natural isomorphism from V to V**

[yifli]

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]....

(Although, I could imagine what you wrote being intended to mean this, but stated awkwardly)