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Let $(M^{2m},h,J)$ be an almost Hermitian manifold with fundamental form $\Omega$.

I want a local expression for this $\Omega (X,Y)=h(JX,Y)$ in this almost-complex manifold, but during my computations I struggle with the almost complex structure $J$. The only thing I know is that $J^2=-Id$.

So my questions so far are:

  1. What does the (1,1)-tensor $J$ do with an arbitrary vector field X (local/global)?

  2. Would it be better to do the computations in a local orthonormal frame?

  3. Does a Riemannian metric in a local orthonormal frame behave like the standard inner product?

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    $\begingroup$ $J$ is the action of $i$ on the tangent space at every point. $\endgroup$
    – jgon
    Commented May 12, 2019 at 16:50
  • $\begingroup$ so I have to complexify the tangent bundle first? In a local chart $(U,\phi )$ with $\phi =(x_1,\dots,x_{2m})$ I can write the metric tensor as $h=\sum_{k,l=1}^{2m}{h_{kl}\cdot dx_k\otimes dx_l}$. As J is the action of $i$, this gives as $\Omega =ih$? How is $\Omega$ then skew-symmetric? $\endgroup$
    – shaiyaiskagga
    Commented May 12, 2019 at 18:39
  • $\begingroup$ Also according to wikipedia, your definition of the fundamental form is wrong, unless you wanted a skew-Hermitian form. To see that $\Omega$ is skew-Hermitian, observe that $h(JX,Y) = ih(X,Y) = i\overline{h(Y,X)} = \overline{-ih(Y,X)} = -\overline{h(JY,X)}$. $\endgroup$
    – jgon
    Commented May 12, 2019 at 18:56
  • $\begingroup$ I think part of what confuses me about this question is that you're asking for a local expression for $\Omega(X,Y)=h(JX,Y)$, but that expression is already valid globally, locally, or even pointwise. Thus I'm not entirely sure what you're looking for in an answer. $\endgroup$
    – jgon
    Commented May 12, 2019 at 19:42
  • $\begingroup$ Also as for question 3, a Riemannian metric behaves like the standard inner product in any frame, but in an orthonormal frame, its matrix is the identity matrix, which is the same matrix as that of the dot product on $\Bbb{R}^n$ with respect to the standard basis. $\endgroup$
    – jgon
    Commented May 12, 2019 at 19:44

1 Answer 1

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Your comments have now led me to a proper understanding of the question, and I can now write an answer.

We have the following set up. $M$ is a $2m$-manifold, with almost complex structure $J$ and a Riemannian metric $h$ that preserves the almost complex structure. I.e., so that $h(Ju,Jv)=h(u,v)$ for all tangent vectors $u$ and $v$ at every point.

You then define the fundamental form $\Omega(X,Y) = h(JX,Y)$, which is skew symmetric, since $$\Omega(Y,X) = h(JY,X) = h(X,JY) = h(JX,J^2Y) = h(JX,-Y) = -\Omega(X,Y).$$

I'll note at this point that your choice of letters is slightly misleading. I would expect $g$ for the Riemannian metric, and wikipedia leads me to expect $\omega$ for the fundamental form. However, now that the meanings have been clarified, it doesn't really matter what letters you picked.

Let $e_i$ be a local orthonormal frame. It's kind of poor notation to regard the vectors $e_i$ as covectors, even though they are canonically identified with such via $h$. Thus let $\theta_i$ be the dual basis, $\theta_i(x) = h(e_i,x)$. Observe that $h(Je_i,-) = -h(e_i,J-)$, so the dual of $Je_i$ is $-\theta_i J$.

Now you want to show that $$\Omega = -\frac{1}{2} \sum_i \theta_i \wedge \theta_i J.$$

First, since $e_i$ are orthonormal, observe that the entries of the matrix for $J$ with respect to the $e_i$ are $J_{ij} = h(e_i,Je_j)$. Also since $h(Jx,Jy)=h(x,y)$, $J$ is orthonormal, so $J^T = J^{-1}$.

To do so, observe that $Je_i$ is also an orthonormal frame, so it suffices to show that both sides agree for all pairs $(e_i,Je_j)$ that we evaluate them at. Evaluating $\Omega$ first, we get $$\Omega(e_i,Je_j) = h(Je_i,Je_j) = \delta_{ij}.$$ Now evaluating the right hand side, we have $$-\frac{1}{2} \sum_k \theta_k(e_i)\theta_k(J^2e_j) - \theta_k(Je_j)\theta_k(Je_i)$$ $$=-\frac{1}{2} \sum_k -\delta_{ik}\delta_{jk} - h(e_k,Je_j)h(e_k,Je_i)$$ $$ = \frac{1}{2}\sum_k \delta_{ik}\delta_{jk} + \frac{1}{2}\sum_k J_{kj}J_{ki}$$ $$ = \frac{1}{2}\delta_{ij} + \frac{1}{2} \sum_k J^T_{ik}J_{kj}$$ $$ = \frac{1}{2}\delta_{ij} + \frac{1}{2} \delta_{ij} = \delta_{ij},$$ as desired.

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