Differential of a 2-form: How to Compute df Using the Wedge Product?

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SUMMARY

The discussion focuses on computing the differential of the 2-form given by \( f = e^{xy} dy \wedge dz \). The correct approach to find \( df \) involves applying the formula \( df = \left(\frac{\partial f}{\partial x}\right)dx + \left(\frac{\partial f}{\partial y}\right)dy + \left(\frac{\partial f}{\partial z}\right)dz \). The user correctly identifies that the wedge product is anti-symmetric and that the dual does not play a role in this computation. The final expression for \( df \) is \( df = y e^{xy} dx \wedge dy \wedge dz + x e^{xy} dy \wedge dy \wedge dz \).

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Homework Statement



Compute the differentials of the following form:

Homework Equations



f = exy dy ^ dz

The Attempt at a Solution



I'm a little confused on how to work with the wedge product. If I'm looking for df, should I start by calculating the dual f*? Or does the dual not come into play?
 
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No, there is no reason to worry about the dual. For any f, [itex]df= (\partial f/\partial x)dx+ (\partial f/\partial y)dy+ (\partial f/\partial z)dz[/itex].

If
[itex]f= e^{xy} dy^dz[/itex] then
[tex]df= (\partial e^{xy}/\partial x)dx\^dy\^dz+ (\partial e^{xy}/\partial y)dy\^dy\^dz+ (\partial e^{xy}/\partial z)dz\^dy\^dz[/tex]
Remember, of course, that the wedge product is "anti-symmetric".
 
So

df = yexy dx ^ dy ^ dz + xexy dy ^ dy ^ dz ? Is that the answer?
 
Last edited:

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