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

#### Peter

##### Well-known member
MHB Site Helper
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 ... ...

In the above proof by Tu we read the following:

" ... ...

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

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

... ... ... "

Can someone please explain/demonstrate how/why we have that

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

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

Help will be much appreciated ... ...

Peter

#### Peter

##### Well-known member
MHB Site Helper
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 ... ...

In the above proof by Tu we read the following:

" ... ...

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

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

... ... ... "

Can someone please explain/demonstrate how/why we have that

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

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

Help will be much appreciated ... ...

Peter

I have been reflecting on my question/problem in the above post ... ... and I think I have resolved the problem ...

Since $$\displaystyle \text{ sgn } \tau$$ is $$\displaystyle +1$$ or $$\displaystyle -1$$ then $$\displaystyle \text{ sgn } \tau \tau = 1$$ ...

Therefore we have

$$\displaystyle \text{ sgn } \sigma = \text{ sgn } \sigma \tau \tau = \text{ sgn } \tau \text{ sgn } \sigma \tau$$

which answers the question ... ...

Is that correct ... ... ?

Peter