This article contains proof of formulas in Riemannian geometry that involve the Christoffel symbols.
Start with the Bianchi identity
  
    
      
        
          R
          
            a
            b
            m
            n
            ;
            l
          
        
        +
        
          R
          
            a
            b
            l
            m
            ;
            n
          
        
        +
        
          R
          
            a
            b
            n
            l
            ;
            m
          
        
        =
        0
        
        
      
    
    
  .
Contract both sides of the above equation with a pair of metric tensors:
  
    
      
        
          g
          
            b
            n
          
        
        
          g
          
            a
            m
          
        
        (
        
          R
          
            a
            b
            m
            n
            ;
            l
          
        
        +
        
          R
          
            a
            b
            l
            m
            ;
            n
          
        
        +
        
          R
          
            a
            b
            n
            l
            ;
            m
          
        
        )
        =
        0
        ,
        
        
      
    
    
  
  
    
      
        
          g
          
            b
            n
          
        
        (
        
          R
          
            m
          
        
        
          
          
          
            b
            m
            n
            ;
            l
          
        
        −
        
          R
          
            m
          
        
        
          
          
          
            b
            m
            l
            ;
            n
          
        
        +
        
          R
          
            m
          
        
        
          
          
          
            b
            n
            l
            ;
            m
          
        
        )
        =
        0
        ,
        
        
      
    
    
  
  
    
      
        
          g
          
            b
            n
          
        
        (
        
          R
          
            b
            n
            ;
            l
          
        
        −
        
          R
          
            b
            l
            ;
            n
          
        
        −
        
          R
          
            b
          
        
        
          
          
          
            m
          
        
        
          
          
          
            n
            l
            ;
            m
          
        
        )
        =
        0
        ,
        
        
      
    
    
  
  
    
      
        
          R
          
            n
          
        
        
          
          
          
            n
            ;
            l
          
        
        −
        
          R
          
            n
          
        
        
          
          
          
            l
            ;
            n
          
        
        −
        
          R
          
            n
            m
          
        
        
          
          
          
            n
            l
            ;
            m
          
        
        =
        0.
        
        
      
    
    
  
The first term on the left contracts to yield a Ricci scalar, while the third term contracts to yield a mixed Ricci tensor,
  
    
      
        
          R
          
            ;
            l
          
        
        −
        
          R
          
            n
          
        
        
          
          
          
            l
            ;
            n
          
        
        −
        
          R
          
            m
          
        
        
          
          
          
            l
            ;
            m
          
        
        =
        0.
        
        
      
    
    
  
The last two terms are the same (changing dummy index n to m) and can be combined into a single term which shall be moved to the right,
  
    
      
        
          R
          
            ;
            l
          
        
        =
        2
        
          R
          
            m
          
        
        
          
          
          
            l
            ;
            m
          
        
        ,
        
        
      
    
    
  
which is the same as
  
    
      
        
          ∇
          
            m
          
        
        
          R
          
            m
          
        
        
          
          
          
            l
          
        
        =
        
          
            1
            2
          
        
        
          ∇
          
            l
          
        
        R
        
        
      
    
    
  .
Swapping the index labels l and m yields
  
    
      
        
          ∇
          
            l
          
        
        
          R
          
            l
          
        
        
          
          
          
            m
          
        
        =
        
          
            1
            2
          
        
        
          ∇
          
            m
          
        
        R
        
        
      
    
    
  ,      
Q.E.D.     (
return to article)
The last equation in Proof 1 above can be expressed as
  
    
      
        
          ∇
          
            l
          
        
        
          R
          
            l
          
        
        
          
          
          
            m
          
        
        −
        
          
            1
            2
          
        
        
          δ
          
            l
          
        
        
          
          
          
            m
          
        
        
          ∇
          
            l
          
        
        R
        =
        0
        
        
      
    
    
  
where δ is the Kronecker delta. Since the mixed Kronecker delta is equivalent to the mixed metric tensor,
  
    
      
        
          δ
          
            l
          
        
        
          
          
          
            m
          
        
        =
        
          g
          
            l
          
        
        
          
          
          
            m
          
        
        ,
        
        
      
    
    
  
and since the covariant derivative of the metric tensor is zero (so it can be moved in or out of the scope of any such derivative), then
  
    
      
        
          ∇
          
            l
          
        
        
          R
          
            l
          
        
        
          
          
          
            m
          
        
        −
        
          
            1
            2
          
        
        
          ∇
          
            l
          
        
        
          g
          
            l
          
        
        
          
          
          
            m
          
        
        R
        =
        0.
        
        
      
    
    
  
Factor out the covariant derivative
  
    
      
        
          ∇
          
            l
          
        
        
          (
          
            R
            
              l
            
          
          
            
            
            
              m
            
          
          −
          
            
              1
              2
            
          
          
            g
            
              l
            
          
          
            
            
            
              m
            
          
          R
          )
        
        =
        0
        ,
        
        
      
    
    
  
then raise the index m throughout
  
    
      
        
          ∇
          
            l
          
        
        
          (
          
            R
            
              l
              m
            
          
          −
          
            
              1
              2
            
          
          
            g
            
              l
              m
            
          
          R
          )
        
        =
        0.
        
        
      
    
    
  
The expression in parentheses is the Einstein tensor, so 
  
    
      
        
          ∇
          
            l
          
        
        
          G
          
            l
            m
          
        
        =
        0
        ,
        
        
      
    
    
       Q.E.D.    (
return to article)
this means that the covariant divergence of the Einstein tensor vanishes.
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