Mentz114 said:
...there must be an invariant inner product operation for vectors, and this requires a dual vector space.
If V is a vector space, you can define the dual space V* without defining an inner product, and you can define an inner product without defining a dual space.
wdlang said:
but in quantum field theory and relativity, why g_{\mu \nu} F^{\nu} transforms differently than F^{\nu} ? what is the effect of the metric g?
If F^\nu are the components of a tensor field in a coordinate system, then that tensor field is
F^\nu\frac{\partial}{\partial x^\nu}
If g_{\mu \nu} F^{\nu} are the components of a tensor field in a coordinate system, then that tensor field is
g_{\mu \nu} F^{\nu}dx^{\mu}
I would write the values of those two tensor fields at a point p in the manifold as
F^\nu(p)\frac{\partial}{\partial x^\nu}\bigg|_p
and
g_{\mu \nu}(p) F^{\nu}(p)dx^{\mu}|_p
The \frac{\partial}{\partial x^\nu}\bigg|_p are basis vectors for the tangent space at p, T_pM. The dx^{\mu}|_p are basis vectors for its dual space, T_pM^* or T_p^*M. Specifically, they are the members of the dual basis of that particular basis of the tangent space.
See
this post for the definition of "dual space", and click the link in it for the definition of "tangent space at p".If \{e_i\} is a basis of a vector space V, then its dual basis, which I will write as \{e^i\}, is defined by e^i(e_j)=\delta^i_j. Suppose that \{f_i\} is another basis of V, related to the first by f_i=A_i^j e_j. (By definition of "basis", the f_i have to be linear combinations of the e_i). The A_i^j can be thought of as the components of a matrix A (row j, column i).
\delta^i_j=f^i(f_j)=f^i(A_j^k e_k)=A_j^k f^i(e_k)
(A^{-1})^j_l\delta^i_j=(A^{-1})^j_l A_j^k f^i(e_k)
\delta^i_l=(A^{-1}A)^k_l f^i(e_k)=\delta^k_l f^i(e_k)=f^i(e_l)
A^k_i\delta^i_l=A^k_if^i(e_l)
\delta^k_l=A^k_i f^i(e_l)
\Rightarrow e^k=A^k_i f^i
(A^{-1})^j_k e^k=(A^{-1})^j_k A^k_i f^i=(A^{-1}A)^j_i f^i=\delta^j_i f^i=f^j
So when the basis "transform" by a matrix A, the dual basis "transforms" by A
-1.
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