Finding a Unitary to Implement the action in a Minimal Dynamical System
$begingroup$
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
asked Jan 8 at 14:10
ervxervx
10.3k31338
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 14:10
ervxervx
10.3k31338
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 8 at 14:10
ervxervx
10.3k31338
$endgroup$
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
asked Jan 8 at 14:10
ervxervx
10.3k31338
asked Jan 8 at 14:10
ervxervx
10.3k31338
edited Jan 8 at 14:36
ervx
asked Jan 8 at 14:10
ervxervx
10.3k31338
asked Jan 8 at 14:10
ervxervx
10.3k31338
asked Jan 8 at 14:10
ervxervx
10.3k31338
10.3k31338
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered yesterday
MaoWaoMaoWao
2,543617
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add a comment |
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$begingroup$
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.
answered yesterday
MaoWaoMaoWao
2,543617
$endgroup$
add a comment |
$begingroup$
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.
answered yesterday
MaoWaoMaoWao
2,543617
$endgroup$
add a comment |
$begingroup$
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.
answered yesterday
MaoWaoMaoWao
2,543617
$endgroup$
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.
answered yesterday
MaoWaoMaoWao
2,543617
answered yesterday
MaoWaoMaoWao
2,543617
answered yesterday
MaoWaoMaoWao
2,543617
answered yesterday
MaoWaoMaoWao
2,543617
2,543617
add a comment |
add a comment |
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$begingroup$
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