Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What is the natural notion of the Lie-derivative of a tensor field along another tensor field, and where can I find an exposition of that?

share|improve this question
3  
Perhaps you could explain why you want to do this. –  Deane Yang Jan 28 at 19:36

1 Answer 1

This can be done in certain special cases besides the usual Lie derivatives along a vector field. More precisely, let $X$ and $Y$ be tensor fields over a manifold $\mathscr{M}$. The Lie derivative $\mathscr{L}_X Y$ of $Y$ along $X$ can be defined in the following cases:

1.) The usual one when $X$ is a vector field. If $Y$ is also a vector field, $\mathscr{L}_X Y=[X,Y]$ is the Lie bracket of $X$ and $Y$. All other extensions can be defined by analogy with the Lie bracket with the aid of the Leibniz rule for derivations;

2.) When both $X$ and $Y$ are completely anti-symmetric contravariant tensor fields (i.e. $p$-vector fields), $\mathscr{L}_X Y=[X,Y]$, where $[\cdot,\cdot]$ is the Schouten-Nijenhuis bracket;

3.) When $X$ and $Y$ are vector-valued differential forms, $\mathscr{L}_X Y=[X,Y]$, where $[\cdot,\cdot]$ is the Frölicher-Nijenhuis bracket.

In complete generality, your problem should be seen as (part of) the problem of finding all bilinear, graded-symmetric natural operators from tensor fields of a certain rank to tensor fields of an appropriate rank. The above special cases of this problem are treated from this point of view in the book by I. Kolár, P. W. Michor and Jan Slovák, Natural Operations in Differential Geometry (Springer-Verlag, 1993), Section 30, pp. 250ff.

share|improve this answer
    
Thank you very much, I will consult that book. –  shuhalo Jan 28 at 19:40
    
This mathematical physics paper might also be interesting. "Do Killing-Yano tensors form a Lie algebra?" (arxiv.org/abs/0705.0535) –  José Figueroa-O'Farrill Jan 28 at 19:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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