Problems on exercise 7.G in the book “K-Theory and C*-Algebras”
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I have a lot problems on exercise 7.G in the book K-Theory and C*-Algebras by Wegge-Olsen.
$newcommand{C}{mathbb{C}}$
$Xsubset mathbb{C}$? As I know the character space of $C^*(u_1,u_2)$ is homeomorphic to a subset of $C^2$: ${(tau(u_1),tau(u_2))|tau mbox{ is a character of } C^*(u_1,u_2)}$.
The example for a standard unitary should be $tmapsto exp(frac{2pi it}{1+|t|})$ as my tutor points out.
When $A$ is unital, $(SA)^sim={fin C(mathbb{T}to A)|f(1)inC}$. Why does $u_1:=1otimes u in M_n((SA)^sim)$?
Most important, what does the author want to tell us?
tensor-products c-star-algebras k-theory
asked Nov 30 '18 at 10:11
C.DingC.Ding
1,3931321
$endgroup$
add a comment |
$begingroup$
I have a lot problems on exercise 7.G in the book K-Theory and C*-Algebras by Wegge-Olsen.
$newcommand{C}{mathbb{C}}$
$Xsubset mathbb{C}$? As I know the character space of $C^*(u_1,u_2)$ is homeomorphic to a subset of $C^2$: ${(tau(u_1),tau(u_2))|tau mbox{ is a character of } C^*(u_1,u_2)}$.
The example for a standard unitary should be $tmapsto exp(frac{2pi it}{1+|t|})$ as my tutor points out.
When $A$ is unital, $(SA)^sim={fin C(mathbb{T}to A)|f(1)inC}$. Why does $u_1:=1otimes u in M_n((SA)^sim)$?
Most important, what does the author want to tell us?
tensor-products c-star-algebras k-theory
asked Nov 30 '18 at 10:11
C.DingC.Ding
1,3931321
$endgroup$
$begingroup$
Ad 1.) Indeed, the C*-algebra generated by two commuting unitaries is $C(mathbb T^2)$, where $mathbb T^2 = S^1 times S^1$.
$endgroup$
– André S.
Dec 1 '18 at 15:09
$begingroup$
@AndréS.: sometimes, but not always. If you take $u_1=u_2$, you will get $mathbb T$ or a subset of it. And even if $u_1$ and $U_2$ are free, they may still have discrete spectrum.
$endgroup$
– Martin Argerami
Dec 1 '18 at 17:00
$begingroup$
Right. So then I mean: Is a quotient of.
$endgroup$
– André S.
Dec 2 '18 at 8:55
add a comment |
$begingroup$
I have a lot problems on exercise 7.G in the book K-Theory and C*-Algebras by Wegge-Olsen.
$newcommand{C}{mathbb{C}}$
$Xsubset mathbb{C}$? As I know the character space of $C^*(u_1,u_2)$ is homeomorphic to a subset of $C^2$: ${(tau(u_1),tau(u_2))|tau mbox{ is a character of } C^*(u_1,u_2)}$.
The example for a standard unitary should be $tmapsto exp(frac{2pi it}{1+|t|})$ as my tutor points out.
When $A$ is unital, $(SA)^sim={fin C(mathbb{T}to A)|f(1)inC}$. Why does $u_1:=1otimes u in M_n((SA)^sim)$?
Most important, what does the author want to tell us?
tensor-products c-star-algebras k-theory
asked Nov 30 '18 at 10:11
C.DingC.Ding
1,3931321
$endgroup$
I have a lot problems on exercise 7.G in the book K-Theory and C*-Algebras by Wegge-Olsen.
$newcommand{C}{mathbb{C}}$
$Xsubset mathbb{C}$? As I know the character space of $C^*(u_1,u_2)$ is homeomorphic to a subset of $C^2$: ${(tau(u_1),tau(u_2))|tau mbox{ is a character of } C^*(u_1,u_2)}$.
The example for a standard unitary should be $tmapsto exp(frac{2pi it}{1+|t|})$ as my tutor points out.
When $A$ is unital, $(SA)^sim={fin C(mathbb{T}to A)|f(1)inC}$. Why does $u_1:=1otimes u in M_n((SA)^sim)$?
Most important, what does the author want to tell us?
tensor-products c-star-algebras k-theory
tensor-products c-star-algebras k-theory
asked Nov 30 '18 at 10:11
C.DingC.Ding
1,3931321
asked Nov 30 '18 at 10:11
C.DingC.Ding
1,3931321
edited Nov 30 '18 at 11:50
C.Ding
asked Nov 30 '18 at 10:11
C.DingC.Ding
1,3931321
asked Nov 30 '18 at 10:11
C.DingC.Ding
1,3931321
asked Nov 30 '18 at 10:11
C.DingC.Ding
1,3931321
1,3931321
$begingroup$
Ad 1.) Indeed, the C*-algebra generated by two commuting unitaries is $C(mathbb T^2)$, where $mathbb T^2 = S^1 times S^1$.
$endgroup$
– André S.
Dec 1 '18 at 15:09
$begingroup$
@AndréS.: sometimes, but not always. If you take $u_1=u_2$, you will get $mathbb T$ or a subset of it. And even if $u_1$ and $U_2$ are free, they may still have discrete spectrum.
$endgroup$
– Martin Argerami
Dec 1 '18 at 17:00
$begingroup$
Right. So then I mean: Is a quotient of.
$endgroup$
– André S.
Dec 2 '18 at 8:55
add a comment |
$begingroup$
Ad 1.) Indeed, the C*-algebra generated by two commuting unitaries is $C(mathbb T^2)$, where $mathbb T^2 = S^1 times S^1$.
$endgroup$
– André S.
Dec 1 '18 at 15:09
$begingroup$
@AndréS.: sometimes, but not always. If you take $u_1=u_2$, you will get $mathbb T$ or a subset of it. And even if $u_1$ and $U_2$ are free, they may still have discrete spectrum.
$endgroup$
– Martin Argerami
Dec 1 '18 at 17:00
$begingroup$
Right. So then I mean: Is a quotient of.
$endgroup$
– André S.
Dec 2 '18 at 8:55
$begingroup$
Ad 1.) Indeed, the C*-algebra generated by two commuting unitaries is $C(mathbb T^2)$, where $mathbb T^2 = S^1 times S^1$.
$endgroup$
– André S.
Dec 1 '18 at 15:09
$begingroup$
Ad 1.) Indeed, the C*-algebra generated by two commuting unitaries is $C(mathbb T^2)$, where $mathbb T^2 = S^1 times S^1$.
$endgroup$
– André S.
Dec 1 '18 at 15:09
$begingroup$
@AndréS.: sometimes, but not always. If you take $u_1=u_2$, you will get $mathbb T$ or a subset of it. And even if $u_1$ and $U_2$ are free, they may still have discrete spectrum.
$endgroup$
– Martin Argerami
Dec 1 '18 at 17:00
$begingroup$
@AndréS.: sometimes, but not always. If you take $u_1=u_2$, you will get $mathbb T$ or a subset of it. And even if $u_1$ and $U_2$ are free, they may still have discrete spectrum.
$endgroup$
– Martin Argerami
Dec 1 '18 at 17:00
$begingroup$
Right. So then I mean: Is a quotient of.
$endgroup$
– André S.
Dec 2 '18 at 8:55
$begingroup$
Right. So then I mean: Is a quotient of.
$endgroup$
– André S.
Dec 2 '18 at 8:55
add a comment |
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$begingroup$
Ad 1.) Indeed, the C*-algebra generated by two commuting unitaries is $C(mathbb T^2)$, where $mathbb T^2 = S^1 times S^1$.
$endgroup$
– André S.
Dec 1 '18 at 15:09
$begingroup$
@AndréS.: sometimes, but not always. If you take $u_1=u_2$, you will get $mathbb T$ or a subset of it. And even if $u_1$ and $U_2$ are free, they may still have discrete spectrum.
$endgroup$
– Martin Argerami
Dec 1 '18 at 17:00
$begingroup$
Right. So then I mean: Is a quotient of.
$endgroup$
– André S.
Dec 2 '18 at 8:55