# Show that $C(S^n)$ is the universal $C^*$-Algebra of selfadjoint, commutative $x_0,\ldots,x_n$ with $\sum x_i^2 = 1$

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

We show that $C(S^n)$ is the the universal algebra by showing that

1. The projections $\pi_0,\ldots,\pi_n\in C(S^n)$ satisfy the relations
2. $C(S^n)=C^*(\pi_0,\ldots,\pi_n)$
3. 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 $$\|\rho(p(\pi_1,\ldots,\pi_n))\|=\|p(x_1,\ldots,x_n)\|\leq\|p\|_{S^n}=\|p(\pi_1,\ldots,\pi_n)\|_{S^n}.$$
• Thanks, that's exactly the inequality i wasn't able to show! I'm not sure what you mean by $n$-variable Gelfand transform, I assume you mean the Gelfand transform of $p(x_1,\ldots,x_n)$? – Kalua May 5 '15 at 9:46
• What I mean is that the inequality $\|p(x_1,\ldots,x_n)\|\leq\|p\|$ depends on doing functional calculus (i.e. Gelfand transform). But here you are doing "several variable" functional calculus. The ideas work as in the single variable case, but this is not usually mentioned in the textbooks. – Martin Argerami May 5 '15 at 15:06
• That sounds interesting, but I really can't find anything about the $n$-variable transform. Can you briefly tell me what it is? I thought you meant the Gelfand transform that is defined for elements of $C^*(x_1,\ldots,x_n)$, which exists because this $C^*$-Algebra is commutative (see my edit for how i used it). Did you have something else in mind? – Kalua May 5 '15 at 16:19
• When you have a single normal operator, the Gelfand transform lets you write $C^*(x)=\{f(x):\ f\in C(\sigma(X))\}$. Here you want to do the same for $n$ commuting selfadjoint operators. The key problem is having the right generalization of $\sigma(x)$, i.e. a "joint spectrum"; see the first paragraph of section 2.1 (and the one before) in this paper. – Martin Argerami May 5 '15 at 16:37