Is $(V_1otimescdotsotimes V_k)^ast simeq V_1^astotimes cdots otimes V_k^ast$ true for infinite dimensional spaces?
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Suppose $V_1,dots,V_k$ are vector spaces of finite dimension. Then I could prove easily that $(V_1otimescdotsotimes V_k)^astsimeq V_1^astotimescdotsotimes V_k^ast$. My proof was like that: first of all, I've shown that if for each $i$ we have $W_i$ another vector space such that $V_isimeq W_i$ then $V_1otimescdotsotimes V_k simeq W_1otimescdotsotimes W_k$.
Then, since I'm supposing each $V_i$ finite dimensional, each $V_isimeq V_i^ast$ and also, we have $V_1otimescdotsotimes V_k$ also finite dimensional, so that
$$(V_1otimescdotsotimes V_k)^ast simeq V_1otimescdotsotimes V_ksimeq V_1^astotimes cdots otimes V_k^ast$$
and so it is proved. Now, if the spaces are not finite dimensional this proof cannot be used. In that case, the property still holds? Is it possible to prove for infinite dimensional spaces?
I've tried to prove it directly, constructing an isomorphism. I've picked first the mapping $psi : V_1^asttimescdotstimes V_k^ast to mathcalL(V_1,dots,V_k;mathbbK)$ given by
$$psi(f_1,dots,f_k)(v_1,dots,v_k) = f_1(v_1)cdots f_k(v_k)$$
this map is multilinear and hence by the universal property there corresponds a unique linear mapping $phi : V_1^astotimes cdots otimes V_k^ast to mathcalL(V_1,dots,V_k;mathbbK)$ such that:
$$phi(f_1otimescdotsotimes f_k)(v_1,dots,v_k) = f_1(v_1)cdots f_k(v_k)$$
To show that this $phi$ is isomorphis I would need to find an inverse, but I didn't have any idea. Is it possible to complete this proof?
Thanks very much in advance.
linear-algebra multilinear-algebra
asked Feb 4 '14 at 19:25
user1620696
11k339104
add a comment |Â
up vote
11
down vote
favorite
Suppose $V_1,dots,V_k$ are vector spaces of finite dimension. Then I could prove easily that $(V_1otimescdotsotimes V_k)^astsimeq V_1^astotimescdotsotimes V_k^ast$. My proof was like that: first of all, I've shown that if for each $i$ we have $W_i$ another vector space such that $V_isimeq W_i$ then $V_1otimescdotsotimes V_k simeq W_1otimescdotsotimes W_k$.
Then, since I'm supposing each $V_i$ finite dimensional, each $V_isimeq V_i^ast$ and also, we have $V_1otimescdotsotimes V_k$ also finite dimensional, so that
$$(V_1otimescdotsotimes V_k)^ast simeq V_1otimescdotsotimes V_ksimeq V_1^astotimes cdots otimes V_k^ast$$
and so it is proved. Now, if the spaces are not finite dimensional this proof cannot be used. In that case, the property still holds? Is it possible to prove for infinite dimensional spaces?
I've tried to prove it directly, constructing an isomorphism. I've picked first the mapping $psi : V_1^asttimescdotstimes V_k^ast to mathcalL(V_1,dots,V_k;mathbbK)$ given by
$$psi(f_1,dots,f_k)(v_1,dots,v_k) = f_1(v_1)cdots f_k(v_k)$$
this map is multilinear and hence by the universal property there corresponds a unique linear mapping $phi : V_1^astotimes cdots otimes V_k^ast to mathcalL(V_1,dots,V_k;mathbbK)$ such that:
$$phi(f_1otimescdotsotimes f_k)(v_1,dots,v_k) = f_1(v_1)cdots f_k(v_k)$$
To show that this $phi$ is isomorphis I would need to find an inverse, but I didn't have any idea. Is it possible to complete this proof?
Thanks very much in advance.
linear-algebra multilinear-algebra
asked Feb 4 '14 at 19:25
user1620696
11k339104
5
1. The infinite-dimensional case is false. A counterexample is the element $e_1^ast otimes e_1^ast + e_2^ast otimes e_2^ast + e_3^ast otimes e_3^ast + ...$ in $left(Votimes Vright)^ast$, where $V$ is the free vector space with basis $left(e_1,e_2,e_3,...right)$, and where $left(e_1^ast, e_2^ast, e_3^ast, ...right)$ denotes the corresponding dual basis of $V^ast$. This is well-defined (check why! it's not like every infinite sum automatically makes sense, but this one does) but manifestly not in $V^ast otimes V^ast$.
– darij grinberg
Feb 4 '14 at 20:04
3
2. Your proof of the finite-dimensional case is correct, but it proves the letter of the statement, not the spirit. What you should be proving is that the canonical injection $V_1^ast otimes V_2^ast otimes ... otimes V_k^ast to left(V_1 otimes V_2 otimes ... otimes V_kright)^ast$ (which sends $f_1 otimes f_2 otimes ... otimes f_k$ to $f_1 otimes f_2 otimes ... otimes f_k$ with the understanding that the tensor signs in these two terms have completely different meanings!) is an isomorphism. This is a stronger claim!
– darij grinberg
Feb 4 '14 at 20:07
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
Suppose $V_1,dots,V_k$ are vector spaces of finite dimension. Then I could prove easily that $(V_1otimescdotsotimes V_k)^astsimeq V_1^astotimescdotsotimes V_k^ast$. My proof was like that: first of all, I've shown that if for each $i$ we have $W_i$ another vector space such that $V_isimeq W_i$ then $V_1otimescdotsotimes V_k simeq W_1otimescdotsotimes W_k$.
Then, since I'm supposing each $V_i$ finite dimensional, each $V_isimeq V_i^ast$ and also, we have $V_1otimescdotsotimes V_k$ also finite dimensional, so that
$$(V_1otimescdotsotimes V_k)^ast simeq V_1otimescdotsotimes V_ksimeq V_1^astotimes cdots otimes V_k^ast$$
and so it is proved. Now, if the spaces are not finite dimensional this proof cannot be used. In that case, the property still holds? Is it possible to prove for infinite dimensional spaces?
I've tried to prove it directly, constructing an isomorphism. I've picked first the mapping $psi : V_1^asttimescdotstimes V_k^ast to mathcalL(V_1,dots,V_k;mathbbK)$ given by
$$psi(f_1,dots,f_k)(v_1,dots,v_k) = f_1(v_1)cdots f_k(v_k)$$
this map is multilinear and hence by the universal property there corresponds a unique linear mapping $phi : V_1^astotimes cdots otimes V_k^ast to mathcalL(V_1,dots,V_k;mathbbK)$ such that:
$$phi(f_1otimescdotsotimes f_k)(v_1,dots,v_k) = f_1(v_1)cdots f_k(v_k)$$
To show that this $phi$ is isomorphis I would need to find an inverse, but I didn't have any idea. Is it possible to complete this proof?
Thanks very much in advance.
linear-algebra multilinear-algebra
asked Feb 4 '14 at 19:25
user1620696
11k339104
Suppose $V_1,dots,V_k$ are vector spaces of finite dimension. Then I could prove easily that $(V_1otimescdotsotimes V_k)^astsimeq V_1^astotimescdotsotimes V_k^ast$. My proof was like that: first of all, I've shown that if for each $i$ we have $W_i$ another vector space such that $V_isimeq W_i$ then $V_1otimescdotsotimes V_k simeq W_1otimescdotsotimes W_k$.
Then, since I'm supposing each $V_i$ finite dimensional, each $V_isimeq V_i^ast$ and also, we have $V_1otimescdotsotimes V_k$ also finite dimensional, so that
$$(V_1otimescdotsotimes V_k)^ast simeq V_1otimescdotsotimes V_ksimeq V_1^astotimes cdots otimes V_k^ast$$
and so it is proved. Now, if the spaces are not finite dimensional this proof cannot be used. In that case, the property still holds? Is it possible to prove for infinite dimensional spaces?
I've tried to prove it directly, constructing an isomorphism. I've picked first the mapping $psi : V_1^asttimescdotstimes V_k^ast to mathcalL(V_1,dots,V_k;mathbbK)$ given by
$$psi(f_1,dots,f_k)(v_1,dots,v_k) = f_1(v_1)cdots f_k(v_k)$$
this map is multilinear and hence by the universal property there corresponds a unique linear mapping $phi : V_1^astotimes cdots otimes V_k^ast to mathcalL(V_1,dots,V_k;mathbbK)$ such that:
$$phi(f_1otimescdotsotimes f_k)(v_1,dots,v_k) = f_1(v_1)cdots f_k(v_k)$$
To show that this $phi$ is isomorphis I would need to find an inverse, but I didn't have any idea. Is it possible to complete this proof?
Thanks very much in advance.
linear-algebra multilinear-algebra
asked Feb 4 '14 at 19:25
user1620696
11k339104
edited Feb 4 '14 at 19:48
asked Feb 4 '14 at 19:25
user1620696
11k339104
asked Feb 4 '14 at 19:25
user1620696
11k339104
asked Feb 4 '14 at 19:25
user1620696
11k339104
11k339104
5
1. The infinite-dimensional case is false. A counterexample is the element $e_1^ast otimes e_1^ast + e_2^ast otimes e_2^ast + e_3^ast otimes e_3^ast + ...$ in $left(Votimes Vright)^ast$, where $V$ is the free vector space with basis $left(e_1,e_2,e_3,...right)$, and where $left(e_1^ast, e_2^ast, e_3^ast, ...right)$ denotes the corresponding dual basis of $V^ast$. This is well-defined (check why! it's not like every infinite sum automatically makes sense, but this one does) but manifestly not in $V^ast otimes V^ast$.
– darij grinberg
Feb 4 '14 at 20:04
3
2. Your proof of the finite-dimensional case is correct, but it proves the letter of the statement, not the spirit. What you should be proving is that the canonical injection $V_1^ast otimes V_2^ast otimes ... otimes V_k^ast to left(V_1 otimes V_2 otimes ... otimes V_kright)^ast$ (which sends $f_1 otimes f_2 otimes ... otimes f_k$ to $f_1 otimes f_2 otimes ... otimes f_k$ with the understanding that the tensor signs in these two terms have completely different meanings!) is an isomorphism. This is a stronger claim!
– darij grinberg
Feb 4 '14 at 20:07
add a comment |Â
5
1. The infinite-dimensional case is false. A counterexample is the element $e_1^ast otimes e_1^ast + e_2^ast otimes e_2^ast + e_3^ast otimes e_3^ast + ...$ in $left(Votimes Vright)^ast$, where $V$ is the free vector space with basis $left(e_1,e_2,e_3,...right)$, and where $left(e_1^ast, e_2^ast, e_3^ast, ...right)$ denotes the corresponding dual basis of $V^ast$. This is well-defined (check why! it's not like every infinite sum automatically makes sense, but this one does) but manifestly not in $V^ast otimes V^ast$.
– darij grinberg
Feb 4 '14 at 20:04
3
2. Your proof of the finite-dimensional case is correct, but it proves the letter of the statement, not the spirit. What you should be proving is that the canonical injection $V_1^ast otimes V_2^ast otimes ... otimes V_k^ast to left(V_1 otimes V_2 otimes ... otimes V_kright)^ast$ (which sends $f_1 otimes f_2 otimes ... otimes f_k$ to $f_1 otimes f_2 otimes ... otimes f_k$ with the understanding that the tensor signs in these two terms have completely different meanings!) is an isomorphism. This is a stronger claim!
– darij grinberg
Feb 4 '14 at 20:07
5
5
1. The infinite-dimensional case is false. A counterexample is the element $e_1^ast otimes e_1^ast + e_2^ast otimes e_2^ast + e_3^ast otimes e_3^ast + ...$ in $left(Votimes Vright)^ast$, where $V$ is the free vector space with basis $left(e_1,e_2,e_3,...right)$, and where $left(e_1^ast, e_2^ast, e_3^ast, ...right)$ denotes the corresponding dual basis of $V^ast$. This is well-defined (check why! it's not like every infinite sum automatically makes sense, but this one does) but manifestly not in $V^ast otimes V^ast$.
– darij grinberg
Feb 4 '14 at 20:04
1. The infinite-dimensional case is false. A counterexample is the element $e_1^ast otimes e_1^ast + e_2^ast otimes e_2^ast + e_3^ast otimes e_3^ast + ...$ in $left(Votimes Vright)^ast$, where $V$ is the free vector space with basis $left(e_1,e_2,e_3,...right)$, and where $left(e_1^ast, e_2^ast, e_3^ast, ...right)$ denotes the corresponding dual basis of $V^ast$. This is well-defined (check why! it's not like every infinite sum automatically makes sense, but this one does) but manifestly not in $V^ast otimes V^ast$.
– darij grinberg
Feb 4 '14 at 20:04
3
3
2. Your proof of the finite-dimensional case is correct, but it proves the letter of the statement, not the spirit. What you should be proving is that the canonical injection $V_1^ast otimes V_2^ast otimes ... otimes V_k^ast to left(V_1 otimes V_2 otimes ... otimes V_kright)^ast$ (which sends $f_1 otimes f_2 otimes ... otimes f_k$ to $f_1 otimes f_2 otimes ... otimes f_k$ with the understanding that the tensor signs in these two terms have completely different meanings!) is an isomorphism. This is a stronger claim!
– darij grinberg
Feb 4 '14 at 20:07
2. Your proof of the finite-dimensional case is correct, but it proves the letter of the statement, not the spirit. What you should be proving is that the canonical injection $V_1^ast otimes V_2^ast otimes ... otimes V_k^ast to left(V_1 otimes V_2 otimes ... otimes V_kright)^ast$ (which sends $f_1 otimes f_2 otimes ... otimes f_k$ to $f_1 otimes f_2 otimes ... otimes f_k$ with the understanding that the tensor signs in these two terms have completely different meanings!) is an isomorphism. This is a stronger claim!
– darij grinberg
Feb 4 '14 at 20:07
add a comment |Â
1 Answer
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The situation here is a little subtle.
The canonical inclusion $(V_1otimesdotsotimes V_k)^*hookrightarrow(V^*_1otimesdotsotimes V_k^*)$ isn't an isomorphism (it's a strict injection), but the spaces are nevertheless isomorphic by some other (non-canonical) map.
The proof that the canonical map isn't an isomorphism is explain in detail in this answer to another question.
The proof that they are isomorphic works by just calculating the dimensions on both sides and finding that they are the same. The dimensions are infinite cardinals. That proof is given here.
answered Feb 12 '15 at 19:40
Oscar Cunningham
9,49422760
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The situation here is a little subtle.
The canonical inclusion $(V_1otimesdotsotimes V_k)^*hookrightarrow(V^*_1otimesdotsotimes V_k^*)$ isn't an isomorphism (it's a strict injection), but the spaces are nevertheless isomorphic by some other (non-canonical) map.
The proof that the canonical map isn't an isomorphism is explain in detail in this answer to another question.
The proof that they are isomorphic works by just calculating the dimensions on both sides and finding that they are the same. The dimensions are infinite cardinals. That proof is given here.
answered Feb 12 '15 at 19:40
Oscar Cunningham
9,49422760
add a comment |Â
up vote
5
down vote
The situation here is a little subtle.
The canonical inclusion $(V_1otimesdotsotimes V_k)^*hookrightarrow(V^*_1otimesdotsotimes V_k^*)$ isn't an isomorphism (it's a strict injection), but the spaces are nevertheless isomorphic by some other (non-canonical) map.
The proof that the canonical map isn't an isomorphism is explain in detail in this answer to another question.
The proof that they are isomorphic works by just calculating the dimensions on both sides and finding that they are the same. The dimensions are infinite cardinals. That proof is given here.
answered Feb 12 '15 at 19:40
Oscar Cunningham
9,49422760
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The situation here is a little subtle.
The canonical inclusion $(V_1otimesdotsotimes V_k)^*hookrightarrow(V^*_1otimesdotsotimes V_k^*)$ isn't an isomorphism (it's a strict injection), but the spaces are nevertheless isomorphic by some other (non-canonical) map.
The proof that the canonical map isn't an isomorphism is explain in detail in this answer to another question.
The proof that they are isomorphic works by just calculating the dimensions on both sides and finding that they are the same. The dimensions are infinite cardinals. That proof is given here.
answered Feb 12 '15 at 19:40
Oscar Cunningham
9,49422760
The situation here is a little subtle.
The canonical inclusion $(V_1otimesdotsotimes V_k)^*hookrightarrow(V^*_1otimesdotsotimes V_k^*)$ isn't an isomorphism (it's a strict injection), but the spaces are nevertheless isomorphic by some other (non-canonical) map.
The proof that the canonical map isn't an isomorphism is explain in detail in this answer to another question.
The proof that they are isomorphic works by just calculating the dimensions on both sides and finding that they are the same. The dimensions are infinite cardinals. That proof is given here.
answered Feb 12 '15 at 19:40
Oscar Cunningham
9,49422760
edited Aug 22 at 8:14
answered Feb 12 '15 at 19:40
Oscar Cunningham
9,49422760
answered Feb 12 '15 at 19:40
Oscar Cunningham
9,49422760
answered Feb 12 '15 at 19:40
Oscar Cunningham
9,49422760
9,49422760
add a comment |Â
add a comment |Â
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5
1. The infinite-dimensional case is false. A counterexample is the element $e_1^ast otimes e_1^ast + e_2^ast otimes e_2^ast + e_3^ast otimes e_3^ast + ...$ in $left(Votimes Vright)^ast$, where $V$ is the free vector space with basis $left(e_1,e_2,e_3,...right)$, and where $left(e_1^ast, e_2^ast, e_3^ast, ...right)$ denotes the corresponding dual basis of $V^ast$. This is well-defined (check why! it's not like every infinite sum automatically makes sense, but this one does) but manifestly not in $V^ast otimes V^ast$.
– darij grinberg
Feb 4 '14 at 20:04
3
2. Your proof of the finite-dimensional case is correct, but it proves the letter of the statement, not the spirit. What you should be proving is that the canonical injection $V_1^ast otimes V_2^ast otimes ... otimes V_k^ast to left(V_1 otimes V_2 otimes ... otimes V_kright)^ast$ (which sends $f_1 otimes f_2 otimes ... otimes f_k$ to $f_1 otimes f_2 otimes ... otimes f_k$ with the understanding that the tensor signs in these two terms have completely different meanings!) is an isomorphism. This is a stronger claim!
– darij grinberg
Feb 4 '14 at 20:07