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A complex valued 1-form $\alpha$ is defined as $\alpha \in C^{\infty}(T^*X\otimes\mathbb{C}$); that is if I understand it correctly a smooth function on the tensor product of the cotangent bundle with $\mathbb{C}$.

What is the factor $\mathbb{C}$ needed, what is it doing in that definition? And while we are at it, why is this 1-form defined on the cotangent bundle? Naively imagining 1-forms as some kind of co-vectors, I would rather have expected it do be defined on the tangent bundle $TX$ ...

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    I don't want to sound pedantic but it a complex valued 1-form is not a section of the tensor product of $T^{*} X$ with a **vector space** $\mathbb{C}$ (this is not even well-defined) but a section of the tensor product of the vector bundle $T^{*} X$ with the **vector bundle** $\underline{\mathbb{C}}$ which is the trivial product bundle $M \times \mathbb{C} \rightarrow M$. This amount to taking the tensor product of each **fiber** $T_p^{*}M$ (the space of linear functionals $T_pM \rightarrow \mathbb{R}$ with the vector space $\mathbb{C}$. – 2017-01-18

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When we write $C^\infty(T^*M\otimes\mathbb{C})$ in this context, that doesn't mean we have functions defined on $T^*M\otimes\mathbb{C}$. Rather, it means we mean certain functions which take values in $T^*M\otimes\mathbb{C}$, specifically smooth sections of the projection map $T^*M\otimes\mathbb{C}\to M$. More generally, if $E$ is a vector bundle on $M$, it is common to write $C^\infty(E)$ for the space of smooth sections of $E$. (This is unfortunately ambiguous with the notation $C^\infty(M)$ for the ring of smooth functions on $M$. A better notation is $C^\infty(M,E)$ for the space of sections of $E$ on $M$, but people sometimes abbreviate this to just $C^\infty(E)$.)

So that is why we have the cotangent bundle--at each point, we are picking a cotangent vector. The factor of $\mathbb{C}$ is because the cotangent space $T^*M_p$ at a point $p$ is the space of real linear functionals on the tangent space: they take in a tangent vector, and spit out a real number. By tensoring with $\mathbb{C}$, you get (up to isomorphism) the space of complex linear functionals on the tangent space, which spit out complex numbers instead. More generally, if $V$ is a real vector space, there is a natural isomorphism $\operatorname{Hom}(V,\mathbb{R})\otimes\mathbb{C}\to\operatorname{Hom}(V,\mathbb{C})$ (induced by the bilinear map which sends a pair $(f,c)\in(\operatorname{Hom}(V,\mathbb{R}),\mathbb{C})$ to the linear map $g(v)=cf(v)$).

So to sum up, a section of $T^*M\otimes\mathbb{C}$ is a function that at each point of $M$ gives you a complex-valued linear functional on the tangent space. That's a complex-valued 1-form.

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The cotangent space at each point is a real vector space, and tensoring it with $C$ turns it into a complex vector space with the same dimension, corresponding to taking a basis and allowing the real coefficients to be complex.

1-forms are the sections on cotangent space, so complex-valued 1-forms are the sections of the complexified cotangent space.

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    Ok thanks, now I understand the factor $\mathbb{C}$. But a real 1-form at a point $x$ is the linear mapping $\alpha_x: T_x X \mapsto \mathbb{C}$, so still dont understand why conversely a complex 1-form maps at a point $x$ is $\alpha_x: T^*_x X \mapsto \mathbb{C}$?Why is the domain of a real 1-form in the tangent space whereas for a complex 1-form the domain is in the cotangent space? – 2017-01-18
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    A real 1-form maps vector fields (in the tangent space) to the reals, and a complex 1-form maps vector fields (in the tangent space) to the complex numbers. But 1-forms are _elements_ of the space of sections of cotangent space, not of tangent space. – 2017-01-18