Upgrading Injectivity of a *-homomorphism From a Dense Subalgebra
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In the proof of Lemma A.4 of this document, the author proves that a C$^*$-algebra is simple by showing that every one of its representations is faithful. They proceed by taking a representation and show it is injective on a (fixed) dense $^*$-subalgebra. Then they claim that the restriction of the representation to this subalgebra is faithful (obvious) and that the extension back to the original algebra is faithful (valid conclusion?).
Perusing other questions on here it would appear that there is a problem with the final claim, if we just look at extending continuous injective functions then you can't just conclude that the extension is injective, and if we look at extending injective $^*$-homomorphisms on dense $^*$-subalgebras, then there are still problems in the general case.
I suppose that my question would be the following; what extra conditions may we impose on an injective $^*$-homomorphism $varphi:mathcalDrightarrow B$ between a dense $^*$-subalgebra $mathcalD$ of a C$^*$-algebra $A$, and another C$^*$-algebra $B$ so that it's extension to $A$ is injective?
The only thing which springs to mind is that if we insist that $varphi$ is an isometry, then its extension will also be an isometry, hence injective.
operator-algebras c-star-algebras
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In the proof of Lemma A.4 of this document, the author proves that a C$^*$-algebra is simple by showing that every one of its representations is faithful. They proceed by taking a representation and show it is injective on a (fixed) dense $^*$-subalgebra. Then they claim that the restriction of the representation to this subalgebra is faithful (obvious) and that the extension back to the original algebra is faithful (valid conclusion?).
Perusing other questions on here it would appear that there is a problem with the final claim, if we just look at extending continuous injective functions then you can't just conclude that the extension is injective, and if we look at extending injective $^*$-homomorphisms on dense $^*$-subalgebras, then there are still problems in the general case.
I suppose that my question would be the following; what extra conditions may we impose on an injective $^*$-homomorphism $varphi:mathcalDrightarrow B$ between a dense $^*$-subalgebra $mathcalD$ of a C$^*$-algebra $A$, and another C$^*$-algebra $B$ so that it's extension to $A$ is injective?
The only thing which springs to mind is that if we insist that $varphi$ is an isometry, then its extension will also be an isometry, hence injective.
operator-algebras c-star-algebras
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
In the proof of Lemma A.4 of this document, the author proves that a C$^*$-algebra is simple by showing that every one of its representations is faithful. They proceed by taking a representation and show it is injective on a (fixed) dense $^*$-subalgebra. Then they claim that the restriction of the representation to this subalgebra is faithful (obvious) and that the extension back to the original algebra is faithful (valid conclusion?).
Perusing other questions on here it would appear that there is a problem with the final claim, if we just look at extending continuous injective functions then you can't just conclude that the extension is injective, and if we look at extending injective $^*$-homomorphisms on dense $^*$-subalgebras, then there are still problems in the general case.
I suppose that my question would be the following; what extra conditions may we impose on an injective $^*$-homomorphism $varphi:mathcalDrightarrow B$ between a dense $^*$-subalgebra $mathcalD$ of a C$^*$-algebra $A$, and another C$^*$-algebra $B$ so that it's extension to $A$ is injective?
The only thing which springs to mind is that if we insist that $varphi$ is an isometry, then its extension will also be an isometry, hence injective.
operator-algebras c-star-algebras
In the proof of Lemma A.4 of this document, the author proves that a C$^*$-algebra is simple by showing that every one of its representations is faithful. They proceed by taking a representation and show it is injective on a (fixed) dense $^*$-subalgebra. Then they claim that the restriction of the representation to this subalgebra is faithful (obvious) and that the extension back to the original algebra is faithful (valid conclusion?).
Perusing other questions on here it would appear that there is a problem with the final claim, if we just look at extending continuous injective functions then you can't just conclude that the extension is injective, and if we look at extending injective $^*$-homomorphisms on dense $^*$-subalgebras, then there are still problems in the general case.
I suppose that my question would be the following; what extra conditions may we impose on an injective $^*$-homomorphism $varphi:mathcalDrightarrow B$ between a dense $^*$-subalgebra $mathcalD$ of a C$^*$-algebra $A$, and another C$^*$-algebra $B$ so that it's extension to $A$ is injective?
The only thing which springs to mind is that if we insist that $varphi$ is an isometry, then its extension will also be an isometry, hence injective.
operator-algebras c-star-algebras
asked Aug 9 at 19:02
user505379
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First, as you say it is not true that a faithful representation on a dense subalgebra extends to a faithful representation. For an example, let $A=C[0,1]$, $D$ the polynomials, and let $pi:Dto ell^infty$ be given by
$$
pi(p)=(p(1),p(1/2),p(1/3),ldots).
$$
Then $pi$ is faithful, but its (unique) extension to $C[0,1]$ is not (as there exist many functions $f$ with $f(1/n)=0$ for all $ninmathbb N$).
As for a sufficient condition, one possibility is to require $D$ to have the property that for any positive $ain A$ there exists positive $din D$ with $dleq a$. Then $pi(a)geq pi(d)$, and so $pi(a)$ cannot be zero for any positive $ain A$.
The polynomials aren't dense in $C_0(mathbb R)$, in fact, the only polynomial in $C_0(mathbb R)$ is the constant $0$ polynomial.
â Aweygan
Aug 9 at 22:49
But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
â Aweygan
Aug 9 at 23:01
You are right, I didn't use $[0,1]$ to go for the easy sequence, but it doesn't work. I'll edit.
â Martin Argerami
Aug 10 at 2:31
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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
First, as you say it is not true that a faithful representation on a dense subalgebra extends to a faithful representation. For an example, let $A=C[0,1]$, $D$ the polynomials, and let $pi:Dto ell^infty$ be given by
$$
pi(p)=(p(1),p(1/2),p(1/3),ldots).
$$
Then $pi$ is faithful, but its (unique) extension to $C[0,1]$ is not (as there exist many functions $f$ with $f(1/n)=0$ for all $ninmathbb N$).
As for a sufficient condition, one possibility is to require $D$ to have the property that for any positive $ain A$ there exists positive $din D$ with $dleq a$. Then $pi(a)geq pi(d)$, and so $pi(a)$ cannot be zero for any positive $ain A$.
The polynomials aren't dense in $C_0(mathbb R)$, in fact, the only polynomial in $C_0(mathbb R)$ is the constant $0$ polynomial.
â Aweygan
Aug 9 at 22:49
But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
â Aweygan
Aug 9 at 23:01
You are right, I didn't use $[0,1]$ to go for the easy sequence, but it doesn't work. I'll edit.
â Martin Argerami
Aug 10 at 2:31
add a comment |Â
up vote
1
down vote
First, as you say it is not true that a faithful representation on a dense subalgebra extends to a faithful representation. For an example, let $A=C[0,1]$, $D$ the polynomials, and let $pi:Dto ell^infty$ be given by
$$
pi(p)=(p(1),p(1/2),p(1/3),ldots).
$$
Then $pi$ is faithful, but its (unique) extension to $C[0,1]$ is not (as there exist many functions $f$ with $f(1/n)=0$ for all $ninmathbb N$).
As for a sufficient condition, one possibility is to require $D$ to have the property that for any positive $ain A$ there exists positive $din D$ with $dleq a$. Then $pi(a)geq pi(d)$, and so $pi(a)$ cannot be zero for any positive $ain A$.
The polynomials aren't dense in $C_0(mathbb R)$, in fact, the only polynomial in $C_0(mathbb R)$ is the constant $0$ polynomial.
â Aweygan
Aug 9 at 22:49
But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
â Aweygan
Aug 9 at 23:01
You are right, I didn't use $[0,1]$ to go for the easy sequence, but it doesn't work. I'll edit.
â Martin Argerami
Aug 10 at 2:31
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First, as you say it is not true that a faithful representation on a dense subalgebra extends to a faithful representation. For an example, let $A=C[0,1]$, $D$ the polynomials, and let $pi:Dto ell^infty$ be given by
$$
pi(p)=(p(1),p(1/2),p(1/3),ldots).
$$
Then $pi$ is faithful, but its (unique) extension to $C[0,1]$ is not (as there exist many functions $f$ with $f(1/n)=0$ for all $ninmathbb N$).
As for a sufficient condition, one possibility is to require $D$ to have the property that for any positive $ain A$ there exists positive $din D$ with $dleq a$. Then $pi(a)geq pi(d)$, and so $pi(a)$ cannot be zero for any positive $ain A$.
First, as you say it is not true that a faithful representation on a dense subalgebra extends to a faithful representation. For an example, let $A=C[0,1]$, $D$ the polynomials, and let $pi:Dto ell^infty$ be given by
$$
pi(p)=(p(1),p(1/2),p(1/3),ldots).
$$
Then $pi$ is faithful, but its (unique) extension to $C[0,1]$ is not (as there exist many functions $f$ with $f(1/n)=0$ for all $ninmathbb N$).
As for a sufficient condition, one possibility is to require $D$ to have the property that for any positive $ain A$ there exists positive $din D$ with $dleq a$. Then $pi(a)geq pi(d)$, and so $pi(a)$ cannot be zero for any positive $ain A$.
edited Aug 10 at 2:32
answered Aug 9 at 21:37
Martin Argerami
116k1071164
116k1071164
The polynomials aren't dense in $C_0(mathbb R)$, in fact, the only polynomial in $C_0(mathbb R)$ is the constant $0$ polynomial.
â Aweygan
Aug 9 at 22:49
But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
â Aweygan
Aug 9 at 23:01
You are right, I didn't use $[0,1]$ to go for the easy sequence, but it doesn't work. I'll edit.
â Martin Argerami
Aug 10 at 2:31
add a comment |Â
The polynomials aren't dense in $C_0(mathbb R)$, in fact, the only polynomial in $C_0(mathbb R)$ is the constant $0$ polynomial.
â Aweygan
Aug 9 at 22:49
But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
â Aweygan
Aug 9 at 23:01
You are right, I didn't use $[0,1]$ to go for the easy sequence, but it doesn't work. I'll edit.
â Martin Argerami
Aug 10 at 2:31
The polynomials aren't dense in $C_0(mathbb R)$, in fact, the only polynomial in $C_0(mathbb R)$ is the constant $0$ polynomial.
â Aweygan
Aug 9 at 22:49
The polynomials aren't dense in $C_0(mathbb R)$, in fact, the only polynomial in $C_0(mathbb R)$ is the constant $0$ polynomial.
â Aweygan
Aug 9 at 22:49
But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
â Aweygan
Aug 9 at 23:01
But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
â Aweygan
Aug 9 at 23:01
You are right, I didn't use $[0,1]$ to go for the easy sequence, but it doesn't work. I'll edit.
â Martin Argerami
Aug 10 at 2:31
You are right, I didn't use $[0,1]$ to go for the easy sequence, but it doesn't work. I'll edit.
â Martin Argerami
Aug 10 at 2:31
add a comment |Â
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