Leo algknt at 04:42, 28 June 2018

2018-06-28T04:42:07Z

Leo algknt: Created page with "The set of differential $k$ -forms on a manifold $M$ (example $\mathbb{R}^3$ ) is a vector space $\Omega^k(M)$ and when $k=0$
2018-06-28T04:38:45Z

Created page with "The set of differential <math>k</math>-forms on a manifold <math>M</math> (example <math>\mathbb{R}^3</math>) is a vector space <math>\Omega^k(M)</math> and when <math>k=0</ma..."

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The set of differential <math>k</math>-forms on a manifold <math>M</math> (example <math>\mathbb{R}^3</math>) is a vector space <math>\Omega^k(M)</math> and when <math>k=0</math> then <math>\Omega^0(M)</math> is the set of smooth functions. Thus smooth functions are 0-forms. Now <math>k</math>-forms are integrated on <math>k</math>-manifolds. For example, a 1-form <math> f(x,y) \mathrm{d}x + g(x,y) \mathrm{d}y</math> can be integrated on a curve <math>\gamma</math>. Also differential forms can be differentiated using the operator d called the exterior operator where <math>d</math> acts on a <math>k</math>-form to produce a <math>k+1</math>-form and that <math>\mathrm{d}\circ \mathrm{d} =0</math>.

Now

1. if <math>f \in \Omega^0(\mathbb{R}^3)</math>, then <math>\mathrm{d}f = \sum_i^3 \frac{\partial{f}}{\partial{x_i}}\mathrm{d}x_i</math> is a 1-form so that <math>\mathrm{d}f \in \Omega^1(M)</math>. Thus <math>d: \mathrm{d} \Omega^0(\mathbb{R}^3) \rightarrow \Omega^1(\mathbb{R}^3)</math> is the gradient operator <math>\mathrm{grad}</math>.

2. If we have a 1-form <math>v = v_x\mathrm{d}x + v_y\mathrm{d}y + v_z\mathrm{d}z</math>, then <math>\mathrm{d}v = \left( \frac{\partial{v_z}}{\partial{y}}- \frac{\partial{v_y}}{\partial{z}}\right)\mathrm{d}y\wedge \mathrm{d}z + \left( \frac{\partial{v_x}}{\partial{z}}- \frac{\partial{v_z}}{\partial{x}}\right)\mathrm{d}x\wedge \mathrm{d}z + \left( \frac{\partial{v_x}}{\partial{y}} - \frac{\partial{v_y}}{\partial{x}}\right)\mathrm{d}x\wedge \mathrm{d}y</math> which is a two form. In this case we have <math>d: \mathrm{d} \Omega^1(\mathbb{R}^3) \rightarrow \Omega^2(\mathbb{R}^3)</math> is the <math>\mathrm{curl}</math> operator.

3. If we have 2-form <math>\omega = (\omega_x, \omega_y, \omega_z)</math> then again get a 3-form <math>\mathrm{d}\omega = \left( \frac{\partial{\omega_x}}{\partial{x}} + \frac{\partial{\omega_y}}{\partial{y}} + \frac{\partial{\omega_z}}{\partial{z}} \right)\mathrm{d}x\wedge\mathrm{d}y\wedge \mathrm{d}z</math>. If we think of <math>\mathrm{d}x\wedge\mathrm{d}y\wedge \mathrm{d}z</math> as a function <math>f</math>, then again we get <math>d: \mathrm{d} \Omega^2(\mathbb{R}^3) \rightarrow \Omega^3(\mathbb{R}^3)</math> is the <math>\mathrm{div}</math> operator.

Notes for AKT-140214/0:08:40 - Revision history

Leo algknt at 04:42, 28 June 2018