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Let $J\in\operatorname{end}(TM)=\Gamma(TM\otimes T^*M)$ with $J^2=-\operatorname{id}$ and for $X,Y\in TM$, let $$N(X,Y):=[JX,JY]-J\big([JX,Y]-[X,JY]\big)-[X,Y].$$ Prove that $N$ is a tensor field of type (1,2).

Since I heard $N$ is the Nijenhuis tensor and saw its component formula, I have tried proving it by starting with the component formula for $N^k_{ij}$ showing that for $X=X^i\frac{\partial}{\partial x^i}$ and $Y=Y^i\frac{\partial}{\partial x^i}$

$$ N(X,Y)=N^k_{ij}\frac{\partial}{\partial x^k}\otimes\mathrm{d}x^i\otimes\mathrm{d}x^j\left(X,Y\right).$$ But I am only able to show \begin{align} N(X,Y) &=\underbrace{\big(J^m_iX^i\frac{\partial}{\partial x^i}(J^k_jY^j)-J^m_jY^j\frac{\partial}{\partial x^m}(J^k_iX^i)\big)\frac{\partial}{\partial x^k}}_{=[JX,JY]}-\underbrace{J^k_m\left(\frac{\partial}{\partial x^i}J^m_j-\frac{\partial}{\partial x^j}J^m_i\right)X^iY^j\frac{\partial}{\partial x^k}}_{=?}. \end{align}

Anyway, I assume this way to be pretty dumb (although I would like to know if it is correct so far and how to go on from there), so I would like to know if there is a more elegant (and for general (k,l)-tensor fields better) way to prove the type of $N$ than writing it locally.

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1 Answer

up vote 4 down vote accepted

Being a tensor field means that you have to show $C^{\infty}(M)$-Linearity. (1,2)-Tensor just means, that it eats two vector fields at spits out another vector field which is obvious from the definition. So look at $N(X,fY)$ for $f \in C^{\infty}(M)$ and use the properties of the Lie-bracket, to show $N(X,fY)=fN(X,Y)$.

Note: $C^{\infty}$-linearity induces, that $N(X,Y)|_p$ just depends on $X_p$ and $Y_p$ in contrast to $\nabla_YX|_p$ in the upper entry.

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