Banach Algebra Isomorphism
$begingroup$
Let $X=l^1$ with coordinate multiplication - it is commutative Banach algebra without unit. The Gelfand transformation is defined as $widehat{x}(e_n)=x_n$ for $x in l^1$. I would like to prove, that this algebra is isomorphic to $C^*$-algebra. Any ideas ? Thanks.
group-isomorphism banach-algebras gelfand-representation
asked Jan 4 at 21:48
lojdmojlojdmoj
877
$endgroup$
add a comment |
$begingroup$
Let $X=l^1$ with coordinate multiplication - it is commutative Banach algebra without unit. The Gelfand transformation is defined as $widehat{x}(e_n)=x_n$ for $x in l^1$. I would like to prove, that this algebra is isomorphic to $C^*$-algebra. Any ideas ? Thanks.
group-isomorphism banach-algebras gelfand-representation
asked Jan 4 at 21:48
lojdmojlojdmoj
877
$endgroup$
add a comment |
$begingroup$
Let $X=l^1$ with coordinate multiplication - it is commutative Banach algebra without unit. The Gelfand transformation is defined as $widehat{x}(e_n)=x_n$ for $x in l^1$. I would like to prove, that this algebra is isomorphic to $C^*$-algebra. Any ideas ? Thanks.
group-isomorphism banach-algebras gelfand-representation
asked Jan 4 at 21:48
lojdmojlojdmoj
877
$endgroup$
Let $X=l^1$ with coordinate multiplication - it is commutative Banach algebra without unit. The Gelfand transformation is defined as $widehat{x}(e_n)=x_n$ for $x in l^1$. I would like to prove, that this algebra is isomorphic to $C^*$-algebra. Any ideas ? Thanks.
group-isomorphism banach-algebras gelfand-representation
group-isomorphism banach-algebras gelfand-representation
asked Jan 4 at 21:48
lojdmojlojdmoj
877
asked Jan 4 at 21:48
lojdmojlojdmoj
877
edited Jan 7 at 22:41
lojdmoj
asked Jan 4 at 21:48
lojdmojlojdmoj
877
asked Jan 4 at 21:48
lojdmojlojdmoj
877
asked Jan 4 at 21:48
lojdmojlojdmoj
877
877
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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It is definitely not, even as a Banach space. Commutative C*-algebras are isomorphic to $C_0(X)$ for some locally compact Hausdorff space, which is infinite when the algebra is infinite-dimensional. In such case, you will find a subalgebra isomorphic to $c_0$. The space $c_0$ does not embed isomorphically into $ell_1$ because the latter space is weakly sequentially complete.
Actually, it is easy to see that $ell_1$ does not satisfy the C*-condition under the original norm. Take $x=e_1 + e_2$. Then $xx^* = x$ but $|xx^*|=2neq |x|^2$. (In other words, observe that non-zero self-adjoint projections in C*-algebras have norm 1.)
answered Jan 4 at 21:59
Tomek KaniaTomek Kania
12.2k11944
$endgroup$
$begingroup$
Is it hold for general $l^p$ space for finite $p$ ?
$endgroup$
– lojdmoj
Jan 4 at 22:15
$begingroup$
@lojdmoj, yes. Why don't you try to modify my arguments to see that?
$endgroup$
– Tomek Kania
Jan 4 at 22:16
$begingroup$
Well, $l^1$ does not satisfy the $C^{*}$-condition, but I would like to show, that there is no isomorphism between $l^{1}$ algebra and $C^{*}$ algebra. We have that $C_{0}(l^{1})=c_0$ but $c_0$ does not embed isomorphically into $l^{1}$. Is is true, that $C_{0}(l^{p})=c_0$ for $p>1$ ?
$endgroup$
– lojdmoj
Jan 7 at 13:23
$begingroup$
@lodmoj, I have a feeling that you don't read my answers. I've shown that there is even no linear isomorphism between $ell_1$ and a C*-algebra...
$endgroup$
– Tomek Kania
Jan 7 at 13:45
$begingroup$
Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
$endgroup$
– lojdmoj
Jan 7 at 22:43
|
show 1 more comment
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1 Answer
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active
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1 Answer
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active
oldest
votes
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active
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votes
$begingroup$
It is definitely not, even as a Banach space. Commutative C*-algebras are isomorphic to $C_0(X)$ for some locally compact Hausdorff space, which is infinite when the algebra is infinite-dimensional. In such case, you will find a subalgebra isomorphic to $c_0$. The space $c_0$ does not embed isomorphically into $ell_1$ because the latter space is weakly sequentially complete.
Actually, it is easy to see that $ell_1$ does not satisfy the C*-condition under the original norm. Take $x=e_1 + e_2$. Then $xx^* = x$ but $|xx^*|=2neq |x|^2$. (In other words, observe that non-zero self-adjoint projections in C*-algebras have norm 1.)
answered Jan 4 at 21:59
Tomek KaniaTomek Kania
12.2k11944
$endgroup$
$begingroup$
Is it hold for general $l^p$ space for finite $p$ ?
$endgroup$
– lojdmoj
Jan 4 at 22:15
$begingroup$
@lojdmoj, yes. Why don't you try to modify my arguments to see that?
$endgroup$
– Tomek Kania
Jan 4 at 22:16
$begingroup$
Well, $l^1$ does not satisfy the $C^{*}$-condition, but I would like to show, that there is no isomorphism between $l^{1}$ algebra and $C^{*}$ algebra. We have that $C_{0}(l^{1})=c_0$ but $c_0$ does not embed isomorphically into $l^{1}$. Is is true, that $C_{0}(l^{p})=c_0$ for $p>1$ ?
$endgroup$
– lojdmoj
Jan 7 at 13:23
$begingroup$
@lodmoj, I have a feeling that you don't read my answers. I've shown that there is even no linear isomorphism between $ell_1$ and a C*-algebra...
$endgroup$
– Tomek Kania
Jan 7 at 13:45
$begingroup$
Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
$endgroup$
– lojdmoj
Jan 7 at 22:43
|
show 1 more comment
$begingroup$
It is definitely not, even as a Banach space. Commutative C*-algebras are isomorphic to $C_0(X)$ for some locally compact Hausdorff space, which is infinite when the algebra is infinite-dimensional. In such case, you will find a subalgebra isomorphic to $c_0$. The space $c_0$ does not embed isomorphically into $ell_1$ because the latter space is weakly sequentially complete.
Actually, it is easy to see that $ell_1$ does not satisfy the C*-condition under the original norm. Take $x=e_1 + e_2$. Then $xx^* = x$ but $|xx^*|=2neq |x|^2$. (In other words, observe that non-zero self-adjoint projections in C*-algebras have norm 1.)
answered Jan 4 at 21:59
Tomek KaniaTomek Kania
12.2k11944
$endgroup$
$begingroup$
Is it hold for general $l^p$ space for finite $p$ ?
$endgroup$
– lojdmoj
Jan 4 at 22:15
$begingroup$
@lojdmoj, yes. Why don't you try to modify my arguments to see that?
$endgroup$
– Tomek Kania
Jan 4 at 22:16
$begingroup$
Well, $l^1$ does not satisfy the $C^{*}$-condition, but I would like to show, that there is no isomorphism between $l^{1}$ algebra and $C^{*}$ algebra. We have that $C_{0}(l^{1})=c_0$ but $c_0$ does not embed isomorphically into $l^{1}$. Is is true, that $C_{0}(l^{p})=c_0$ for $p>1$ ?
$endgroup$
– lojdmoj
Jan 7 at 13:23
$begingroup$
@lodmoj, I have a feeling that you don't read my answers. I've shown that there is even no linear isomorphism between $ell_1$ and a C*-algebra...
$endgroup$
– Tomek Kania
Jan 7 at 13:45
$begingroup$
Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
$endgroup$
– lojdmoj
Jan 7 at 22:43
|
show 1 more comment
$begingroup$
It is definitely not, even as a Banach space. Commutative C*-algebras are isomorphic to $C_0(X)$ for some locally compact Hausdorff space, which is infinite when the algebra is infinite-dimensional. In such case, you will find a subalgebra isomorphic to $c_0$. The space $c_0$ does not embed isomorphically into $ell_1$ because the latter space is weakly sequentially complete.
Actually, it is easy to see that $ell_1$ does not satisfy the C*-condition under the original norm. Take $x=e_1 + e_2$. Then $xx^* = x$ but $|xx^*|=2neq |x|^2$. (In other words, observe that non-zero self-adjoint projections in C*-algebras have norm 1.)
answered Jan 4 at 21:59
Tomek KaniaTomek Kania
12.2k11944
$endgroup$
It is definitely not, even as a Banach space. Commutative C*-algebras are isomorphic to $C_0(X)$ for some locally compact Hausdorff space, which is infinite when the algebra is infinite-dimensional. In such case, you will find a subalgebra isomorphic to $c_0$. The space $c_0$ does not embed isomorphically into $ell_1$ because the latter space is weakly sequentially complete.
Actually, it is easy to see that $ell_1$ does not satisfy the C*-condition under the original norm. Take $x=e_1 + e_2$. Then $xx^* = x$ but $|xx^*|=2neq |x|^2$. (In other words, observe that non-zero self-adjoint projections in C*-algebras have norm 1.)
answered Jan 4 at 21:59
Tomek KaniaTomek Kania
12.2k11944
answered Jan 4 at 21:59
Tomek KaniaTomek Kania
12.2k11944
answered Jan 4 at 21:59
Tomek KaniaTomek Kania
12.2k11944
answered Jan 4 at 21:59
Tomek KaniaTomek Kania
12.2k11944
12.2k11944
$begingroup$
Is it hold for general $l^p$ space for finite $p$ ?
$endgroup$
– lojdmoj
Jan 4 at 22:15
$begingroup$
@lojdmoj, yes. Why don't you try to modify my arguments to see that?
$endgroup$
– Tomek Kania
Jan 4 at 22:16
$begingroup$
Well, $l^1$ does not satisfy the $C^{*}$-condition, but I would like to show, that there is no isomorphism between $l^{1}$ algebra and $C^{*}$ algebra. We have that $C_{0}(l^{1})=c_0$ but $c_0$ does not embed isomorphically into $l^{1}$. Is is true, that $C_{0}(l^{p})=c_0$ for $p>1$ ?
$endgroup$
– lojdmoj
Jan 7 at 13:23
$begingroup$
@lodmoj, I have a feeling that you don't read my answers. I've shown that there is even no linear isomorphism between $ell_1$ and a C*-algebra...
$endgroup$
– Tomek Kania
Jan 7 at 13:45
$begingroup$
Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
$endgroup$
– lojdmoj
Jan 7 at 22:43
|
show 1 more comment
$begingroup$
Is it hold for general $l^p$ space for finite $p$ ?
$endgroup$
– lojdmoj
Jan 4 at 22:15
$begingroup$
@lojdmoj, yes. Why don't you try to modify my arguments to see that?
$endgroup$
– Tomek Kania
Jan 4 at 22:16
$begingroup$
Well, $l^1$ does not satisfy the $C^{*}$-condition, but I would like to show, that there is no isomorphism between $l^{1}$ algebra and $C^{*}$ algebra. We have that $C_{0}(l^{1})=c_0$ but $c_0$ does not embed isomorphically into $l^{1}$. Is is true, that $C_{0}(l^{p})=c_0$ for $p>1$ ?
$endgroup$
– lojdmoj
Jan 7 at 13:23
$begingroup$
@lodmoj, I have a feeling that you don't read my answers. I've shown that there is even no linear isomorphism between $ell_1$ and a C*-algebra...
$endgroup$
– Tomek Kania
Jan 7 at 13:45
$begingroup$
Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
$endgroup$
– lojdmoj
Jan 7 at 22:43
$begingroup$
Is it hold for general $l^p$ space for finite $p$ ?
$endgroup$
– lojdmoj
Jan 4 at 22:15
$begingroup$
Is it hold for general $l^p$ space for finite $p$ ?
$endgroup$
– lojdmoj
Jan 4 at 22:15
$begingroup$
@lojdmoj, yes. Why don't you try to modify my arguments to see that?
$endgroup$
– Tomek Kania
Jan 4 at 22:16
$begingroup$
@lojdmoj, yes. Why don't you try to modify my arguments to see that?
$endgroup$
– Tomek Kania
Jan 4 at 22:16
$begingroup$
Well, $l^1$ does not satisfy the $C^{*}$-condition, but I would like to show, that there is no isomorphism between $l^{1}$ algebra and $C^{*}$ algebra. We have that $C_{0}(l^{1})=c_0$ but $c_0$ does not embed isomorphically into $l^{1}$. Is is true, that $C_{0}(l^{p})=c_0$ for $p>1$ ?
$endgroup$
– lojdmoj
Jan 7 at 13:23
$begingroup$
Well, $l^1$ does not satisfy the $C^{*}$-condition, but I would like to show, that there is no isomorphism between $l^{1}$ algebra and $C^{*}$ algebra. We have that $C_{0}(l^{1})=c_0$ but $c_0$ does not embed isomorphically into $l^{1}$. Is is true, that $C_{0}(l^{p})=c_0$ for $p>1$ ?
$endgroup$
– lojdmoj
Jan 7 at 13:23
$begingroup$
@lodmoj, I have a feeling that you don't read my answers. I've shown that there is even no linear isomorphism between $ell_1$ and a C*-algebra...
$endgroup$
– Tomek Kania
Jan 7 at 13:45
$begingroup$
@lodmoj, I have a feeling that you don't read my answers. I've shown that there is even no linear isomorphism between $ell_1$ and a C*-algebra...
$endgroup$
– Tomek Kania
Jan 7 at 13:45
$begingroup$
Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
$endgroup$
– lojdmoj
Jan 7 at 22:43
$begingroup$
Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
$endgroup$
– lojdmoj
Jan 7 at 22:43
|
show 1 more comment
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