# Why is the intersection of complex submanifolds always positive.?

Hi, everyone: I was finally able to show that all complex manifolds are orientable, by generalizing to many variables the fact that , for a single complex variable, the Jacobian matrix is of the form (after using Cauchy-Riemann to substitute). ( This is my first post here. I read the FAQ's, but I apologize if I am not following protocol correctly.Please let me know if so.)

(a b)
(-b a)


which has non-zero determinant a2+b2 . We can induct, to show something similar holds for higher dimensions, i.e., the Jacobian ( of overlapping charts will necessarily be positive. Now, a couple of questions, please:

1)Is there a more topological proof of the orientability.?.I thought of using Lie theory, that Gl(n;C) is connected, may work, but I don't see how to rigorize this argument; in Milnor and Stasheff's book , it is stated that(paraphrase) this path connectedness allows one basis to be deformed into another homotopically, so that orientation is preserved. But, AFAIK, the bases are just elements in Cn. Any ideas on this direction.?. I know that in Cn, connectedness implies path-connectedness. I also know that Gl(n;C) can be embedded in Gl+(2n;IR), one of the connected components of Gl(2n;IR). I think this helps, but I don't know how to "rigorize" this idea. Any suggestions, please.?

I also wonder if one can generalize the CW-decomposition of CPn, where we can see that there is only one cell in the top dimension, to other complex manifolds.

2)Any suggestions, please, for showing that complex submanifolds S1,S2( of the "right dimensions, to make sure they can intersect) of a manifold M ,have positive intersection (self- or otherwise).? . I understand that, at every point p of intersection, we append the tangent spaces, defining:

Tp(S1/\S2)=Tp(S1)(+)Tp(S2)

and then the intersection is positive if there is an even permutation from the basis of TpM to the basis of Tp(S1/\S2). But I have no idea of how to show that, for complex submanifolds, the intersection is always positive.

Thanks For Any Help/Suggestions.

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Hint: For two complex operators A,B consider a complex affine line L passing through A,B. L consists of operators zA+(1-z)B, z is a complex number. Show, that if A and B are non-degenerate then only finite number of operators from L are degenerate. Deduce from it that Gl(n,C) is connected. – Petya May 6 '10 at 5:12
well, take a look at Griffiths and Harris, pag 62...hope that helps. – Csar Lozano Huerta May 6 '10 at 6:32
$GL_n(\mathbb{C})$ is path-connected because Gaussian elimination can be accomplished through a sequence of continuous shears and rotations. – S. Carnahan May 6 '10 at 6:51
Alternatively, $GL_n(\mathbb{C})$ is connected because it is $\mathbb{C}^{n^2} \setminus \Delta$, where $\Delta$ has real codimension $2$. – David Speyer May 6 '10 at 12:08
Jordan normal form is yet another reason that $GL_n(C)$ is connected. – Keivan Karai May 6 '10 at 17:54

You can answer both your questions with the following remark: if $x_1,\ldots,x_n$ is a complex basis of $\mathbb{C}^n$, then $x_1,ix_1,x_2,ix_2,\ldots,x_n,ix_n$ is a real basis of $\mathbb{C}^n$ whose orientation does not depend upon $x_1,\ldots,x_n$ (in particular it does not depend upon their order).
Now you can give an orientation to a complex manifold $M$ by locally choosing a complex basis and use the subsequent local orientation (which will be globally defined thanks to the remark), and if you endow two complex submanifolds $N_1,N_2$ of complementary dimensions with the orientations defined in the same way, then automatically at each (transversal) intersection point $x$ the orientation of $T_x N_1\oplus T_xN_2$ obtained by concatenation is the same than that of $T_x M$. This exactly means that the intersection is positive.
Your main problem for (2) is your definition of sign of intersection- the sign of the determinant of the unique matrix mapping your direct sum basis to your overarching basis includes your definition as a special case (since $det(v_\sigma(1) v_\sigma(2)...v_\sigma(n))= sign(\sigma)det(v_1v_2...v_n)$-by definition in some books! and $det(I)=1$). Just observe that your unique matrix is in $Gl(n,\mathbb{C}) \subset Gl(2n,\mathbb{R})$ and look at the real determinant.
$\mathrm{GL}(2n,\mathbb R)$, you mean. – Mariano Suárez-Alvarez May 6 '10 at 17:31