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.


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 $C(X\otimes\omega)$.

  • $\begingroup$ 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

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