Anticommutativity of Wedge Product .... Tu, Proposition 3.21

[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 Proposition 3.21 ... ...

Proposition 3.21 reads as follows:

In the above proof by Tu we read the following:

" ... ...

... $= \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma ) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) }) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) })$$= ( \text{ sgn } \tau ) \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma \tau ) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) }) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) })$

... ... ... "
Can someone please explain/demonstrate how/why we have that$\sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma ) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) }) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) })$$= ( \text{ sgn } \tau ) \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma \tau ) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) }) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) })$Help will be much appreciated ... ...

Peter

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*** EDIT ***I have been reflecting on my question/problem in the above post ... ... and I think I have resolved the problem ...Since $\text{ sgn } \tau$ is $+1$ or $-1$ then $\text{ sgn } \tau \tau = 1$ ...Therefore we have$\text{ sgn } \sigma = \text{ sgn } \sigma \tau \tau = \text{ sgn } \tau \text{ sgn } \sigma \tau$which answers the question ... ...Is that correct ... ... ?

Peter

 

[andrewkirk]

Yes, that is correct. Well done!

 

[math771]

Yes, your explanation is correct. The sign of a permutation is equal to the product of the signs of its disjoint cycles. Since $\tau$ is a permutation of length $l + k$, it can be decomposed into disjoint cycles of lengths $l$ and $k$. Therefore, the sign of $\tau$ is equal to the product of the signs of these cycles, which is equal to $\text{sgn } \tau \text{ sgn } \sigma \tau$. This is why we can rewrite the sum in the proof as $( \text{ sgn } \tau ) \sum_{ \sigma_{ k + l } } ( \text{ sgn } \sigma \tau ) g ( v_{ \sigma \tau (1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l) }) f( v_{ \sigma \tau (l+1) }, \cdot \cdot \cdot , v_{ \sigma \tau (l+k) })$.