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The Wedge Product ... Tu, Section 3.7 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,887
Hobart, Tasmania
I am reading Loring W.Tu's book: "An Introduction to Manifolds" (Second Edition) ...

I need help in order to fully understand Tu's section on the wedge product (Section 3.7 ... ) ... ...

The start of Section 3.7 reads as follows:






Tu - Start of Section 3.7 ... Wedge Product ... ... .png







In the above text from Tu we read the following:

" ... ... for every permutation \(\displaystyle \sigma \in S_{ k + l }\), there are \(\displaystyle k!\) permutations \(\displaystyle \tau\) in \(\displaystyle S_k\) that permute the first \(\displaystyle k\) arguments \(\displaystyle v_{ \sigma (1) }, \cdot \cdot \cdot , v_{ \sigma (k) }\) and leave the arguments of \(\displaystyle g\) alone; for all \(\displaystyle \tau\) in \(\displaystyle S_k\), the resulting permutations \(\displaystyle \sigma \tau\) in \(\displaystyle S_{ k + l }\) contribute the same term to the sum, ... ... "


Since I did not completely understand the above quoted statement I developed an example with \(\displaystyle f \in A_2 (V)\) and \(\displaystyle g \in A_3 (V)\) ... ... so we have


\(\displaystyle f \wedge g ( v_1 , \cdot \cdot \cdot , v_5) \)

\(\displaystyle = \frac{1}{2!} \frac{1}{3!} \sum_{ \sigma \in S_5 } f ( v_{ \sigma (1) }, v_{ \sigma (2) } ) g( v_{ \sigma (3) }, v_{ \sigma (4) }, v_{ \sigma (5) } )\)


Now ... translating Tu's quoted statement above into the terms of the example we have ... ...

" ... for every permutation \(\displaystyle \sigma \in S_{ 2 + 3 }\) there are \(\displaystyle 2! = 2\) permutations \(\displaystyle \tau\) in \(\displaystyle S_2\) that permute the first \(\displaystyle 2\) arguments \(\displaystyle v_{ \sigma (1) }, v_{ \sigma (2) }\) and leave the arguments of \(\displaystyle g\) alone ... ... "

Now, following the above quoted text ... consider a specific permutation \(\displaystyle \sigma\) ... say


\(\displaystyle \sigma_1 = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}\)


Now there are \(\displaystyle 2!= 2\) permutations \(\displaystyle \tau\) in \(\displaystyle S_2\) that permute the first \(\displaystyle k = 2\) arguments \(\displaystyle ( v_{ \sigma (1) }, v_{ \sigma (2) } ) = ( v_2, v_3 )\) ...

[Note ... one of the \(\displaystyle k!\) permutations is essentially \(\displaystyle \sigma_1\) itself ... ]

These permutations may be represented by (is this correct?)

... in \(\displaystyle S_2\) ...

\(\displaystyle \tau_1 = \begin{bmatrix} 2 & 3 \\ 2 & 3 \end{bmatrix}\)

\(\displaystyle \tau_2 = \begin{bmatrix} 2 & 3 \\ 3 & 2 \end{bmatrix}\)


and in \(\displaystyle S_{ 2 + 3 }\) ...


\(\displaystyle \tau_1 = \begin{bmatrix} 2 & 3 & 4 & 5 & 1 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}\)

\(\displaystyle \tau_2 = \begin{bmatrix} 2 & 3 & 4 & 5 & 1 \\ 3 & 2 & 4 & 5 & 1 \end{bmatrix}\)


Now the resulting permutations \(\displaystyle \sigma \tau\) are supposed to contribute the same term to the sum ...


... ... \(\displaystyle \sigma_1 \tau_1 = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 & 5 & 1 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}\)


and


\(\displaystyle \sigma_1 \tau_2 = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 & 5 & 1 \\ 3 & 2 & 4 & 5 & 1 \end{bmatrix}\)


\(\displaystyle = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 4 & 3 & 5 & 1 \end{bmatrix} \)



Now the two permutations are not identical ... but ... maybe the difference is accounted for in the \(\displaystyle ( \text{ sgn } )\) function somehow ... but how ... ?



Can someone please clarify the above ...

Peter
 
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