# (Non)Isomorphism of V with V** in Infinite-Dimension Case

1. Jun 8, 2014

### WWGD

Hi, please give me some leeway for my laziness here:

We have that , in the finite-dimensional case for vector spaces, V~V** in a natural, i.e.,
basis-independent way ; one way of proving this is by showing that the identity functor is
naturally-isomorphic to the double-dual functor. An easy way of showing that this is not the
case for V~V* is that the two functors ( identity and 1st-dual ) go in opposite directions.

Questions:
1) Is there a canonical bilinear form B:V-->V** here that gives us the isomorphism V~V**? How do we show there is no such form from V to V*?

2) How/why does the functor argument showing V~V** fail when V is infinite-dimensional?

2. Jun 9, 2014

### mathman

Example of failure. Let V be space of continuous functions on [0,1], V* is L1[0,1], V** is L∞[0,1].

3. Jun 9, 2014

### WWGD

Thanks, Mathman, I understand; only Reflexive spaces have this property; but my confusion is about:

1) How/where the argument of the natural (functor) isomorphism between the identity and the double dual fails in the infinite-dimensional case.

2) The existence of a non-degenerate bilinear form B(x,y) in V implies that there is a natural isomorphism between V and V**. My question is: if there exists a natural isomorphism, is there necessarily an associated non-degenerate form in V? Specifically for the case of V, V** in the finite-dimensional case: since V~V** is natural: is there a bilinear form in V that gives us a natural isomorphism between V, V**?

4. Jun 9, 2014

### micromass

Staff Emeritus
I'm dense, but how can a function from $V$ to $V^{**}$ be a bilinear form? What does this have to do with anything?

The functor argument will likely use some basis for $V$ and the associated basis for $V^{**}$. However, if $V$ is finite-dimensional, then you cannot show that the associated "basis" in fact does span the space. It works in the finite-dimensional case because you can just say a linear independent set with a given cardinality is a basis.

You are thinking of the continuous dual, the OP is asking about the algebraic dual.

5. Jun 9, 2014

### micromass

Staff Emeritus
No, in fact no infinite dimensional space has the property that $V\cong V^{**}$ if you mean the algebraic dual.

Last edited: Jun 9, 2014
6. Jun 9, 2014

### micromass

Staff Emeritus
You can equip any vector space with an inner product, so there is always a non-degenerate bilinear form. Just pick a basis $\{e_i~\vert~i\in I\}$ for the vector space in the Hamel sense (every element is the unique linear combination of basis elements). Then define

$$B\left(\sum_{\alpha\in A} \alpha e_\alpha,\sum_{\beta \in B}\beta e_\beta\right) = \sum_{\alpha\in A, \beta\in B} \alpha\beta$$

The problem is that this inner product only provides you with continuous linear functionals (continuous with respect to the inner product),

7. Jun 9, 2014

### WWGD

Yes, of course, the continuous dual; I'm aware that by cardinality reasons alone they cannot be isomorphic, i.e., the algebraic dual is a product, compared to a sum ( finite support ).

8. Jun 9, 2014

### micromass

Staff Emeritus
So let me get this clear. In your posts you denote with $V^*$ the continous dual? Then what structure is there on $V$ that allows you to speak of continuity?

9. Jun 9, 2014

### WWGD

Why can't you define a bilinear form between any two vector spaces? The existence of a non-degenerate form determines a natural (basis-free ) isomorphism between a space and its dual. I was wondering if the converse is also true, i.e ., if there is a natural isomorphism, is there a non-degenerate form that gives rise to this isomorphism?

And, thanks, I got the issue with the natural isomorphism in the infinite-dimensional case, but (at least ) my version of the proof is messy (not Messi ).

10. Jun 9, 2014

### WWGD

Of course, but there must be some naturality/categorical choice of this non-degenerate form, otherwise any vector space would be isomorphic with its dual by just choosing/defining any non-degenerate form on the space.

11. Jun 9, 2014

### WWGD

O.K, I think I know where I was confused: I thought, given any inner-product space (V, <,>) , why doesn't <,> give us a natural isomorphism with the continuous V*? Well, if we use the norm that originates from an inner-product (as a non-degenerate bilinear form), then we have a Hilbert space; but there are many cases in which the given norm does not originate from an inner-product (e.g., in the standard products in the $L^p$ spaces) and this gives the necessary compatibility condition between the norm and the bilinear map, otherwise any inner-product space would be isomorphic to its (continuous dual) .I know this is true when V is a Hilbert space, so not just any inner-product will do. But I got it through writing out my confusions; thanks for letting me get rid of my confusions by writing things here, and sorry if I was not too clear; feel free to delete this if it seemed confusing, and thanks again for letting me clear my confusions, and sorry for the confusion.

Please remind me from time-to-time that others cannot read my mind and that I need to explain myself instead :) .

Last edited: Jun 9, 2014
12. Jun 10, 2014

### mathman

I am not sure what your point is. However Lp and Lq are dual for p, q > 1 and 1/p + 1/q = 1.
Therefore V = V** in all these cases.

13. Jun 10, 2014

### WWGD

Mathman, I just meant to ask whether there is a pairing V-->V** like the pairing v-><v,.> in V-->V* given when there is an inner-product that allows us to define a natural isomorphism between V, V**, or, more generally, if V,W are any two naturally-isomorphic vector spaces, can this isomorphism always be described using a non-degenerate form, i.e., by assigning ?

14. Jun 11, 2014

### WWGD

QUOTE=micromass;4769684]I'm dense, but how can a function from $V$ to $V^{**}$ be a bilinear form? What does this have to do with anything?

The functor argument will likely use some basis for $V$ and the associated basis for $V^{**}$. However, if $V$ is finite-dimensional, then you cannot show that the associated "basis" in fact does span the space. It works in the finite-dimensional case because you can just say a linear independent set with a given cardinality is a basis.

You're right (and, of course, I am wrong here), it is not really a bilinear form

( unless maybe one may construct one by playing

around with pairings) , but an isomorphism from V to V** , i.e., an assignment v-->v**

maybe of a sort similar to the one we have between V and V** when we have an associated

non-degenerate form < , >, where the assignment v-->< v, .> gives us a natural isomorphism.

In this case, the isomorphism V-->V** is given by: v-->v**:=v*(v); I am curious as to

what type of algebraic object this is; is it a bilinear form, a general tensor

15. Jun 11, 2014

### micromass

Staff Emeritus
16. Jun 11, 2014

### WWGD

Yes; that is the map, thanks. I think this comment fits-in here somewhere; in the finite dimensional case, a linear map is described by a matrix A; its dual map (i.e., the image of A in a non-canonical isomorphism between A and A*; in this case the transpose $A^T$is the adjoint map) is given by $A^T$, and so its double dual is given by $(A^T)^T=A$ ( I just got paid \$100 for product-placement by AT&T !) , which suggests the natural isomorphism