§ Canonical bundle over RP2 is not trivial

• Recall that RP2 can be defined as the set of all lines in R3.
• The trivial bundle is the bundle RP2xR -> RP2. We shall call it T.
• The canonical bundle L is defined as follows: It is a subspace of RP2xR3, which has elements { (l, v) : l ∈ v }. That is, it is all lines, and elements on these lines.

§ Intuition: Canonical bundle is not trivial

• One might try to construct an isomorphism from the trivial bundle T to the canonical bundle Lby sending (l ∈ RP2, t ∈ R) to (l, tv_l) where v_l is the unit vector along l, or the intersection of l with the upper hemisphere of S2.
• Think of RP2 as the upper hemisphere. See that on the equator, we only need half of the equator, since the "other half" is already given by the first half of the equator.
• Unfortunately, this does not work for the following subtle reason.
• Let us use the above """isomorphism""" to consider what happens to the trivial section RP2 x {1} (ie, the scalar field that is 1 everywhere on RP2.
• We will assign to each line, the unit vector along that line.
• On the equator, for the "front half", we will get unit vectors emenating from the origin into the boundary.
• For the "back half", we will get unit vectors pointing towards the origin of the sphere .
• But for all points "just above" the sphere, we will have unit vectors emenating out from the sphere.
• So, it will be discontinuous at the "back half" of the equator!

§ Formal proof: Canonical bundle is not trivial

• Consider a section of s of L, the canonical line bundle.
• s : (l : RP2) → l is a function that sends a line l ∈ RP2 to a point on the line v ∈ l.
• This can be seen as a map that sends, for points in the upper hemisphere, s(l ∈ RP2) = ((λ(l) x) ∈ l) for some function λ: RP2 → R, where x is the intersection of l with the upper hemisphere. λ must be continuous in l.
• We will create a new function s' : S2 → R3, such that the map s'(x ∈ S2) = s(l) for all points x, where l is the line that contains x. So s'(x) is going to be a point on l, where l is the line that contains x.
• We will write s'(x ∈ S2) = λ'(x) x for some function λ': S2 → R.
• We now calculuate what λ' is like.
• We must have that s'(x) = s'(-x), since both x and -x lie on the same line l, and thus s'(x) = s(l) = s'(-x).
• Let us now calculate this in terms of λ'.
• We see that s'(x) = λ'(x)x, and s'(-x) = λ'(-x)(-x) = -λ'(x) x.
• But since s'(x) = s'(-x), we must have that λ'(x) x = - λ'(x) x, or λ'(x) = - λ'(-x).
• Intuitively, the line l is determined only by the positive vector, but the function λ is given as input the vector on the sphere. Thus, on input -x, it must rescale the vector negatively , to push the vector in the positive direction! (as s needs us to do).
• So this means that we have a continuous odd function λ' : S2 → R, such that λ'(x) = - λ'(-x).
• By the intermediate value theorem, this function must vanish.
• Said differently, take a curve c : [0, 1] → S2 on the sphere connecting x to -x. Then λ' . c : [0, 1] → R gives us a continous function which is positive at 0 and negative at 1, and thus must have crossed 0 at some point by the intermediate value theorem.
• This means that the vector field s must have vanished at some point, since there is a point where λ' equals 0, which means there is a point where s' vanishes (on S2), which means there is a line where s vanishes (on RP2).
• Thus, there cannot exist a canonical line bundle on RP2, which means that the canonical line bundle is not trivial.

• Now see that we need
• Now RP2 is this vector field restricted to the top hemisphere.
• If λ is zero somewhere, then we are done.
• We will show that λ must be zero somewhere.
• consider the scalar field on S^2 that sends x to λ(x).