# Complexification of Tangent Bundle

I am currently reading a book where the author says that the tangent and cotangent bundles $TM$ and $T^*M$ of a manifold $M$ are complexified.

I am not familiar with Complex Manifolds so looked it up on Wikipedia.

Now I just would like to check whether my impression is correct that if $(x^1, \dots, x^n)$ is a local coordinate system and so $\frac{\partial}{\partial x^i} , i = 1, \dots n$ forms a basis for the corresponing local region in the tangent bundle then a vector $v$ in the complexified tangent space $T^C_pM$ at $p \in M$ can be written as \begin{align} v \,\otimes_\mathbb{R} \mathbb{C} &= (\sum_j v^j \frac{\partial}{\partial x^j} \! |_p) \otimes _\mathbb{R} \mathbb{C} = (\sum_j v^j \frac{\partial}{\partial x^j} \! |_p) \otimes _\mathbb{R} (a + ib) \\ &= \sum_j (av^j \frac{\partial}{\partial x^j} \! |_p \otimes 1 + bv^j \frac{\partial}{\partial x^j} \! |_p \otimes i) \\ &= (\, \sum_j av^j \frac{\partial}{\partial x^j} \! |_p ) \otimes 1 + (\, \sum_j bv^j \frac{\partial}{\partial x^j} \! |_p) \otimes i \\ &= av \otimes 1 + bv \otimes i \end{align}

Is this description correct ? I am very new to tensor products so I hope I could use this as a way to check whether I actually understandt how extension of scalars work, thanks for any feedback!

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As far as formal manipulation of variables go, it looks okay to me. – Willie Wong Jan 16 '12 at 15:04
That's great! Hopefully this means I am finally getting a little fluency in dealing with tensor products. Tks again for the help! – harlekin Jan 16 '12 at 15:29

The result you get is correct, but the calculation indicates to me that you have not really understood how the tensor product works. Personally I would not accept this computation from an undergraduate student. For example, i could ask you, what exactly the term $v\otimes\mathbb{C}$ means, i.e. in which set lives it?
Depending on your concrete definition of the tensor product, you will sooner or later proof the following fundamental fact: If $V,W$ are vector spaces over $k$ with $k$-bases $\{v_i\}_{i\in I}$ and $\{w_j\}_{j\in J}$, then $\{v_i\otimes w_j\}_{(i,j)\in I\times J}$ is a $k$-basis of $V\otimes_k W$. In your concrete example you have $V=T_pM$ with the basis $$\left.\frac{\partial}{\partial x^1}\right|_p,\dots\left.\frac{\partial}{\partial x^n}\right|_p$$ and $W=\mathbb{C}$ with the basis (over $\mathbb{R}$) $1,i$. Then by the above considerations you know that a generic element $v\in T_p^CM=(T_pM)\otimes_\mathbb{R}\mathbb{C}$ is of the form $$v=\sum_i a_i\left.\frac{\partial}{\partial x^i}\right|_p\otimes 1\quad+\quad\sum_j b_j \left.\frac{\partial}{\partial x^j}\right|_p\otimes i$$ where $a_i,b_j\in\mathbb{R}$. Of course, this is a pain if you have to do concrete calculations, so one often lights the notation by writing $i\cdot v$ for $v\otimes i\in V\otimes\mathbb{C}$ so that a generic element in $V\otimes\mathbb{C}$ has the form $v+i\cdot w$, with $v,w\in V$. This is also the way one should think (in my opinion) about the complexification of a real vector space $V$: you have the elements $v$ of $V$ carrying complex coefficients, such that the product $\mathbb{C}\times V\ni (z,v)\mapsto z\cdot v\in V_\mathbb{C}$ is $\mathbb{R}$-bilinear. In this sense, for example a complex vector field $X\in\Gamma(T^CM)$ can b written locally as $$X=\sum_i X^i\frac{\partial}{\partial x^i}$$ with complex component functions $X^i\in C(U,\mathbb{C})$. Cheers, Robert