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We discuss here, following , the direct approach to Theorem 10.21, with purely algebraic techniques. We will use semigroup algebras, jointly generalizing the main models that we have, namely group algebras, and free Fock spaces. Let us start with:

Definition

We call “semigroup” a unital semigroup, embeddable into a group:

[[math]] M\subset G [[/math]]

For such a semigroup [math]M[/math], we use the notation

[[math]] M^{-1}=\left\{m^{-1}\Big|m\in M\right\} [[/math]]

regarded as a subset of some group [math]G[/math] containing [math]M[/math], as above.

As a first observation, the above embeddability assumption [math]M\subset G[/math] tells us that the usual group cancellation rules hold in [math]M[/math], namely:

[[math]] ab=ac\implies b=c [[/math]]

[[math]] ba=ca\implies b=c [[/math]]

Regarding the precise relation between [math]M[/math] and the various groups [math]G[/math] containing it, it is possible to talk here about the Grothendieck group [math]G[/math] associated to such a semigroup [math]M[/math]. However, we will not need this in what follows, and use Definition 10.22 as such.

With the above definition in hand, we have the following construction, which unifies the main models that we have, namely the group algebras, and the free Fock spaces:

Proposition

Let [math]M[/math] be a semigroup. By using the left simplifiability of [math]M[/math] we can define, as for the discrete groups, an embedding of semigroups, as follows:

[[math]] (M,\cdot)\to (B(l^2(M)),\circ) [[/math]]

[[math]] m\to\lambda_M(m)=[\delta _n\to\delta_{mn}] [[/math]]

Via this embedding, the [math]C^*[/math]-algebra [math]C^*(M)\subset B(l^2(M))[/math] generated by [math]\lambda_M(M)[/math], together with the following canonical state, is a noncommutative random variable algebra:

[[math]] \tau_M(T)= \lt T\delta_e,\delta_e \gt [[/math]]

Also, the operators in [math]\lambda _M(M)[/math] are isometries, but not necessarily unitaries.

Show Proof

Everything here is standard, as for the usual group algebras, with the only subtlety appearing at the level of the isometry property of the operators [math]\lambda_M(m)[/math]. To be more precise, for every [math]m\in M[/math], the adjoint operator [math]\lambda_M(m)^*[/math] is given by:

[[math]] \lambda_M(m)^*(\delta_n) =\sum_{x\in M} \lt \lambda_M(m)^*\delta_n,\delta_x \gt \delta_x =\sum_{x\in M}\delta_{n ,mx}\delta_x [[/math]]

Thus we have indeed the isometry property for these operators, namely:

[[math]] \lambda_M(m)^*\lambda_M(m)=1 [[/math]]

As for the unitarity propety of the such operators, this definitely holds in the usual discrete group case, [math]M=G[/math], but not in general. As a basic example here, for the semigroup [math]M=\mathbb N[/math], which satisfies of course the assumptions in Definition 10.22, the operator [math]\lambda_M(m)[/math] associated to the element [math]m=1\in\mathbb N[/math] is the usual shift:

[[math]] \lambda_\mathbb N(1)=S\in B(l^2(\mathbb N)) [[/math]]

But this shift [math]S[/math], that we know well from the above, is an isometry which is not a unitary. Thus, we are led to the conclusions in the statement.

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At the level of examples now, as announced above, we have:

Proposition

The construction [math]M\to C^*(M)[/math] is as follows:

• For the discrete groups, [math]M=G[/math], we obtain in this way the usual discrete group algebras [math]C^*(G)[/math], as previously constructed in the above.
• For a free semigroup, [math]M=\mathbb N^{*I}[/math], we obtain the algebra of creation operators over the full Fock space over [math]\mathbb R^I[/math], with the state associated to the vacuum vector.

Show Proof

All this is clear from definitions, with (1) being obvious, and (2) coming via our usual identifications for the free Fock spaces and related algebras.

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As a key observation now, enabling us to do some probability, we have:

Proposition

If [math]M\subset N[/math] are semigroups satisfying the condition

[[math]] M(N-M)=N-M [[/math]]

then for every family [math]\{a_i\}_{i\in I}[/math] of elements in [math]M[/math], we have the formula

[[math]] \{\lambda_N(a_i)\}_{i\in I}\sim\{\lambda_M(a_i)\}_{i\in I} [[/math]]

as an equality of joint distributions, with respect to the canonical states.

Show Proof

Assuming [math]M\subset N[/math] we have [math]l^2(M)\subset l^2(N)[/math], and for [math]m,m'\in M[/math] we have:

[[math]] \lambda_M(m)\delta_{m'}=\lambda_N(m)\delta_{m'} [[/math]]

Thus if we suppose [math]M(N-M)=N-M[/math], as in the statement, then we have:

[[math]] \begin{eqnarray*} \lambda_M(m)^*\delta_{m'} &=&\sum_{x\in M}\delta_{m',mx}\delta_x\\ &=&\sum_{x\in N}\delta_{m',mx}\delta_x\\ &=&\lambda_N(m)^*\delta_{m'} \end{eqnarray*} [[/math]]

In particular, if [math]m_1,\ldots,m_k\in M[/math], and [math]\alpha_1,\ldots,\alpha_k[/math] are exponents in [math] \{1,*\}[/math], then:

[[math]] \lambda_M(m_1)^{\alpha_1}\ldots\lambda_M(m_k)^{\alpha_k}\delta_e =\lambda_N(m_1)^{\alpha_1}\ldots\lambda_N(m_k)^{\alpha_k}\delta_e [[/math]]

Thus, we are led to the conclusion in the statement.

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Following , let us introduce the following technical notion:

Definition

Let [math]N[/math] be a semigroup. Consider the following order on it:

[[math]] a\preceq_Nb\iff b\in aN [[/math]]

We say that [math]N[/math] is in the class [math]E[/math] if it satisfies one of the following equivalent conditions:

• For [math]\preceq_N[/math] every bounded subset is totally ordered.
• [math]a\preceq c,b\preceq c\implies a\preceq b[/math] or [math]b\preceq a[/math].
• [math]aN\cap bN\neq\emptyset\implies aN\subset bN[/math] or [math]bN\subset aN[/math].
• [math]NN^{-1}\cap N^{-1}N=N\cup N^{-1}[/math].

Also by following , let us introduce as well the following notion, which is something standard in the combinatorial theory of semigroups:

Definition

Let [math](a_i)_{i\in I}[/math] be a family of elements in a semigroup [math]N[/math].

• We say that [math](a_i)_{i\in I}[/math] is a code if the semigroup [math]M\subset N[/math] generated by the [math]a_i[/math] is isomorphic to [math]\mathbb N^{*I}[/math], via [math]a_i\to e_i[/math], and satisfies [math]M(N-M)=N-M[/math].
• We say that [math](a_i)_{i\in I}[/math] is a prefix if [math]a_i\in a_jN\implies i=j[/math], which means that the elements [math]a_i[/math] are not comparable via the order relation [math]\preceq_N[/math].

In our probabilistic setting, the notion of code is of interest, due to:

Proposition

Assuming that [math](a_i,b_i)_{i\in I}[/math] is a code, the family

[[math]] \left(\frac{1}{2}(\lambda_N(a_i)+\lambda_N(b_i)^*)\right)_{i\in I} [[/math]]

is a circular family, in the sense of free probability theory.

Show Proof

Let [math](a_i,b_i)_{i\in I}[/math] be a code, and consider the following family:

[[math]] \Big(\lambda_N(a_i),\lambda_N(b_i)\Big)_{i\in I}\in B(l^2(N)) [[/math]]

By using Proposition 10.25, this family has the same distribution as a family of creation operators associated to a family of [math]2I[/math] orthonormal vectors, on the free Fock space:

[[math]] \Big(\lambda_{\mathbb N^{*I}}(e_i),\lambda_{\mathbb N^{*I}}(f_i)\Big)_{i\in I}\in B(l^2(N^{*I})) [[/math]]

Thus, we obtain the result, via the standard facts about the circular systems on free Fock spaces, that we know from chapter 9.

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In view of this, the following result provides us with a criterion for finding circular systems in the algebras of the semigroups in the class [math]E[/math], from Definition 10.26:

Proposition

For a semigroup [math]N\in E[/math], a family

[[math]] (a_i)_{i\in I}\subset N [[/math]]

having at least two elements is a prefix if and only if it is a code.

Show Proof

We have two implications to be proved, as follows:

(1) Let first [math](a_i)_{i\in I}[/math] be a code which is not a prefix, for instance because we have [math]a_i=a_jn[/math] with [math]i\neq j,n\in N[/math]. Then [math]n[/math] is in the semigroup [math]M[/math] generated by the [math]a_k[/math] and [math]a_i=a_jn[/math] with [math]i\neq j[/math], so [math]M[/math] cannot be free, and this is a contradiction, as desired.

(2) Conversely, suppose now that [math](a_i)_{i\in I}[/math] is a prefix and let, with [math]m\in N[/math]:

[[math]] A=a_{i_1}^{{\alpha}_1}\ldots a_{i_n}^{{\alpha}_n}m=a_{j_ 1}^{{\beta}_1}\ldots a_{j_s}^{ {\beta}_s} [[/math]]

We have then [math]a_{i_1}\preceq A[/math], [math]a _{j_1}\preceq A[/math], and so [math]i_1=j_1[/math]. We can therefore simplify [math]A[/math] to the left by [math]a_{i_1}[/math]. A reccurence on [math]\sum\alpha_ i[/math] shows then that we have [math]n\leq s[/math] and:

[[math]] a_{i_k}=a_{j_k}\quad,\quad\forall k\leq n [[/math]]

[[math]] {\alpha}_k={\beta}_k\quad,\quad\forall k \lt n [[/math]]

[[math]] {\alpha}_n\leq{\beta}_n [[/math]]

[[math]] m=a_{j_n}^{{\beta}_n-{\alpha}_n}a_{j_{n+1}}^{{\beta}_{n+1}}\ldots a_{j_s }^{{\beta}_s} [[/math]]

Finally, we know that [math]m[/math] is in the semigroup generated by the [math]a_i[/math], so we have a code. Moreover, for [math]m=e[/math] we obtain that we have [math]n=s[/math], [math]a_{j_k}=a_{i_k}[/math] and [math]{\alpha}_k={\beta}_k[/math] for any [math]k\leq n[/math]. Thus the variables [math]a_i[/math] freely generate the semigroup [math]M[/math], and so the family [math](a_i)_{i\in I}[/math] is a code. Thus, we are led to the conclusion in the statement.

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Summarizing, we have some good freeness results, for our semigroups. Before getting into applications, let us discuss now the examples. We have here the following result:

Proposition

The class [math]E[/math] has the following properties:

• All the groups are in [math]E[/math].
• The positive parts of totally ordered abelian groups are in [math]E[/math].
• If [math]G[/math] is a group and [math]M\in E[/math], then [math]M\times G\in E[/math].
• If [math]A_1[/math], [math]A_2[/math] are in [math]E[/math], then the free product [math]A_1*A_2[/math] is in [math]E[/math].

Show Proof

This is something elementary, whose proof goes as follows:

(1) This is obvious, coming from definitions.

(2) This is obvious as well, because [math]M[/math] is here totally ordered by [math]\preceq_M[/math].

(3) Let [math]G[/math] be a group and [math]M\in E[/math]. We have then, as desired:

[[math]] \begin{eqnarray*} &&(M\times G)(M\times G)^{-1}\cap (M\times G)^{-1}(M\times G)\\ &=&(M\times G)(M^{-1}\times G)\cap (M^{-1}\times G)(M\times G)\\ &=&(MM^{-1}\times G)\cap (M^{-1}M\times G)\\ &=&(MM^{-1}\cap M^{-1}M)\times G\\ &=&(M\cup M^{-1})\times G\\ &=&(M\times G)\cup (M^{-1}\times G)\\ &=&(M\times G)\cup (M\times G)^{-1} \end{eqnarray*} [[/math]]

(4) Let [math]a,b,c\in A_1*A_2[/math] such that [math]ab=c[/math]. We write, as reduced words:

[[math]] a=x_1\ldots x_n\quad,\quad b=y_1\ldots y_m\quad,\quad c=z_1\ldots z_p [[/math]]

Now let [math]s[/math] be such that the following equalities happen:

[[math]] x_ny_1=1\quad,\quad \ldots\quad,\quad x_{n-s+1}y_s=1\quad,\quad x_{n-s}y_{s+1}\neq 1 [[/math]]

Consider now the following element:

[[math]] u =x_{n-s+1}\ldots x_n =(y_1\ldots y_s)^{-1} [[/math]]

We have then the following computation:

[[math]] c =ab =x_1\ldots x_{n-s}y_{s+1}\ldots y_m [[/math]]

Now let [math]i\in\{ 1,2\}[/math] be such that [math]z_{n-s}\in A_i[/math]. There are two cases:

-- If [math]x_{n-s}\in A_1[/math] and [math]y_{s+1}\in A_2[/math] or if [math]x_{n-s}\in A_2[/math] and [math]y_{s+1}\in A_1[/math], then [math]x_1\ldots x_{n-s}y_{s+1}\ldots y_m[/math] is a reduced word. In particular, we have [math]x_1=z_1[/math], [math]x_2=z_2[/math], and so on up to [math]x_{n-s}=z_{n-s}[/math]. Thus we have [math]a=z_1\ldots z_{n-s}u[/math], with [math]u[/math] invertible.

-- If [math]x_{n-s},y_{s+1}\in A_i[/math] then [math]x_1=z_1[/math] and so on, up to [math]x_{n-s-1}=z_{n-s-1}[/math] and [math]x_{n-s}y_{s+1}=z_{n-s}[/math]. In this case we have [math]a=z_1\ldots z_{n-s-1}x_{n-s}u[/math], with [math]u[/math] invertible.

Now observe that in both cases we obtained that [math]a[/math] is of the form [math]z_1\ldots z_fxu[/math] for some [math]f[/math], with [math]u[/math] invertible and such that if [math]z_{f+1}\in A_i[/math], then there exists [math]y\in A_i[/math] such that:

[[math]] xy=z_{f+1} [[/math]]

Indeed, we can take [math]f=n-s-1[/math] and [math]x=z_{n-s},y=1[/math] in the first case, and [math]x=x_{n-s},y=y_{s+1}[/math] in the second one. Suppose now that [math]A_1,A_2\in E[/math] and let [math]a,b,a',b'\in A_1*A_2[/math] such that [math]ab=a'b'[/math]. Let [math]z_1\ldots z_p[/math] be the decomposition of [math]ab=a'b'[/math] as a reduced word. Then we can decompose our words, as above, in the following way:

[[math]] a=z_1\ldots z_fxu\quad,\quad a'=z_1\ldots z_{f'}x'u' [[/math]]

We have to show that [math]a=a'm[/math] or that [math]a'=am[/math] for some [math]m\in A_1*A_2[/math]. But this is clear in all three cases that can appear, namely [math]f \lt f'[/math], [math]f' \lt f[/math], [math]f=f'[/math].

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We can now formulate a main result about semigroup freeness, as follows:

Theorem

The following happen:

• Given [math]M\subset N[/math], both in the class [math]E[/math], satisfying [math]M(N-M)=N-M[/math], any [math]x[/math] in the [math]*[/math]-algebra generated by [math]\lambda (M)[/math] can be written as follows, with [math]p_i,q_i\in M[/math]:
  [[math]] x=\sum_ia_i\lambda_N(p_i)\lambda_N(q_i)^* [[/math]]
• Asssume [math]A,B\in E[/math], and let [math]x[/math] be an element of the [math]*[/math]-algebra generated by [math]\lambda_{A*B}(A)[/math] such that [math]\tau(x)=0[/math]. If [math]W_A,W_B[/math] are respectively the sets of reduced words beginning by an element of [math]A,B[/math], then [math]x[/math] acts as follows:
  [[math]] l^2(W_B\cup\{e\})\to l^2(W_A) [[/math]]
• Let [math]A,B\in E[/math]. Then [math]\lambda_{A*B}(A)[/math] and [math]\lambda_{A*B}(B)[/math] are free.

Show Proof

This follows from our results so far, the idea being is as follows:

(1) It is enough to prove this for elements of the form [math]x=\lambda(m)^*\lambda(n)[/math] with [math]m,n\in M[/math], because the general case will follow easily from this. In order to do so, observe that [math]x=\lambda(m)^*\lambda(n)[/math] is different from [math]0[/math] precisely when there exist [math]a,b\in N[/math] such that:

[[math]] \lt \lambda(m)^*\lambda(n)\delta_a,\delta_b \gt \neq 0 [[/math]]

That is, the following condition must be satisfied:

[[math]] na=mb [[/math]]

We know that there exists [math]c\in N[/math] with [math]n=mc[/math] or with [math]m=nc[/math]. Moreover, as [math]M(N-M)=N-M[/math], it follows that [math]c\in M[/math]. Thus [math]x=\lambda(m)^*\lambda (n)\neq 0[/math] implies that [math]x=\lambda(c)[/math] or [math]x=\lambda(c)^*[/math] with [math]c\in M[/math], and this finishes the proof.

(2) We apply (1) with [math]M=A[/math] and [math]N=A*B[/math] for writing, with [math]p_i,q_i\in A[/math]:

[[math]] x=\sum_ia_i\lambda(p_i)\lambda (q_i)^* [[/math]]

Consider now the following element:

[[math]] \tau (\lambda(p_i)\lambda(q_i)^*)=\sum_x\delta_{e,p_ix}\delta_{e,q_ix} [[/math]]

This element is nonzero precisely when [math]p_i=q_i[/math] is invertible, and in this case:

[[math]] \lambda(p_i)\lambda(q_i)^*=1 [[/math]]

Now since we assumed [math]\tau (x)=0[/math], it follows that we can write:

[[math]] x=\sum a_i\lambda(p_i)\lambda (q_i)^*\quad,\quad \tau(\lambda(p_i)\lambda (q_i)^*)=0 [[/math]]

By linearity, it is enough to prove the result for [math]x=\lambda(p_i)\lambda (q_i)^*[/math]. Let [math]m\in W_B\cup\{ e\}[/math] and suppose that [math]x\delta_m\neq 0[/math]. Then [math]\lambda(q_i)^*\delta_m\neq0[/math] implies that [math]m=q_ic[/math] for some word [math]c\in A*B[/math]. As [math]q_i\in A[/math] and [math]m\in W_B\cup\{ e\}[/math], it follows that [math]q_i[/math] is invertible. Now observe that:

[[math]] p_iq_i^{-1}=1\implies\tau (x)=1 [[/math]]

It follows that we have, as desired:

[[math]] x\delta_m=\delta_{p_iq_i^{-1}m}\in l^2(W_A) [[/math]]

(3) This follows from (2) above. Indeed, let [math]P=x_n\ldots x_1[/math] be a product of elements in [math]ker(\tau)[/math], such that [math]x_{2k}[/math] is in the [math]*[/math]-algebra generated by [math]\lambda (B)[/math] and [math]x_{2k+1}[/math] is in the [math]*[/math]-algebra generated by [math]\lambda(A)[/math]. Then [math]x_1\delta_e\in l^2(W_A)[/math]. Thus [math]x_2x_1\delta_e\in l^2(W_B)[/math], and so on. By a reccurence, [math]P\delta_e[/math] is in [math]l^2(W_A)[/math] or in [math]l^2(W_B)[/math]. But this implies that [math]\tau (P)=0[/math], as desired.

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As a main application of the above semigroup technology, we have:

Theorem

Consider a Haar unitary [math]u[/math], free from a semicircular [math]s[/math]. Then

[[math]] c=us [[/math]]

is a circular variable.

Show Proof

Denote by [math]z[/math] the image of [math]1\in \mathbb Z[/math] and by [math]n[/math] the image of [math]1\in \mathbb N[/math] by the canonical embeddings into the free product [math]\mathbb Z *\mathbb N[/math]. Let [math]\lambda ={\lambda} _{\mathbb Z *\mathbb N}[/math]. We know that [math]\mathbb Z *\mathbb N\in E[/math]. Also [math](zn,nz^{-1})[/math] is obviously a prefix, so it is a code. Thus, the following variable is circular:

[[math]] c=\frac{1}{2}(\lambda (zn)+\lambda (nz^{-1})^*) [[/math]]

The point now is that we have the following formula:

[[math]] \frac{1}{2}(\lambda (zn)+\lambda (nz^{-1})^*)=us [[/math]]

But this gives the result, in our model and so in general as well, because [math]u={\lambda}(z)[/math] is a Haar-unitary, [math]s=1/2(\lambda (n)+\lambda (n)^*)[/math] is semicircular, and [math]u[/math] and [math]s[/math] are free.

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We can now recover the Voiculescu polar decomposition result for the circular variables, obtained in , by using random matrix techniques, as follows:

Theorem

Consider the polar decomposition of a circular variable, in some von Neumann algebraic probability space with faithful normal state:

[[math]] x=vb [[/math]]

Then [math]v[/math] is Haar unitary, [math]b[/math] is quarter-circular, and [math](v,b)[/math] are free.

Show Proof

This follows by suitably manipulating Theorem 10.32, as to replace the semicircular element there by a quarter-circular. Consider indeed the following group:

[[math]] G=\mathbb Z*(\mathbb Z\times\mathbb Z /2\mathbb Z) [[/math]]

Let [math]z,t,a[/math] be the images of the following elements, into this group [math]G[/math]:

[[math]] 1\in\mathbb Z\quad,\quad (1,\hat{0})\in\mathbb Z\times(\mathbb Z /2\mathbb Z)\quad,\quad (0,\hat{1})\in\mathbb Z\times(\mathbb Z /2\mathbb Z) [[/math]]

Let [math]u=\lambda_G (z)[/math], [math]d=\lambda_G (a)[/math] and choose a quarter-circular [math]q\in C^*(\lambda_G(t))[/math]. Then [math](q,d)[/math] are independent, so [math]dq[/math] is semicircular, and so [math]c=udq[/math] is circular, and:

-- The module of [math]c[/math] is [math]q[/math], which is a quarter-circular.

-- The polar part of [math]c[/math] is [math]ud[/math], which is obviously a Haar unitary.

-- Consider the automorphism of [math]G[/math] which is the identity on [math]\mathbb Z\times\mathbb Z /2\mathbb Z [/math] and maps [math]z\to za[/math]. This extends to a trace-preserving automorphism of [math]C^*(G)[/math] which maps:

[[math]] u\to ud\quad,\quad q\to q [[/math]]

Since [math]u,q[/math] are free, it follows that [math]ud,q[/math] are free too, finishing the proof.

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General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].