7d. Cuntz algebras

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We would like to end this chapter with an interesting class of [math]C^*[/math]-algebras, discovered by Cuntz in , and heavily used since then, for various technical purposes. These algebras are not obviously related to the quantum space program that we have been developing so far, and might even look like some sort of Devil's invention, orthogonal to what is beautiful in operator algebras, but believe me, if planning to do some serious operator algebra work, you will certainly run into them. Their definition is as follows:

Definition

The Cuntz algebra [math]O_n[/math] is the [math]C^*[/math]-algebra generated by isometries [math]S_1,\ldots,S_n[/math] satisfying the following condition:

[[math]] S_1S_1^*+\ldots+S_nS_n^*=1 [[/math]]

That is, [math]O_n\subset B(H)[/math] is generated by [math]n[/math] isometries whose ranges sum up to [math]H[/math].

Observe that [math]H[/math] must be infinite dimensional, in order to have isometries as above. In what follows we will prove that [math]O_n[/math] is independent on the choice of such isometries, and also that this algebra is simple. We will restrict the attention to the case [math]n=2[/math], the proof in general being similar. Let us start with some simple computations, as follows:

Proposition

Given a word [math]i=i_1\ldots i_k[/math] with [math]i_l\in\{1,2\}[/math], we associate to it the element [math]S_i=S_{i_1}\ldots S_{i_k}[/math] of the algebra [math]O_2[/math]. Then [math]S_i[/math] are isometries, and we have

[[math]] S_i^*S_j=\delta_{ij}1 [[/math]]

for any two words [math]i,j[/math] having the same lenght.

Show Proof

We use the relations defining the algebra [math]O_2[/math], namely:

[[math]] S_1^*S_1=S_2^*S_2=1\quad,\quad S_1S_1^*+S_2S_2^*=1 [[/math]]

The fact that [math]S_i[/math] are isometries is clear, here being the check for [math]i=12[/math]:

[[math]] S_{12}^*S_{12}=(S_1S_2)^*(S_1S_2)=S_2^{*}S_1^*S_1S_2=S_2^{*}S_2=1 [[/math]]

Regarding the last assertion, by recurrence we just have to establish the formula there for the words of length 1. That is, we want to prove the following formulae:

[[math]] S_1^*S_2=S_2^*S_1=0 [[/math]]

But these two formulae follow from the fact that the projections [math]P_i=S_iS_i^*[/math] satisfy by definition [math]P_1+P_2=1[/math]. Indeed, we have the following computation:

[[math]] \begin{eqnarray*} P_1+P_2=1 &\implies&P_1P_2=0\\ &\implies&S_1S_1^*S_2S_2^*=0\\ &\implies&S_1^*S_2=S_1^*S_1S_1^*S_2S_2^*S_2=0 \end{eqnarray*} [[/math]]

Thus, we have the first formula, and the proof of the second one is similar.

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We can use the formulae in Proposition 7.38 as follows:

Proposition

Consider words in [math]O_2[/math], meaning products of [math]S_1,S_1^*,S_2,S_2^*[/math].

• Each word in [math]O_2[/math] is of form [math]0[/math] or [math]S_iS_j^*[/math] for some words [math]i,j[/math].
• Words of type [math]S_iS_j^*[/math] with [math]l(i)=l(j)=k[/math] form a system of [math]2^k\times 2^k[/math] matrix units.
• The algebra [math]A_k[/math] generated by matrix units in (2) is a subalgebra of [math]A_{k+1}[/math].

Show Proof

Here the first two assertions follow from the formulae in Proposition 7.38, and for the last assertion, we can use the following formula:

[[math]] S_iS_j^*=S_i1S_j^*=S_i(S_1S_1^*+S_2S_2^*)S_j^* [[/math]]

Thus, we obtain an embedding of algebras [math]A_k[/math], as in the statement.

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Observe now that the embedding constructed in (3) above is compatible with the matrix unit systems in (2). Consider indeed the following diagram:

[[math]] \begin{matrix} A_{k+1}&\simeq &M_{2^{k+1}}(\mathbb C)\\ \ &\ &\ \\ \cup&\ &\cup\\ \ &\ &\ \\ A_k&\simeq&M_{2^k}(\mathbb C) \end{matrix} [[/math]]

With the notation [math]e_{ix,yj}=e_{ij}\otimes e_{xy}[/math], the inclusion on the right is given by:

[[math]] \begin{eqnarray*} e_{ij} &\to&e_{i1,1h}+e_{i2,2j}\\ &=&e_{ij}\otimes e_{11}+e_{ij}\otimes e_{22}\\ &=&e_{ij}\otimes 1 \end{eqnarray*} [[/math]]

Thus, with standard tensor product notations, the inclusion on the right is the canonical inclusion [math]m\to m\otimes1[/math], and so the above diagram becomes:

[[math]] \begin{matrix} A_{k+1}&\simeq &M_2(\mathbb C)^{\otimes k+1}\\ \ &\ &\ \\ \cup&\ &\cup\\ \ &\ &\ \\ A_k&\simeq &M_2(\mathbb C)^{\otimes k} \end{matrix} [[/math]]

The passage from the algebra [math]A=\cup_kA_k\simeq M_2(\mathbb C)^{\otimes\infty}[/math] coming from this observation to the full the algebra [math]O_2[/math] that we are interested in can be done by using:

Proposition

Each element [math]X\in \lt S_1,S_2 \gt \subset O_2[/math] decomposes as a finite sum

[[math]] X=\sum_{i \gt 0}S_1^{*i}X_{-i}+X_0+\sum_{i \gt 0}X_iS_1^i [[/math]]

where each [math]X_i[/math] is in the union [math]A[/math] of algebras [math]A_k[/math].

Show Proof

By linearity and by using Proposition 7.39 we may assume that [math]X[/math] is a nonzero word, say [math]X=S_iS_j^*[/math]. In the case [math]l(i)=l(j)[/math] we can set [math]X_0=X[/math] and we are done. Otherwise, we just have to add at left or at right terms of the form [math]1=S_1^*S_1[/math]. For instance [math]X=S_2[/math] is equal to [math]S_2S_1^*S_1[/math], and we can take [math]X_1=S_2S_1^*\in A_1[/math].

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We must show now that the decomposition [math]X\to (X_i)[/math] found above is unique, and then prove that each application [math]X\to X_i[/math] has good continuity properties. The following formulae show that in both problems we may restrict attention to the case [math]i=0[/math]:

[[math]] X_{i+1}=(XS_1^*)_i\hskip 2cm X_{-i-1}=(S_1X)_i [[/math]]

In order to solve these questions, we use the following fact:

Proposition

If [math]P[/math] is a nonzero projection in [math]\mathcal O_2= \lt S_1,S_2 \gt \subset O_2[/math], its [math]k[/math]-th average, given by the formula

[[math]] Q=\sum_{l(i)=k}S_iPS_i^* [[/math]]

is a nonzero projection in [math]\mathcal O_2[/math] having the property that the linear subspace [math]QA_kQ[/math] is isomorphic to a matrix algebra, and [math]Y\to QYQ[/math] is an isomorphism of [math]A_k[/math] onto it.

Show Proof

We know that the words of form [math]S_iS_j^*[/math] with [math]l(i)=l(j)=k[/math] are a system of matrix units in [math]A_k[/math]. We apply to them the map [math]Y\to QYQ[/math], and we obtain:

[[math]] \begin{eqnarray*} QS_iS_j^*Q &=&\sum_{pq} S_pPS_p^*S_iS_j^*S_qPS_q^*\\ &=&\sum_{pq}\delta_{ip}\delta_{jq}S_pP^2S_q^*\\ &=&S_iPS_j^* \end{eqnarray*} [[/math]]

The output being a system of matrix units, [math]Y\to QYQ[/math] is an isomorphism from the algebra of matrices [math]A_k[/math] to another algebra of matrices [math]QA_kQ[/math], and this gives the result.

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Thus any map [math]Y\to QYQ[/math] behaves well on the [math]i=0[/math] part of the decomposition on [math]X[/math]. It remains to find [math]P[/math] such that [math]Y\to QYQ[/math] destroys all [math]i\neq 0[/math] terms, and we have here:

Proposition

Assuming [math]X_0\in A_k[/math], there is a nonzero projection [math]P\in A[/math] such that [math]QXQ=QX_0Q[/math], where [math]Q[/math] is the [math]k[/math]-th average of [math]P[/math].

Show Proof

We want [math]Y\to QYQ[/math] to map to zero all terms in the decomposition of [math]X[/math], except for [math]X_0[/math]. Let us call [math]M_1,\ldots,M_t\in\mathcal O_2-A[/math] the terms to be destroyed. We want the following equalities to hold, with the sum over all pairs of length [math]k[/math] indices:

[[math]] \sum_{ij}S_iPS_i^*M_qS_jPS_j^*=0 [[/math]]

The simplest way is to look for [math]P[/math] such that all terms of all sums are [math]0[/math]:

[[math]] S_iPS_i^*M_qS_jPS_j^*=0 [[/math]]

By multiplying to the left by [math]S_i^*[/math] and to the right by [math]S_j[/math], we want to have:

[[math]] PS_i^*M_qS_jP=0 [[/math]]

With [math]N_z=S_i^*M_qS_j[/math], where [math]z[/math] belongs to some new index set, we want to have:

[[math]] PN_zP=0 [[/math]]

Since [math]N_z\in\mathcal O_2-A[/math], we can write [math]N_z=S_{m_z}S_{n_z}^*[/math] with [math]l(m_z)\neq l(n_z)[/math], and we want:

[[math]] PS_{m_z}S_{n_z}^*P=0 [[/math]]

In order to do this, we can the projections of form [math]P=S_rS_r^*[/math]. We want:

[[math]] S_rS_r^*S_{m_z}S_{n_z}^*S_rS_r^*=0 [[/math]]

Let [math]K[/math] be the biggest length of all [math]m_z,n_z[/math]. Assume that we have fixed [math]r[/math], of length bigger than [math]K[/math]. If the above product is nonzero then both [math]S_r^*S_{m_z}[/math] and [math]S_{n_z}^*S_r[/math] must be nonzero, which gives the following equalities of words:

[[math]] r_1\ldots r_{l(m_z)}=m_z\quad,\quad r_1\ldots r_{l(n_z)}=n_z [[/math]]

Assuming that these equalities hold indeed, the above product reduces as follows:

[[math]] S_rS_{r_{l(r)}}^*\ldots S_{r_{l(m_z)+1}}^*S_{r_{l(n_z)+1}}\ldots S_{r_{l(r)}}S_r^* [[/math]]

Now if this product is nonzero, the middle term must be nonzero:

[[math]] S_{r_{l(r)}}^*\ldots S_{r_{l(m_z)+1}}^*S_{r_{l(n_z)+1}}\ldots S_{r_{l(r)}}\neq 0 [[/math]]

In order for this for hold, the indices starting from the middle to the right must be equal to the indices starting from the middle to the left. Thus [math]r[/math] must be periodic, of period [math]|l(m_z)-l(n_z)| \gt 0[/math]. But this is certainly possible, because we can take any aperiodic infinite word, and let [math]r[/math] be the sequence of first [math]M[/math] letters, with [math]M[/math] big enough.

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We can now start solving our problems. We first have:

Proposition

The decomposition of [math]X[/math] is unique, and we have

[[math]] ||X_i||\leq||X|| [[/math]]

for any [math]i[/math].

Show Proof

It is enough to do this for [math]i=0[/math]. But this follows from the previous result, via the following sequence of equalities and inequalities:

[[math]] ||X_0||=||QX_0Q||=||QXQ||\leq||X|| [[/math]]

Thus we got the inequality in the statement. As for the uniqueness part, this follows from the fact that [math]X_0\to QX_0Q=QXQ[/math] is an isomorphism.

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Remember now we want to prove that the Cuntz algebra [math]O_2[/math] does not depend on the choice of the isometries [math]S_1,S_2[/math]. In order to do so, let [math]\overline{\mathcal O}_2[/math] be the completion of the [math]*[/math]-algebra [math]\mathcal O_2= \lt S_1,S_2 \gt \subset O_2[/math] with respect to the biggest [math]C^*[/math]-norm. We have:

Proposition

We have the equivalence

[[math]] X=0\iff X_i=0,\forall i [[/math]]

valid for any element [math]X\in\overline{\mathcal O}_2[/math].

Show Proof

Assume [math]X_i=0[/math] for any [math]i[/math], and choose a sequence [math]X^k\to X[/math] with [math]X^k\in\mathcal O_2[/math]. For [math]\lambda\in\mathbb T[/math] we define a representation [math]\rho_\lambda[/math] in the following way:

[[math]] \rho_\lambda :S_i\to\lambda S_i [[/math]]

We have then [math]\rho_\lambda(Y)=Y[/math] for any element [math]Y\in A[/math]. We fix norm one vectors [math]\xi,\eta[/math] and we consider the following continuous functions [math]f:\mathbb T\to\mathbb C[/math]:

[[math]] f^k(\lambda)= \lt \rho_\lambda (X^k)\xi,\eta \gt [[/math]]

From [math]X^k\to X[/math] we get, with respect to the usual sup norm of [math]C(\mathbb T)[/math]:

[[math]] f^k\to f [[/math]]

Each [math]X^k\in\mathcal O_2[/math] can be decomposed, and [math]f^k[/math] is given by the following formula:

[[math]] f^k(\lambda )=\sum_{i \gt 0}\lambda^{-i} \lt S_1^{*i}X^k_{-i}\xi,\eta \gt + \lt X_0\xi,\eta \gt + \sum_{i \gt 0}\lambda^i \lt X_i^kS_1^i\xi,\eta \gt [[/math]]

This is a Fourier type expansion of [math]f^k[/math], that can we write in the following way:

[[math]] f^k(\lambda)=\sum_{j=-\infty}^\infty a_j^k\lambda^j [[/math]]

By using Proposition 7.43 we obtain that with [math]k\to\infty[/math], we have:

[[math]] |a_j^k|\leq||X_j^k||\to||X_j^\infty||=0 [[/math]]

On the other hand we have [math]a_j^k\to a_j[/math] with [math]k\to\infty[/math]. Thus all Fourier coefficients [math]a_j[/math] of [math]f[/math] are zero, so [math]f=0[/math]. With [math]\lambda=1[/math] this gives the following equality:

[[math]] \lt X\xi,\eta \gt =0 [[/math]]

This is true for arbitrary norm one vectors [math]\xi,\eta[/math], so [math]X=0[/math] and we are done.

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We can now formulate the Cuntz theorem, from , as follows:

Theorem (Cuntz)

Let [math]S_1,S_2[/math] be isometries satisfying [math]S_1S_1^*+S_2S_2^*=1[/math].

• The [math]C^*[/math]-algebra [math]O_2[/math] generated by [math]S_1,S_2[/math] does not depend on the choice of [math]S_1,S_2[/math].
• For any nonzero [math]X\in O_2[/math] there are [math]A,B\in O_2[/math] with [math]AXB=1[/math].
• In particular [math]O_2[/math] is simple.

Show Proof

This basically follows from the various results established above:

(1) Consider the canonical projection map [math]\pi:\overline{O}_2\to O_2[/math]. We know that [math]\pi[/math] is surjective, and we will prove now that [math]\pi[/math] is injective. Indeed, if [math]\pi(X)=0[/math] then [math]\pi(X)_i=0[/math] for any [math]i[/math]. But [math]\pi(X)_i[/math] is in the dense [math]*[/math]-algebra [math]A[/math], so it can be regarded as an element of [math]\overline{O}_2[/math], and with this identification, we have [math]\pi(X)_i=X_i[/math] in [math]\overline{O}_2[/math]. Thus [math]X_i=0[/math] for any [math]i[/math], so [math]X=0[/math]. Thus [math]\pi[/math] is an isomorphism. On the other hand [math]\overline{O}_2[/math] depends only on [math]\mathcal O_2[/math], and the above formulae in [math]\mathcal O_2[/math], for algebraic calculus and for decomposition of an arbitrary [math]X\in\mathcal O_2[/math], show that [math]\mathcal O_2[/math] does not depend on the choice of [math]S_1,S_2[/math]. Thus, we obtain the result.

(2) Choose a sequence [math]X^k\to X[/math] with [math]X^k\in\mathcal O_2[/math]. We have the following formula:

[[math]] (X^*X)_0=\lim_{k\to\infty}\left( \sum_{i \gt 0}X^{k*}_{-i}X^k_{-i}+X^{k*}_0X^k_0+ \sum_{i \gt 0}S_1^{*i}X^{k*}_iX^k_iS_1^i\right) [[/math]]

Thus [math]X\neq 0[/math] implies [math](X^*X)_0\neq 0[/math]. By linearity we can assume that we have:

[[math]] ||(X^*X)_0||=1 [[/math]]

Now choose a positive element [math]Y\in\mathcal O_2[/math] which is close enough to [math]X^*X[/math]:

[[math]] ||X^*X-Y|| \lt \varepsilon [[/math]]

Since [math]Z\to Z_0[/math] is norm decreasing, we have the following estimate:

[[math]] ||Y_0|| \gt 1-\varepsilon [[/math]]

We apply Proposition 7.42 to our positive element [math]Y\in\mathcal O_2[/math]. We obtain in this way a certain projection [math]Q[/math] such that [math]QY_0Q=QYQ[/math] belongs to a certain matrix algebra. We have [math]QYQ \gt 0[/math], so we can diagonalize this latter element, as follows:

[[math]] QYQ=\sum\lambda_iR_i [[/math]]

Here [math]\lambda_i[/math] are positive numbers and [math]R_i[/math] are minimal projections in the matrix algebra. Now since [math]||QYQ||=||Y_0||[/math], there must be an eigenvalue greater that [math]1-\varepsilon[/math]:

[[math]] \lambda_0 \gt 1-\varepsilon [[/math]]

By linear algebra, we can pass from a minimal projection to another:

[[math]] U^*U=R_i\quad,\quad UU^*=S_1^kS_1^{*k} [[/math]]

The element [math]B=QU^*S_1^k[/math] has norm [math]\leq 1[/math], and we get the following inequality:

[[math]] \begin{eqnarray*} ||1-B^*X^*XB|| &\leq&||1-B^*YB||+||B^*YB-B^*X^*XB||\\ & \lt &||1-B^*YB||+\varepsilon \end{eqnarray*} [[/math]]

The last term can be computed by using the diagonalization of [math]QYQ[/math], as follows:

[[math]] \begin{eqnarray*} B^*YB &=&S_1^{*k}UQYQU^*S_1^k\\ &=&S_1^{*k}\left(\sum \lambda_i UR_iU^*\right) S_1^k\\ &=&\lambda_0S_1^{*k}S_1^kS_1^{*k}S_1^k\\ &=&\lambda_0 \end{eqnarray*} [[/math]]

From [math]\lambda_0 \gt 1-\varepsilon[/math] we get [math]||1-B^*YB|| \lt \varepsilon[/math], and we obtain the following estimate:

[[math]] ||1-B^*X^*XB|| \lt 2\varepsilon [[/math]]

Thus [math]B^*X^*XB[/math] is invertible, say with inverse [math]C[/math], and we have [math](B^*X^*)X(BC)=1[/math].

(3) This is clear from the formula [math]AXB=1[/math] established in (2).

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General references

Banica, Teo (2024). "Principles of operator algebras". arXiv:2208.03600 [math.OA].