Notes for AKT-140214/0:08:40: Difference between revisions

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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.
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.
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 divergence operator <math>\mathrm{div}</math>.

Latest revision as of 23:42, 27 June 2018

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

Now


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

2. If we have a 1-form [math]\displaystyle{ v = v_x\mathrm{d}x + v_y\mathrm{d}y + v_z\mathrm{d}z }[/math], then [math]\displaystyle{ \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]\displaystyle{ d: \mathrm{d} \Omega^1(\mathbb{R}^3) \rightarrow \Omega^2(\mathbb{R}^3) }[/math] is the [math]\displaystyle{ \mathrm{curl} }[/math] operator.

3. If we have 2-form [math]\displaystyle{ \omega = (\omega_x, \omega_y, \omega_z) }[/math] then again get a 3-form [math]\displaystyle{ \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]\displaystyle{ \mathrm{d}x\wedge\mathrm{d}y\wedge \mathrm{d}z }[/math] as a function [math]\displaystyle{ f }[/math], then again we get [math]\displaystyle{ d: \mathrm{d} \Omega^2(\mathbb{R}^3) \rightarrow \Omega^3(\mathbb{R}^3) }[/math] is the divergence operator [math]\displaystyle{ \mathrm{div} }[/math].