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Given a $2m$-dimensional manifold $M$, an almost complex structure $J$ is equivalent to a $\text{GL}(m,\mathbb C)$-structure on $M$.

I wonder why the intrinsic torsion of the $\text{GL}(m,\mathbb C)$-structure is equivalent to the Nijenhuis tensor of $J$.

Can some one give a suggestion to calculate this or a reference with readable proof.

Thanks in advance.

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  • $\begingroup$ It no one beats me, I will try to give answer tomorrow or the day after. $\endgroup$
    – M.G.
    Commented Oct 20, 2018 at 1:39
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    $\begingroup$ @M.G. I hope no one beats you. $\endgroup$
    – Yikun Qiao
    Commented Oct 22, 2018 at 15:51

2 Answers 2

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Not trying to beat M. G., I would rather provide an elementary explanation of what's essentially going on, trying to convince you that what you've asked is really something one might expect.

Let $\nabla$ be a connection preserving the field of endomorphisms $J$, i. e. $\nabla_XJY = J\nabla_XY$, and $T$ be its torsion. If we change the connection, if would differ by a vector valued 2-tensor $A$ satisfying $A(X,JY) = JA(X,Y)$. The torsion $T'$ of the changed connection $\nabla'$ would be given by $T'(X,Y) = T(X,Y) + A(X,Y) - A(Y,X)$.

Suppose we want to come up with an expression in terms of torsion, which would be invariant w. r. t. adding a $J$-linear in its second entry tensor $A$ to the connection. One needs to find a nontrivial relation satisfied by the skew symmetric vector valued 2-tensor $$\alpha(X,Y) = A(X,Y) - A(Y,X),$$ which would give rise to an expression in terms of torsion depending on the field $J$ alone. This is straightforward:

$$\alpha(X,Y) = A(X,Y) - A(Y,X) = -JA(X,JY) + JA(Y,JX) = \\ = -J(\alpha(X,JY) + A(JY,X) - \alpha(Y,JX)-A(JX,Y)) = \\ = -J(\alpha(X,JY) + \alpha(JX,Y)) + A(JX,JY) - A(JY,JX) =\\ = -J(\alpha(X,JY) + \alpha(JX,Y)) + \alpha(JX,JY).$$

Hence the tensor $\alpha$ satisfies the formula $$\alpha(X,Y) + J\alpha(X,JY) + J\alpha(JX,Y) - \alpha(JX,JY) = 0,$$

and the corresponding expression involving torsion

$$\nu(X,Y) = T(X,Y) + JT(X,JY) + JT(JX,Y) - T(JX,JY)$$

is independent on the choice of the connection.

Let's write the tensor $\nu$ in terms of the connection.

$$\nu(X,Y) = \nabla_XY - \nabla_YX - [X,Y] + \\ + J\nabla_XJY - J\nabla_{JY}X - J[X,JY] + \\ + J\nabla_{JX}Y - J\nabla_YJX - J[JX,Y] - \\ - \nabla_{JX}JY + \nabla_{JY}JX + [JX,JY].$$

The connection terms cancel out as $\nabla_XY+J\nabla_XJY = 0$, $J\nabla_{JX}Y - \nabla_{JX}JY = 0$, $-\nabla_YX - J\nabla_YJX = 0$ and $-J\nabla_{JY}X + \nabla_{JY}JX = 0$. What rests is $$\nu(X,Y) = [JX,JY] - J[X,JY] - J[JX,Y] - [X,Y],$$ i. e. $\nu$ is the Nijenhuis tensor.

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  • $\begingroup$ To me your name commemorates the abolition of Gulag in 1957, thank you for that reminder and for the answer too of course $\endgroup$
    – rych
    Commented Oct 25, 2018 at 12:44
  • $\begingroup$ Excellent answer! One minor nitpicking: it's not that $J\nabla_X Y = \nabla_X JY$ is an almost complex structure, but the connection form $A$ satisfying this is. $\endgroup$ Commented Aug 28, 2023 at 2:44
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The $G$-structure consists of all coframes (or equivalently, frames) by complex linear 1-forms, i.e. its local sections are coframings $\omega^1,\dots,\omega^n$ consisting of locally defined $\mathbb{C}$-linearly independent $(1,0)$-forms. The torsion of the $G$-structure (as for any $G$-structure) is expressed by Cartan's structure equations $d\omega+\gamma\wedge\omega=T\omega\wedge\omega$, where we allow any $\mathfrak{g}$-valued 1-form $\gamma$ here (the pseudoconnection). The intrinsic torsion is the part you can't absorb by choice of $\gamma$. So in our case $G=GL(n,\mathbb{C})$, and we can absorb any complex 1-form multiple of the $\omega^a$ into choice of $\gamma$, i.e. $d\omega^a+\gamma^a_b \wedge \omega^b = \frac{1}{2}T^a_{\bar{b}\bar{c}} \omega^{\bar{b}}\wedge \omega^{\bar{c}}$. A long calculation, using the vector fields $X_a$ dual to the $\omega^a$, reveals that $T^a_{\bar{b}\bar{c}}\omega^{\bar{b}}\wedge\omega^{\bar{c}}X_a$ is the negative of the Nijenhuis tensor.

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