Definition of Dual Vector Space used to define a Tensor
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A variant of this question has been asked before, but not the specific one I had in mind.
A tensor is defined as a multi-linear map from $l$ copies of a dual vector space $V^*$ and $m$ copies of a vector space $V$ to the field $K$ of the vector space. How has the dual vector space been defined? I have seen two definitions for the dual vector space- the set of all linear functionals and the set of bounded linear functionals (which would require a norm to be defined to define a tensor(!?)).
If we were to define a norm on $V$, would the definition of the dual space, and hence a tensor, change?
Link to a similar question: Dual space - bounded functionals?
linear-algebra tensors dual-spaces
asked Sep 6 at 6:34
D12ac
112
add a comment |Â
up vote
2
down vote
favorite
A variant of this question has been asked before, but not the specific one I had in mind.
A tensor is defined as a multi-linear map from $l$ copies of a dual vector space $V^*$ and $m$ copies of a vector space $V$ to the field $K$ of the vector space. How has the dual vector space been defined? I have seen two definitions for the dual vector space- the set of all linear functionals and the set of bounded linear functionals (which would require a norm to be defined to define a tensor(!?)).
If we were to define a norm on $V$, would the definition of the dual space, and hence a tensor, change?
Link to a similar question: Dual space - bounded functionals?
linear-algebra tensors dual-spaces
asked Sep 6 at 6:34
D12ac
112
1
This may have to do with the dimension of $V$ (if I have all my facts straight). Isn't any linear functional on a finite-dimensional (normed) vector space bounded?
– Arthur
Sep 6 at 6:37
@Arthur Yes, on a finite dimensional vector space $V$, every linear functional is bounded. On such a $V$, both definitions of the dual would be equivalent.
– D12ac
Sep 6 at 6:40
So... if we're in a setting where $V$ is assumed to be finite-dimensional (like, say, the tangent space over a point on a manifold) then we don't need to specify boundedness, while in general we do. Maybe that's what's going on with your two definitions?
– Arthur
Sep 6 at 6:50
I believe it depends on the application. In functional analysis the dual is almost always the continuous dual, but I imagine that the algebraic dual is used in other settings. See en.wikipedia.org/wiki/Dual_space#Continuous_dual_space
– Calvin Khor
Sep 6 at 6:55
@Arthur What irks me is (on infinite dimension) the need to defined a norm on a vector space to define a tensor. I don't even know if this statement is correct or not.
– D12ac
Sep 6 at 7:05
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
A variant of this question has been asked before, but not the specific one I had in mind.
A tensor is defined as a multi-linear map from $l$ copies of a dual vector space $V^*$ and $m$ copies of a vector space $V$ to the field $K$ of the vector space. How has the dual vector space been defined? I have seen two definitions for the dual vector space- the set of all linear functionals and the set of bounded linear functionals (which would require a norm to be defined to define a tensor(!?)).
If we were to define a norm on $V$, would the definition of the dual space, and hence a tensor, change?
Link to a similar question: Dual space - bounded functionals?
linear-algebra tensors dual-spaces
asked Sep 6 at 6:34
D12ac
112
A variant of this question has been asked before, but not the specific one I had in mind.
A tensor is defined as a multi-linear map from $l$ copies of a dual vector space $V^*$ and $m$ copies of a vector space $V$ to the field $K$ of the vector space. How has the dual vector space been defined? I have seen two definitions for the dual vector space- the set of all linear functionals and the set of bounded linear functionals (which would require a norm to be defined to define a tensor(!?)).
If we were to define a norm on $V$, would the definition of the dual space, and hence a tensor, change?
Link to a similar question: Dual space - bounded functionals?
linear-algebra tensors dual-spaces
linear-algebra tensors dual-spaces
asked Sep 6 at 6:34
D12ac
112
asked Sep 6 at 6:34
D12ac
112
edited Sep 6 at 6:43
asked Sep 6 at 6:34
D12ac
112
asked Sep 6 at 6:34
D12ac
112
asked Sep 6 at 6:34
D12ac
112
112
1
This may have to do with the dimension of $V$ (if I have all my facts straight). Isn't any linear functional on a finite-dimensional (normed) vector space bounded?
– Arthur
Sep 6 at 6:37
@Arthur Yes, on a finite dimensional vector space $V$, every linear functional is bounded. On such a $V$, both definitions of the dual would be equivalent.
– D12ac
Sep 6 at 6:40
So... if we're in a setting where $V$ is assumed to be finite-dimensional (like, say, the tangent space over a point on a manifold) then we don't need to specify boundedness, while in general we do. Maybe that's what's going on with your two definitions?
– Arthur
Sep 6 at 6:50
I believe it depends on the application. In functional analysis the dual is almost always the continuous dual, but I imagine that the algebraic dual is used in other settings. See en.wikipedia.org/wiki/Dual_space#Continuous_dual_space
– Calvin Khor
Sep 6 at 6:55
@Arthur What irks me is (on infinite dimension) the need to defined a norm on a vector space to define a tensor. I don't even know if this statement is correct or not.
– D12ac
Sep 6 at 7:05
add a comment |Â
1
This may have to do with the dimension of $V$ (if I have all my facts straight). Isn't any linear functional on a finite-dimensional (normed) vector space bounded?
– Arthur
Sep 6 at 6:37
@Arthur Yes, on a finite dimensional vector space $V$, every linear functional is bounded. On such a $V$, both definitions of the dual would be equivalent.
– D12ac
Sep 6 at 6:40
So... if we're in a setting where $V$ is assumed to be finite-dimensional (like, say, the tangent space over a point on a manifold) then we don't need to specify boundedness, while in general we do. Maybe that's what's going on with your two definitions?
– Arthur
Sep 6 at 6:50
I believe it depends on the application. In functional analysis the dual is almost always the continuous dual, but I imagine that the algebraic dual is used in other settings. See en.wikipedia.org/wiki/Dual_space#Continuous_dual_space
– Calvin Khor
Sep 6 at 6:55
@Arthur What irks me is (on infinite dimension) the need to defined a norm on a vector space to define a tensor. I don't even know if this statement is correct or not.
– D12ac
Sep 6 at 7:05
1
1
This may have to do with the dimension of $V$ (if I have all my facts straight). Isn't any linear functional on a finite-dimensional (normed) vector space bounded?
– Arthur
Sep 6 at 6:37
This may have to do with the dimension of $V$ (if I have all my facts straight). Isn't any linear functional on a finite-dimensional (normed) vector space bounded?
– Arthur
Sep 6 at 6:37
@Arthur Yes, on a finite dimensional vector space $V$, every linear functional is bounded. On such a $V$, both definitions of the dual would be equivalent.
– D12ac
Sep 6 at 6:40
@Arthur Yes, on a finite dimensional vector space $V$, every linear functional is bounded. On such a $V$, both definitions of the dual would be equivalent.
– D12ac
Sep 6 at 6:40
So... if we're in a setting where $V$ is assumed to be finite-dimensional (like, say, the tangent space over a point on a manifold) then we don't need to specify boundedness, while in general we do. Maybe that's what's going on with your two definitions?
– Arthur
Sep 6 at 6:50
So... if we're in a setting where $V$ is assumed to be finite-dimensional (like, say, the tangent space over a point on a manifold) then we don't need to specify boundedness, while in general we do. Maybe that's what's going on with your two definitions?
– Arthur
Sep 6 at 6:50
I believe it depends on the application. In functional analysis the dual is almost always the continuous dual, but I imagine that the algebraic dual is used in other settings. See en.wikipedia.org/wiki/Dual_space#Continuous_dual_space
– Calvin Khor
Sep 6 at 6:55
I believe it depends on the application. In functional analysis the dual is almost always the continuous dual, but I imagine that the algebraic dual is used in other settings. See en.wikipedia.org/wiki/Dual_space#Continuous_dual_space
– Calvin Khor
Sep 6 at 6:55
@Arthur What irks me is (on infinite dimension) the need to defined a norm on a vector space to define a tensor. I don't even know if this statement is correct or not.
– D12ac
Sep 6 at 7:05
@Arthur What irks me is (on infinite dimension) the need to defined a norm on a vector space to define a tensor. I don't even know if this statement is correct or not.
– D12ac
Sep 6 at 7:05
add a comment |Â
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1
This may have to do with the dimension of $V$ (if I have all my facts straight). Isn't any linear functional on a finite-dimensional (normed) vector space bounded?
– Arthur
Sep 6 at 6:37
@Arthur Yes, on a finite dimensional vector space $V$, every linear functional is bounded. On such a $V$, both definitions of the dual would be equivalent.
– D12ac
Sep 6 at 6:40
So... if we're in a setting where $V$ is assumed to be finite-dimensional (like, say, the tangent space over a point on a manifold) then we don't need to specify boundedness, while in general we do. Maybe that's what's going on with your two definitions?
– Arthur
Sep 6 at 6:50
I believe it depends on the application. In functional analysis the dual is almost always the continuous dual, but I imagine that the algebraic dual is used in other settings. See en.wikipedia.org/wiki/Dual_space#Continuous_dual_space
– Calvin Khor
Sep 6 at 6:55
@Arthur What irks me is (on infinite dimension) the need to defined a norm on a vector space to define a tensor. I don't even know if this statement is correct or not.
– D12ac
Sep 6 at 7:05