# Lie bracket as a tensor

First of all one comment: I know there is a question titled ''Lie bracket is not a vector field'' which proofs Lie bracket is not $C^\infty(M)$-linear.

Now, I show my question. In the paper Poisson-Nijenhuis Structures, F. Magri and Y. Kosmann-Schwarzbach (1990) (avaible on http://www.numdam.org/item?id=AIHPA_1990__53_1_35_0) one can find:

Let $\mu$ an $TM$-valued 2-form [i.e. a (1-2) tensor skew-symmetric in its covariant indexes] that defines a Lie algebra-structure on $TM$.

But they write the Lie bracket as

$$[X,Y]$$

instead of $\mu(X,Y)$. In addition they distinguish between

$$N.\mu(X,Y) = \mu(N(X),Y)) + \mu(X,N(Y)) - N(\mu(X,Y))$$

and

$$[X,Y]_{N.\mu} = [N(X),Y] + [X,N(Y)] - N([X,Y])$$

when they define the ''deformed Lie bracket''.

So, there is a (''canonical'') way to induce a Lie bracket (being not $C^\infty(M)$-lienar) by means of a vectorial valued 2-form? For example, it is possible with Poisson bracket (setting Schouten-Nijenhuis bracket of the bivector with itself be zero).

Thank you

## 1 Answer

The solution can be found in the Lie algebroids.

Basically, it is a quatern $(M,E,[\cdot,\cdot],\Phi)$ where

• $M$ is a manifold,

• $E$ is a vector bundle over $M$,

• $[\cdot,\cdot]$ is a Lie bracket on the $\mathbb R$-algebra of vector fields over $M$ and

• $\Phi:E\rightarrow TM$ is a bundle morphism that also satisfy the Leibniz rule,

$$[X,fY] = (\Phi(X))(f) Y + f[X,Y] ,$$ for all vector fields $X,Y$ and for all differentiable function $f$ over the manifold.

This structure allows us to generalize the usual case of vector fields.

In addition, with this formalism, every vectorial-valued 2-form which satisfies Jacobi identity can define a Lie bracket (as an operation of a $\mathbb R$-algebra) setting $E=TM$ and $\Phi$ the identity map. In the paper, the authors set $\Phi=N$ the Nijenhuis operator associated to $\mu$ and as Lie bracket the "deformed" bracket $[\cdot,\cdot]_{N.\mu}$. It is $C^{\infty}$-linear, that is true, but the Leibniz rule is also satisfy because we have the morphism.

PD: My definition is essentially the same as in https://en.wikipedia.org/wiki/Lie_algebroid but including the manifold explicitly.