"Adding" a Vector Space and its Dual

[Phinrich]

TL;DR Summary

Hi All

Given a Vector Space V (of dimension n) and its Dual V* (also having dimension n) can we mathematically combine these vector spaces to form a Vector Space of Dimension 2n ? I dont believe the Direct Sum will do because I think to do a Direct Sum we will end up with a Direct Sum still having dimension n and not 2n. I stand to be corrected.

Any help would be greatly appreciated.

Given that the Set of 1-Forms is a Vector Space distinct from, but complimentary to, the Linear Vector Space of Vectors. And given that there is an Isomorphism between the linear space of vectors and the dual vector space of 1-forms, does it make mathematical sense to combine a vector space and its dual space of 1-forms to produce a new space having dimension equal to the sum of the dimension of the vector space plus the dimension of its dual? I believe that if we define V as {v1, v2, 0} and a second vector space W = {0,0,w1} then we can form the Direct Sum Space of the two as Y = {v1, v2, w1}. However since the components of V = {v1, v2,0} and the components of V* = {v1*,v2*,0} there is no way to form a Direct Sum and still satisfy the rules for a Direct Sum. So, in component form, I want to take {v1,v2,0} PLUS {w1,w2,0} to produce Y={v1,v2,0,w1,w2,0}.

Hope this makes sense to someone.

Thanks

Paul

 

[fresh_42]

$\dim (V\oplus V^*) = 2n\, , \,\dim(V\otimes V^*)=n^2$

So you can take the direct sum, which doesn't provide much new structure, as they are only ordered pairs. The tensor product is more interesting as it is isomorphic to the vector space of linear transformations on $V$. One possibility to get back to dimension $2n$ would be to factor out a subspace $U\subseteq V\otimes V^*$ with $\dim U=n^2-2n=n(n-2)$ and consider $V\otimes V^*/U$.

But the entire question is a bit of in reverse. Usually you start with some vector space and ask for the dimension and not the other way around.

Summary:
The direct sum has the requested dimension, factor spaces of the tensor product are more interesting.

 

[Phinrich]

Thank you very much for your reply.

The second option (factoring out) seems a bit complicated to me. The first option, I think, at least as you have written it, seems to be the answer. I say this because you say the Direct Sum will produce a Space of Dim 2n from 2 spaces each of dim n. I thought one condition of a Direct Sum was that the intersection of the Summed spaces had to be zero? That is why I thought the two Summed spaces had (to start with) to have the same Dimension and then the Sum kept that Dimension. Are you saying V can have Dim 3 and V* also Dim 3 and I can form the Direct Sum to have Dim 3 + 3 = 6 because we are not adding an element of V to its corresponding element of V* ?
Thanks
Paul

 

[Phinrich]

I guess I am asking if we can do the following;

V = (v1,v2,v3) and V* = (v1*,v2*,v3*) and the Direct Sum = (v1,v2,v3,v1*,v2*,v3*)

Thanks.

 

[fresh_42]

I thought one condition of a Direct Sum was that the intersection of the Summed spaces had to be zero?

It has to be, only that the zero is $0=(0,0)=\vec{0}+(0\, : \,V\longrightarrow \{\,0\,\})$.

I guess I am asking if we can do the following;

V = (v1,v2,v3) and V* = (v1*,v2*,v3*) and the Direct Sum = (v1,v2,v3,v1*,v2*,v3*)

Thanks.

Yes.

 

[Phinrich]

So the 1-Form 0 maps the 0 vector to 0 ?

 

[fresh_42]

It maps the entire vector space to $0$.

 

[Phinrich]

Maybe I don't need to form any kind of Sum. Maybe I just need to recognise that I have a Tensor V(n,n) consisting of the n contravariant components together with the n covectors (or 1-Forms). The these n 1-Forms Map the n contravariant elements to the set of Real numbers ?

 

[fresh_42]

Yes, but then you have $V\otimes V^* \cong L(V,V) \cong \mathbb{M}(n,\mathbb{R})$ and dimension $n^2$.

 

[Phinrich]

OK that's correct as you say

 

[Phinrich]

So can I form the following then, in any way,

V = (v1,v2,v3) and V* = (v1*,v2*,v3*) and the "Direct Sum" or whatever Sum = (v1,v2,v3,v1*,v2*,v3*)

 

[Phinrich]

OK Fresh_42, you have REALLY got me thinking now about my original question which may well have been misguided by my ignorance.

You said earlier;

Yes, but then you have V⊗V∗≅L(V,V)≅M(n,R)V⊗V∗≅L(V,V)≅M(n,R) and dimension n2n2.

Yes this has dimension nxn rather than 2n. I need to rethink my original purpose. Maybe I was wrong in insisting the result should be 2xn. Sorry but I am a hobbyist mathematician (the worst kind I suspect). Thank you very much for your time.

Paul

 

[fresh_42]

OK Fresh_42, you have REALLY got me thinking now about my original question which may well have been misguided by my ignorance.

You said earlier;

Yes, but then you have V⊗V∗≅L(V,V)≅M(n,R)V⊗V∗≅L(V,V)≅M(n,R) and dimension n2n2.

Yes this has dimension nxn rather than 2n. I need to rethink my original purpose. Maybe I was wrong in insisting the result should be 2xn. Sorry but I am a hobbyist mathematician (the worst kind I suspect). Thank you very much for your time.

Paul

Don't mind. And, yes, you can still consider the direct sum. Direct sums can be formed with any vector spaces, so $V \oplus V^*$ is just an example.

 

[Phinrich]

OK then if V has 3 dim and so does V* does it make sense to define a metric tensor for the Direct Sum Space and what would be its Dimension ?

 

[fresh_42]

I'm not sure what you are heading for. A metric tensor associates a metric to some point: $A \longrightarrow (V\oplus V)^*$ where the image is a non degenerate symmetric bilinear form. Of course you can set $A=V$.

 

[Phinrich]

OK perhaps, to explain, If instead of involving V with V*. If we had two vector spaces V and W being subspaces of R4 we might expect V to have dim 2 and perhaps W to have dim 2. Then, if I am correct, the Direct Sum V and W would give a vector space of dim 4. So if we could have defined a metric tensor for V it would be a 2 x 2 matrix. Similarly a metric tensor for W might have been a 2 x 2 matrix. Now the metric tensor for R4 I would expect to be a 4 x 4 matrix.

No if this is correct then instead of W being W what if W = V* instead ? Would V and V* have separate metrics and would the metric tensor of the Direct Sum be some way of combining the metric tensor for V with that for V* as we did in the case of V and W ?

 

[Phinrich]

Fresh_42. I think you understand this whole thing way better than I do and probably my questions are not making sense to you. The problem is to convince me of my ignorance. You have changed my perspective on this and I need to think deeply on my new perspective. Its late here so I am retiring for the night. Will get back to you perhaps tomorrow after a long deep thinking process. I still think you have convinced me that I am really only talking about vectors (of dim n) and their corresponding 1-Forms (also of Dim n). Rather than any Direct Sums of Spaces.

Thank you again for your time and knowledge.