2
$\begingroup$

This is a content in page 35 of foundation in differential geometry - KN

For a form $r$-form $\omega $ interior product is $$ i_X\omega\doteq C(X\otimes \omega)$$

where $\{ e_i\}$ is ON, notation is $T=T_{i_1\cdots i_r}^{j_1\cdots j_s} e_{i_1}^\ast \otimes \cdots e_{i_r}^\ast \otimes e_{j_1}\cdots \otimes e_{j_s} $ and ${\bf contraction}$ map is $$C(T)_{i_2\cdots i_r}^{j_2\cdots j_s}=T_{ti_2\cdots i_r}^{tj_2\cdots j_s}$$

${\bf Question}$ : This book says that $$ i_X\omega(Y_1,\ \cdots,\ Y_{r-1}) =r\cdot \omega (X,Y_1,\ \cdots,\ Y_{r-1})$$

${\bf Try}$ : If $$\omega = \sum_{i<j} \omega_{ij} e_i^\ast\wedge e_j^\ast = \frac{1}{2}\sum_{i<j}\omega_{ij} \{ e_i^\ast \otimes e_j^\ast - e_j^\ast\otimes e_i^\ast \} $$ is two-form $(r=2)$, then $$(i_{e_1}\omega)(e_n) = (e_1\otimes \omega)_{1n}^1=\omega (e_1,e_n)=\frac{1}{2}\omega_{1n} $$

But $$ 2\cdot \omega(e_1,e_n)=\omega_{1n} $$

Am I right ? Thank you in advance.

$\endgroup$
4
$\begingroup$

In the definition of the contraction map, you have $T=T_{i_1\cdots i_r}^{j_1\cdots j_s} e_{i_1}^\ast \otimes \cdots e_{i_r}^\ast \otimes e_{j_1}\cdots \otimes e_{j_s} $, where the implicit summation is over all multi-indices, not just increasing ones. For a $2$-form $\omega$, this means $\omega = \omega_{ij} e_i^* \otimes e_j^*$ and therefore $\omega(e_1,e_n) = \omega_{1n}$. To relate this to the wedge product, compute: \begin{align*} \omega &= \sum_{i,j} \omega_{ij} e_i^* \otimes e_j^*\\ &=\sum_{i<j} \omega_{ij} e_i^* \otimes e_j^* + \sum_{i>j}\omega_{ij} e_i^* \otimes e_j^*\\ &=\sum_{i<j} \omega_{ij} e_i^* \otimes e_j^* + \sum_{j>i} \omega_{ji} e_j^* \otimes e_j^*\\ &= \sum_{i<j} \omega_{ij}\big(e_i^* \otimes e_j^*- e_j^* \otimes e_j^*\big)\\ &= 2\sum_{i<j} \omega_{ij} e_i^*\wedge e_j^*. \end{align*} (This is based on Kobayashi & Nomizu's convention for the wedge product, which is the alternating projection applied to the tensor product.)

EDIT: Now I've had a chance to look back at Kobayashi-Nomizu, and it appears that their definitions are in fact inconsistent. Here's what they say:

Consider every $r$-form $\omega$ as an element of $\mathfrak T^0_r(M)$ and define $\iota_X\omega = C(X\otimes \omega)$. In other words, $$ (\iota_X\omega) (Y_1,\dots,Y_{r-1}) = r\cdot \omega(X,Y_1,\dots,Y_{r-1})\quad \text{for }Y_i\in \mathfrak X(M). $$

Unfortunately, these two formulas don't match except on $1$-forms. The easiest way to see this is by considering a $2$-form $\omega = \alpha\wedge\beta = \frac12 (\alpha\otimes\beta - \beta\otimes\alpha)$. On the one hand, $$ C(X\otimes\omega) = \tfrac12 C(X\otimes\alpha\otimes\beta - X\otimes\beta\otimes\alpha) = \tfrac12 (\alpha(X)\beta - \beta(X)\alpha), $$ and therefore, $$ C(X\otimes\omega)(Y) = \tfrac12 (\alpha(X)\beta(Y) - \beta(X)\alpha(Y)). $$ On the other hand, $$ 2\cdot\omega(X,Y) = 2\cdot\tfrac12 (\alpha(X)\beta(Y) - \beta(X)\alpha(Y)). $$

Given that their wedge product convention is $\omega\wedge\eta = \text{Alt}(\omega\otimes\eta)$, the correct definition of the interior product is their second one, with the factor of $r$. Without that factor, it would not be an antiderivation ("skew-derivation" in their terminology). If you use the other wedge product convention (the one I call the "determinant convention" in Introduction to Smooth Manifolds), there's no $r$, and in fact $\iota_X\omega$ is equal to $C(X\otimes\omega)$.

$\endgroup$
  • $\begingroup$ Thank you, but there is still a problem : By definition $i_{e_1}\omega (e_n)=C(e_1\otimes \omega)(e_n)=\omega_{1n}$ And $2\cdot \omega (e_1,e_n)=2\omega_{1n}$ $\endgroup$ – HK Lee May 2 '14 at 14:55
  • 1
    $\begingroup$ Oh, right. There does seem to be an inconsistency here. I'll have to wait until I get ahold of my copy of KN and try to understand better what they're claiming. $\endgroup$ – Jack Lee May 2 '14 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.