Equivalent Definition of a Hermitian Metric on an Almost Complex Manifold? For an almost-complex manifold $M$ with almost-complex structure $J$, we say that a metric $h:T_p(M;{\mathbb R}) \times T_p(M;{\mathbb R}) \to {\mathbb R}$ is Hermitian if it holds that 
$$
h(J(v),J(w)) = h(v,w),
$$
for all $v,w \in T_p(M;{\mathbb R})$, and for all $p \in M$. 
Let us denote the complexification of $T_p(M;{\mathbb R})$ by $T_p(M;{\mathbb C})$, i.e.
$$
T_p(M;{\mathbb  C)} := T_p(M;{\mathbb R}) \otimes {\mathbb C}.
$$
As is easy to see, since $J^2 = -$id, we get a decomposition of $T_p(M;{\mathbb  C)}$ into $\pm i$ eigenspaces of the complex linear extension of $J$ to $T_p(M;{\mathbb  C)}$. We denote these spaces by $T_p(M;{\mathbb  C)}^{(1,0))}$, and $T_p(M;{\mathbb  C)}^{(0,1))}$ respectively. Assuming that $g$ is Hermitian easily implies that  $T_p(M;{\mathbb  C)}^{(1,0)}$, and $T_p(M;{\mathbb  C)}^{(0,1))}$ are orthogonal with respect to $g$. But is this an equivalent definition, i.e. is $g$ Hermitian if and only if the $(1,0)$ and $(0,1)$ vector fields are orthogonal?
 A: Let $V=T_pM$, $V_{\mathbb C}=V\otimes\mathbb C$. Then $J$ extends, complex linearly, uniquely, to $V_{\mathbb C}$, with the $\pm i$ eigen-space  decomposition $V_{\mathbb C}=V^{1,0}\oplus V^{0,1}.$ Likewise $V^*_{\mathbb C}=(V^*)^{1,0}\oplus (V^*)^{0,1}.$ A scalar product on $V$ is a symmetric tensor $g\in S^2(V^*)$, and can also be extended complex linearly (uniquely) to $g_{\mathbb C} \in  S^2(V^*_{\mathbb C})=S^2((V^*)^{1,0})\oplus S^2((V^*)^{0,1})\oplus (V^*)^{1,0}\otimes (V^*)^{0,1}$. So $g_{\mathbb C}$ decomposes accordingly as $g= g^{2,0}+g^{0,2}+g^{1,1},$ where 
$g^{0,2}=\overline{g^{2,0}}$ and $\overline{g^{1,1}}=g^{1,1}$. 
Now you take  the condition that  $J$ is $g$-orthogonal, ie $g(Jv_1, Jv_2)=g(v_1, v_2)$ for all $v_1, v_2\in V$, and seperate it into $(p,q)$ types and get that it is equivalent to $g^{0,2}=g^{2,0}=0$, i.e.  $g=g^{1,1}.$ 
In other words, when $J$ is $g$-orthogonal $V^{1,0}$ and $V^{0,1}$ are not orthogonal with respect to $g_{\mathbb C},$ but just the opposite... 
The correct condition for the $g$-orthogonality of the almost complex structure $J$, in terms of the decomposition  $V_{\mathbb C}=V^{1,0}\oplus V^{0,1}$ associated with $J$, is that each summand is isotropic with respect to $g_{\mathbb C}$ (the restriction of the quadratic form to it vanishes). 
