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What is a good geometric way of thinking of complex tangent vectors on a manifold? I can convince myself that I understand tangent vectors by thinking of them as paths on the manifold. Is there a nice way to visualize or think of complex vectors on a manifold?

I mean, I know the definition of a complexified tangent bundle, and if a manifold has an almost complex structure, and I know what it means for the complex vector to be holomorphic, antiholomorphic, and that there is an isomorphism between these eigenspaces and the real tangent bundle...

But I just don't "get" it. I feel like I have no intuitive understanding of what these things are.

If the manifold comes with an almost complex structure, is it correct to think of holomorphic tangent vectors as some kind of germs of J-holomorphic maps from an open set in $\mathbb{C}$ into the manifold?

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sorry to distract from your post; please see the last comment here: math.stackexchange.com/questions/54973/… –  gary Aug 8 '11 at 20:49
The germ of a $J$-holomorphic map from $\mathbb{C}$ to an almost complex manifold is determined by the image of $\mathbb{R}$: if $v$ is a tangent vector for $\mathbb{R}$, the tangent vectors for $\mathbb{C}$ are spanned by $v$ and $Jv$, and if $f:\mathbb C\to M$ is $J$-holomorphic, then $df(Jv)=J(df(v))$, and so everything is determined completely by the image of $df(v)$. Germs of such maps should be in correspondence with regular tangent vectors. You say that you have no intuitive feeling for what complex tangent vectors are. Do you have intuition about general vector bundles? –  Aaron Aug 8 '11 at 22:04
@Aaron - I have some intuitive notion for general vector bundles. But for general vector bundles, a priori there is no special relationship between the fibers and the base space, so complexification of an abstract real vector bundle doesn't feel like it's such a bizarre thing to do. –  Braindead Aug 9 '11 at 1:56
The thing that really drives me insane about this formalism is when you start out with a vector space that is already a vector space over complex numbers. –  Braindead Aug 9 '11 at 2:13

1 Answer 1

I'm not entirely sure what you are looking for, but complexifying a vector space just means that you take $\mathbb{C}$-linear combinations instead of $\mathbb{R}$-linear ones. In general, they are nothing more and nothing less.

For the tangent bundle, there is a decent interpretation, but it requires changing how you think about vectors (and in particular, shifting your attention to vector fields).

While there is a correspondence between vectors at a point and equivalence classes of paths through the point, we can think of a vector field $V$ as being an $\mathbb{R}$-linear functional $V:C^{\infty}(M)\to C^{\infty}(M)$ that satisfies $V(fg)(p)=V(f)g+fV(g)$. You should verify that every $V$ with this property actually is a vector field.

A section of of the complexified bundle are the same thing, except we are looking at $\mathbb C$ linear derivations on the smooth, complex valued functions. This can be seen in local coordinates, where normal vector fields will be $C^{\infty}(M)$-linear combinations of the coordinate vector fields $\partial_{x_i}$ while the complex vector fields will be $C^{\infty}_{\mathbb C}(M)$ linear combinations.

Unfortunately, this still doesn't give the geometric interpretation of individual vectors that I feel you are looking for, and you are left with the harder to understand "a complexified vector at $p$ is an element of the stalk (at $p$) of the sheaf of complex vector fields."

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