Is There a Natural Isomorphism from V to V**?

[yifli]

natural isomorphism from V to V**

It is known that there is a natural isomorphism $$\epsilon \rightleftharpoons \omega^\epsilon$$ from V to V**, where $$\omega: V \times V* \rightarrow R$$ is a bilinear mapping.

So given a certain $$\epsilon \in V$$, its image under the isomorphism is actually a set of values $$\left\{f(\epsilon),f \in V^*\right\}$$, i.e., a vector is mapped to a set of numbers

Is my understanding correct?

Thanks

 

[lavinia]

It is known that there is a natural isomorphism $$\epsilon \rightleftharpoons \omega^\epsilon$$ from V to V**, where $$\omega: V \times V* \rightarrow R$$ is a bilinear mapping.

So given a certain $$\epsilon \in V$$, its image under the isomorphism is actually a set of values $$\left\{f(\epsilon),f \in V^*\right\}$$, i.e., a vector is mapped to a set of numbers

Is my understanding correct?

Thanks

There is no natural isomorphism between V and V*. What isomorphism are you thinking of?

Generally, any isomorphism requires the choice of an inner product where a vector in V is identified with a linear map from V into the base field. You can think of a linear map as a set of numbers but it is more than that because it is not just any map but is a linear map.

 

[HallsofIvy]

You are misreading what he said. There is no natural isomorphism from V to V*, the dual space, (if V has infinite dimension) but there is from V to V**, the dual of the dual.

 

[yifli]

There is no natural isomorphism between V and V*. What isomorphism are you thinking of?

it 's true that there's no natural isomorphism between V and V*, but I'm talking about the natural isomorphism between V and V**

 

[Hurkyl]

So given a certain $$\epsilon \in V$$, its image under the isomorphism is actually a set of values $$\left\{f(\epsilon),f \in V^*\right\}$$, i.e., a vector is mapped to a set of numbers

Is my understanding correct?

Not really. The image of $\epsilon$ is the function that takes $f \in V^*$ and maps it to $f(\epsilon)$. I.E. $\omega^\epsilon$ the function defined by

$$\omega^\epsilon(f) = f(\epsilon)$$​

Incidentally, I assume by $\omega$ you mean not a general bilinear mapping, but instead the specific map [itex]\omega(x,f) = f(x)[/tex]...<br /> <br /> (Although, I could imagine what you wrote being intended to mean this, but stated awkwardly)[/itex]