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Topic: How do we Evaluate This Form on S^1?
Replies: 1   Last Post: Jan 21, 2013 6:05 AM

 W. Dale Hall Posts: 71 Registered: 2/11/05
Re: How do we Evaluate This Form on S^1?
Posted: Jan 21, 2013 6:05 AM

hbertaz@gmail.com wrote:
> On Sunday, January 20, 2013 4:28:53 PM UTC-8, W. Dale Hall wrote:
>> hbe...@gmail.com wrote:
>>

>>> Hi, All:
>>
>>>
>>
>>> Just curious:
>>
>>>
>>
>>> How does one evaluate the form d(theta) on the circle?
>>
>>>
>>
>>> I know a form takes a tangent vector X and spits out a real number ,
>>
>> > in a linear way. The tangent space to S^1 is R^1, so that a tangent
>>
>> > vector is of the form c ; any non-zero real number can be a basis.
>>
>>>
>>
>>> But, how do we get a number from d(theta)[c]?
>>
>>>
>>
>>> I tried using a change of coordinate, changing from cartesian to
>>
>> > polar, but I don't see much difference.
>>
>>>
>>
>>
>>>
>>
>>> Thanks.
>>
>>>
>>
>>
>>
>> At any point x on the circle S^1, d(theta) is the linear function on the
>>
>> tangent space T_x(S^1) that sends the vector \partial_theta in
>>
>> T_x(S^1) to 1.
>>
>>
>>
>> In order to identify the reals R with the tangent space T_x(S^1), you
>>
>> need to map r in R to the vector r*\partial_theta. A tangent vector
>>
>> isn't *really* a real number [after all, tangent vectors to different
>>
>> points on S^1 aren't quite identifiable with one another since they
>>
>> belong to different tangent spaces], but is uniquely expressible as a
>>
>> basis element times a real number (where the basis element is chosen
>>
>> freely, but once chosen, it's fixed). So, if you've chosen the vector
>>
>> \partial_theta as the basis element for T_x(S^1), the map d(theta) sends
>>
>> r*\partial_theta to r in R.
>>
>>
>>
>> Dale

>
> Thanks, Dale Hall. I guess d\theta is a basis for the dual of d\theta, i.e.,
> the basis so that v*(v)=1 (general case, given a basis B_V={v1,..,vn} for V,
> the dual basis would be B_V*={v1*,..,vn*} , with v_i*(vj):=Del^i_j ). I guess
> this is part of why all 1-manifolds are orientable: we can replicate this 1-form
> in any 1-manifold.
> Now, a follow up, please. Using the fact that tensors are multi-linear maps,
> and that vectors are 1-tensors. How do we describe a vector as a linear map?
> Sorry, I know this is simple, but I'm stuck. We know there is a
> vector space isomorphism (non-natural iso.) between v and v*. I
> tried to use a choice h of isomorphism to understand the image of v
> in V* under h, but I seem to be going in circles (on top of the
> confusing issue that the image seems to depend on h, since the iso.
> between V and V* is not natural); I'm kind of new to all this.
>

For finite dimensional vector spaces over a given field k, the dimension
is a complete invariant. Since the dual V* of the (finite-dimensional)
vector space V has the same dimension as V, they are
isomorphic. However, as you note, the isomorphism isn't natural:
given a basis {v1 ... vk} of V, the canonical duals {v1* ... vk*},
where vi* maps vj to the Kronecker delta d(i,j), form a basis for V*.

To view V as the V**, the dual of V*, just note that the pairing

p: VxV* ------> R
by

p:(v,w) |-----> w(v)

that is, evaluation of the linear function w at v, can equally be viewed
as evaluation of v at w. Linearity of the mapping p with
respect to v is guaranteed, and linearity with respect to w can
easily be established. The pairing itself *is* natural; it's just
the isomorphism between V and V* that depends on choosing a basis.