- 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.

- One might try to construct an isomorphism from the trivial bundle
`T`

to the canonical bundle`L`

by 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!

- 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)`

.