# How many differential forms on the complex plane?

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?

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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 $(\Bbb C^\ast)^{0,1}$.

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Dear Henry, the statement " $\Bbb C$ is a real vector space of dimension $2$, with basis $\{x,y\}$" does not make sense: a basis of $\mathbb C$ consists of two complex numbers and $x,y$ are not complex numbers (they are real linear forms). The basis you want is $(1,i)$. Also, $dz = dx + i\, dy$ and $d\bar{z} = dx + i\, dy$; in other words the signs of the imaginary parts are wrong. – Georges Elencwajg May 17 '14 at 7:07
Your question helped me immensely in focusing on the right mathematical framework. However, I still have a small doubt. Once you have the complxified space $V\otimes_{\mathbb{R}}\mathbb{C}$, you can act on it both by multiplication by $i$, both by applying $J$. Are those two operations the same thing? Precisely, is it true that $$i(\mathbb{v}\otimes z)=J(\mathbb{v}\otimes z)?$$ – Giuseppe Negro May 18 '14 at 23:07
(PS: I wrote "question" where, of course, I meant "answer") – Giuseppe Negro May 18 '14 at 23:55
@GiuseppeNegro They are different. We can complexify any real vector space $V$ my taking $V \otimes_{\Bbb R} \Bbb C$, and complexification makes no reference to a complex structure on $J$. For example, $J(dz - d\bar{z}) = i(dz + d\bar{z}) = 2\, dx \neq i(dz - d\bar{z}) = -2\, dy$, so we explicitly see that $J$ is not the same as multiplication by $i$. – Henry T. Horton May 19 '14 at 21:13
It is a bit confusing indeed. Anyway, I found this Wikipedia page: en.wikipedia.org/wiki/… which says something about the construction you suggested. I am leaving it here for future reference – Giuseppe Negro May 20 '14 at 11:42

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.

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So according to this interpretation the objects $dz$ and $d\overline{z}$ are real-linear, complex-valued differential forms, if I understand well. But then, what do you mean by "complexify the space"? Can you give some more detail on this, please? – Giuseppe Negro May 17 '14 at 1:12
Ja I mean $V\otimes \mathbb{C}$. – Rene Schipperus May 17 '14 at 1:31
Maybe it helps not to think of them as forms but just as elements of a vector space, where we are now allowing complex coefficients, but they remain independent. – Rene Schipperus May 17 '14 at 1:33
Here's a thing that I fail to understand. You say "the underlying real vector space is 2 dimensional", so we are considering $\mathbb{R}^2$. Then we complexify it, that is, we take $\mathbb{R}^2\otimes_{\mathbb{R}} \mathbb{C}$. Now, shouldn't this last space be $2\times 2=4$ dimensional? – Giuseppe Negro May 18 '14 at 10:46
You have to be careful about what the base field is. Yes it is 4 dimensional over $\mathbb{R}$ but only 2 dimensional over $\mathbb{C}$ – Rene Schipperus May 20 '14 at 0:24