Natural isomorphisms between dual spaces

[wisvuze]

EDIT: finite dimensional only!

Hello, I would like to ask a question; I understand that the cannonical "evaluation map" ( (p(v))(f) = f(v) , f is a functional, v in V ) from V -> V** is a "natural" isomorphism ( we don't have to select any bases, the isomorphism relies on no choices ), so V and V** are naturally isomorphic. However, we usually establish an isomorphism from V to V* by the use of dual bases, which involves a selection of bases and is not "natural".
I've been told that V and V* are not naturally isomorphic, can we not find an isomorphism from V to V* that requires no choice? I can't think of one on my own, but that is certainly no proof.

Thanks

 

[Deveno]

if V is not finite-dimensional, V* is not isomorphic to V, it's "bigger"

 

[wisvuze]

Sorry, I meant finite dimensional only. And it's also true that V** and V won't be isomorphic in the infinite dimensional case

 

[Deveno]

think of it this way: suppose you have V, and V*. given v in V, how can you unambiguously choose an f in V* that goes with v? you can't just use f(v), because there are multiple f in V* wih f(v) all the same. you have to compare how f acts on a basis to distinguish it from all the other linear functionals (and a basis is the smallest set of vectors that will do).

in other words, to figure out which linear functional you have, you need n independent "test-values". this is because f collapses n values to 1 value. the reason we DON'T need to do this for V**, is that if we have n^2 values collapsing down to n, that is the same as having n values (a vector in F^n) to start with.

 

[Landau]

I've been told that V and V* are not naturally isomorphic, can we not find an isomorphism from V to V* that requires no choice? I can't think of one on my own, but that is certainly no proof.

To actually prove this, one of course needs to make the notion of 'natural isomorphism' precise, which has been done in category theory. See here for a proof of the fact there there is no natural isomorphism between the identity functor and the dual functor.

 

[wisvuze]

think of it this way: suppose you have V, and V*. given v in V, how can you unambiguously choose an f in V* that goes with v? you can't just use f(v), because there are multiple f in V* wih f(v) all the same. you have to compare how f acts on a basis to distinguish it from all the other linear functionals (and a basis is the smallest set of vectors that will do).

in other words, to figure out which linear functional you have, you need n independent "test-values". this is because f collapses n values to 1 value. the reason we DON'T need to do this for V**, is that if we have n^2 values collapsing down to n, that is the same as having n values (a vector in F^n) to start with.

great reply! Thanks! That's a cool way of thinking about it

 

[wisvuze]

To actually prove this, one of course needs to make the notion of 'natural isomorphism' precise, which has been done in category theory. See here for a proof of the fact there there is no natural isomorphism between the identity functor and the dual functor.

thanks, I was just looking at this in my algebra book