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We say that the $C^*$-algebra $A$ generated by $a_1,...,a_n$ is universal subject to relations $R_1,...,R_m$ if for every $C^*$-algebra $B$ with elements $b_1,...,b_n$ satisfying relations $R_1,...,R_m$ there is $C^*$-epimorphism $\varphi: A \to B$ such that $\varphi(a_i)=b_i$. One of the basic examples is the $C^*$-algebra of complex valued function on three sphere $C(S^3)$ which is the universal commutative unital $C^*$-algebra generated by $a,b$ with relation $a^*a+b^*b=1$. My question is the following: what kind of relations can we impose on our $C^*$-algebra? In all examples which I saw the relations were algebraic and were of the form: $f(a_1,...,a_n,a_1^*,...,a_n^*)=0$ where $f$ was some polynomial. In particular do we admit:

  • quantification and referring to other elements not being the generators

  • order properties of $C^*$-algebras

  • functions which are no longer polynomials (continuous functions, Borel functions etc.)

If the answer is positive I would be grateful to know some (known in literature) examples of universal $C^*$-algebras arising in such a way.

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    $\begingroup$ If you require your generators to be unitaries, and your relations to be multiplicative words in the generators and their inverses, then the universal $C^*$-algebra is exactly the maximal $C^*$-algebra of the finitely presented group with the same presentation. $\endgroup$ – Alain Valette Dec 15 '13 at 21:22
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Since free C*-algebras don't exist, we can't give a concrete description of all relations that are allowed. Instead, we need to give axioms that determine what collections of n-tuples $(a_1,\dots,a_n)$ in $A$ are allowed, where $A$ varies over all C*-algebras. It is important that the elements not "know about" the ambient C*-algebra, so ``a is in a separable C*-algebra'' is not allowed.

Some of what is allowed is conditions like $0 \leq x \leq 1$ and $$ 0\leq\left[\begin{array}{cc} \mathbf{1} & x\\ x^{*} & \mathbf{1} \end{array}\right]\leq 1 $$ where $\mathbf{1}$ is in the unitization of $A$. Another fun example is ``x is hermitian and has spectrum contained in the Cantor set''. We can't use Borel functional calculus, but we can take a continuous function on $\mathbb R$ that is bounded and use as a relation "$x$ is normal and $f(x)=0$." See section 5 of "From Matrix to Operator Inequalities" by me, Canad. Math. Bull. 55(2012), 339--350 for a use of analytic functional calculus.

I could drone on forever here, but I did already: ``C*-Algebra Relations'' in Mathematica Scandinavica 107, 43--72, 2010.

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This a long comment rather than a complete answer.

Let me point out a paper of Bruce Blackadar

B. Blackadar, Shape theory for C* -algebras, Math. Scand. 56 (1985), 249-275.

where slightly more general conditions, which can be imposed in a natural manner on the generating relations, are considered. More specifically, in this setting the relations considered in the paper have the form

$$\|p(x_1, \ldots, x_n, x_1^*, \ldots, x_n^*)\|\leqslant \eta,$$

where $p$ is a polynomial of $2n$ non-commuting variables and $\eta\geqslant 0$. I am quite sure that this is not what you are looking for, though.

When the functions allowed in the generating relations are no longer polynomials but arbitrary Borel functions, it is difficult to talk about any kind of universality of such creatures. Indeed, in this case $h(f(a))$ need not be the same as $f(h(a))$ where $h$ is some *-homomorphism (these are not even well-defined a priori).

It is also possible to talk about C*-algebras generated by order-zero c.p.c maps from matrix algebras $M_n$ etc (this is perhaps the order ingredient you have in mind). However those maps correspond precisely to ${}^*$-homomorphisms from $C([0,1], M_n)$ so this is the old notion of universality in disguise. An important example of a C*-algebra which can be expressed in terms of (rather messy) relations involving order zero c.p.c maps is the Jiang-Su algebra $\mathcal{Z}$:

B. Jacelon and W. Winter, $\mathcal{Z}$ is universal, to appear in J. Noncommut. Geom., arXiv version.

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    $\begingroup$ Just a correction on a detail: a c.p.c. order zero map with domain $A$ corresponds to a *-homomorphism with domain $C_0( (0,1], A )$, i.e. the cone over $A$. $\endgroup$ – Gabor Szabo Jan 29 '14 at 22:42
  • $\begingroup$ Yes Gabor, you're right. $\endgroup$ – Tomek Kania Jan 30 '14 at 10:40
  • $\begingroup$ @TomekKania "When the functions allowed in the generating relations are no longer polynomials but arbitrary Borel functions, it is difficult to talk about any kind of universality of such creatures." What about when "$f$" is more than Borel function, say Holomorphic function? In particular is there a universal algebra with two generators a, b with $sin(ab)=sin(ba)$? $\endgroup$ – Ali Taghavi Apr 8 '18 at 7:53
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This is not a complete answer, just a well-known example of relation we can't impose to a $C^*$-algebra.

Let $\mathcal{A}$ be a $C^*$-algebra, then, the relation $ab-ba = 1$ is impossible for $a, b \in \mathcal{A}$ :

Lemma : If $ab-ba = 1$ then $ab^n-b^na = nb^{n-1}$.
Proof by induction : for $n=1$ it's ok.
Now if it's true for $n$, then $ab^{n+1}-b^{n+1}a = ab^nb -b^{n+1}a = (b^na+ nb^{n-1})b-b^{n+1}a $ $ = b^nab+nb^n-b^{n+1}a = b^n(1+ba)+nb^n-b^{n+1}a = (n+1)b^n $. $\square$

Corollary : The relation $ab-ba = 1$ is impossible in a Banach algebra.
Proof : If it's possible, then $ab^n-b^na = nb^{n-1}$. Next $\Vert nb^{n-1} \Vert = n \Vert b^{n-1} \Vert = \Vert ab^n-b^na \Vert $ $ \le \Vert ab^n \Vert + \Vert b^na \Vert \le \Vert ab \Vert \Vert b^{n-1} \Vert + \Vert b^{n-1} \Vert \Vert ba \Vert$.
Conclusion $\Vert ab \Vert + \Vert ba \Vert \ge n$ $\forall n$, contradiction. $\square$

Remark : This (Heisenberg) relation is realized by unbounded operators.
For example let the Hilbert space $H = l^2(\mathbb{Z})$, $a: e_n \to ne_{n-1}$ and $b: e_n \to e_{n+1}$

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    $\begingroup$ As an application of this in Quantum Mechanics, one can consider the position and momentum operators $x$ and $p$, which, according to the standard quantization, must satisfy $xp - px \subset 1$ (assuming natural units, where $\hbar = 1$). Then the above argument shows that $x$ and $p$ cannot be bounded operators. To deal with bounded operators one can do the Weil trick to take the exponentials $e^{i\xi p}$ and $e^{i\eta x}$ in order to get unitaries through Borel functional calculus. $\endgroup$ – Phoenix87 Jul 15 '14 at 13:30

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