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Let $X$ be a vector field, $L_X$ the Lie derivative, $i_X$ the interior derivative, $A$ the alternation operation (that assign for each covariant tensor field $K$, a differential form $AK$).

The definitions I am following are:

$$(AK)(X_1,\cdots, X_r)= \frac{1}{r!}\sum_{\sigma\in S_r}{\rm sgn}(\sigma)K(X_{\sigma(1)},\cdots,X_{\sigma(r)}) $$ $$L_X(\omega)(X_1,\cdots,X_k)=X(\omega(X_1,\cdots,X_k)) -\sum_{i=1}^k \omega(X_1,\cdots, [X,X_i],\cdots, X_k),$$

$$(i_X\omega)(X_1,\cdots, X_{k-1})=k\omega(X,X_1,\cdots,X_{k-1}),$$ for $X_i\in \mathfrak{X}(M)$; $1\leq i\leq k$.

It is easy to see that the operators $L_X$ and $A$ commute with each other; that is, for a covariant tensor field $K$, we have $$L_X(AK)=A(L_XK).$$

This commutativity condition fails in case of $i_X$ and $A$.

Let $K$ be a covariant tensor field of degree $r$. Let $X_2,\cdots,X_r$ be vector fields on $M$. We have

$$\begin{align}i_{X_1}(AK)(X_2,\cdots,X_r)=&r (AK)(X_1,X_2,\cdots,X_r)\\ =&r\frac{1}{r!}\sum_{\sigma\in S_r}{\rm sgn}(\sigma) K(X_{\sigma(1)},\cdots,X_{\sigma(r)})\\ =&r\frac{1}{r!}\sum_{\sigma\in S_r|\sigma(1)=1}{\rm sgn}(\sigma) K(X_{\sigma(1)},\cdots,X_{\sigma(r)})+r\frac{1}{r!}\sum_{\sigma\in S_r|\sigma(1)\neq1}{\rm sgn}(\sigma) K(X_{\sigma(1)},\cdots,X_{\sigma(r)})\\ =&\frac{1}{(r-1)!}\sum_{\sigma\in S_{r-1}}{\rm sgn}(\sigma) K(X_1,X_{\sigma(2)},\cdots,X_{\sigma(r)})+({\rm something})\\ =&\frac{1}{(r-1)!}\sum_{\sigma\in S_{r-1}}{\rm sgn}(\sigma) i_{X_1}K(X_{\sigma(2)},\cdots,X_{\sigma(r)})+({\rm something})\\ =&A(i_{X_1}K)(X_2,\cdots,X_r)+({\rm something}) \end{align}$$

So, $i_X$ and $A$ does not really commute. We have $$i_X(AK)(X_1,\cdots,X_r)=A(i_XK)(X_1,\cdots,X_r)+({\rm something})$$

I am trying to see if there is a nice expression for the ``something" part. It would just be an appropriate rearrangement of terms, but, I am not able to do it in a clever way.

Any help is appreciated.

Praphulla Koushik
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  • Maybe I am missing something here, but isn't the interior product only defined for differential forms and not for general covariant tensor fields? – AlexD Sep 18 '21 at 11:13
  • @AlexD Yes. interior product is defined only for differential forms... I took the liberty of extending the interior derivative to general covariant tensor fields :) Just put the vector field in first place :) – Praphulla Koushik Sep 18 '21 at 12:51

1 Answers1

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I'll give it a try:

$$ (i_{X_1}(AK))(X_2,\dots,X_r) $$ $$ = \frac{1}{(r-1)!}\sum_{\sigma\in S_r}\mathrm{sgn}(\sigma)K(X_{\sigma(1)},\dots,X_{\sigma(r)})$$ $$ = \frac{1}{(r-1)!}\sum_{\alpha=1}^r\sum_{\sigma\in S_{r-1}}(-1)^{\alpha+1}\mathrm{sgn}(\sigma)K(X_\alpha,X_{\sigma(2)},\dots,X_{\sigma(r)})$$ $$ = \sum_{\alpha=1}^r(-1)^{\alpha+1}\frac{1}{(r-1)!}\sum_{\sigma\in S_{r-1}}\mathrm{sgn}(\sigma)(i_{X_\alpha}K)(X_{\sigma(2)},\dots,X_{\sigma(r)})$$ $$ = \sum_{\alpha=1}^r(-1)^{\alpha+1}A(i_{X_\alpha}K)(X_1,\dots,\hat{X}_\alpha,\dots,X_r)$$

Here $\hat{X}_\alpha$ means that $X_\alpha$ is omitted.

AlexD
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  • This look very nice except that I could not figure out how you got $(-1) ^{\alpha+1}$ is coming.. Please explain... – Praphulla Koushik Sep 18 '21 at 17:11
  • I think you have to pick up that factor from the permutation. For each $\alpha$ you fix $\sigma(1)=\alpha$, so I think for a $\sigma\in S_r$ with $\sigma(1)=\alpha$, you have $\mathrm{sgn}(\sigma) = (-1)^{\alpha}\mathrm{\tau}$ for some $\tau\in S_{r-1}$. Think of it as dividing the permutation $\sigma$ into two parts: One part moves $\alpha$ to position $1$. This part contributes $(-1)^{\alpha+1}$ to the sign of $\sigma$. The other part is a permutation $\tau$ of the remaining indices. – AlexD Sep 18 '21 at 17:34
  • I am still not sure if I understand it correctly.. we are writing $S_r$ as disjoint union of sets; each of which is of the form ${\sigma\in S_r:\sigma(1)=i}$ for $1\leq i\leq r$.. I see ${\sigma\in S_r:\sigma(1)=1}$ can be identified with $S_{r-1}$.. not sure what to do with the other subsets… – Praphulla Koushik Sep 18 '21 at 17:46