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$\newcommand{\CA}{{\mathcal{A}}} \newcommand{\CG}{{\mathcal{G}}} \newcommand{\BR}{{\mathbb{R}}} \newcommand{\Fm}{{\mathfrak{m}}} \newcommand{\smint}{{(-\varepsilon,\varepsilon)}} \newcommand{\seq}{{\subseteq}}$

I know many equivalent definitions of the tangent space of a smooth manifold at a point, and I would like to understand whether they all apply to complex manifolds as well. Of course I could try to just go through the proofs and see whether they can be translated into the holomorphic setting, but first I would like to understand if I get the general picture right. I am going to explain how I understand the picture so far and I would appreciate any comments about it.

Let $M$ be a smooth manifold and $x$ a point of $M$. Let $\CA_x$ be the local $\BR-$algebra of germs of smooth functions at $x$ and $\Fm_x$ its maximal ideal, consisting of germs vanishing at $x$. Similarly, let $\CG_x$ be the set of germs of smooth paths based at $x$ (i.e., $\gamma(0)=x$). We write $\gamma$ both for a smooth path and for its germ, and the same for functions. There are four definitions of the tangent space at $x$ and they all give isomorphic vector spaces (the canonical isomorphisms are indicated below):

  1. $\{ (U,\phi, \xi) \mid (U, \phi) \text{ chart at }x, \xi \in \BR^n\} \ / \ (U, \phi, \xi) \sim (V, \psi, D_{\phi(x)}(\psi \circ \phi^{-1})(\xi))$

  2. $\left\{ \gamma \in \CG_x \right\} / \left\{ (\phi \circ \gamma)'(0)=0 \text{ for some chart } (U, \phi) \right\}$

  3. $\left\{ D \in {\CA_x}^\vee \mid D(fg) = D(f)g(x) + f(x)D(g)\right\}$

  4. $\left(\Fm_x \ / \ {\Fm_x}^2\right)^\vee$

Here are the isomorphisms of vector spaces between the different descriptions:

1.$\to$2.: $[\gamma] \mapsto [U, \phi, (\phi \circ \gamma)'(0)]$

1.$\to$3.: $v \mapsto D_v: D_v(f):= \frac{d}{dt}((f\circ \phi^{-1})(\phi(x)+tv))\vert_{t=0}$

2.$\to$3.: $[\gamma] \mapsto D_{[\gamma]}: D_{[\gamma]}(f):= (f \circ \gamma)'(0)$

3.$\to$4.: $D \mapsto D \vert_{\Fm_x}$

4.$\to$3.: $R \mapsto D: D(f):=R([f-f(x)])$

Remark: We could also consider smooth functions instead of germs. The argument still works because of the existence of bump functions. But if we want the definitions to apply to complex manifolds as well, then we need germs.

First question: is all of the above correct?

Second question: if we replace "smooth" by "holomorphic" throughout, does the same work for complex manifolds and complex tangent spaces?

Third question: what is the relation between the real and complex tangent spaces of a complex manifold at a point?

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57Jimmy
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  • For the third question, the complex tangent space is the complexification of the real tangent space, i.e. $T_{p,\mathbb C}M=T_{p,\mathbb R}M\otimes_{\mathbb R}\mathbb C$. A real tangent vector can be thought of as a real derivation of real smooth functions, similar a complex tangent vector can be seen as a complex derivation of complex-valued smooth functions. – lEm Nov 20 '19 at 07:16
  • @IEm I think that there could be two meaningful definitions of "complex tangent space". One is yours, and the other one consists of complex derivations of holomorphic functions. As mentioned here, the latter is a direct summand of the former. – 57Jimmy Nov 20 '19 at 10:25
  • Yes, indeed, $T^{1,0}_p$ is known as the holomorphic tangent space, the bundle associated to which is the holomorphic tangent bundle, which is isomorphic to the real tangent bundle as real vector bundles. – lEm Nov 20 '19 at 10:32
  • I am not really familiar with different definitions of the complex/holomorphic tangent spaces. But for practical purposes, the complex/holomorphic tangent spaces have a neat description in terms of local coordinates. – lEm Nov 20 '19 at 10:34
  • @IEm When you say "which is isomorphic...", did I get it right: are you saying that the complex derivations of holomorphic functions on $M$ are the same as real derivations of smooth real-valued functions on $M$ (as real vector spaces)? That sounds strange – 57Jimmy Nov 20 '19 at 15:09
  • No, not in this sense. I was talking about the vector bundle structure. Roughly speaking, they are the same because if you try to change the coordinates, the "extra factors" for each bundle correspond to the Jacobian matrix of the transition functions. Anyway it was just a remark, which isn't really relevant to your question here. I am not really sure if one can canonically construct an isomorphism between two vector spaces of deviations though. By the way, I believe the definition (4) extends to holomorphic setting, by taking the maximal ideal of $\mathbcal O_{M, x}$ . – lEm Nov 20 '19 at 15:31
  • Oh, sorry, I indeed misread vector bundle as vector space! Thanks for the explanation – 57Jimmy Nov 20 '19 at 16:27

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