In a previous post Lie bracket as a tensor I asked about the formalism of deformed Lie brackets, how a tensor could satisfy the Leibniz rule. This is question is solved and now I want a proof of the next result:

Let $M$ be a finite-dimensional manifold and let $[\cdot,\cdot]$ be the usual Lie bracket of the $\mathbb R$-algebra of vector fields over $M$. Consider an endomorphism $N\in \Gamma^{\infty}(TM\otimes T^*M)$. Then, the bracket

$$ [X,Y]_N = [N(X),Y] + [X,N(Y)] - N([X,Y]) $$

satisfies the Jacobi identity (and then defines a Lie bracket) if and only if $N$ is a Nijenhuis tensor, i.e.

$$ \frac{1}{2}[N,N]_{FN}(X,Y) = [N(X),N(Y)]-N([N(X),Y]) - N([X,N(Y)]) + N^2([X,Y]) = 0 $$

for all $X,Y$ in $\Gamma^{\infty}(TM)$.

One proof can be found in Deformations on Lie algebra structures, of A. Nijenhuis and R. W. Richardson, (1967), publised on Journal of Mathematics and Mechanics, T. 17, No. 1, pp. 89-105.

However, I have been looking for at Internet and i have not found anything. And also, the library of my university neither have it.


After big efforts and many computations finally I founded this equality:

$$ \oint_{X,Y,Z} [X,[Y,Z]_N]_N = -\frac{1}{2} \oint_{X,Y,Z} \Bigg(T_N(X,[Y,Z]) + [X, T_N(Y,Z)] \Bigg) , $$

where I am naming

$$ T_N(X,Y) = \frac{1}{2} [N,N]_{FN}(X,Y) $$


$$ \oint_{X,Y,Z} f(X,Y,Z) = f(X,Y,Z) + f(Y,Z,X) + f(Z,X,Y) $$

the cyclic sum that appears in the Jacobi identity.

So one direction is clear. But, I am not sure

$$ \oint_{X,Y,Z} \Bigg(T_N(X,[Y,Z]) + [X, T_N(Y,Z)] \Bigg) = 0 $$


$$ T_N(X,Y)=0 $$

for all vector fields $X$ and $Y$.

Can anybody helps me?


  • $\begingroup$ Another perfect answer would be a photo (or pdf) with the original proof of the paper or a link to download the pdf $\endgroup$ – Dog_69 Apr 28 '17 at 22:35
  • $\begingroup$ You can just compute the Jacobi identity for $[X,Y]_N$. $\endgroup$ – Dietrich Burde Apr 29 '17 at 19:55
  • $\begingroup$ @DietrichBurde the computation was hard and it is not enough. Some extra idea is necessary. See my edit please. $\endgroup$ – Dog_69 May 2 '17 at 22:51

Well... as is usual, I will answer my own question. The statement is more general than my question. In fact, $(TM,[\cdot,\cdot]_N,N)$ is a Lie algebroid if and only if $N$ is a Nijenhuis operatoor. The proof is as follows.

First, check $N$ is a homomorphism if and only if $T_N=0$:

$$ N([X,Y]_N)= N([N(X),Y]) + N([X,N(Y)]) - N^2([X,Y]) = [N(X),N(Y)] \, , \quad \forall X,Y\in{\frak X}(M) \, . $$ This expresion is equivaent to say $T_N=0$.

After, consider the derivation $D$ defined on $T^*M$ by means of this new bracket and the anchor map $N$:

$$ D \alpha \, (X_1,\dots, X_{p+1}) = \sum_{i=1}^{p+1} (-1)^{i+1} N(X_i)(\alpha(X_1,\dots,\hat{X}_i,\dots,X_{p+1}) + \sum_{i<j} (-1)^{i+j} \alpha([X_i,X_j]_N,X_1,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_{p+1}) \, . $$

As a derivation of degree 1, it can be write as (Frölicher-Nijenhuis decomposition theorem)

$$ D = \mathcal{L}_K + i_L $$

where $K$ is $TM$-valued 1-form and $L$ is a $TM$-valued 2-form. Studing the behavior of $D$ over functions and 1-forms one can see that $L=0$ and $K=N$, so

$$ D=\mathcal{L}_N \, . $$

Finally we have to study the operator $\mathcal{L}^2_N$. We know

$$ \mathcal{L}^2_N = 0 \Longleftrightarrow T_N = 0 \, , $$


$$ T_N=\frac{1}{2} [N,N]_{FN} = \frac{1}{2} [\mathcal{L}_N,\mathcal{L}_N](f) = \mathcal{L}_N^2 (f) \, . $$

A short computation shows

$$ D^2\alpha (X,Y,Z) = \mathcal{L}^2_N \alpha(X,Y,Z) = \oint_{X,Y,Z} N([X,Y])(\alpha(Z) ) - \oint_{X,Y,Z} [N(X),N(Y)](\alpha(Z)) + \oint_{X,Y,Z} \alpha([[X,Y]_N,Z]_N) \, , $$


$$ \oint_{X,Y,Z} \alpha([[X,Y]_N,Z]_N) = 0 \Longleftrightarrow N([X,Y]_N)= N([N(X),Y]) + N([X,N(Y)]) - N^2([X,Y]) = [N(X),N(Y)] \Longleftrightarrow T_N = 0 $$


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