non isomorphic finite dimensional $C*$ algebras
$begingroup$
How many non isomorphic finite dimensional $C^*$ algebras if the dimensions without a bound? Is it countable or uncountable?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 11 '18 at 8:42
mathrookiemathrookie
875512
$endgroup$
add a comment |
$begingroup$
How many non isomorphic finite dimensional $C^*$ algebras if the dimensions without a bound? Is it countable or uncountable?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 11 '18 at 8:42
mathrookiemathrookie
875512
$endgroup$
$begingroup$
Finite-dimensional C*-algebras are finite products of matrix algebras.
$endgroup$
– Qiaochu Yuan
Dec 11 '18 at 9:13
1
$begingroup$
This would be a more interesting question for a fixed dimension $d$ (although completely unrelated to $C^ast$-algebras)
$endgroup$
– Adrián González-Pérez
Dec 11 '18 at 11:36
add a comment |
$begingroup$
How many non isomorphic finite dimensional $C^*$ algebras if the dimensions without a bound? Is it countable or uncountable?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 11 '18 at 8:42
mathrookiemathrookie
875512
$endgroup$
How many non isomorphic finite dimensional $C^*$ algebras if the dimensions without a bound? Is it countable or uncountable?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 11 '18 at 8:42
mathrookiemathrookie
875512
asked Dec 11 '18 at 8:42
mathrookiemathrookie
875512
asked Dec 11 '18 at 8:42
mathrookiemathrookie
875512
asked Dec 11 '18 at 8:42
mathrookiemathrookie
875512
asked Dec 11 '18 at 8:42
mathrookiemathrookie
875512
875512
$begingroup$
Finite-dimensional C*-algebras are finite products of matrix algebras.
$endgroup$
– Qiaochu Yuan
Dec 11 '18 at 9:13
1
$begingroup$
This would be a more interesting question for a fixed dimension $d$ (although completely unrelated to $C^ast$-algebras)
$endgroup$
– Adrián González-Pérez
Dec 11 '18 at 11:36
add a comment |
$begingroup$
Finite-dimensional C*-algebras are finite products of matrix algebras.
$endgroup$
– Qiaochu Yuan
Dec 11 '18 at 9:13
1
$begingroup$
This would be a more interesting question for a fixed dimension $d$ (although completely unrelated to $C^ast$-algebras)
$endgroup$
– Adrián González-Pérez
Dec 11 '18 at 11:36
$begingroup$
Finite-dimensional C*-algebras are finite products of matrix algebras.
$endgroup$
– Qiaochu Yuan
Dec 11 '18 at 9:13
$begingroup$
Finite-dimensional C*-algebras are finite products of matrix algebras.
$endgroup$
– Qiaochu Yuan
Dec 11 '18 at 9:13
1
1
$begingroup$
This would be a more interesting question for a fixed dimension $d$ (although completely unrelated to $C^ast$-algebras)
$endgroup$
– Adrián González-Pérez
Dec 11 '18 at 11:36
$begingroup$
This would be a more interesting question for a fixed dimension $d$ (although completely unrelated to $C^ast$-algebras)
$endgroup$
– Adrián González-Pérez
Dec 11 '18 at 11:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For each $n$, the set of (isomorphism classes) of $C^*$-algebras of dimension $leq n$ is finite (being the number of collections ${(n_1,m_1),ldots,(n_k,m_k)}$ of pairs of natural numbers with the $m_j$ distinct such that $n_1cdot m_1^2+cdots+n_kcdot m_k^2leq n$), hence the collection of (isomorphism classes) of finite-dimensional $C^*$-algebras is countable.
answered Dec 11 '18 at 19:09
AweyganAweygan
14.1k21441
$endgroup$
$begingroup$
why shuould we multiply $n_k$?each finite dimensional $C^*$ algebra is $*$ isomorphic to direct sum of matrix algebras
$endgroup$
– mathrookie
Dec 11 '18 at 20:54
$begingroup$
I meant to write that the $m_j$ were distinct. This just allows for multiple $M_{m_j}(mathbb C)$ summands in the same algebra.
$endgroup$
– Aweygan
Dec 11 '18 at 21:49
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
For each $n$, the set of (isomorphism classes) of $C^*$-algebras of dimension $leq n$ is finite (being the number of collections ${(n_1,m_1),ldots,(n_k,m_k)}$ of pairs of natural numbers with the $m_j$ distinct such that $n_1cdot m_1^2+cdots+n_kcdot m_k^2leq n$), hence the collection of (isomorphism classes) of finite-dimensional $C^*$-algebras is countable.
answered Dec 11 '18 at 19:09
AweyganAweygan
14.1k21441
$endgroup$
$begingroup$
why shuould we multiply $n_k$?each finite dimensional $C^*$ algebra is $*$ isomorphic to direct sum of matrix algebras
$endgroup$
– mathrookie
Dec 11 '18 at 20:54
$begingroup$
I meant to write that the $m_j$ were distinct. This just allows for multiple $M_{m_j}(mathbb C)$ summands in the same algebra.
$endgroup$
– Aweygan
Dec 11 '18 at 21:49
add a comment |
$begingroup$
For each $n$, the set of (isomorphism classes) of $C^*$-algebras of dimension $leq n$ is finite (being the number of collections ${(n_1,m_1),ldots,(n_k,m_k)}$ of pairs of natural numbers with the $m_j$ distinct such that $n_1cdot m_1^2+cdots+n_kcdot m_k^2leq n$), hence the collection of (isomorphism classes) of finite-dimensional $C^*$-algebras is countable.
answered Dec 11 '18 at 19:09
AweyganAweygan
14.1k21441
$endgroup$
$begingroup$
why shuould we multiply $n_k$?each finite dimensional $C^*$ algebra is $*$ isomorphic to direct sum of matrix algebras
$endgroup$
– mathrookie
Dec 11 '18 at 20:54
$begingroup$
I meant to write that the $m_j$ were distinct. This just allows for multiple $M_{m_j}(mathbb C)$ summands in the same algebra.
$endgroup$
– Aweygan
Dec 11 '18 at 21:49
add a comment |
$begingroup$
For each $n$, the set of (isomorphism classes) of $C^*$-algebras of dimension $leq n$ is finite (being the number of collections ${(n_1,m_1),ldots,(n_k,m_k)}$ of pairs of natural numbers with the $m_j$ distinct such that $n_1cdot m_1^2+cdots+n_kcdot m_k^2leq n$), hence the collection of (isomorphism classes) of finite-dimensional $C^*$-algebras is countable.
answered Dec 11 '18 at 19:09
AweyganAweygan
14.1k21441
$endgroup$
For each $n$, the set of (isomorphism classes) of $C^*$-algebras of dimension $leq n$ is finite (being the number of collections ${(n_1,m_1),ldots,(n_k,m_k)}$ of pairs of natural numbers with the $m_j$ distinct such that $n_1cdot m_1^2+cdots+n_kcdot m_k^2leq n$), hence the collection of (isomorphism classes) of finite-dimensional $C^*$-algebras is countable.
answered Dec 11 '18 at 19:09
AweyganAweygan
14.1k21441
edited Dec 11 '18 at 21:48
answered Dec 11 '18 at 19:09
AweyganAweygan
14.1k21441
answered Dec 11 '18 at 19:09
AweyganAweygan
14.1k21441
answered Dec 11 '18 at 19:09
AweyganAweygan
14.1k21441
14.1k21441
$begingroup$
why shuould we multiply $n_k$?each finite dimensional $C^*$ algebra is $*$ isomorphic to direct sum of matrix algebras
$endgroup$
– mathrookie
Dec 11 '18 at 20:54
$begingroup$
I meant to write that the $m_j$ were distinct. This just allows for multiple $M_{m_j}(mathbb C)$ summands in the same algebra.
$endgroup$
– Aweygan
Dec 11 '18 at 21:49
add a comment |
$begingroup$
why shuould we multiply $n_k$?each finite dimensional $C^*$ algebra is $*$ isomorphic to direct sum of matrix algebras
$endgroup$
– mathrookie
Dec 11 '18 at 20:54
$begingroup$
I meant to write that the $m_j$ were distinct. This just allows for multiple $M_{m_j}(mathbb C)$ summands in the same algebra.
$endgroup$
– Aweygan
Dec 11 '18 at 21:49
$begingroup$
why shuould we multiply $n_k$?each finite dimensional $C^*$ algebra is $*$ isomorphic to direct sum of matrix algebras
$endgroup$
– mathrookie
Dec 11 '18 at 20:54
$begingroup$
why shuould we multiply $n_k$?each finite dimensional $C^*$ algebra is $*$ isomorphic to direct sum of matrix algebras
$endgroup$
– mathrookie
Dec 11 '18 at 20:54
$begingroup$
I meant to write that the $m_j$ were distinct. This just allows for multiple $M_{m_j}(mathbb C)$ summands in the same algebra.
$endgroup$
– Aweygan
Dec 11 '18 at 21:49
$begingroup$
I meant to write that the $m_j$ were distinct. This just allows for multiple $M_{m_j}(mathbb C)$ summands in the same algebra.
$endgroup$
– Aweygan
Dec 11 '18 at 21:49
add a comment |
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$begingroup$
Finite-dimensional C*-algebras are finite products of matrix algebras.
$endgroup$
– Qiaochu Yuan
Dec 11 '18 at 9:13
1
$begingroup$
This would be a more interesting question for a fixed dimension $d$ (although completely unrelated to $C^ast$-algebras)
$endgroup$
– Adrián González-Pérez
Dec 11 '18 at 11:36