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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.

1 Answers 1

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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.

  • 0
    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.2012-02-26
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    But compactness + contractibility doesn't imply the fixed point property. Why here it implies?2012-02-26
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    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.2012-02-26
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    There is a famous example due to Kinoshita of a compact, contractible space without the topological fixed point property: http://matwbn.icm.edu.pl/ksiazki/fm/fm40/fm4019.pdf2012-02-26
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    @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! :) )2012-02-26
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    @GregMartin, if you prefer, the composition of the retraction, a rotation of the circle and the inclusion of the circle in the big space :) (but not *conjugation*, as an interesting retraction rarely has an inverse)2012-02-26
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    @Nimza, see my comment above to Michael.2012-02-26
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    Thank you, @MarianoSuárez-Alvarez. And how to show that "If X is such a space which is contractible and f:X→X is continuous, then the Lefschetz number of f is 1"?2012-02-27
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    @Nimza: Since all the higher homology groups are zero, the Lefschetz number is simply the trace of the map induced on $H_0$. You can easily compute that!2012-02-27