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.

Consider the English alphabet in this font with serifs

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

Consider any letter in this font as a topological space (assume that letters don't have weight and are genersted by lines) and consider a continuous mapping from any letter to itself. For which letters any such mapping have a fixed point and for which it has not? For O the answer is negative, for C, Z, S it is positive.

The main difficulty for me is serifs. For the sans-serif font like Arial the problem is much easier. Here we can't even easily divide letters into topologically equal groups and these groups aren't obvious: G ~ J, T ~ I ~ U ~ W, E ~ F, C ~ Z ~ S etc.

share|improve this question
add comment

1 Answer 1

up vote 7 down vote accepted

Each of the letters deformation retracts to either a point, a circle, or a figure 8.

  • If a letter deformation retracts to a point, Lefschetz fixed point theorem tells us that every map to itself has a fixed point.

  • If a letter retracts to a circle, then the composition of the retraction with a rotation gives us a map which does not have a fixed point.

  • If a letter retracts to a figure 8, the composition of the retraction with the map that collapses one of the loops to a point and rotates the other some angle does not have any fixed points.

In particular, the serifs play no role here.

share|improve this answer
    
If the letters do not have a weight, the figure 8 will be rather an 8 of a digital clock —this does not change anything, though. –  Mariano Suárez-Alvarez Feb 26 '12 at 10:13
1  
But compactness + contractibility doesn't imply the fixed point property. Why here it implies? –  Nimza Feb 26 '12 at 11:01
    
I might say "the conjugation of a rotation of the circle by the retraction" instead of "the composition of the retraction with a rotation", since we're looking for a map from the letter-space to itself rather than from the letter-space to the circle. –  Greg Martin Feb 26 '12 at 11:01
    
There is a famous example due to Kinoshita of a compact, contractible space without the topological fixed point property: matwbn.icm.edu.pl/ksiazki/fm/fm40/fm4019.pdf –  Michael Greinecker Feb 26 '12 at 11:51
    
@MichaelGreinecker, I am not familiar with the example (I'll look later) but the Lefschetz fixed point theorem applies to compact triangulable spaces. If $X$ is such a space which is contractible and $f:X\to X$ is continuous, then the Lefschetz number of $f$ is $1$. (And the letters are obviously triangulable! :) ) –  Mariano Suárez-Alvarez Feb 26 '12 at 20:33
show 4 more comments

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.