HW 07 - Mathematics 218: Analysis in Several Variables...

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Unformatted text preview: Mathematics 218: Analysis in Several Variables Professor Robert Gunning Fall Term 2009 Seventh Problem Set Rik Sengupta December 4, 2009 1 Group 1 1. In R 4 consider the differential forms ω 1 = x 1 dx 2 + x 3 dx 4 ω 2 = ( x 2 1 + x 2 3 ) dx 1 ∧ dx 2 + ( x 2 2 + x 2 4 ) dx 3 ∧ dx 4 ω 3 = dx 1 ∧ dx 3 + x 2 dx 1 ∧ dx 4 + x 1 dx 2 ∧ dx 4 . (a) Find the exterior products ω 1 ∧ ω 1 , ω 1 ∧ ω 2 and ω 2 ∧ ω 2 in reduced form. (b) Find the exterior derivatives dω 1 , dω 2 and dω 3 in reduced form. (c) Which of these three differential forms are closed, that is, satisfy dω = 0? (d) For each of the differentials ω which is closed find a differential form σ such that ω = dσ . Having found one such form σ , what is the most general such form? (Answer this part of the question without giving explicit formulas for all possible solutions, but just by characterizing all possible solutions using your general knowledge of the properties of differential forms). Solution. We will use the basic ideas of exterior algebra. (a) We have 1 ω 1 ∧ ω 1 = ( x 1 dx 2 + x 3 dx 4 ) ∧ ( x 1 dx 2 + x 3 dx 4 ) = x 2 1 dx 2 ∧ dx 2 + x 1 x 3 dx 2 ∧ dx 4 + x 1 x 3 dx 4 ∧ dx 2 + x 2 3 dx 4 ∧ dx 4 = x 1 x 3 dx 2 ∧ dx 4- x 1 x 3 dx 2 ∧ dx 4 = 0 . Next, we have ω 1 ∧ ω 2 = ( x 1 dx 2 + x 3 dx 4 ) ∧ (( x 2 1 + x 2 3 ) dx 1 ∧ dx 2 + ( x 2 2 + x 2 4 ) dx 3 ∧ dx 4 ) = x 1 ( x 2 1 + x 2 3 ) dx 2 ∧ dx 1 ∧ dx 2 + x 1 ( x 2 2 + x 2 4 ) dx 2 ∧ dx 3 ∧ dx 4 + x 3 ( x 2 1 + x 2 3 ) dx 4 ∧ dx 1 ∧ dx 2 + x 3 ( x 2 2 + x 2 4 ) dx 4 ∧ dx 3 ∧ dx 4 =- x 1 ( x 2 1 + x 2 3 ) dx 2 ∧ dx 2 ∧ dx 1 + x 1 ( x 2 2 + x 2 4 ) dx 2 ∧ dx 3 ∧ dx 4 + x 3 ( x 2 1 + x 2 3 ) dx 1 ∧ dx 2 ∧ dx 4- x 3 ( x 2 2 + x 2 4 ) dx 4 ∧ dx 4 ∧ dx 3 = x 1 ( x 2 2 + x 2 4 ) dx 2 ∧ dx 3 ∧ dx 4 + x 3 ( x 2 1 + x 2 3 ) dx 1 ∧ dx 2 ∧ dx 4 . Finally, we get ω 2 ∧ ω 2 = (( x 2 1 + x 2 3 ) dx 1 ∧ dx 2 + ( x 2 2 + x 2 4 ) dx 3 ∧ dx 4 ) ∧ (( x 2 1 + x 2 3 ) dx 1 ∧ dx 2 + ( x 2 2 + x 2 4 ) dx 3 ∧ dx 4 ) = ( x 2 1 + x 2 3 ) 2 dx 1 ∧ dx 2 ∧ dx 1 ∧ dx 2 + ( x 2 1 + x 2 3 )( x 2 2 + x 2 4 ) dx 1 ∧ dx 2 ∧ dx 3 ∧ dx 4 + ( x 2 1 + x 2 3 )( x 2 2 + x 2 4 ) dx 3 ∧ dx 4 ∧ dx 1 ∧ dx 2 + ( x 2 2 + x 2 4 ) 2 dx 3 ∧ dx 4 ∧ dx 3 ∧ dx 4 =- ( x 2 1 + x 2 3 ) 2 dx 1 ∧ dx 1 ∧ dx 2 ∧ dx 2 + ( x 2 1 + x 2 3 )( x 2 2 + x 2 4 ) dx 1 ∧ dx 2 ∧ dx 3 ∧ dx 4 + ( x 2 1 + x 2 3 )( x 2 2 + x 2 4 ) dx 1 ∧ dx 2 ∧ dx 3 ∧ dx 4- ( x 2 2 + x 2 4 ) 2 dx 3 ∧ dx 3 ∧ dx 4 ∧ dx 4 = 2( x 2 1 + x 2 3 )( x 2 2 + x 2 4 ) dx 1 ∧ dx 2 ∧ dx 3 ∧ dx 4 . (b) The exterior derivatives can be found more easily. We have dω 1 = d ( x 1 dx 2 + x 3 dx 4 ) = dx 1 ∧ dx 2 + x 1 ∧ d ( dx 2 ) + dx 3 ∧ dx 4 + x 3 ∧ d ( dx 4 ) = dx 1 ∧ dx 2 + dx 3 ∧ dx 4 ....
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HW 07 - Mathematics 218: Analysis in Several Variables...

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