How many differential forms on the complex plane?

Question

I am puzzled by the fact that the two differential forms $$\begin{array}{cc} dz=dx+i dy , & d\overline{z}=dx-i dy \end{array} $$ are $\mathbb{C}$-linearly independent, even if the underlying manifold $\mathbb{C}$ has complex dimension $1$. Of course there is nothing wrong: the second "differential form" is not $\mathbb{C}$-linear. Hence, it is not a true complex differential form. However it is not real valued, so it is not a real differential form either. So what it is?

Question. Can you clarify the nature of the objects $dz, d\overline{z}$ above? Which abstract notion, relative to complex manifolds, includes and generalizes them?

Answer 1

A complex structure $J$ on a real vector space $V$ is a endomorphism $J: V \longrightarrow V$ that squares to minus the identity: $J^2 = -\text{Id}_{V}$. Such a complex structure allows us to define scalar multiplication of elements $v \in V$ by complex numbers $a + bi$: $$(a + bi)\cdot v = av + b Jv.$$ But since $V$ is a real vector space, in order to actually use this scalar multiplication by complex numbers and consider $V$ as a complex vector space, we need to extend our field of scalars from $\Bbb R$ to $\Bbb C$ via complexification and consider $V \otimes_{\Bbb R} \Bbb C$. This will be where the "extra" forms come from.

Note that a complex structure $J$ has eigenvalues $\pm i$. Then we have a direct sum decomposition $$V \otimes_{\Bbb R} \Bbb C = V^{1,0} \oplus V^{0,1},$$ where $V^{1,0}$ is the $+i$-eigenspace of $J$ and $V^{0,1}$ is the $-i$-eigenspace of $J$. For $v \in V \otimes_{\Bbb R} \Bbb C$, $v - iJv \in V^{1,0}$ and $v + iJv \in V^{0,1}$.

The above relates to your question as follows. $\Bbb C$ is a real vector space of dimension $2$, with basis $\{1,i\}$ and complex structure $J = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. The dual space $\Bbb C^\ast$ is also real of dimension $2$ with a complex structure $J$ as above and basis $\{dx, dy\}$. By the above, we have a direct sum decomposition $$\Bbb C^\ast \otimes_{\Bbb R} \Bbb C = (\Bbb C^\ast)^{1,0} \oplus (\Bbb C^\ast)^{0,1}.$$ We see that $$dz = dx + i\, dy$$ is a generator for $(\Bbb C^\ast)^{1,0}$ and $$d\bar{z} = dx - i\, dy$$ is a generator for $(\Bbb C^\ast)^{0,1}$.

Answer 2

The underlying real vector space is 2 dimensional, spanned by $dx, dy$. We can complexify the space by taking the tensor product with $\mathbb{C}$ but it remains 2 dimensional. Then we are just choosing a different basis. Of course $d\overline{z}$ is not holomorphic but is real differentiable. That is one way of looking at it. All complex analysis can be reduced to real analysis using harmonic functions and Cauchy-Riemann.