Ex 3.2 Verification, 2.1

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Functional Differential Geometry

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Verify that the coefficients of a one-form field transform as described in equation (3.56). You should use equation (3.44) in your derivation.

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Eq. (3.56):

a(\chi(\mathbf{m})) = a'(\chi'(\mathbf{m})) \left[D(\chi \circ (\chi')^{-1})(\chi'(\mathbf{m}))\right]^{-1}

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The general one-form \boldsymbol{\omega} is a linear combination of coordinate-basis one-form fields:

Eq. (3.15):

\begin{aligned} \text{X}_i(\text{f})(\mathbf{m}) &= D(\text{f} \circ \chi^{-1})(\chi(\mathbf{m})) u_i(\chi(\mathbf{m})) \\ &= \partial_i(\text{f} \circ \chi^{-1})(\chi(\mathbf{m})) \end{aligned}

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Given a vector field \mathbf{v}, we define the coordinate-basis one-form fields as

Eq. (3.40):

\tilde{\text{X}} (\mathbf{v}) (\mathbf{m}) = \mathbf{v}(\chi)(\mathbf{m})

So

\begin{aligned} \tilde{\text{X}}^i (\text{X}_j) (\mathbf{m}) &= \text{X}_j (\chi^i)(\mathbf{m}) \\ &= \text{X}_j (\chi^i \circ \chi^{-1})(\chi (\mathbf{m})) \\ &= \partial_j (\chi^i \circ \chi^{-1})(\chi (\mathbf{m})) \end{aligned}

Consider also that

\displaystyle{ \begin{aligned} x^i &= \chi^i (\mathbf{m}) \\ &= \chi^i (\chi^{-1} (\chi (\mathbf{m}))) \\ &= (\chi^i \circ \chi^{-1}) (x) \\ \end{aligned}}

Then, we have Eq. (3.41):

\begin{aligned} \tilde{\text{X}}^i (\text{X}_j) (\mathbf{m}) &= \delta^i_{~j} \ \end{aligned}

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Eq. (3.42):

\begin{aligned} \boldsymbol{\omega}(\mathbf{v})(\mathbf{m}) &= a(\chi(\mathbf{m})) \tilde{\text{X}}(\mathbf{v})(\mathbf{m}) \\ &= \sum_{i} a_i(\chi(\mathbf{m})) \tilde{\text{X}}^i(\mathbf{v})(\mathbf{m}) \\ \boldsymbol{\omega}(\mathbf{v}) &= (a \circ \chi) \tilde{\text{X}} (\mathbf{v}) \\ \end{aligned}

Therefore,

\begin{aligned} \boldsymbol{\omega}(\text{X}_i)(\mathbf{m}) &= \sum_{j} a_j(\chi(\mathbf{m})) \tilde{\text{X}}^j(\text{X}_i)(\mathbf{m}) \\ &= a_i(\chi(\mathbf{m})) \\ \boldsymbol{\omega}(\text{X}_i)(\chi^{-1}(x)) &= a_i(x) \\ \end{aligned}

Eq. (3.44):

\begin{aligned} a_i(x) &= \boldsymbol{\omega}(\text{X}_i)(\chi^{-1}(x)) \\ &= (a' \circ \chi') \tilde{\text{X}'}(\text{X}_i) (\chi^{-1}(x)) \\ \end{aligned}

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— Me@2024-09-18 06:38:02 AM

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