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finite dimensional representation of $C^*$ representation
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According to the standard definition of representation of a $C^*$ algebra $A$ ,we need to construct a $*$ homomorphism $pi$ from $A$ to $B(H)$,where $H$ is a Hilbert space.My question is : if we have a $*$ homomorphism $phi$ from $A$ to $B$,where B is a finite dimensional $C^*$ algebra.Can we call $phi$ the finite dimensional representation of $A$?
operator-theory operator-algebras c-star-algebras
edited Jul 21 at 22:37
Martin Argerami
116k1071164
asked Jul 20 at 0:00
mathrookie
437211
add a comment |Â
up vote
1
down vote
favorite
According to the standard definition of representation of a $C^*$ algebra $A$ ,we need to construct a $*$ homomorphism $pi$ from $A$ to $B(H)$,where $H$ is a Hilbert space.My question is : if we have a $*$ homomorphism $phi$ from $A$ to $B$,where B is a finite dimensional $C^*$ algebra.Can we call $phi$ the finite dimensional representation of $A$?
operator-theory operator-algebras c-star-algebras
edited Jul 21 at 22:37
Martin Argerami
116k1071164
asked Jul 20 at 0:00
mathrookie
437211
Let $C(X)$ be the unital $C^ast$-algebra of continuous complex-valued functions on some compact Hausdorff space $X$. Then evaluation at a point $x in X$, i.e., the map $hatx : C(X) to mathbbC$ defined by $hatx(f) := f(x)$ gives you a $1$-dimensional representation, and the moment $X$ has at least two points, different points will give you different $1$-dimensional representations.
– Branimir Ćaćić
Jul 20 at 8:59
I mean "when talking about the represemtation,should we construct a Hilbert space?"In my opinition,constructing a $*$ homomorphism is enough,since a $C^*$ algebra can be embeded in $B(H)$ for some Hilbert space $H$.
– mathrookie
Jul 21 at 2:05
Ah, sorry, I misunderstood your question!
– Branimir Ćaćić
Jul 21 at 6:09
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
According to the standard definition of representation of a $C^*$ algebra $A$ ,we need to construct a $*$ homomorphism $pi$ from $A$ to $B(H)$,where $H$ is a Hilbert space.My question is : if we have a $*$ homomorphism $phi$ from $A$ to $B$,where B is a finite dimensional $C^*$ algebra.Can we call $phi$ the finite dimensional representation of $A$?
operator-theory operator-algebras c-star-algebras
edited Jul 21 at 22:37
Martin Argerami
116k1071164
asked Jul 20 at 0:00
mathrookie
437211
According to the standard definition of representation of a $C^*$ algebra $A$ ,we need to construct a $*$ homomorphism $pi$ from $A$ to $B(H)$,where $H$ is a Hilbert space.My question is : if we have a $*$ homomorphism $phi$ from $A$ to $B$,where B is a finite dimensional $C^*$ algebra.Can we call $phi$ the finite dimensional representation of $A$?
operator-theory operator-algebras c-star-algebras
edited Jul 21 at 22:37
Martin Argerami
116k1071164
asked Jul 20 at 0:00
mathrookie
437211
edited Jul 21 at 22:37
Martin Argerami
116k1071164
edited Jul 21 at 22:37
Martin Argerami
116k1071164
edited Jul 21 at 22:37
Martin Argerami
116k1071164
116k1071164
asked Jul 20 at 0:00
mathrookie
437211
asked Jul 20 at 0:00
mathrookie
437211
asked Jul 20 at 0:00
mathrookie
437211
437211
Let $C(X)$ be the unital $C^ast$-algebra of continuous complex-valued functions on some compact Hausdorff space $X$. Then evaluation at a point $x in X$, i.e., the map $hatx : C(X) to mathbbC$ defined by $hatx(f) := f(x)$ gives you a $1$-dimensional representation, and the moment $X$ has at least two points, different points will give you different $1$-dimensional representations.
– Branimir Ćaćić
Jul 20 at 8:59
I mean "when talking about the represemtation,should we construct a Hilbert space?"In my opinition,constructing a $*$ homomorphism is enough,since a $C^*$ algebra can be embeded in $B(H)$ for some Hilbert space $H$.
– mathrookie
Jul 21 at 2:05
Ah, sorry, I misunderstood your question!
– Branimir Ćaćić
Jul 21 at 6:09
add a comment |Â
Let $C(X)$ be the unital $C^ast$-algebra of continuous complex-valued functions on some compact Hausdorff space $X$. Then evaluation at a point $x in X$, i.e., the map $hatx : C(X) to mathbbC$ defined by $hatx(f) := f(x)$ gives you a $1$-dimensional representation, and the moment $X$ has at least two points, different points will give you different $1$-dimensional representations.
– Branimir Ćaćić
Jul 20 at 8:59
I mean "when talking about the represemtation,should we construct a Hilbert space?"In my opinition,constructing a $*$ homomorphism is enough,since a $C^*$ algebra can be embeded in $B(H)$ for some Hilbert space $H$.
– mathrookie
Jul 21 at 2:05
Ah, sorry, I misunderstood your question!
– Branimir Ćaćić
Jul 21 at 6:09
Let $C(X)$ be the unital $C^ast$-algebra of continuous complex-valued functions on some compact Hausdorff space $X$. Then evaluation at a point $x in X$, i.e., the map $hatx : C(X) to mathbbC$ defined by $hatx(f) := f(x)$ gives you a $1$-dimensional representation, and the moment $X$ has at least two points, different points will give you different $1$-dimensional representations.
– Branimir Ćaćić
Jul 20 at 8:59
Let $C(X)$ be the unital $C^ast$-algebra of continuous complex-valued functions on some compact Hausdorff space $X$. Then evaluation at a point $x in X$, i.e., the map $hatx : C(X) to mathbbC$ defined by $hatx(f) := f(x)$ gives you a $1$-dimensional representation, and the moment $X$ has at least two points, different points will give you different $1$-dimensional representations.
– Branimir Ćaćić
Jul 20 at 8:59
I mean "when talking about the represemtation,should we construct a Hilbert space?"In my opinition,constructing a $*$ homomorphism is enough,since a $C^*$ algebra can be embeded in $B(H)$ for some Hilbert space $H$.
– mathrookie
Jul 21 at 2:05
I mean "when talking about the represemtation,should we construct a Hilbert space?"In my opinition,constructing a $*$ homomorphism is enough,since a $C^*$ algebra can be embeded in $B(H)$ for some Hilbert space $H$.
– mathrookie
Jul 21 at 2:05
Ah, sorry, I misunderstood your question!
– Branimir Ćaćić
Jul 21 at 6:09
Ah, sorry, I misunderstood your question!
– Branimir Ćaćić
Jul 21 at 6:09
add a comment |Â
2 Answers
2
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oldest
votes
up vote
1
down vote
accepted
The answer is yes, because if $B$ is finite-dimensional, then it admits a representation $Bsubset M_n(mathbb C)$. So via this representation for $B$, we may see $phi:Ato M_n(mathbb C)=B(mathbb C^n)$.
answered Jul 21 at 22:38
Martin Argerami
116k1071164
add a comment |Â
up vote
1
down vote
The answer is absolutely yes. Suppose that $phi : A to B$ is a $ast$-homomorphism, where $B$ is finite-dimensional. By the structure theorem for finite-dimensional $C^ast$-algebras (i.e., the appropriate refinement of Wedderburn's theorem), the $C^ast$-algebra $B$ necessarily takes the form
$$
B cong M_n_1(mathbbC) oplus cdots oplus M_n_k(mathbbC) subseteq M_n_1 + cdots + n_k(mathbbC)
$$
for some positive integers $n_1,dotsc,n_k$. Now, let $operatornametr$ be the normalised matrix trace on $M_n_1 + cdots + n_k(mathbbC)$, which is, in fact, a faithful trace, satisfying $operatornameTr(x^ast x) geq 0$ for any matrix $x$, with
$$
forall x in M_n_1 + cdots + n_k(mathbbC), quad operatornametr(x^ast x) = 0 iff x = 0.
$$
We can therefore restrict $operatornametr$ to a faithful trace on $B$, and use it to construct a positive-definite inner product on the finite-dimensional complex vector space $B$, by
$$
forall b_1, b_2 in B, quad langle b_1,b_2 rangle := operatornametr(b_1^ast b_2).
$$
Since $operatornametr$ is faithful and $B$ is finite-dimensional,
$$
L^2(B,operatornametr) := (B,langle cdot,cdot rangle)
$$
becomes a faithful finite-dimensional $ast$-representation of $B$ by left multiplication—indeed, it's precisely the GNS representation of $B$ induced by the faithful tracial state $operatornametrvert_B$. As a result, the $ast$-homomorphism $phi : A to B subseteq B(L^2(B,operatornametr))$ really does definite a finite-dimensional $ast$-representation on $L^2(B,operatornametr)$.
answered Jul 21 at 6:23
Branimir Ćaćić
9,69221846
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The answer is yes, because if $B$ is finite-dimensional, then it admits a representation $Bsubset M_n(mathbb C)$. So via this representation for $B$, we may see $phi:Ato M_n(mathbb C)=B(mathbb C^n)$.
answered Jul 21 at 22:38
Martin Argerami
116k1071164
add a comment |Â
up vote
1
down vote
accepted
The answer is yes, because if $B$ is finite-dimensional, then it admits a representation $Bsubset M_n(mathbb C)$. So via this representation for $B$, we may see $phi:Ato M_n(mathbb C)=B(mathbb C^n)$.
answered Jul 21 at 22:38
Martin Argerami
116k1071164
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The answer is yes, because if $B$ is finite-dimensional, then it admits a representation $Bsubset M_n(mathbb C)$. So via this representation for $B$, we may see $phi:Ato M_n(mathbb C)=B(mathbb C^n)$.
answered Jul 21 at 22:38
Martin Argerami
116k1071164
The answer is yes, because if $B$ is finite-dimensional, then it admits a representation $Bsubset M_n(mathbb C)$. So via this representation for $B$, we may see $phi:Ato M_n(mathbb C)=B(mathbb C^n)$.
answered Jul 21 at 22:38
Martin Argerami
116k1071164
answered Jul 21 at 22:38
Martin Argerami
116k1071164
answered Jul 21 at 22:38
Martin Argerami
116k1071164
answered Jul 21 at 22:38
Martin Argerami
116k1071164
116k1071164
add a comment |Â
add a comment |Â
up vote
1
down vote
The answer is absolutely yes. Suppose that $phi : A to B$ is a $ast$-homomorphism, where $B$ is finite-dimensional. By the structure theorem for finite-dimensional $C^ast$-algebras (i.e., the appropriate refinement of Wedderburn's theorem), the $C^ast$-algebra $B$ necessarily takes the form
$$
B cong M_n_1(mathbbC) oplus cdots oplus M_n_k(mathbbC) subseteq M_n_1 + cdots + n_k(mathbbC)
$$
for some positive integers $n_1,dotsc,n_k$. Now, let $operatornametr$ be the normalised matrix trace on $M_n_1 + cdots + n_k(mathbbC)$, which is, in fact, a faithful trace, satisfying $operatornameTr(x^ast x) geq 0$ for any matrix $x$, with
$$
forall x in M_n_1 + cdots + n_k(mathbbC), quad operatornametr(x^ast x) = 0 iff x = 0.
$$
We can therefore restrict $operatornametr$ to a faithful trace on $B$, and use it to construct a positive-definite inner product on the finite-dimensional complex vector space $B$, by
$$
forall b_1, b_2 in B, quad langle b_1,b_2 rangle := operatornametr(b_1^ast b_2).
$$
Since $operatornametr$ is faithful and $B$ is finite-dimensional,
$$
L^2(B,operatornametr) := (B,langle cdot,cdot rangle)
$$
becomes a faithful finite-dimensional $ast$-representation of $B$ by left multiplication—indeed, it's precisely the GNS representation of $B$ induced by the faithful tracial state $operatornametrvert_B$. As a result, the $ast$-homomorphism $phi : A to B subseteq B(L^2(B,operatornametr))$ really does definite a finite-dimensional $ast$-representation on $L^2(B,operatornametr)$.
answered Jul 21 at 6:23
Branimir Ćaćić
9,69221846
add a comment |Â
up vote
1
down vote
The answer is absolutely yes. Suppose that $phi : A to B$ is a $ast$-homomorphism, where $B$ is finite-dimensional. By the structure theorem for finite-dimensional $C^ast$-algebras (i.e., the appropriate refinement of Wedderburn's theorem), the $C^ast$-algebra $B$ necessarily takes the form
$$
B cong M_n_1(mathbbC) oplus cdots oplus M_n_k(mathbbC) subseteq M_n_1 + cdots + n_k(mathbbC)
$$
for some positive integers $n_1,dotsc,n_k$. Now, let $operatornametr$ be the normalised matrix trace on $M_n_1 + cdots + n_k(mathbbC)$, which is, in fact, a faithful trace, satisfying $operatornameTr(x^ast x) geq 0$ for any matrix $x$, with
$$
forall x in M_n_1 + cdots + n_k(mathbbC), quad operatornametr(x^ast x) = 0 iff x = 0.
$$
We can therefore restrict $operatornametr$ to a faithful trace on $B$, and use it to construct a positive-definite inner product on the finite-dimensional complex vector space $B$, by
$$
forall b_1, b_2 in B, quad langle b_1,b_2 rangle := operatornametr(b_1^ast b_2).
$$
Since $operatornametr$ is faithful and $B$ is finite-dimensional,
$$
L^2(B,operatornametr) := (B,langle cdot,cdot rangle)
$$
becomes a faithful finite-dimensional $ast$-representation of $B$ by left multiplication—indeed, it's precisely the GNS representation of $B$ induced by the faithful tracial state $operatornametrvert_B$. As a result, the $ast$-homomorphism $phi : A to B subseteq B(L^2(B,operatornametr))$ really does definite a finite-dimensional $ast$-representation on $L^2(B,operatornametr)$.
answered Jul 21 at 6:23
Branimir Ćaćić
9,69221846
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The answer is absolutely yes. Suppose that $phi : A to B$ is a $ast$-homomorphism, where $B$ is finite-dimensional. By the structure theorem for finite-dimensional $C^ast$-algebras (i.e., the appropriate refinement of Wedderburn's theorem), the $C^ast$-algebra $B$ necessarily takes the form
$$
B cong M_n_1(mathbbC) oplus cdots oplus M_n_k(mathbbC) subseteq M_n_1 + cdots + n_k(mathbbC)
$$
for some positive integers $n_1,dotsc,n_k$. Now, let $operatornametr$ be the normalised matrix trace on $M_n_1 + cdots + n_k(mathbbC)$, which is, in fact, a faithful trace, satisfying $operatornameTr(x^ast x) geq 0$ for any matrix $x$, with
$$
forall x in M_n_1 + cdots + n_k(mathbbC), quad operatornametr(x^ast x) = 0 iff x = 0.
$$
We can therefore restrict $operatornametr$ to a faithful trace on $B$, and use it to construct a positive-definite inner product on the finite-dimensional complex vector space $B$, by
$$
forall b_1, b_2 in B, quad langle b_1,b_2 rangle := operatornametr(b_1^ast b_2).
$$
Since $operatornametr$ is faithful and $B$ is finite-dimensional,
$$
L^2(B,operatornametr) := (B,langle cdot,cdot rangle)
$$
becomes a faithful finite-dimensional $ast$-representation of $B$ by left multiplication—indeed, it's precisely the GNS representation of $B$ induced by the faithful tracial state $operatornametrvert_B$. As a result, the $ast$-homomorphism $phi : A to B subseteq B(L^2(B,operatornametr))$ really does definite a finite-dimensional $ast$-representation on $L^2(B,operatornametr)$.
answered Jul 21 at 6:23
Branimir Ćaćić
9,69221846
The answer is absolutely yes. Suppose that $phi : A to B$ is a $ast$-homomorphism, where $B$ is finite-dimensional. By the structure theorem for finite-dimensional $C^ast$-algebras (i.e., the appropriate refinement of Wedderburn's theorem), the $C^ast$-algebra $B$ necessarily takes the form
$$
B cong M_n_1(mathbbC) oplus cdots oplus M_n_k(mathbbC) subseteq M_n_1 + cdots + n_k(mathbbC)
$$
for some positive integers $n_1,dotsc,n_k$. Now, let $operatornametr$ be the normalised matrix trace on $M_n_1 + cdots + n_k(mathbbC)$, which is, in fact, a faithful trace, satisfying $operatornameTr(x^ast x) geq 0$ for any matrix $x$, with
$$
forall x in M_n_1 + cdots + n_k(mathbbC), quad operatornametr(x^ast x) = 0 iff x = 0.
$$
We can therefore restrict $operatornametr$ to a faithful trace on $B$, and use it to construct a positive-definite inner product on the finite-dimensional complex vector space $B$, by
$$
forall b_1, b_2 in B, quad langle b_1,b_2 rangle := operatornametr(b_1^ast b_2).
$$
Since $operatornametr$ is faithful and $B$ is finite-dimensional,
$$
L^2(B,operatornametr) := (B,langle cdot,cdot rangle)
$$
becomes a faithful finite-dimensional $ast$-representation of $B$ by left multiplication—indeed, it's precisely the GNS representation of $B$ induced by the faithful tracial state $operatornametrvert_B$. As a result, the $ast$-homomorphism $phi : A to B subseteq B(L^2(B,operatornametr))$ really does definite a finite-dimensional $ast$-representation on $L^2(B,operatornametr)$.
answered Jul 21 at 6:23
Branimir Ćaćić
9,69221846
answered Jul 21 at 6:23
Branimir Ćaćić
9,69221846
answered Jul 21 at 6:23
Branimir Ćaćić
9,69221846
answered Jul 21 at 6:23
Branimir Ćaćić
9,69221846
9,69221846
add a comment |Â
add a comment |Â
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Let $C(X)$ be the unital $C^ast$-algebra of continuous complex-valued functions on some compact Hausdorff space $X$. Then evaluation at a point $x in X$, i.e., the map $hatx : C(X) to mathbbC$ defined by $hatx(f) := f(x)$ gives you a $1$-dimensional representation, and the moment $X$ has at least two points, different points will give you different $1$-dimensional representations.
– Branimir Ćaćić
Jul 20 at 8:59
I mean "when talking about the represemtation,should we construct a Hilbert space?"In my opinition,constructing a $*$ homomorphism is enough,since a $C^*$ algebra can be embeded in $B(H)$ for some Hilbert space $H$.
– mathrookie
Jul 21 at 2:05
Ah, sorry, I misunderstood your question!
– Branimir Ćaćić
Jul 21 at 6:09