Mixing
infinite representation of a $C^*$ algebra
Clash Royale CLAN TAG#URR8PPP
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Suppose $pi$ is a finite dimensional representation of $C^*$ algebra $A$ on a Hilbert space $H$,then $pi$ is the direct sum of finite dimensional irreducible subrepresentations.
My quesion is :If $pi$ is an infinite dimensional representation of $C^*$ algebra $A$ on a Hilbert space $H$ with dim($H$)=$infty$.Does there exsit a finite dimensional representation of A on $H_0$,where $H_0$ is a subspace of $H$?Can we decompose the infinite dimensional representation?
operator-theory operator-algebras c-star-algebras
asked Jul 22 at 19:01
mathrookie
437211
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up vote
1
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Suppose $pi$ is a finite dimensional representation of $C^*$ algebra $A$ on a Hilbert space $H$,then $pi$ is the direct sum of finite dimensional irreducible subrepresentations.
My quesion is :If $pi$ is an infinite dimensional representation of $C^*$ algebra $A$ on a Hilbert space $H$ with dim($H$)=$infty$.Does there exsit a finite dimensional representation of A on $H_0$,where $H_0$ is a subspace of $H$?Can we decompose the infinite dimensional representation?
operator-theory operator-algebras c-star-algebras
asked Jul 22 at 19:01
mathrookie
437211
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $pi$ is a finite dimensional representation of $C^*$ algebra $A$ on a Hilbert space $H$,then $pi$ is the direct sum of finite dimensional irreducible subrepresentations.
My quesion is :If $pi$ is an infinite dimensional representation of $C^*$ algebra $A$ on a Hilbert space $H$ with dim($H$)=$infty$.Does there exsit a finite dimensional representation of A on $H_0$,where $H_0$ is a subspace of $H$?Can we decompose the infinite dimensional representation?
operator-theory operator-algebras c-star-algebras
asked Jul 22 at 19:01
mathrookie
437211
Suppose $pi$ is a finite dimensional representation of $C^*$ algebra $A$ on a Hilbert space $H$,then $pi$ is the direct sum of finite dimensional irreducible subrepresentations.
My quesion is :If $pi$ is an infinite dimensional representation of $C^*$ algebra $A$ on a Hilbert space $H$ with dim($H$)=$infty$.Does there exsit a finite dimensional representation of A on $H_0$,where $H_0$ is a subspace of $H$?Can we decompose the infinite dimensional representation?
operator-theory operator-algebras c-star-algebras
asked Jul 22 at 19:01
mathrookie
437211
asked Jul 22 at 19:01
mathrookie
437211
asked Jul 22 at 19:01
mathrookie
437211
asked Jul 22 at 19:01
mathrookie
437211
437211
add a comment |Â
add a comment |Â
1 Answer
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Not necessarily. Take $A=B(H)$, and $pi$ the identity representation.
More generally, take $A$ any C$^*$-algebra with $pi$ an infinite-dimensional irreducible representation (easy examples of such $A$ are $K(H)$ and UHF$(2^infty)$). This $pi$ cannot have summands, since $pi(A)'=mathbb C$.
answered Jul 22 at 21:43
Martin Argerami
115k1071164
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Not necessarily. Take $A=B(H)$, and $pi$ the identity representation.
More generally, take $A$ any C$^*$-algebra with $pi$ an infinite-dimensional irreducible representation (easy examples of such $A$ are $K(H)$ and UHF$(2^infty)$). This $pi$ cannot have summands, since $pi(A)'=mathbb C$.
answered Jul 22 at 21:43
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
Not necessarily. Take $A=B(H)$, and $pi$ the identity representation.
More generally, take $A$ any C$^*$-algebra with $pi$ an infinite-dimensional irreducible representation (easy examples of such $A$ are $K(H)$ and UHF$(2^infty)$). This $pi$ cannot have summands, since $pi(A)'=mathbb C$.
answered Jul 22 at 21:43
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Not necessarily. Take $A=B(H)$, and $pi$ the identity representation.
More generally, take $A$ any C$^*$-algebra with $pi$ an infinite-dimensional irreducible representation (easy examples of such $A$ are $K(H)$ and UHF$(2^infty)$). This $pi$ cannot have summands, since $pi(A)'=mathbb C$.
answered Jul 22 at 21:43
Martin Argerami
115k1071164
Not necessarily. Take $A=B(H)$, and $pi$ the identity representation.
More generally, take $A$ any C$^*$-algebra with $pi$ an infinite-dimensional irreducible representation (easy examples of such $A$ are $K(H)$ and UHF$(2^infty)$). This $pi$ cannot have summands, since $pi(A)'=mathbb C$.
answered Jul 22 at 21:43
Martin Argerami
115k1071164
answered Jul 22 at 21:43
Martin Argerami
115k1071164
answered Jul 22 at 21:43
Martin Argerami
115k1071164
answered Jul 22 at 21:43
Martin Argerami
115k1071164
115k1071164
add a comment |Â
add a comment |Â
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