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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 UTC8, 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 nonzero 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. >> >>> >> >>> Any suggestions, please? >> >>> >> >>> 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 1manifolds are orientable: we can replicate this 1form > in any 1manifold. > Now, a follow up, please. Using the fact that tensors are multilinear maps, > and that vectors are 1tensors. 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 (nonnatural 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 (finitedimensional) 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.



