differential forms

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- Tags: - #geomtop/differential-geometry - Refs: - #todo/add-references - Links: - Stokes theorem

Differential Forms

  • An \(n{\hbox{-}}\)form on \({\mathbf{R}}^n\): \(\omega=f(\mathbf{x}) d x_{1} \wedge \cdots \wedge d x_{n}\)
  • A \(k{\hbox{-}}\)form on \({\mathbf{R}}^n\) for \(k<n\): \(= \sum _{i_{1}<\cdots<i_{k}} f_{I} \,\, d x_{i_{1}} \wedge \cdots \wedge d x_{i_{k}}\) where \(f_I\) are smooth functions
  • Differentiating a function: \begin{align*}d f=\sum_{j=1}^{n} \frac{\partial f}{\partial x_{j}} d x_{j}\end{align*}
  • Differentiating a form: \begin{align*}d \omega=\sum_{I} d f_{I} \wedge d \mathbf{x}_{I}=\sum_{I} \sum_{j=1}^{n} \frac{\partial f_{I}}{\partial x_{j}} d x_{j} \wedge d x_{i_{1}} \wedge \cdots \wedge d x_{i_{k}}\end{align*}

Example computations

  • Differentiating a form: \begin{align*}\begin{aligned} d(y\,dx+ x\,dy) &= df_{0, 1} \wedge dx_{0, 1} + df_{1,0} \wedge dx_{1,0}\\ &= d(x) \wedge dy + d(y)\wedge dx \\ &= \qty{{\frac{\partial }{\partial x}\,}(y)\,dx+ {\frac{\partial }{\partial y}\,} (y)\,dy} \wedge dx + \qty{{\frac{\partial }{\partial x}\,} (x) \,dx+ {\frac{\partial }{\partial y}\,} (x) \,dy}\wedge dy \\ &= \qty{0 + 1\,dy}\wedge \,dx+ (1\,dx+ 0)\wedge \,dy\\ &= 1\,dy\wedge dx + 1\,dx\wedge \,dy\\ &= (-1)\,dx\wedge\,dy+ 1\,dx\wedge\,dy\\ &= 0 \end{aligned}\end{align*}
  • Differentiating a form: \begin{align*}\begin{aligned} d(-y\,dx+ x\,dy) &= df_{0, 1} \wedge dx_{0, 1} + df_{1,0} \wedge dx_{1,0}\\ &= d(x) \wedge dy + d(-y)\wedge dx \\ &= \qty{{\frac{\partial }{\partial x}\,}(-y)\,dx+ {\frac{\partial }{\partial y}\,} (-y)\,dy} \wedge dx + \qty{{\frac{\partial }{\partial x}\,} (x) \,dx+ {\frac{\partial }{\partial y}\,} (x) \,dy}\wedge dy \\ &= \qty{0 + (-1) \,dy}\wedge \,dx+ (1\,dx+ 0)\wedge \,dy\\ &= (-1)\,dy\wedge dx + (1)\,dx\wedge \,dy\\ &= 2 dx\wedge \,dy \end{aligned}\end{align*}
  • Differentiating a function: \begin{align*}\begin{aligned} d(\arctan(y/x)) &= {\frac{\partial }{\partial x}\,} \arctan(y/x) \,dx+ {\frac{\partial }{\partial y}\,}\arctan(y/x) \,dy\\ &= {1\over 1+ (y/x)^2} {-y\over x^2} \,dx+ {1\over 1+ (y/x)^2} {1\over x}\,dy\\ &= {x^2 \over x^2 + y^2} {-y\over x^2} \,dx+ {x^2\over x^2 + y^2} {1\over x}\,dx\\ &= \qty{-{y\over x^2 + y^2} \,dx+ {x\over x^2 + y^2}\,dy} \end{aligned}\end{align*}
    • Checking that \(d^2 = 0\): \begin{align*}\begin{aligned} d^2(\arctan(y/x)) &= d\qty{-{y\over x^2 + y^2} \,dx+ {x\over x^2 + y^2}\,dy} \\ &\coloneqq d\qty{f_{1, 0} \,dx+ f_{0, 1} \,dy} \\ &\coloneqq\qty{{\frac{\partial }{\partial x}\,}f_{1, 0} \,dx+ {\frac{\partial }{\partial y}\,}f_{1, 0} \,dy} \wedge \,dx+ \qty{{\frac{\partial }{\partial x}\,}f_{0, 1} \,dx+ {\frac{\partial }{\partial y}\,}f_{0, 1} \,dy} \wedge \,dy\\ &= \qty{{2xy \over (x^2+y^2)^2}\,dx+ { y^2-x^2 \over (x^2+y^2)^2}\,dy}\wedge dx + \qty{{y^2-x^2 \over (x^2+y^2)^2}\,dx+ {-2xy\over (x^2+y^2)^2}\,dy}\wedge dy \\ &= {y^2-x^2 \over (x^2+y^2)^2}\,dy\wedge \,dx+ {y^2-x^2 \over (x^2+y^2)^2}\,dx\wedge \,dy\\ &= -{y^2-x^2 \over (x^2+y^2)^2}\,dx\wedge \,dy+ {y^2-x^2 \over (x^2+y^2)^2}\,dx\wedge \,dy\\ &= 0 \end{aligned}\end{align*}
  • Differentiating higher \(k{\hbox{-}}\)forms: \begin{align*}\begin{aligned} d(x_1\,dx_2 + x_3\,dx_4 + x_5\,dx_6) &= \qty{\sum {\frac{\partial }{\partial x_i}\,} (x_1) \,dx_i}\wedge dx_2\\ &\qquad +\qty{\sum {\frac{\partial }{\partial x_i}\,} (x_3) \,dx_i}\wedge dx_4 \\ &\qquad + \qty{\sum {\frac{\partial }{\partial x_i}\,} (x_5) \,dx_i}\wedge dx_6 \\ &= \,dx_1\wedge \,dx_2 + \,dx_3 \wedge \,dx_4 + \,dx_5 \wedge \,dx_6 \end{aligned}\end{align*}
  • Wedging two 2-forms:

\begin{align*} \begin{aligned} \omega&=a_{1} d x_{1}+a_{2} d x_{2} \text { and } \\ \eta&=b_{1} d x_{1}+b_{2} d x_{2} \in \Lambda^{1}\left(\mathbb{R}^{2}\right)^{*}=\left(\mathbb{R}^{2}\right)^{*} \implies \\ \\ \omega \wedge \eta &=\left(a_{1} d x_{1}+a_{2} d x_{2}\right) \wedge\left(b_{1} d x_{1}+b_{2} d x_{2}\right) \\ &=a_{1} b_{1} d x_{1} \wedge d x_{1}+a_{2} b_{1} d x_{2} \wedge d x_{1}+a_{1} b_{2} d x_{1} \wedge d x_{2}+a_{2} b_{2} d x_{2} \wedge d x_{2} \\ &=a_{1} b_{1} d \mathbf{x}_{(1,1)}+a_{2} b_{1} d \mathbf{x}_{(2,1)}+a_{1} b_{2} d \mathbf{x}_{(1,2)}+a_{2} b_{2} d \mathbf{x}_{(2,2)} \\ &=\left(a_{1} b_{2}-a_{2} b_{1}\right) d \mathbf{x}_{(1,2)} \\ &= \operatorname{det}{ \, {}^{t}{ \left( { \begin{bmatrix} {\omega} & {\eta} \\ {\mathrel{\Big|}} & {\mathrel{\Big|}} \end{bmatrix} } \right) } } = \operatorname{det}{ \begin{bmatrix} {a_{1}} & {a_2} \\ {b_1} & {b_2} \end{bmatrix} } . \end{aligned} \end{align*} - Integrating a form: let \(\gamma(t) = {\left[ {1+t, -1+3t, 2t} \right]}\) for \(t\in [0, 1]\) and \(\omega = xy\,dz\). Then \begin{align*} \int_\gamma \omega = \int_0^1 \gamma^*\omega = \int_0^1 (1+t)(-1+3t)(2\,dt) = 2 \end{align*}

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#geomtop/differential-geometry #todo/add-references