# Is there a natural Lie bracket for $\mathfrak X (M) \times C^\infty(M)$ (pairs of vector fields and smooth functions)?

Space of smooth vector fields $\mathfrak X(M)$ on a manifold $M$ has a structure of Lie algebra with the bracket being a commutator of two vector fields.

Does cartesian product $\mathfrak X (M) \times C^\infty(M)$ of a space of smooth vector fields $\mathfrak X(M)$ with a space of smooth functions $C^\infty(M)$ on the same manifold naturally form a Lie algebra?

What about $\mathfrak X (M) \times C^\infty(M) \times \ldots \times C^\infty(M)$ ?

• Do you have any reason to believe the answer to your question is positive? Could you write something about such a reason? Sep 28 '14 at 17:15
• @AmitaiYuval If you are still interested, the reason for the question comes from fluid dynamics. Ideal incompressible fluid is described solely by its (divergence-free) velocity field. This system is Hamiltonian with a Lie-Posson bracket based on a Lie bracket of vector fields. If there are additional variables, like energy, the reversible part of evolution is still described by a Poisson bracket. I wondered, if this is still a Lie-Poisson bracket and what is the Lie bracket then for velocity fields coupled with this additional scalar fields. Sep 29 '14 at 7:08

If $\mathfrak g$ is a Lie algebra, and $V$ a representation of $\mathfrak g$, then you can define a Lie algebra structure on $\mathfrak g\oplus V$ by the formula $$[(X,u),(Y,v)]=([X,Y],X\cdot v-Y\cdot u)$$ This applies to your example via the standard action of vector fields on functions by derivation.
Here are the details. The bracket is clearly antisymmetric. We only need to check the Jocaobi identity. If $X,X',X''\in\mathfrak g$ and $v,v',v''\in V$, then \begin{align} [(X,v),[(X',v'),(X'',v'')]]&=[(X,v),([X',X''],X'\cdot v''-X''\cdot v')\big]\\ &=([X,[X',X'']],X\cdot (X'\cdot v''- X''\cdot v')-[X',X'']\cdot v)\\ &=([X,[X',X'']],X\cdot X'\cdot v''- X\cdot X''\cdot v'\\ &\qquad\qquad -X'\cdot X''\cdot v+X''\cdot X'\cdot v) \end{align} So for $X,Y,Z\in\mathfrak g$ and $u,v,w\in V$. Then \begin{align} [(X,u),[(Y,v),(Z,w)]]\\ +\:[(Y,v),[(Z,w),(X,u)]]\\ +\:[(Z,w),[(X,u),(Y,v)]] \end{align} has its first coordinate equal to $$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$$ and its second coordinate equal to \begin{align} &\phantom{+} X\cdot Y\cdot w- X\cdot Z\cdot v -Y\cdot Z\cdot u+Z\cdot Y\cdot u\\ &+Y\cdot Z\cdot u- Y\cdot X\cdot w -Z\cdot X\cdot v+X\cdot Z\cdot v\\ &+Z\cdot X\cdot v- Z\cdot Y\cdot u -X\cdot Y\cdot w+Y\cdot X\cdot w \end{align} which equals $0$, as every term appears along with its opposite.