# Orientability of Complex Manifolds.

Hi, everyone: I am trying to show that any complex manifold is orientable.

I know this has to see with properties of Gl(n;C) (C complexes, of course.) ;

specifically, with Gl(n;C) being connected (as a Lie Group.). Now this means

that the determinant map must be either always pos. or always negative, but

I am not clear on why it is not always negative.

Also, I am confused about the fact that the determinant may be complex-valued,

so that it does not make sense to say it is positive or negative.

Any Ideas.?

Thanks.

Office_Shredder
Staff Emeritus
Gold Member
With regard to GL(n,C):

The determinant still can be anything except for 0. Except in this case 'anything' means any complex number instead of any real number like you're probably more familiar with. So with real numbers, if you remove zero you get two disconnected sets. If you just remove 0 from the complex plane you're left with a set that is still connected

Thanks. So how can I then show that complex manifolds are orientable.?.

I had been told that properties of Gl(n;C) gave the answer, but I cannot see

why/how.

Manifold is orientable if and only if the chart transitions can be chosen to be orientation-preserving (i.e. Jacobian positive). Holomorphic maps certainly do that.

I can see that for one complex variable, where we can may be use conformality,
but it does not seem so clear for many complex variables.

I also wonder if we're given a complex n-manifold N, if the orientability of N is
equivalent to the orientability of the equivalent real 2n-manifold that we get by
"decomplexifying" N.

Think of the standard embedding of GL(n,C) into GL(2n,R).

lavinia
Gold Member
I think this works.

- the tangent bundle of a complex manifold has a complex structure. This is because the coordinate charts lie in GL(n:C)

any complex vector space has a canonical orientation. Choose any basis x1, x2, ..., xn and extend it to a real basis, x1 ix1 x2 ix2, ...xn ixn. This ordering defines an orientation of the underlying real vector space. The choice is independent of the choice of basis x1 ... xn because GL(n;C) is path connected.