Banach Algebra Isomorphism












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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.






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    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







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      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







      share|cite|improve this question















      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






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      edited Jan 7 at 22:41







      lojdmoj

















      asked Jan 4 at 21:48









      lojdmojlojdmoj

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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.)






          share|cite|improve this answer





















          • Is it hold for general $l^p$ space for finite $p$ ?
            – lojdmoj
            Jan 4 at 22:15










          • @lojdmoj, yes. Why don't you try to modify my arguments to see that?
            – Tomek Kania
            Jan 4 at 22:16










          • 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 $p>1$ ?
            – lojdmoj
            Jan 7 at 13:23












          • @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...
            – Tomek Kania
            Jan 7 at 13:45










          • Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
            – lojdmoj
            Jan 7 at 22:43













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          1 Answer
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          1 Answer
          1






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          active

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          active

          oldest

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          2














          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.)






          share|cite|improve this answer





















          • Is it hold for general $l^p$ space for finite $p$ ?
            – lojdmoj
            Jan 4 at 22:15










          • @lojdmoj, yes. Why don't you try to modify my arguments to see that?
            – Tomek Kania
            Jan 4 at 22:16










          • 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 $p>1$ ?
            – lojdmoj
            Jan 7 at 13:23












          • @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...
            – Tomek Kania
            Jan 7 at 13:45










          • Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
            – lojdmoj
            Jan 7 at 22:43


















          2














          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.)






          share|cite|improve this answer





















          • Is it hold for general $l^p$ space for finite $p$ ?
            – lojdmoj
            Jan 4 at 22:15










          • @lojdmoj, yes. Why don't you try to modify my arguments to see that?
            – Tomek Kania
            Jan 4 at 22:16










          • 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 $p>1$ ?
            – lojdmoj
            Jan 7 at 13:23












          • @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...
            – Tomek Kania
            Jan 7 at 13:45










          • Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
            – lojdmoj
            Jan 7 at 22:43
















          2












          2








          2






          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.)






          share|cite|improve this answer












          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.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 21:59









          Tomek KaniaTomek Kania

          12.1k11943




          12.1k11943












          • Is it hold for general $l^p$ space for finite $p$ ?
            – lojdmoj
            Jan 4 at 22:15










          • @lojdmoj, yes. Why don't you try to modify my arguments to see that?
            – Tomek Kania
            Jan 4 at 22:16










          • 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 $p>1$ ?
            – lojdmoj
            Jan 7 at 13:23












          • @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...
            – Tomek Kania
            Jan 7 at 13:45










          • Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
            – lojdmoj
            Jan 7 at 22:43




















          • Is it hold for general $l^p$ space for finite $p$ ?
            – lojdmoj
            Jan 4 at 22:15










          • @lojdmoj, yes. Why don't you try to modify my arguments to see that?
            – Tomek Kania
            Jan 4 at 22:16










          • 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 $p>1$ ?
            – lojdmoj
            Jan 7 at 13:23












          • @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...
            – Tomek Kania
            Jan 7 at 13:45










          • Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
            – lojdmoj
            Jan 7 at 22:43


















          Is it hold for general $l^p$ space for finite $p$ ?
          – lojdmoj
          Jan 4 at 22:15




          Is it hold for general $l^p$ space for finite $p$ ?
          – lojdmoj
          Jan 4 at 22:15












          @lojdmoj, yes. Why don't you try to modify my arguments to see that?
          – Tomek Kania
          Jan 4 at 22:16




          @lojdmoj, yes. Why don't you try to modify my arguments to see that?
          – Tomek Kania
          Jan 4 at 22:16












          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 $p>1$ ?
          – lojdmoj
          Jan 7 at 13:23






          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 $p>1$ ?
          – lojdmoj
          Jan 7 at 13:23














          @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...
          – Tomek Kania
          Jan 7 at 13:45




          @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...
          – Tomek Kania
          Jan 7 at 13:45












          Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
          – lojdmoj
          Jan 7 at 22:43






          Yes, but I am confused. I still do not understand if algebra $l^p$ is isomorphic to $C^{*}$ algebra.
          – lojdmoj
          Jan 7 at 22:43




















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