Yahoo Answers: Answers and Comments for Line in CP_2 [數學]
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Tue, 21 Aug 2007 14:39:34 +0000
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Yahoo Answers: Answers and Comments for Line in CP_2 [數學]
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https://answers.yahoo.com/question/index?qid=20070821000051KK03545
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From Ivan: First let us see see what is CP_2
as complex m...
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Fri, 24 Aug 2007 21:04:13 +0000
First let us see see what is CP_2
as complex manifold, it is 2 dim
as real manifold, it is 4 dim
1) Project line in CP_2 always mean complex line  which is a (2 dim over C) plane define on C^3 by the homogenous polynomial ax+by+cz=0, and project to CP_2. (2dim become 1 dim over C). We need it to be homogenous, because we want (cx,cy,cz) satisfies this equation for all complex number c, so that we can well definedly identify all the point.
2) Hyperplane of some ndim space always mean the space that is (n1)dim. So, in this case, CP_2 is 2dim, hyperplane is 1dim  the same as 1)
3) you say it is a real line. So in this case we need to consider CP_2 as a real manifold.
before we have c*(z1,z2,z3) = (cz1, cz2, cz3). in terms of real number, it becomes
(c,d)*(u,v,w,x,y,z) = (cudv, cv+du, cwdx, cx+dw, cydz, cz+dy)
then, it is R^6 quotient out the relation: (c,d)*(u,v,w,x,y,z)~(u,v,w,x,y,z)
Now there is no "straight" line in CP_2, because this is not a subspace, it is a curved manifold. There are 2 other ideas.
First, a "straight" line may be a geodesic line, i.e. the shortest distance between two points.
another idea is, take a 3dim(over R) space on R^6. This looks like C x R. Under identification, if the direction of R is chosen in a right way (so that you have no extra idenfication) then you will get a nice 1dim manifold in CP_2, this will be your real line.
However, I am not too sure about this, may be you can try a plane like (u,v,w,0,0,0), and see if under the above identifaction, it becomes something that can be parametrized by a real number t.
If your parameter is a complex number, then it is a complex line, so the same as 1)
4) (complex) plane in CP_2 is just CP_2 itself: a 2dim thing in 2dim thing ma~
if it is (real) plane, then it is dim2 over R. The problem is similar to 3). If you extend back to C^3 (or R^6), it becomes something 4 dimensional over R. Even if it is a very nice R^4 in C^3 = R^6, however, there are "less direction" in C^3 than R^6, so that even though it is a R^4, it may not be viewed as C^2. If luckily it is also a complex subspace, then it will be the same as 1).
So, to conclude the real case, I will say:
A) In general, there are more real plane than complex line
B) a real straight line (resp. plane) = {S subset of CP_2 : p^1(S) = R^3 (resp. R^4) in C^6}
where p is the projection C^3 > CP_2.
Anyway, talking about CP_2, people just study the complex line, i.e. the object in 1)
20070828 13:44:00 補充：
Let me give a smaller example. Consider C^2 ~ R^4write the coordinate as (z1,z2) = (x,y,z,u)then consider the plane of the form (0,y,z,0) in R^4
20070828 13:44:11 補充：
Now this is a real plane: real combination of them is still in this plane.But complex combination is not, for example:(a ib) (0,1,0,0) = (b, 0, 0, 0) is not in this plane.So this is a real plane, but not a complex line.