$\mathbb{C}-$linear extension of 2-forms to (1,1)-forms

Question

I am trying to analyze a bit how we can extend a differential form to the complexification $V\otimes\mathbb{C}=V_{\mathbb{C}}$ of the vector space. Of course, you can do this for a general $k$-form, but I think the simpler the example the better I can grasp the idea.

So let $V=\{<v_1,\ldots,v_m>\}$ be a real vector space of even dimension and $V^{1,0},V^{0,1}$ the usual eigenspace related to the almost complex structure $I.$ Consider the $2$ form on $\bigwedge ^2V,$ $a=\sum_{i<j} f_{ij}v_i\wedge v_j$. Then $a$ is a map $V\times V\to \mathbb{R}$ such that $ a(u,w)=\sum_{i<j}f_{ij}(v_i\otimes v_j-v_j\otimes v_i)(u\otimes w)=\sum_{i<j}f_{ij}(v_i(u) v_j(w)-v_j(u) v_i(w))$.

Now we wish to expand $a$ in a "linear" way so it becomes a mapping $V_{\mathbb{C}}\times V_{\mathbb{C}}\to \mathbb{C}.$ And this is where I need help. I have looked a few sources, but none of them seems to address this. (so from here and beyond maybe nothing makes sense)

So I am only trying to do what it feels "natural". And it feels like there are $2$ ways we can go

1. Just feed complex values via $a(u\otimes z_1,w\otimes z_2)=\sum_{i<j}f_{ij}(z_1v_i(u) v_j(w)-z_2v_j(u) v_i(w))$, where $(u\otimes z_1,w\otimes z_2)\in V_{\mathbb{C}}\times V_{\mathbb{C}}$

2. Use the eigenspaces, $V^{1,0},V^{0,1}$ writing $v=x+iy\in V^{1,0}$ and $\bar{v}=x-iy\in V^{0,1}$ where $x,y\in V.$ And pass to

$(1,1)-$forms $a=\sum_{i<j} f_{ij}v_i\wedge \bar{v_j}$ and I assume when we compute $a(u,w)$ here we mean to expand $v_i\wedge \bar{v_j}=dx_i\wedge dx_j-idx_i\wedge dy_j+idy_i\wedge dx_j+dy_i\wedge dy_j$ and do the same as before $dx_i\wedge dx_j(u\otimes z_1,w\otimes z_2)\ldots$

I'm maybe mixing up some stuff, but the subject is fairly new to me. Can you give me an insight how to make a bit of sense of these objects and how we properly extend the differential forms?

Answer 1

A differential $2$-form $\alpha$ on a smooth manifold $M$ is a smooth section of $\bigwedge^2T^*M$. For each $p \in M$, we see that $\alpha_p \in \bigwedge^2T_p^*M$ and hence can be viewed as a skew-symmetric map $\alpha_p : T_pM\times T_pM \to \mathbb{R}$.

So the relevant linear algebra is as follows. Given a real vector space $V$, an element $a \in \bigwedge^2V^*$ can be viewed as a skew-symmetric map $a : V^2 \to \mathbb{R}$. We can extend $a$ to a map $V_{\mathbb{C}}^2 \to \mathbb{C}$ by extending $a$ complex multilinearly. That is, if $u, w \in V$ and $z_1, z_2 \in \mathbb{C}$, then $\alpha(u\otimes z_1, w\otimes z_2) := z_1z_2\alpha(u, w)$.

If $\{v_1, \dots, v_n\}$ is a basis for $V$ with dual basis $\{v^1, \dots, v^n\}$ of $V^*$, then we can write $a = \sum_{i<j}f_{ij}v^i\wedge v^j$ for some $f_{ij} \in \mathbb{R}$, where $(v^i\wedge v^j)(u, w) = v^i(u)v^j(w) - v^i(w)v^j(u)$. Extending as above, we still have $a = \sum_{i<j}f_{ij}v^i\wedge v^j$ as before, but now $(v^i\wedge v^j)(u\otimes z_1, w\otimes z_2) = z_1z_2(v^i\wedge v^j)(u, w)$.