Finding a Unitary to Implement the action in a Minimal Dynamical System












1












$begingroup$


Let $X$ be an infinite compact Hausdorff space and let $sigmacolon Xto X$ be a minimal homeomorphism thereof. Then $sigma$ gives rise to an automorphism $sigma'$ of $C(X)$ defined by $sigma'(f):=fcircsigma^{-1}$. While reading a paper, I then come across a line that says that there is a unitary $u$ such that $ufu^{*}=sigma'(f)$ for all $fin C(X)$.




Is $uin C(X)$? And how does one obtain such a unitary?







dynamical-systems operator-algebras c-star-algebras







share|cite|improve this question











$endgroup$












  • $begingroup$
    A unitary operator is a map $u colon C(X) to C(X)$, so no, $u$ is not in $C(X)$. What does $sigma(f)$ mean here? $sigma$ is a map $X to X$ and $f$ is a map $X to mathbb{C}$, so it's not clear what this composition could mean.
    $endgroup$
    – Dan Rust
    Jan 8 at 14:23










  • $begingroup$
    If $uin C(X)$, then $ufu^ast=uu^ast f=f$. So no, you usually won't find such a $uin C(X)$. Representing $sigma$ by a unitary is exactly what the crossed product construction does, but without context, it's hard to say more.
    $endgroup$
    – MaoWao
    Jan 8 at 14:24












  • $begingroup$
    @Dan Rust: Thank you for pointing out the typos; I have fixed them.
    $endgroup$
    – ervx
    Jan 8 at 14:40






  • 1




    $begingroup$
    Depending on the context, you might be implicitly dealing with a representation of $C(X)$ on some Hilbert space $H$, and $u$ is a unitary in $B(H)$.
    $endgroup$
    – Aweygan
    Jan 8 at 14:41
















1












$begingroup$


Let $X$ be an infinite compact Hausdorff space and let $sigmacolon Xto X$ be a minimal homeomorphism thereof. Then $sigma$ gives rise to an automorphism $sigma'$ of $C(X)$ defined by $sigma'(f):=fcircsigma^{-1}$. While reading a paper, I then come across a line that says that there is a unitary $u$ such that $ufu^{*}=sigma'(f)$ for all $fin C(X)$.




Is $uin C(X)$? And how does one obtain such a unitary?







dynamical-systems operator-algebras c-star-algebras







share|cite|improve this question











$endgroup$












  • $begingroup$
    A unitary operator is a map $u colon C(X) to C(X)$, so no, $u$ is not in $C(X)$. What does $sigma(f)$ mean here? $sigma$ is a map $X to X$ and $f$ is a map $X to mathbb{C}$, so it's not clear what this composition could mean.
    $endgroup$
    – Dan Rust
    Jan 8 at 14:23










  • $begingroup$
    If $uin C(X)$, then $ufu^ast=uu^ast f=f$. So no, you usually won't find such a $uin C(X)$. Representing $sigma$ by a unitary is exactly what the crossed product construction does, but without context, it's hard to say more.
    $endgroup$
    – MaoWao
    Jan 8 at 14:24












  • $begingroup$
    @Dan Rust: Thank you for pointing out the typos; I have fixed them.
    $endgroup$
    – ervx
    Jan 8 at 14:40






  • 1




    $begingroup$
    Depending on the context, you might be implicitly dealing with a representation of $C(X)$ on some Hilbert space $H$, and $u$ is a unitary in $B(H)$.
    $endgroup$
    – Aweygan
    Jan 8 at 14:41














1












1








1





$begingroup$


Let $X$ be an infinite compact Hausdorff space and let $sigmacolon Xto X$ be a minimal homeomorphism thereof. Then $sigma$ gives rise to an automorphism $sigma'$ of $C(X)$ defined by $sigma'(f):=fcircsigma^{-1}$. While reading a paper, I then come across a line that says that there is a unitary $u$ such that $ufu^{*}=sigma'(f)$ for all $fin C(X)$.




Is $uin C(X)$? And how does one obtain such a unitary?







dynamical-systems operator-algebras c-star-algebras







share|cite|improve this question











$endgroup$




Let $X$ be an infinite compact Hausdorff space and let $sigmacolon Xto X$ be a minimal homeomorphism thereof. Then $sigma$ gives rise to an automorphism $sigma'$ of $C(X)$ defined by $sigma'(f):=fcircsigma^{-1}$. While reading a paper, I then come across a line that says that there is a unitary $u$ such that $ufu^{*}=sigma'(f)$ for all $fin C(X)$.




Is $uin C(X)$? And how does one obtain such a unitary?







dynamical-systems operator-algebras c-star-algebras




dynamical-systems operator-algebras c-star-algebras






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 14:36







ervx

















asked Jan 8 at 14:10









ervxervx

10.3k31438




10.3k31438












  • $begingroup$
    A unitary operator is a map $u colon C(X) to C(X)$, so no, $u$ is not in $C(X)$. What does $sigma(f)$ mean here? $sigma$ is a map $X to X$ and $f$ is a map $X to mathbb{C}$, so it's not clear what this composition could mean.
    $endgroup$
    – Dan Rust
    Jan 8 at 14:23










  • $begingroup$
    If $uin C(X)$, then $ufu^ast=uu^ast f=f$. So no, you usually won't find such a $uin C(X)$. Representing $sigma$ by a unitary is exactly what the crossed product construction does, but without context, it's hard to say more.
    $endgroup$
    – MaoWao
    Jan 8 at 14:24












  • $begingroup$
    @Dan Rust: Thank you for pointing out the typos; I have fixed them.
    $endgroup$
    – ervx
    Jan 8 at 14:40






  • 1




    $begingroup$
    Depending on the context, you might be implicitly dealing with a representation of $C(X)$ on some Hilbert space $H$, and $u$ is a unitary in $B(H)$.
    $endgroup$
    – Aweygan
    Jan 8 at 14:41


















  • $begingroup$
    A unitary operator is a map $u colon C(X) to C(X)$, so no, $u$ is not in $C(X)$. What does $sigma(f)$ mean here? $sigma$ is a map $X to X$ and $f$ is a map $X to mathbb{C}$, so it's not clear what this composition could mean.
    $endgroup$
    – Dan Rust
    Jan 8 at 14:23










  • $begingroup$
    If $uin C(X)$, then $ufu^ast=uu^ast f=f$. So no, you usually won't find such a $uin C(X)$. Representing $sigma$ by a unitary is exactly what the crossed product construction does, but without context, it's hard to say more.
    $endgroup$
    – MaoWao
    Jan 8 at 14:24












  • $begingroup$
    @Dan Rust: Thank you for pointing out the typos; I have fixed them.
    $endgroup$
    – ervx
    Jan 8 at 14:40






  • 1




    $begingroup$
    Depending on the context, you might be implicitly dealing with a representation of $C(X)$ on some Hilbert space $H$, and $u$ is a unitary in $B(H)$.
    $endgroup$
    – Aweygan
    Jan 8 at 14:41
















$begingroup$
A unitary operator is a map $u colon C(X) to C(X)$, so no, $u$ is not in $C(X)$. What does $sigma(f)$ mean here? $sigma$ is a map $X to X$ and $f$ is a map $X to mathbb{C}$, so it's not clear what this composition could mean.
$endgroup$
– Dan Rust
Jan 8 at 14:23




$begingroup$
A unitary operator is a map $u colon C(X) to C(X)$, so no, $u$ is not in $C(X)$. What does $sigma(f)$ mean here? $sigma$ is a map $X to X$ and $f$ is a map $X to mathbb{C}$, so it's not clear what this composition could mean.
$endgroup$
– Dan Rust
Jan 8 at 14:23












$begingroup$
If $uin C(X)$, then $ufu^ast=uu^ast f=f$. So no, you usually won't find such a $uin C(X)$. Representing $sigma$ by a unitary is exactly what the crossed product construction does, but without context, it's hard to say more.
$endgroup$
– MaoWao
Jan 8 at 14:24






$begingroup$
If $uin C(X)$, then $ufu^ast=uu^ast f=f$. So no, you usually won't find such a $uin C(X)$. Representing $sigma$ by a unitary is exactly what the crossed product construction does, but without context, it's hard to say more.
$endgroup$
– MaoWao
Jan 8 at 14:24














$begingroup$
@Dan Rust: Thank you for pointing out the typos; I have fixed them.
$endgroup$
– ervx
Jan 8 at 14:40




$begingroup$
@Dan Rust: Thank you for pointing out the typos; I have fixed them.
$endgroup$
– ervx
Jan 8 at 14:40




1




1




$begingroup$
Depending on the context, you might be implicitly dealing with a representation of $C(X)$ on some Hilbert space $H$, and $u$ is a unitary in $B(H)$.
$endgroup$
– Aweygan
Jan 8 at 14:41




$begingroup$
Depending on the context, you might be implicitly dealing with a representation of $C(X)$ on some Hilbert space $H$, and $u$ is a unitary in $B(H)$.
$endgroup$
– Aweygan
Jan 8 at 14:41










1 Answer
1






active

oldest

votes


















1












$begingroup$

The crossed product is a construction that implements a group action on a $C^ast$-algebra as an action by unitaries. Similar to group $C^ast$-algebras, there are two variants, the reduced and the full (universal) crossed product.



Let $A$ be a $C^ast$-algebra, $G$ a (discrete) group and $alphacolon Glongrightarrow mathrm{Aut}(A)$ a group homomorphism. The full crossed product $Artimes_alpha G$ is the universal $C^ast$-algebra generated by $A$ and $C^ast(G)$ subject to the relation $u_g a u_g^ast=alpha_g(a)$.



The reduced crossed product is constructed as follows. Assume for simplicity that $Asubset B(H)$. Define representations of $A$ and $G$ on $Hotimesell^2(G)$ by
begin{align*}
&picolon Alongrightarrow B(Hotimes ell^2(G)),,pi(a)(xiotimes delta_g)=alpha_{g^{-1}}(a)xiotimesdelta_g\
&lambdacolon Glongrightarrow B(Hotimesell^2(G)),,lambda(h)(xiotimesdelta_g)=xiotimes delta_{hg}
end{align*}
The reduced crossed product $Artimes_{alpha,mathrm{r}}G$ is the $C^ast$-algebra generated by $pi(A)$ and $lambda(G)$.



The elements $pi(a)$ and $lambda(g)$ satisfiy the relation $lambda(g)pi(a)lambda(g)^ast=pi(alpha_{g}(a))$. Thus there exists a (surjective) $ast$-homomorphism $Lambdacolon Artimes_alpha Glongrightarrow Artimes_{alpha,mathrm{r}} G$ such that $Lambda(a)=pi(a)$ and $Lambda(u_g)=lambda(g)$.



In your example, $A=C(X)$, $G=mathbb{Z}$, $alpha(n)=(sigma')^n$, and $C^ast(mathbb{Z})$ is the universal $C^ast$-algebra generated by a unitary $u_1$. Since $mathbb{Z}$ is amenable, the map $Lambda$ is injective, that is, the full and reduced crossed product coincide.



What is even more, if the homeomorphism $sigma$ is not only minimal, but also free, then the crossed product $Artimes G$ is simple. Thus whenever you have representation $pi'$ of $C(X)$ on a Hilbert space $K$ and a unitary $uin B(K)$ such that $upi'(f)u^ast=pi'(sigma'(f))$, there exists a unique $ast$-isomorphism from $C(X)rtimes mathbb{Z}$ onto the $C^ast$-algebra generated by $pi'(C(X))$ and $u$ that is the identity on $C(X)$ and maps $u_1$ to $u$.



All these statements should be contained in Dana Williams's book on crossed products of $C^ast$-algebras.






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    1












    $begingroup$

    The crossed product is a construction that implements a group action on a $C^ast$-algebra as an action by unitaries. Similar to group $C^ast$-algebras, there are two variants, the reduced and the full (universal) crossed product.



    Let $A$ be a $C^ast$-algebra, $G$ a (discrete) group and $alphacolon Glongrightarrow mathrm{Aut}(A)$ a group homomorphism. The full crossed product $Artimes_alpha G$ is the universal $C^ast$-algebra generated by $A$ and $C^ast(G)$ subject to the relation $u_g a u_g^ast=alpha_g(a)$.



    The reduced crossed product is constructed as follows. Assume for simplicity that $Asubset B(H)$. Define representations of $A$ and $G$ on $Hotimesell^2(G)$ by
    begin{align*}
    &picolon Alongrightarrow B(Hotimes ell^2(G)),,pi(a)(xiotimes delta_g)=alpha_{g^{-1}}(a)xiotimesdelta_g\
    &lambdacolon Glongrightarrow B(Hotimesell^2(G)),,lambda(h)(xiotimesdelta_g)=xiotimes delta_{hg}
    end{align*}
    The reduced crossed product $Artimes_{alpha,mathrm{r}}G$ is the $C^ast$-algebra generated by $pi(A)$ and $lambda(G)$.



    The elements $pi(a)$ and $lambda(g)$ satisfiy the relation $lambda(g)pi(a)lambda(g)^ast=pi(alpha_{g}(a))$. Thus there exists a (surjective) $ast$-homomorphism $Lambdacolon Artimes_alpha Glongrightarrow Artimes_{alpha,mathrm{r}} G$ such that $Lambda(a)=pi(a)$ and $Lambda(u_g)=lambda(g)$.



    In your example, $A=C(X)$, $G=mathbb{Z}$, $alpha(n)=(sigma')^n$, and $C^ast(mathbb{Z})$ is the universal $C^ast$-algebra generated by a unitary $u_1$. Since $mathbb{Z}$ is amenable, the map $Lambda$ is injective, that is, the full and reduced crossed product coincide.



    What is even more, if the homeomorphism $sigma$ is not only minimal, but also free, then the crossed product $Artimes G$ is simple. Thus whenever you have representation $pi'$ of $C(X)$ on a Hilbert space $K$ and a unitary $uin B(K)$ such that $upi'(f)u^ast=pi'(sigma'(f))$, there exists a unique $ast$-isomorphism from $C(X)rtimes mathbb{Z}$ onto the $C^ast$-algebra generated by $pi'(C(X))$ and $u$ that is the identity on $C(X)$ and maps $u_1$ to $u$.



    All these statements should be contained in Dana Williams's book on crossed products of $C^ast$-algebras.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The crossed product is a construction that implements a group action on a $C^ast$-algebra as an action by unitaries. Similar to group $C^ast$-algebras, there are two variants, the reduced and the full (universal) crossed product.



      Let $A$ be a $C^ast$-algebra, $G$ a (discrete) group and $alphacolon Glongrightarrow mathrm{Aut}(A)$ a group homomorphism. The full crossed product $Artimes_alpha G$ is the universal $C^ast$-algebra generated by $A$ and $C^ast(G)$ subject to the relation $u_g a u_g^ast=alpha_g(a)$.



      The reduced crossed product is constructed as follows. Assume for simplicity that $Asubset B(H)$. Define representations of $A$ and $G$ on $Hotimesell^2(G)$ by
      begin{align*}
      &picolon Alongrightarrow B(Hotimes ell^2(G)),,pi(a)(xiotimes delta_g)=alpha_{g^{-1}}(a)xiotimesdelta_g\
      &lambdacolon Glongrightarrow B(Hotimesell^2(G)),,lambda(h)(xiotimesdelta_g)=xiotimes delta_{hg}
      end{align*}
      The reduced crossed product $Artimes_{alpha,mathrm{r}}G$ is the $C^ast$-algebra generated by $pi(A)$ and $lambda(G)$.



      The elements $pi(a)$ and $lambda(g)$ satisfiy the relation $lambda(g)pi(a)lambda(g)^ast=pi(alpha_{g}(a))$. Thus there exists a (surjective) $ast$-homomorphism $Lambdacolon Artimes_alpha Glongrightarrow Artimes_{alpha,mathrm{r}} G$ such that $Lambda(a)=pi(a)$ and $Lambda(u_g)=lambda(g)$.



      In your example, $A=C(X)$, $G=mathbb{Z}$, $alpha(n)=(sigma')^n$, and $C^ast(mathbb{Z})$ is the universal $C^ast$-algebra generated by a unitary $u_1$. Since $mathbb{Z}$ is amenable, the map $Lambda$ is injective, that is, the full and reduced crossed product coincide.



      What is even more, if the homeomorphism $sigma$ is not only minimal, but also free, then the crossed product $Artimes G$ is simple. Thus whenever you have representation $pi'$ of $C(X)$ on a Hilbert space $K$ and a unitary $uin B(K)$ such that $upi'(f)u^ast=pi'(sigma'(f))$, there exists a unique $ast$-isomorphism from $C(X)rtimes mathbb{Z}$ onto the $C^ast$-algebra generated by $pi'(C(X))$ and $u$ that is the identity on $C(X)$ and maps $u_1$ to $u$.



      All these statements should be contained in Dana Williams's book on crossed products of $C^ast$-algebras.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The crossed product is a construction that implements a group action on a $C^ast$-algebra as an action by unitaries. Similar to group $C^ast$-algebras, there are two variants, the reduced and the full (universal) crossed product.



        Let $A$ be a $C^ast$-algebra, $G$ a (discrete) group and $alphacolon Glongrightarrow mathrm{Aut}(A)$ a group homomorphism. The full crossed product $Artimes_alpha G$ is the universal $C^ast$-algebra generated by $A$ and $C^ast(G)$ subject to the relation $u_g a u_g^ast=alpha_g(a)$.



        The reduced crossed product is constructed as follows. Assume for simplicity that $Asubset B(H)$. Define representations of $A$ and $G$ on $Hotimesell^2(G)$ by
        begin{align*}
        &picolon Alongrightarrow B(Hotimes ell^2(G)),,pi(a)(xiotimes delta_g)=alpha_{g^{-1}}(a)xiotimesdelta_g\
        &lambdacolon Glongrightarrow B(Hotimesell^2(G)),,lambda(h)(xiotimesdelta_g)=xiotimes delta_{hg}
        end{align*}
        The reduced crossed product $Artimes_{alpha,mathrm{r}}G$ is the $C^ast$-algebra generated by $pi(A)$ and $lambda(G)$.



        The elements $pi(a)$ and $lambda(g)$ satisfiy the relation $lambda(g)pi(a)lambda(g)^ast=pi(alpha_{g}(a))$. Thus there exists a (surjective) $ast$-homomorphism $Lambdacolon Artimes_alpha Glongrightarrow Artimes_{alpha,mathrm{r}} G$ such that $Lambda(a)=pi(a)$ and $Lambda(u_g)=lambda(g)$.



        In your example, $A=C(X)$, $G=mathbb{Z}$, $alpha(n)=(sigma')^n$, and $C^ast(mathbb{Z})$ is the universal $C^ast$-algebra generated by a unitary $u_1$. Since $mathbb{Z}$ is amenable, the map $Lambda$ is injective, that is, the full and reduced crossed product coincide.



        What is even more, if the homeomorphism $sigma$ is not only minimal, but also free, then the crossed product $Artimes G$ is simple. Thus whenever you have representation $pi'$ of $C(X)$ on a Hilbert space $K$ and a unitary $uin B(K)$ such that $upi'(f)u^ast=pi'(sigma'(f))$, there exists a unique $ast$-isomorphism from $C(X)rtimes mathbb{Z}$ onto the $C^ast$-algebra generated by $pi'(C(X))$ and $u$ that is the identity on $C(X)$ and maps $u_1$ to $u$.



        All these statements should be contained in Dana Williams's book on crossed products of $C^ast$-algebras.






        share|cite|improve this answer









        $endgroup$



        The crossed product is a construction that implements a group action on a $C^ast$-algebra as an action by unitaries. Similar to group $C^ast$-algebras, there are two variants, the reduced and the full (universal) crossed product.



        Let $A$ be a $C^ast$-algebra, $G$ a (discrete) group and $alphacolon Glongrightarrow mathrm{Aut}(A)$ a group homomorphism. The full crossed product $Artimes_alpha G$ is the universal $C^ast$-algebra generated by $A$ and $C^ast(G)$ subject to the relation $u_g a u_g^ast=alpha_g(a)$.



        The reduced crossed product is constructed as follows. Assume for simplicity that $Asubset B(H)$. Define representations of $A$ and $G$ on $Hotimesell^2(G)$ by
        begin{align*}
        &picolon Alongrightarrow B(Hotimes ell^2(G)),,pi(a)(xiotimes delta_g)=alpha_{g^{-1}}(a)xiotimesdelta_g\
        &lambdacolon Glongrightarrow B(Hotimesell^2(G)),,lambda(h)(xiotimesdelta_g)=xiotimes delta_{hg}
        end{align*}
        The reduced crossed product $Artimes_{alpha,mathrm{r}}G$ is the $C^ast$-algebra generated by $pi(A)$ and $lambda(G)$.



        The elements $pi(a)$ and $lambda(g)$ satisfiy the relation $lambda(g)pi(a)lambda(g)^ast=pi(alpha_{g}(a))$. Thus there exists a (surjective) $ast$-homomorphism $Lambdacolon Artimes_alpha Glongrightarrow Artimes_{alpha,mathrm{r}} G$ such that $Lambda(a)=pi(a)$ and $Lambda(u_g)=lambda(g)$.



        In your example, $A=C(X)$, $G=mathbb{Z}$, $alpha(n)=(sigma')^n$, and $C^ast(mathbb{Z})$ is the universal $C^ast$-algebra generated by a unitary $u_1$. Since $mathbb{Z}$ is amenable, the map $Lambda$ is injective, that is, the full and reduced crossed product coincide.



        What is even more, if the homeomorphism $sigma$ is not only minimal, but also free, then the crossed product $Artimes G$ is simple. Thus whenever you have representation $pi'$ of $C(X)$ on a Hilbert space $K$ and a unitary $uin B(K)$ such that $upi'(f)u^ast=pi'(sigma'(f))$, there exists a unique $ast$-isomorphism from $C(X)rtimes mathbb{Z}$ onto the $C^ast$-algebra generated by $pi'(C(X))$ and $u$ that is the identity on $C(X)$ and maps $u_1$ to $u$.



        All these statements should be contained in Dana Williams's book on crossed products of $C^ast$-algebras.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 15:12









        MaoWaoMaoWao

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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            Gyllenstierna

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