Fixed Points and Degree Theory 


    
    
      
        
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      The degree of a constant map is 0.
    
    
  
        
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      \(f\simeq g \iff \deg f = \deg g\), since this implies \(f_* = g_*\).
    
    
  
        
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      If \(f\) is a homotopy equivalence, \({\left\lvert {\deg f} \right\rvert} = 1\).
    
    
  
    
      
      
        
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          This is because \(f\simeq g \implies H_*(f) = H_*(g)\).
        
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      \(\deg\operatorname{id}_{S^n} = 1\)
    
    
  
        
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      \(\text{deg} (f\circ g) = \deg f \cdot \deg g\)
    
    
  
        
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      \(\deg(H_{x_i}) = -1\) for \(H_{x_i}\) any rotation about the hyperplane \(x_i = 0\), i.e.  

\begin{align*}
H_{x_i}: {\mathbf{R}}^{n+1} &\to {\mathbf{R}}^{n+1} \\
{\left[ {x_1, \cdots, x_i, \cdots, x_{n+1}} \right]} 
&\mapsto
{\left[ {x_1, \cdots, - x_i, \cdots, x_{n+1}} \right]} 
.\end{align*}

    
    
  
        
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      The antipodal map on \(S^n\subset {\mathbf{R}}^{n+1}\) is the composition of \(n+1\) hyperplane reflections, so \(\deg\alpha = (-1)^{n+1}\).
    
    
  
    
      
      
        
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          As a consequence, if \(\deg f\) is even then \(f\) is not homotopic to the antipodal map.
        
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      Show that if \(f: S^n\to S^n\) has no fixed points \(\iff \deg f = (-1)^{n+1}\) and \(f\) is homotopic to the antipodal map.
    

  

    
    
      
        
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      If \(f(x)\neq x\), the line segment \(L(-x, f(x))\) does not contain \(0\).
    
    
  
        
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      If \(f(x) \neq -x\), the line segment \(L(x, f(x))\) does not contain \(0\).
    
    
  
        
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      The straight line homotopy \(H(t, x) = (1-t)f(x) + t(-x)\) is a homotopy between \(f\) and the antipodal map in \({\mathbf{R}}^{n+1}\). Use that \(H(t, x) = 0 \iff t=1/2 \iff f(x) = x\), so \(H(t, {-})\) is always a line from \(x\) to \(f(x)\) not passing through \(0\), so \(H(t, {-}) \neq 0\). So this descends to a homotopy of \({\mathbf{R}}^{n+1}\setminus\left\{{0}\right\}\simeq S^n\), or explicitly one can project 

\begin{align*}
H: I\times S^n &\to S^n \\
(t, x) &\mapsto {H(t, x) \over {\left\lVert {H(t, x)} \right\rVert}}
.\end{align*}

    

  
    

      Less explicitly: if \(f(x) \neq x\) for all \(x\in S^n\), there is a unique geodesic through these two points, so let each point flow along its corresponding geodesic.
    

  

    

      Show that if \(f\) is not surjective then \(\deg f = 0\).
    

  

    

      In this case \(f\) factors as \(S^n \xrightarrow{f_1} S^n\setminus\left\{{ {\operatorname{pt}} }\right\}\xrightarrow{f_2} S_n\). But \(S^n\setminus\left\{{ {\operatorname{pt}} }\right\} \simeq{\mathbf{R}}^n\), to \(H_*(f_1) = 0\) and \(\deg f_1 = 0\). Then apply \(\deg f = \qty{\deg f_1} \qty{\deg f_2}\).
    

  

    

      For \(f:X\to X\), define the trace of \(f\) to be 

\begin{align*}
\Lambda_f \coloneqq\sum_{k \geq 0} (-1)^k ~\mathrm{Tr}(f_* \mathrel{\Big|}H_k(X; {\mathbf{Q}}))
\end{align*}
 where \(f_*: H_k(X; {\mathbf{Q}}) \to H_k(X; {\mathbf{Q}})\) is the induced map on homology. If \(\Lambda_f \neq 0\) then \(f\) has a fixed point.
    

  

    

      Every \(f: B^n \to B^n\) has a fixed point.
    

  
\todo[inline]{Proof}


    

      There is no non-vanishing tangent vector field on even dimensional spheres \(S^{2n}\).
    

  

    

      For every \(S^n \xrightarrow{f} {\mathbf{R}}^n \exists x\in S^n\) such that \(f(x) = f(-x)\).