I did all explanation. Can you just teach me how to calculate this interior product?

182 Views Asked by At

Orient the unit sphere $S^n$ in $\Bbb R^{n+1}$ as the boundary of the closed unit ball.

Show that an orientation form on $S^n$ is

$w=\sum _{i=1}^{n+1}(-1)^{i-1}x^i dx^1∧...∧dx^i∧...∧dx^{n+1}$

I think that

I need to calculate i$_{X}(dx^1 ∧···∧dx^{n+1})$.

Let's write $X=\sum x^i\frac{\partial}{\partial x^i}$

Then I obtain that $i_{\sum x^i\frac{\partial}{\partial x^i}}(dx^1 ∧···∧dx^{n+1})$

After here, How to calculate this? I have my final exam tommorow. So I need to learn this calculation. Please show me. Thank you so much:)

1

There are 1 best solutions below

6
On BEST ANSWER

Using the results that for 1 form $ \omega \in \Omega^1(M) $ you have $i_X(\omega) = \omega(X)$ and in general for any two forms $\omega,\eta $ you have $ i_X(\omega\wedge \eta) = (i_X\omega)\wedge \eta + (-1)^{\deg(\omega)}\omega\wedge (i_X\eta) $. Using these two it is easy to prove for 1 forms $ \omega_1,\omega_2,...,\omega_k \in \Omega^1(M) $ you have (the hat implies that is omitted) $$ i_X(\omega_1\wedge\omega_2\wedge...\wedge \omega_k) = \sum_{i=1}^k (-1)^{i-1} \omega_i(X)(\omega_1\wedge\omega_2\wedge...\hat{\omega_i}...\wedge \omega_k) $$ So you have your answer.