Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have been trying to solve this exercise from the book Fundamental Group and Covering Spaces written by Elon Lages Lima, chapter 3. It says that given $f,g : B^{2} \to S^{2}$ continuous such that for $(x,y) \in S^{1}$, $f(x,y) = (x,y,0)$ and $g(x,y) = (-y,x,0)$, there exists $(x,y) \in B^{2}$ with $f(x,y) = g(x,y)$ or $f(x,y) = -g(x,y)$.

I'm not really sure how to proceed. The definition of $f,g$ makes me think that I should try inner product or projective space. Or maybe i should simply proceed by contradiction and try to construct a function in contradiction with some application like Borsuk-Ulam theorem or that doesn't exists a not null tangent vectorial field on S^{2}. Neither of those have worked to me.

Any ideas, suggestions or answers? :)

share|improve this question

1 Answer 1

If neither $f(x,y)=g(x,y)$ nor $f(x,y)=-g(x,y)$ then there is a well defined nonzero tangent vector to the sphere pointing along the great circle containing $f(x,y)$ and $g(x,y)$ in the direction from $f(x,y)$ toward $g(x,y)$. This uses that $f(x,y) \neq g(x,y)$ to guarantee nonzero length, and $f(x,y) \neq -g(x,y)$ to guarantee there is only one such direction.

So if you can fill in from your assumptions about $f,g$ on the equator of the sphere to this vector field, you can finish with Borsuk-Ulam as you suggest. Thus the idea you had might be made to work.

EDIT: I think the way is to define $F(x,y,z)=f(x,y)$ and $G(x,y,z)=g(x,y),$ so that now $F,G$ are maps from the sphere $S^2$ to itself, where the sphere is considered as points in three space where $x^2+y^2+z^2=1$. In other words to get the image of a point on the sphere under $F$ or $G$ one projects onto the $xy$ plane and reads off the value of $f$ or $g$ respectively. This done we now have the functions $F,G$ defined on $S_2$ and I think the above idea produces a nonvanishing vector field.

share|improve this answer
    
It's seems like a great idea!! But I don't see where are we using that the functions are already defined on the equator. I think is for the continuity of the vector field but I'm not really sure. –  Frank Feb 2 '13 at 18:33
    
Now I have a doubt, when you find the direction from $f(x,y)$ towards $g(x,y)$ that tangent vector makes sense when you think of it like a vector starting at $f(x,y)$ but, how do we move it to the position $(x,y,z)$ ? –  Frank Feb 2 '13 at 18:38
    
For your first comment: The use on the equator is that, given the definitions there, the points $f(x,y)=(x,y,0)$ and $g(x,y)=(-y,x,0)$ are neither equal nor antipodal. (One is rotated 90 degrees from the other.) I'll think more about second comment... –  coffeemath Feb 3 '13 at 4:57
    
Yes but we are not really using that definition, we are only using that they are neither equal or antipodal, so the condition can be weakened, I think that condition must play an important role in the proof :S –  Frank Feb 3 '13 at 14:28
    
@Frank : The more I think about it, my idea might not be right. The two given maps $f,g$ are defined on the 2-disk, and when "lifted" to the sphere as I suggested they might not work as intended. Because if I'm at say $(x,y,z)$ with $z>0$ all I have to work with is two points $f(x,y),g(x,y)$ somewhere on the sphere, not really related to the specific point $(x,y,z)$ where I'm located, but could be anywhere else. Using these I now don't see how to define a direction at the given point $(x,y,z)$. –  coffeemath Feb 4 '13 at 0:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.