_i=x_i^*$ and
$$\sum_i \chi(x_i)^2 = \chi(\sum_i x_i^2) = \chi(1)=1$$
so $(\chi(x_i))_i\in S^n$, hence $\sigma(x_1,\ldots,x_n) \subset S^n$. Let $U:S^n\to S^n$ be a rotation. Let $\tilde x_i := U_i(x_1,\ldots,x_n)$, then $\tilde x_1,\ldots,\tilde x_n$ satisfy the same relations as $x_1,\ldots,x_n$, because:
- $U$ can be approximated by polynomials with real valued coefficients, hence $U(x_1,\ldots,x_n)=\overline U(x_1,\ldots,x_n)=\overline U(x_1^*,\ldots,x_n^*) = (U(x_1,\ldots,x_n))^*$
- $A$ is commutative, so $\tilde x_i\tilde x_j=\tilde x_j\tilde x_i$
- $\sum_i U_i(x_1,\ldots,x_n)^2 = (\sum_i U_i^2)(x_1,\ldots,x_n) = 1(x_1,\ldots,x_n) = 1$.
By universality of $A=C^*(x_1,\ldots,x_n)$ there is a $*$-homomorphisms $\varphi:A\to A$ with $\varphi(x_i)=\tilde x_i$. By universality of $A\cong(\tilde x_1,\ldots,\tilde x_n)$ there is a $*$-homomorphisms $\tilde\varphi:A\to A$ with $\tilde\varphi(\tilde x_i)=x_i$. Then $\varphi\circ\tilde\varphi=\tilde\varphi\circ\varphi=\text{id}_A$, so $\varphi:A\to A$ is a $*$-automorphism, hence it preserves the spectrum, ie.
\begin{align*} \sigma(x_1,\ldots,x_n) &= \sigma(\varphi(x_1),\ldots,\varphi(x_n)) \\ &= \sigma(\tilde x_1,\ldots,\tilde x_n) \\ &= \sigma(U(x_1,\ldots,x_n)) \\ &= U(\sigma(x_1,\ldots,x_n)) \end{align*}
Because this is true for all rotations $U$, we must have $\sigma(x_1,\ldots,x_n)=S^n$.
I'm not sure if this approach works, but it looks too complicated. The usual way to show what the universal algebra is, is to show that you have the $*$-homomorphism onto any C$^*$-algebra with the desired generators and relations.
In this case, if $x_1,\ldots,x_n\in B(H)$ are as you postulate, you want to construct a $*$-homomorphism $\rho:C(S^n)\to B(H)$ with $\rho(\pi_j)=x_j$. The only issue is whether this is bounded. The key observation here is that because the $x_j$ commute, you can do an $n$-variable Gelfand transform. Then, for any polynomial $p$ you have $$ \|\rho(p(\pi_1,\ldots,\pi_n))\|=\|p(x_1,\ldots,x_n)\|\leq\|p\|_{S^n}=\|p(\pi_1,\ldots,\pi_n)\|_{S^n}. $$