Mixing
Verapoulous Algebra $C(K) mathbin{hatotimes} C(L)$ is a subalgebra of $C(Ktimes L)$?
$begingroup$
Let $K$ and $L$ be compact spaces. Consider the Banach algebra $V(K,L)=C(K)mathbin{hatotimes} C(L)$ , which is the completion of the $C(K)otimes C(L)$ with respect to the projective tensor norm. It is known that $V(K,L)$ is a subalgebra of $C(Ktimes L)$ , not isometrically. I am trying to establish this result . Heres what I have figured out.
Due to the universal property of tensor product , there is an algebra homomorphism $theta : C(K)otimes C(L)to C(Ktimes L)$ such that $$theta(sum_{i=1}^nf_iotimes g_i)(x,y)=sum_{i=1}^nf_i(x)g_i(y)$$
I want to show that this map is injective. Once this is done then we can extend it to its completion.
functional-analysis tensor-products operator-algebras c-star-algebras banach-algebras
edited Jan 14 at 10:39
Andreas Caranti
56.6k34395
asked Jan 14 at 10:30
Lav KumarLav Kumar
219112
$endgroup$
add a comment |
$begingroup$
Let $K$ and $L$ be compact spaces. Consider the Banach algebra $V(K,L)=C(K)mathbin{hatotimes} C(L)$ , which is the completion of the $C(K)otimes C(L)$ with respect to the projective tensor norm. It is known that $V(K,L)$ is a subalgebra of $C(Ktimes L)$ , not isometrically. I am trying to establish this result . Heres what I have figured out.
Due to the universal property of tensor product , there is an algebra homomorphism $theta : C(K)otimes C(L)to C(Ktimes L)$ such that $$theta(sum_{i=1}^nf_iotimes g_i)(x,y)=sum_{i=1}^nf_i(x)g_i(y)$$
I want to show that this map is injective. Once this is done then we can extend it to its completion.
functional-analysis tensor-products operator-algebras c-star-algebras banach-algebras
edited Jan 14 at 10:39
Andreas Caranti
56.6k34395
asked Jan 14 at 10:30
Lav KumarLav Kumar
219112
$endgroup$
$begingroup$
But be careful: The extension of an injective uniformly continuous map to the completion needn't be injective! Consider a Banach space $(X,|cdot|)$ and a strictly finer norm $||cdot||$ on it (e.g., $|x|+|f(x)|$ for a discontinuous linear functional). Then the identity from $(X,||cdot||)$ to $(X,|cdot|)$ is continuous and injective but its extension to the completion is not injective (because otherwise its inverse would be continuous by the closed graph theorem).
$endgroup$
– Jochen
Jan 15 at 13:47
$begingroup$
@jochen Thanks for reminding , I was actually thinking the same .
$endgroup$
– Lav Kumar
Jan 16 at 13:55
add a comment |
$begingroup$
Let $K$ and $L$ be compact spaces. Consider the Banach algebra $V(K,L)=C(K)mathbin{hatotimes} C(L)$ , which is the completion of the $C(K)otimes C(L)$ with respect to the projective tensor norm. It is known that $V(K,L)$ is a subalgebra of $C(Ktimes L)$ , not isometrically. I am trying to establish this result . Heres what I have figured out.
Due to the universal property of tensor product , there is an algebra homomorphism $theta : C(K)otimes C(L)to C(Ktimes L)$ such that $$theta(sum_{i=1}^nf_iotimes g_i)(x,y)=sum_{i=1}^nf_i(x)g_i(y)$$
I want to show that this map is injective. Once this is done then we can extend it to its completion.
functional-analysis tensor-products operator-algebras c-star-algebras banach-algebras
edited Jan 14 at 10:39
Andreas Caranti
56.6k34395
asked Jan 14 at 10:30
Lav KumarLav Kumar
219112
$endgroup$
Let $K$ and $L$ be compact spaces. Consider the Banach algebra $V(K,L)=C(K)mathbin{hatotimes} C(L)$ , which is the completion of the $C(K)otimes C(L)$ with respect to the projective tensor norm. It is known that $V(K,L)$ is a subalgebra of $C(Ktimes L)$ , not isometrically. I am trying to establish this result . Heres what I have figured out.
Due to the universal property of tensor product , there is an algebra homomorphism $theta : C(K)otimes C(L)to C(Ktimes L)$ such that $$theta(sum_{i=1}^nf_iotimes g_i)(x,y)=sum_{i=1}^nf_i(x)g_i(y)$$
I want to show that this map is injective. Once this is done then we can extend it to its completion.
functional-analysis tensor-products operator-algebras c-star-algebras banach-algebras
functional-analysis tensor-products operator-algebras c-star-algebras banach-algebras
edited Jan 14 at 10:39
Andreas Caranti
56.6k34395
asked Jan 14 at 10:30
Lav KumarLav Kumar
219112
edited Jan 14 at 10:39
Andreas Caranti
56.6k34395
asked Jan 14 at 10:30
Lav KumarLav Kumar
219112
edited Jan 14 at 10:39
Andreas Caranti
56.6k34395
edited Jan 14 at 10:39
Andreas Caranti
56.6k34395
edited Jan 14 at 10:39
Andreas Caranti
56.6k34395
56.6k34395
asked Jan 14 at 10:30
Lav KumarLav Kumar
219112
asked Jan 14 at 10:30
Lav KumarLav Kumar
219112
asked Jan 14 at 10:30
Lav KumarLav Kumar
219112
219112
$begingroup$
But be careful: The extension of an injective uniformly continuous map to the completion needn't be injective! Consider a Banach space $(X,|cdot|)$ and a strictly finer norm $||cdot||$ on it (e.g., $|x|+|f(x)|$ for a discontinuous linear functional). Then the identity from $(X,||cdot||)$ to $(X,|cdot|)$ is continuous and injective but its extension to the completion is not injective (because otherwise its inverse would be continuous by the closed graph theorem).
$endgroup$
– Jochen
Jan 15 at 13:47
$begingroup$
@jochen Thanks for reminding , I was actually thinking the same .
$endgroup$
– Lav Kumar
Jan 16 at 13:55
add a comment |
$begingroup$
But be careful: The extension of an injective uniformly continuous map to the completion needn't be injective! Consider a Banach space $(X,|cdot|)$ and a strictly finer norm $||cdot||$ on it (e.g., $|x|+|f(x)|$ for a discontinuous linear functional). Then the identity from $(X,||cdot||)$ to $(X,|cdot|)$ is continuous and injective but its extension to the completion is not injective (because otherwise its inverse would be continuous by the closed graph theorem).
$endgroup$
– Jochen
Jan 15 at 13:47
$begingroup$
@jochen Thanks for reminding , I was actually thinking the same .
$endgroup$
– Lav Kumar
Jan 16 at 13:55
$begingroup$
But be careful: The extension of an injective uniformly continuous map to the completion needn't be injective! Consider a Banach space $(X,|cdot|)$ and a strictly finer norm $||cdot||$ on it (e.g., $|x|+|f(x)|$ for a discontinuous linear functional). Then the identity from $(X,||cdot||)$ to $(X,|cdot|)$ is continuous and injective but its extension to the completion is not injective (because otherwise its inverse would be continuous by the closed graph theorem).
$endgroup$
– Jochen
Jan 15 at 13:47
$begingroup$
But be careful: The extension of an injective uniformly continuous map to the completion needn't be injective! Consider a Banach space $(X,|cdot|)$ and a strictly finer norm $||cdot||$ on it (e.g., $|x|+|f(x)|$ for a discontinuous linear functional). Then the identity from $(X,||cdot||)$ to $(X,|cdot|)$ is continuous and injective but its extension to the completion is not injective (because otherwise its inverse would be continuous by the closed graph theorem).
$endgroup$
– Jochen
Jan 15 at 13:47
$begingroup$
@jochen Thanks for reminding , I was actually thinking the same .
$endgroup$
– Lav Kumar
Jan 16 at 13:55
$begingroup$
@jochen Thanks for reminding , I was actually thinking the same .
$endgroup$
– Lav Kumar
Jan 16 at 13:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $hin C(K)otimes C(L)$ and $theta(h)=0$. Write $h=sum_{i=1}^nf_iotimes g_i$, where the $g_i$ are linearly independent. But then this implies that for all $xin K$,
$$0=theta(h)(x,cdot)=sum_{i=1}^nf_i(x)g_i(cdot),$$
whence $f_i=0$ for all $i$, and thus $h=0$.
answered Jan 14 at 14:52
AweyganAweygan
14.2k21441
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Lav Kumar
Jan 14 at 16:59
1
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 17:00
$begingroup$
Is the choice of representation of h with g's linearly independent comes from smallest representation of h as sums of elementary tensors ?
$endgroup$
– Lav Kumar
Jan 14 at 17:02
$begingroup$
I wouldn't think so, but possibly. You might be able to do it with the $f_i$ linearly independent and obtain a representation with less elementary tensors.
$endgroup$
– Aweygan
Jan 14 at 17:09
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Suppose $hin C(K)otimes C(L)$ and $theta(h)=0$. Write $h=sum_{i=1}^nf_iotimes g_i$, where the $g_i$ are linearly independent. But then this implies that for all $xin K$,
$$0=theta(h)(x,cdot)=sum_{i=1}^nf_i(x)g_i(cdot),$$
whence $f_i=0$ for all $i$, and thus $h=0$.
answered Jan 14 at 14:52
AweyganAweygan
14.2k21441
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Lav Kumar
Jan 14 at 16:59
1
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 17:00
$begingroup$
Is the choice of representation of h with g's linearly independent comes from smallest representation of h as sums of elementary tensors ?
$endgroup$
– Lav Kumar
Jan 14 at 17:02
$begingroup$
I wouldn't think so, but possibly. You might be able to do it with the $f_i$ linearly independent and obtain a representation with less elementary tensors.
$endgroup$
– Aweygan
Jan 14 at 17:09
add a comment |
$begingroup$
Suppose $hin C(K)otimes C(L)$ and $theta(h)=0$. Write $h=sum_{i=1}^nf_iotimes g_i$, where the $g_i$ are linearly independent. But then this implies that for all $xin K$,
$$0=theta(h)(x,cdot)=sum_{i=1}^nf_i(x)g_i(cdot),$$
whence $f_i=0$ for all $i$, and thus $h=0$.
answered Jan 14 at 14:52
AweyganAweygan
14.2k21441
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Lav Kumar
Jan 14 at 16:59
1
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 17:00
$begingroup$
Is the choice of representation of h with g's linearly independent comes from smallest representation of h as sums of elementary tensors ?
$endgroup$
– Lav Kumar
Jan 14 at 17:02
$begingroup$
I wouldn't think so, but possibly. You might be able to do it with the $f_i$ linearly independent and obtain a representation with less elementary tensors.
$endgroup$
– Aweygan
Jan 14 at 17:09
add a comment |
$begingroup$
Suppose $hin C(K)otimes C(L)$ and $theta(h)=0$. Write $h=sum_{i=1}^nf_iotimes g_i$, where the $g_i$ are linearly independent. But then this implies that for all $xin K$,
$$0=theta(h)(x,cdot)=sum_{i=1}^nf_i(x)g_i(cdot),$$
whence $f_i=0$ for all $i$, and thus $h=0$.
answered Jan 14 at 14:52
AweyganAweygan
14.2k21441
$endgroup$
Suppose $hin C(K)otimes C(L)$ and $theta(h)=0$. Write $h=sum_{i=1}^nf_iotimes g_i$, where the $g_i$ are linearly independent. But then this implies that for all $xin K$,
$$0=theta(h)(x,cdot)=sum_{i=1}^nf_i(x)g_i(cdot),$$
whence $f_i=0$ for all $i$, and thus $h=0$.
answered Jan 14 at 14:52
AweyganAweygan
14.2k21441
answered Jan 14 at 14:52
AweyganAweygan
14.2k21441
answered Jan 14 at 14:52
AweyganAweygan
14.2k21441
answered Jan 14 at 14:52
AweyganAweygan
14.2k21441
14.2k21441
$begingroup$
Thank you so much.
$endgroup$
– Lav Kumar
Jan 14 at 16:59
1
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 17:00
$begingroup$
Is the choice of representation of h with g's linearly independent comes from smallest representation of h as sums of elementary tensors ?
$endgroup$
– Lav Kumar
Jan 14 at 17:02
$begingroup$
I wouldn't think so, but possibly. You might be able to do it with the $f_i$ linearly independent and obtain a representation with less elementary tensors.
$endgroup$
– Aweygan
Jan 14 at 17:09
add a comment |
$begingroup$
Thank you so much.
$endgroup$
– Lav Kumar
Jan 14 at 16:59
1
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 17:00
$begingroup$
Is the choice of representation of h with g's linearly independent comes from smallest representation of h as sums of elementary tensors ?
$endgroup$
– Lav Kumar
Jan 14 at 17:02
$begingroup$
I wouldn't think so, but possibly. You might be able to do it with the $f_i$ linearly independent and obtain a representation with less elementary tensors.
$endgroup$
– Aweygan
Jan 14 at 17:09
$begingroup$
Thank you so much.
$endgroup$
– Lav Kumar
Jan 14 at 16:59
$begingroup$
Thank you so much.
$endgroup$
– Lav Kumar
Jan 14 at 16:59
1
1
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 17:00
$begingroup$
You're welcome. Glad to help!
$endgroup$
– Aweygan
Jan 14 at 17:00
$begingroup$
Is the choice of representation of h with g's linearly independent comes from smallest representation of h as sums of elementary tensors ?
$endgroup$
– Lav Kumar
Jan 14 at 17:02
$begingroup$
Is the choice of representation of h with g's linearly independent comes from smallest representation of h as sums of elementary tensors ?
$endgroup$
– Lav Kumar
Jan 14 at 17:02
$begingroup$
I wouldn't think so, but possibly. You might be able to do it with the $f_i$ linearly independent and obtain a representation with less elementary tensors.
$endgroup$
– Aweygan
Jan 14 at 17:09
$begingroup$
I wouldn't think so, but possibly. You might be able to do it with the $f_i$ linearly independent and obtain a representation with less elementary tensors.
$endgroup$
– Aweygan
Jan 14 at 17:09
add a comment |
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$begingroup$
But be careful: The extension of an injective uniformly continuous map to the completion needn't be injective! Consider a Banach space $(X,|cdot|)$ and a strictly finer norm $||cdot||$ on it (e.g., $|x|+|f(x)|$ for a discontinuous linear functional). Then the identity from $(X,||cdot||)$ to $(X,|cdot|)$ is continuous and injective but its extension to the completion is not injective (because otherwise its inverse would be continuous by the closed graph theorem).
$endgroup$
– Jochen
Jan 15 at 13:47
$begingroup$
@jochen Thanks for reminding , I was actually thinking the same .
$endgroup$
– Lav Kumar
Jan 16 at 13:55