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Let $M$ be a complex manifold with almost complex structure $J$, and $TM$ it's real tangent space, $TM^{\mathbb{C}}=TM \otimes \mathbb{C}$ its complexified tangent space.

Now I'm getting confused with the metrics on this space, and so I will try and list what I think is going on and where I'm confused.

• A Riemannian metric on $M$ is a real , symmetric, positive definite sesquilinear form $$g : TM \times TM \to \mathbb{R}$$ such that it is $J$-invariant (i.e. $g(X,Y)=g(JX, JY), \, \forall X,Y \in \Gamma (TM$)).

• A Hermitian structure on M is a map $$h: TM^{\mathbb{C}} \times TM^{\mathbb{C}} \to \mathbb{C}$$ such that it is a $J$-invariant, symmetric (i.e. $h(X,Y)=\overline{h(Y,X)}, \, \forall X, Y \in \Gamma(TM^{\mathbb{C}})$ ), positive definite, sesquilinear form.

• Now the texts usually say something like: given a Riemannian metric g, there exists a Hermitian structure $$h(X,Y) = g(X,Y)-ig(X,JY)$$

But this is only defined on $TM$ and we need h to be defined on $TM^{\mathbb{C}}$? Do we then extend it linearly in the first argument and anti-linearly on the second to define it on all of $TM^{\mathbb{C}}$ and thus defining h as a Hermitian structure? Also, why can't we define $h(X,Y) = g(X,Y)$? I don't see why this would not define a Hermitian structure.

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Your problem is local, so let me just work over one point, or in other words in a real vector space $V$ with a symmetric bilinear form $g$ (note: the term sesquilinear does not make sense in this case).

There are two ways to associate a complex vector space to $V$ that we have to discuss separately:

(1) Complexification: $V^\mathbb{C}:=V\otimes_\mathbb{R}\mathbb{C}$ becomes a complex vectorspace by defining $$ \lambda \cdot (v\otimes \mu) = v \otimes(\lambda \mu) \quad \forall \lambda \in \mathbb{C}, v\otimes\mu \in V\otimes_\mathbb{R} C=V^\mathbb{C} $$ For the dimension you have $\dim_\mathbb{C}V^\mathbb{C}=\dim_\mathbb{R}V$. Now you can extend $g$ to a bilinear form $G$ or to a sesquilinear form $H$ on $V^\mathbb{C}$. They are defined as follows: $$G(v\otimes \lambda , w\otimes \mu)=\lambda \mu g(v,w) \quad \text{and} \quad H(v\otimes \lambda , w\otimes \mu)=\lambda \bar{\mu} g(v,w) $$ Note that $G$ will be symmetric again (i.e. $G(X,Y)=G(Y,X))$, whereas $H$ turns out to be hermitian (i.e. $H(X,Y)=\overline{H(Y,X)}$). In geometry one usually works with $H$, because it naturally provides us with a norm on $V^\mathbb{C}$.

Note that what we have done in the preceeding paragraph does not depend on a complex structure $J$, it always works.

(2) Complex Structure: Assume now that you have an (real!) endomorphism $J$ on V with $J^2=-I$. (In the vector space case this is called a linear complex structure and in in geometry usually an almost complex structure, while the term complex structure is reserved for those (global) endomorphisms which are integrable.)

Now you can turn $V$ itseld into a complex vector space that I will denote with $V_\mathbb{C}$. Complex multiplication is defined as $$ \lambda \cdot v = \Re \lambda \cdot v + \Im\lambda \cdot Jv \quad \forall \lambda\in\mathbb{C}, v\in V_\mathbb{C}=V $$ For the dimensions you have $\dim_\mathbb{C} V_\mathbb{C} = \dim_\mathbb{R}/2$.

Note that $g$ is still a function on $V_\mathbb{C}\otimes_\mathbb{C}V_\mathbb{C}$, but in general there is no reason to assume that it will be sesquilinear. This is really a stronger statement, because it involves the endomorphism $J$ and even $J-$invariance is usually not enough. However if $g$ happens to be $J$ invariant, then we can define a hermitian sesquilinear form $h$ on $V_\mathbb{C}$ in the way you have mentioned: $$ h(v,w)=g(v,w)-ig(v,Jw) $$

A counterexample: In the second case I have claimed that we cannot expect $g$ to be sequilinear and indeed there is an easy counterexample: Take $V=\mathbb{R}^2$ with the usual euclidean inner product $g$ and let $J$ be the rotation by $\pi/2$. This is an isometry so $g$ is clearly $J-$invariant. This $J$ induces a complex multiplication on $V$ that is identical to the usual one for the complex numbers, so $V_\mathbb{C}=\mathbb{C}$.

Let $v=(1,0)$, then $i\cdot v = (0,1)$, but $$ g(i \cdot v,v) = 0 \neq i = i g(v,v), $$ hence $g$ is not sesquilinear.

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    $\begingroup$ This is exactly the answer I was wanting! So we are considering the second case, but I was mixing it up with the first. Thanks! $\endgroup$
    – user500074
    Commented Jan 7, 2018 at 18:17
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    $\begingroup$ You're welcome. Maybe I should add something: Actually it is impossible that a symmetric $\mathbb{R}-$bilinear form $g:V\otimes V \rightarrow \mathbb{R}$ is already sesquilinear (w.r.t. some $J$) itself, because sesquilinear forms necessarily also assume imaginary values whereas $g$ is always real. $\endgroup$
    – Jan Bohr
    Commented Jan 7, 2018 at 20:55
  • $\begingroup$ In fact, the $h$ defined here is not sesquilinear in the sense of $\mathbb{C}$ linear in the first argument and $\overline{\mathbb{C}}$ linear in the second argument - it has the opposite symmetries. For example, $h(Ju,v) = -i h(u,v)$ under this definition. $\endgroup$
    – Triangle
    Commented Feb 16 at 19:30

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