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.







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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.







    share|cite|improve this question






















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      1
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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.







      share|cite|improve this question












      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.









      share|cite|improve this question











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      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 $ain A$.






          share|cite|improve this answer






















          • 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.
            – Aweygan
            Aug 9 at 22:49










          • But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
            – 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.
            – Martin Argerami
            Aug 10 at 2:31











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          up vote
          1
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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 $ain A$.






          share|cite|improve this answer






















          • 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.
            – Aweygan
            Aug 9 at 22:49










          • But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
            – 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.
            – Martin Argerami
            Aug 10 at 2:31















          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 $ain A$.






          share|cite|improve this answer






















          • 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.
            – Aweygan
            Aug 9 at 22:49










          • But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
            – 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.
            – Martin Argerami
            Aug 10 at 2:31













          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 $ain A$.






          share|cite|improve this answer














          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 $ain A$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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.
            – Aweygan
            Aug 9 at 22:49










          • But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
            – 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.
            – 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.
            – Aweygan
            Aug 9 at 22:49










          • But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
            – 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.
            – 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.
          – 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.
          – Aweygan
          Aug 9 at 22:49












          But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
          – Aweygan
          Aug 9 at 23:01




          But this can be modified to $A=C([0,1])$, $D$ the polynomials, $pi: Dtoell^infty$ by $pi(p)=(p(1),p(frac12),p(frac13),ldots)$.
          – 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.
          – 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.
          – Martin Argerami
          Aug 10 at 2:31













           

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