Simple/Basic Example on Wedge Products

  • #1
Math Amateur
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I am reading Barrett O'Neil's book: Elementary Differential Geometry ...

I need help with fully understanding the example on wedge products of differential forms in O'Neill's text on page 31 in Section 1.6 ..


The example reads as follows:


?temp_hash=61bf9d43e782a3c8ffa0ec43c4bd722e.png



I do not follow the computations in this example ... can someone please explain (in very simple terms) how the example "works" ... essentially how do we justify each step in the reasoning ...


Particularly puzzling for me is the following ... ...


... in the first equation (equation *) we find

[itex]d(fg \ dx \ dy ) = \partial (fg) / \partial z \ dz \ dx \ dy [/itex]

How is this step justified?


... then I find all the steps in the second and third equations puzzling as well ...

Can someone explain and justify the steps in the example in terms of O'Neill's Definitions, Theorems and Lemma's ... ...


To give readers of this post the context, notation and necessary Definitions, Theorems and Lemma's, I am providing the text of Section 1.6: Differential Forms ... ... as follows ... ...


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?temp_hash=61bf9d43e782a3c8ffa0ec43c4bd722e.png

?temp_hash=61bf9d43e782a3c8ffa0ec43c4bd722e.png

?temp_hash=61bf9d43e782a3c8ffa0ec43c4bd722e.png
 

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  • O'Neill - 1 - Section 1.6 - page 28 - Differential Forms - PART 1     .png
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  • O'Neill - 2 - Section 1.6 - page 29 - Differential Forms - PART 2     .png
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  • O'Neill - 3 - Section 1.6 - page 30 - Differential Forms - PART 3     .png
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  • O'Neill - 4 - Section 1.6 - page 31 - Differential Forms - PART 4      .png
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Answers and Replies

  • #2
RUber
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Thanks for the interesting post. I have never worked with wedge products before, but I think I can help you tear this one apart.
Using the definition of the exterior derivative see 6.3 and definition 5.2 (not included):
##d(fg \partial x \partial y ) = d(fg) \wedge \partial x \partial y##
and
##d(fg) = \frac{\partial (fg)}{\partial x } \partial x + \frac{\partial (fg)}{\partial y } \partial y+ \frac{\partial (fg)}{\partial z} \partial z.##
Using the rule about repeats being zero, you can drop the ##\partial x## and ## \partial y## terms.
This gives you:
##d(fg \partial x \partial y ) = \frac{\partial (fg)}{\partial z} \partial z \wedge \partial x \partial y.##

For the second equation:
##d(f \partial x ) =df \wedge \partial x\\
\quad = \frac{\partial (f)}{\partial x } \partial x + \frac{\partial (f)}{\partial y } \partial y+ \frac{\partial (f)}{\partial z} \partial z \wedge \partial x \\
\quad =\frac{\partial (f)}{\partial y } \partial y \partial x+ \frac{\partial (f)}{\partial z} \partial z \partial x ##
Then looking at
##d(f \partial x )\wedge g\partial y ##
gives you:
##d(f \partial x )\wedge g\partial y = \frac{\partial (f)}{\partial y } \partial y \partial x+ \frac{\partial (f)}{\partial z} \partial z \partial x \wedge g\partial y\\
\quad = \frac{\partial (f)}{\partial y } g \partial y \partial x \partial y + \frac{\partial (f)}{\partial z} g \partial z \partial x \partial y ##
Eliminating the duplicate partial y, you get:
##d(f \partial x )\wedge g\partial y= \frac{\partial (f)}{\partial z} g \partial z \partial x \partial y\\
\quad = - \frac{\partial (f)}{\partial z} g \partial x \partial z \partial y\\
\quad = \frac{\partial (f)}{\partial z} g \partial x \partial y\partial z##

For the third equation, the process is about the same.
 
  • #3
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Hi RUber ... thanks for the help ... appreciate it ...

Just a question about the justification for

[itex] d(fg \ \partial x \partial y ) = d(fg) \ \wedge \ \partial x \partial y [/itex]

or, in the notation of O'Neil ... ...

[itex] d(fg \ dx dy ) = d(fg) \ \wedge \ dx dy [/itex]

Now you quote Definition 6.3 ( I cannot see the relevance of Definition 5.2 ? maybe you can help ...)

Definition 6.3 says that if we have a 1-form [itex] \phi = \sum f_i \ dx_i [/itex] then [itex] d \phi = \sum df_i \ \wedge dx_i [/itex]

Now how do we get 6.3 about 1-forms to apply to a situation about 1-forms ... I'm not sure ...

It does not solve my problem ... but to get closer to apply 6.3 we can put [itex] fg = F [/itex] and then try to apply 6.3 ...but how do we deal with the fact that we are dealing with [itex] dx \ dy[/itex] and not just [itex] dx [/itex] ... I guess we could write [itex] dX = dx \ dy [/itex] ... but could we then legitimately use 6.3 ...

Can you help?

Peter
 
  • #4
lavinia
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The general definition of the exterior derivative of a wedge product of two differential forms is

##d(α∧β) = dα∧β =
+ (-1)^{p}α∧dβ## where ##α## is a ##p##-form.

For a zero form - i.e. a function - the wedge is omitted since it is just scalar multiplication for zero forms.

##dxdy## is really ##dx∧dy## so

##d(fdxdy) = d(f∧(dx∧dy)) = df∧(dx∧dy) + (-1)^{0}f∧d(dx∧dy) =df∧(dx∧dy) + f(d^{2}x∧dy - x∧d^{2}y)##
## = df∧(dx∧dy) = df∧dxdy##

since ##d^2## is zero always.
 
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  • #5
Math Amateur
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Thanks for you help, Lavinia ... most helpful ...

... BUT ... just a clarification ... (perhaps just a detail. but anyway ...)

How exactly do you get [itex] d(f \ dx \ dy ) = d ( f \ \wedge \ (dx \ \wedge \ dy ) [/itex] ... ... ?

I can see that if [itex] dx \ dy = dx \ \wedge \ dy [/itex] that we would have

[itex] d ( f \ dx \ dy ) = d ( f \ dx \ \wedge \ dy ) [/itex] ... ...
 
  • #6
lavinia
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Wedge product with a zero form is just scalar multiplication - by definition. The wedge is usually left out for zero forms but this is just a notational convenience.
 
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