I 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.