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I'm in the following situation: let's assume I am given the following vector fields in $\mathbb{C}^2$, not identically zero: $$ X=X_1\partial _x+X_2\partial _y $$ $$ Y=Y_1\partial _x+Y_2\partial _y $$ $$ S=px\partial _x+qy\partial _y $$ where $p$ and $q$ are natural numbers which are relatively prime, and $X_1, X_2, Y_1, Y_2$ are holomorphic functions in $\mathbb{C}^2$. Suppose that $$ X\wedge Y=f\partial_x\wedge\partial_y,S\wedge X=g\partial_x\wedge\partial_y,S\wedge Y=h\partial_x\wedge\partial_y $$ where $f,g,h$ are holomorphic in $\mathbb{C}^2$. Also suppose that $$ [S,X]=mX, [S,Y]=nY, [X,Y]=0 $$ where $m$ and $n$ are non-zero natural numbers. I need to show the following identities: $$ S(f)=(m+n+p+q)f $$ $$ S(g)=(m+p+q)g $$ $$ S(h)=(n+p+q)h $$

where a field $F=F_1\partial_x+S_2\partial_y$ applied to a function $\delta$ is $$ F(\delta)=F_1\dfrac{\partial \delta}{\partial x}+F_2\dfrac{\partial \delta}{\partial y} $$

I have tried a lot of things, even writting the Lie brackets in local coordinates and hoped for the best but nothing came out. I need help! Thanks in advance.

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Gustavo,

Are you sure this is all correct? The right approach is to use the fact that Lie derivative is a derivation and compute $L_S(X\wedge Y)$, etc. But I don't get your answers, and I think the hypotheses are mutually contradictory. In particular, the only way $[S,X]=mX$ can hold is to have $m=p$ and $X_2=0$. What is going on? :)

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I'm trying to check that $[S,X]=mX$ can only check in the conditions you mentioned, but I think that there is no problem here (there could be elsewhere!). Under the conditions for $S$ (it's called a linear diagonalizable vector field which under a variable change can be written as mentioned), when $[S,X]=mX$, we say that $X$ is quasi-homogeneous with respect to $S$. It's also possible to show that this $m$ is a non-zero natural and that $X$ must be a polynomial vector field to be quasi-homogeneous. –  Marra May 4 '13 at 2:39
    
I will follow your advice and see if $L_S(X\wedge Y)$ gives me the answer I need :) –  Marra May 4 '13 at 2:41
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