Linear Maps and Fixed Points in RPn

[mathshelp]

Regard the n-dimensional real projective space RPⁿ as the
space of lines in ^Rn+1 through {0}, i.e.

RPⁿ = (Rⁿ⁺¹ − {0}) /~ with x ~ y if y = λx for λ not equal to 0 ∈ R ;with the equivalence class of x denoted by [x].


(i) Work out the necessary and sufficient condition on a linear map
f : Rⁿ⁺¹ → R ^m+1 for the formula [f][x] = [f(x)] to define a map
[f]: RPⁿ → RP^m ; [x] → [f(x)] :

(ii) For a linear map f : Rⁿ⁺¹ → Rⁿ⁺¹ satisfying the condition of (i)
prove that the fixed point set
Fix([f]) = {[x] ∈ RPⁿ | [x] = [f(x)] ∈ RPⁿ}
consists of the equivalence classes of the lines in Rⁿ⁺¹ through {0} which contain eigenvectors of f.

(iii) Construct examples of linear maps f : R³ → R³ satisfying the condition of (i) such that

(a) Fix([f]) is a point.
(b) Fix([f]) is the disjoint union of a point and a circle.
(c) Fix([f]) is a projective plane.

 

[Mikemaths]

for part (ii)
if [x]=[f(x)] in RPn
then by the definition
λx=f(x) so that they are in the same equivalence class therefore since f is linear x is an eigenvector and λ is the eigenvalue

 

[mathshelp]

Yea that makes sense. Does anyone know about question iii? Thats the part I'm not sure about really

 

[mrbohn1]

Here are some suggestions off the top of my head (read: check the details as they might be wrong!):

(a) Any map which rotates R³ around the origin: then the only fixed point is the origin.

(b) Project R³ onto R² via the map f:(x,y,z)-->(x,y,0). Then compose this with the map g which fixes zero and takes every other point x-->x/|x|.

This will fix the unit circle: {(x,y): x²+y²=1} and the origin.

(c) Any map f:x-->ax where a is a scalar should fix RP².

 

[blue_raver22]

(i) The necessary and sufficient condition for the formula [f][x] = [f(x)] to define a map [f]: RPn → RPm ; [x] → [f(x)] is that f must be a well-defined map on the quotient space (Rn+1 - {0})/~. This means that for any [x] in RPn, the result [f(x)] must be independent of the representative of [x] chosen. In other words, if [x] = [y], then [f(x)] = [f(y)]. This is equivalent to saying that f(x) and f(y) must lie on the same line through {0} in Rm+1. Therefore, the necessary and sufficient condition is that f must map lines through {0} in Rn+1 to lines through {0} in Rm+1.

(ii) To prove that the fixed point set Fix([f]) consists of the equivalence classes of lines in Rn+1 through {0} which contain eigenvectors of f, we first note that any fixed point [x] in RPn must satisfy [x] = [f(x)]. This means that f(x) and x must lie on the same line through {0} in Rn+1. Since x is a point in Rn+1, this line must be the span of x, which is the line through {0} in Rn+1 containing x as an eigenvector of f. Therefore, Fix([f]) consists of the equivalence classes of lines in Rn+1 through {0} which contain eigenvectors of f.

(iii)

(a) One example of a linear map f: R3 → R3 satisfying the condition of (i) such that Fix([f]) is a point is the identity map. In this case, Fix([f]) = RP3, which is just a single point in R3.

(b) Another example is the map f(x,y,z) = (x+y, y+z, z), which maps lines through {0} in R3 to lines through {0} in R3. The fixed point set in this case is the disjoint union of the point [0,0,0] and the circle [x,y,z] = [x,0,z] for all x, z ∈ R.

(c) An example of a linear map f: R3 → R