The Wedge Product .... Tu, Section 3.7

[Math Amateur]

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:

In the above text from Tu we read the following:

" ... ... for every permutation $\sigma \in S_{ k + l }$, there are $k!$ permutations $\tau$ in $S_k$ that permute the first $k$ arguments $v_{ \sigma (1) }, \cdot \cdot \cdot , v_{ \sigma (k) }$ and leave the arguments of $g$ alone; for all $\tau$ in $S_k$, the resulting permutations $\sigma \tau$ in $S_{ k + l }$ contribute the same term to the sum, ... ... "Since I did not completely understand the above quoted statement I developed an example with $f \in A_2 (V)$ and $g \in A_3 (V)$ ... ... so we have$f \wedge g ( v_1 \cdot \cdot \cdot v_5)$

$= \frac{1}{2!} \frac{1}{3!} \sum_{ \sigma \in S_5 } f ( v_{ \sigma (1) }, v_{ \sigma (2) } ) g( v_{ \sigma (3) }, v_{ \sigma (4) }, v_{ \sigma (5) } )$Now ... translating Tu's quoted statement above into the terms of the example we have ... ...

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

Now, following the above quoted text ... consider a specific permutation $\sigma$ ... say$\sigma_1 = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}$Now there are $2!= 2$ permutations $\tau$ in $S_2$ that permute the first $k = 2$ arguments $( v_{ \sigma (1) }, v_{ \sigma (2) } ) = ( v_2, v_3 )$ ...

[Note ... one of the $k!$ permutations is essentially $\sigma_1$ itself ... ]

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

... in $S_2$ ...

$\tau_1 = \begin{bmatrix} 2 & 3 \\ 2 & 3 \end{bmatrix}$

$\tau_2 = \begin{bmatrix} 2 & 3 \\ 3 & 2 \end{bmatrix}$and in $S_{ 2 + 3 }$ ...$\tau_1 = \begin{bmatrix} 2 & 3 & 4 & 5 & 1 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}$

$\tau_2 = \begin{bmatrix} 2 & 3 & 4 & 5 & 1 \\ 3 & 2 & 4 & 5 & 1 \end{bmatrix}$Now the resulting permutations $\sigma \tau$ are supposed to contribute the same term to the sum ...... ... $\sigma_1 \tau_1 = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 & 5 & 1 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}$

$= \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}$and$\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}$$= \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 $( \text{ sgn } )$ function somehow ... but how ... ?
Can someone please clarify the above ...

... ... could someone also confirm that the above analysis is basically correct ... ?

Peter

 

[andrewkirk]

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

I think there is a bit of abuse of notation from the author there. What he means that there is a perm $\tau\in S_{k+l}$ (not $S_k$) that permutes the first $k$ elements and is the identity map on the last $l$ elements. So the two possible values for $\tau$ are $\tau_1 =$(1 2 3 4 5) and$\tau_2=$ (2 1 3 4 5).
If $\sigma$ is (5 1 3 4 2) then
$\sigma\tau_1=$(5 1 3 4 2)(1 2 3 4 5) = (5 1 3 4 2)
and
$\sigma\tau_2=$(5 1 3 4 2)(2 1 3 4 5) = (1 5 3 4 2)

In this case the two terms in the sum (3.5) from this $\sigma$ are:

$\textrm{sgn} (\sigma\tau_1) f(v_5, v_1) g(v_3,v_4,v_2) = \textrm{sgn} (\sigma)\textrm{sgn}(\tau_1) f(v_5, v_1) g(v_3,v_4,v_2)$
and
$\textrm{sgn} (\sigma\tau_2) f(v_1, v_5) g(v_3,v_4,v_2) = \textrm{sgn} (\sigma)\textrm{sgn}(\tau_2) f(v_1, v_5) g(v_3,v_4,v_2)$
and the signs of both the second 'sgn' piece and the $f$ piece change between the two, cancelling each other out and making the values the same.

 

[math771]

Hello Peter,

Your example and analysis seem to be correct. The difference between the two permutations is accounted for by the sign function, also known as the signature or parity function. This function assigns a value of either +1 or -1 to each permutation, depending on whether it is an even or odd permutation. In your example, the first permutation, $\sigma_1 \tau_1$, has a sign of +1 while the second permutation, $\sigma_1 \tau_2$, has a sign of -1. This is because the second permutation involves an odd number of swaps, while the first permutation does not involve any swaps.

In general, when we are dealing with wedge products of differential forms, the sign function comes into play because we are dealing with alternating multilinear maps. This means that the order of the inputs matters and swapping inputs changes the sign of the output. The sign function helps us keep track of these changes in sign.

I hope this helps clarify the concept of the sign function in the context of wedge products. Let me know if you have any further questions.