- #1
jojo12345
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When considering finite dimensional vector spaces V and W over a field K, there exists a natural isomorphism between their tensor product and the space of bilinear maps from the cartesian product of the dual spaces to the underlying field. However, the text I'm reading asserts that if V and W are infinite dimensional, then the two spaces are not isomorphic.
In the finite dimensional case, one natural isomorphism between the two spaces is given by
T:V\otimes W\rightarrow (V^{*},W^{*})^{*}
where
T(A)=\bar{A},\forall A\in V\otimes W
and
\bar{A}(\rho,\psi)=(\bar{a^{r}v_{r}\otimes w_{r}})(\rho,\psi)=a^{r}\rho(v_{r})\psi(w_{r}) where I'm using the summation convention. There isn't some fixed range for the sums- they must only be finite.
What I am trying to show is that this mapping is not surjective when V and W are not finite dimensional. I have an argument to suggest this is the case, but I would appreciate being corrected or having my thoughts confirmed.
Let S_{1}=\{ \rho{}_{1},\rho{}_{2},...\} be a countable subset of a basis for V^{*} and S_{2}=\{ \psi{}_{1},\psi{}_{2},...\} be a countable subset of a basis for W^{*}. Now let F\in(V^{*},W^{*})^{*} be a bilinear map such that F(\rho{}_{i},\psi{}_{j})=k_{i,j}. There will not always be a tensor that maps to this function because every tensor is specified by a finite number of scalar components (they are vectors after all) and the system of equations one would have to solve in order to determine the components is \displaystyle very overdetermined.
The things I'm most unsure about are my ability to restrict F as I have and how the fact that a tensor can be represented by an arbitrarily large number of components affects the last sentence in my argument.
As a side note, I am also wondering about how to go about proving no isomorphism exists in the case of infinite dimensional spaces for even if what I tried to prove above is true, it only rules out a single map from being an isomorphism. It seems to me that perhaps my argument applies to any linear map between the two spaces. However, if my argument above doesn't hold water, then I have a bit more thinking to do.
In the finite dimensional case, one natural isomorphism between the two spaces is given by
T:V\otimes W\rightarrow (V^{*},W^{*})^{*}
where
T(A)=\bar{A},\forall A\in V\otimes W
and
\bar{A}(\rho,\psi)=(\bar{a^{r}v_{r}\otimes w_{r}})(\rho,\psi)=a^{r}\rho(v_{r})\psi(w_{r}) where I'm using the summation convention. There isn't some fixed range for the sums- they must only be finite.
What I am trying to show is that this mapping is not surjective when V and W are not finite dimensional. I have an argument to suggest this is the case, but I would appreciate being corrected or having my thoughts confirmed.
Let S_{1}=\{ \rho{}_{1},\rho{}_{2},...\} be a countable subset of a basis for V^{*} and S_{2}=\{ \psi{}_{1},\psi{}_{2},...\} be a countable subset of a basis for W^{*}. Now let F\in(V^{*},W^{*})^{*} be a bilinear map such that F(\rho{}_{i},\psi{}_{j})=k_{i,j}. There will not always be a tensor that maps to this function because every tensor is specified by a finite number of scalar components (they are vectors after all) and the system of equations one would have to solve in order to determine the components is \displaystyle very overdetermined.
The things I'm most unsure about are my ability to restrict F as I have and how the fact that a tensor can be represented by an arbitrarily large number of components affects the last sentence in my argument.
As a side note, I am also wondering about how to go about proving no isomorphism exists in the case of infinite dimensional spaces for even if what I tried to prove above is true, it only rules out a single map from being an isomorphism. It seems to me that perhaps my argument applies to any linear map between the two spaces. However, if my argument above doesn't hold water, then I have a bit more thinking to do.