Difference between revisions of "Curvature is tensorial"
(→Proof) 

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 2  <math>f\nabla_X\nabla_Y  (Yf)\nabla_X  f \nabla_Y\nabla_X  \nabla_{[fX,Y]}</math>  Fact (3), the Leibnizlike axiom for connection.  <math>\nabla_Y(f \nabla_X) \to (Yf)\nabla_X + f\nabla_Y\nabla_X</math>. To understand this more clearly imagine an input <math>Z</math> to the whole expression, so that the rewrite becomes <math>\nabla_Y(f \nabla_X(Z)) \to (Yf)\nabla_X(Z) + f\nabla_Y\nabla_X(Z)</math>. In the notation of fact (3), <math>A = Y</math>, <math>f = f</math>, and <math>B = \nabla_X(Z)</math>.   2  <math>f\nabla_X\nabla_Y  (Yf)\nabla_X  f \nabla_Y\nabla_X  \nabla_{[fX,Y]}</math>  Fact (3), the Leibnizlike axiom for connection.  <math>\nabla_Y(f \nabla_X) \to (Yf)\nabla_X + f\nabla_Y\nabla_X</math>. To understand this more clearly imagine an input <math>Z</math> to the whole expression, so that the rewrite becomes <math>\nabla_Y(f \nabla_X(Z)) \to (Yf)\nabla_X(Z) + f\nabla_Y\nabla_X(Z)</math>. In the notation of fact (3), <math>A = Y</math>, <math>f = f</math>, and <math>B = \nabla_X(Z)</math>.  
    
−   3  <math>f(\nabla_X\nabla_Y  \nabla_Y\nabla_X)  \nabla_{(Yf)X  +   3  <math>f(\nabla_X\nabla_Y  \nabla_Y\nabla_X)  \nabla_{(Yf)X}  \nabla_{[fX,Y]}</math>  <math>\nabla</math> is <math>C^\infty</math>linear in its subscript argument  <math>(Yf)\nabla_X \to \nabla_{(Yf)X}</math> 
    
−   4  <math>f(\nabla_X\nabla_Y  \nabla_Y\nabla_X  \nabla_{[X,Y]})</math>  Fact (2)  <math>[fX,Y] \to f[X,Y]  +   4  <math>f(\nabla_X\nabla_Y  \nabla_Y\nabla_X)  \nabla_{(Yf)X + [fX,Y]}</math>  <math>\nabla</math> is additive in its subscript argument  <math>\nabla_{(Yf)}X} + \nabla_{[fX,Y]} = \nabla_{(Yf)X + [fX,Y]}</math> 
+    
+   5  <math>f(\nabla_X\nabla_Y  \nabla_Y\nabla_X  \nabla_{[X,Y]})</math>  Fact (2)  <math>[fX,Y] + (Yf)X \to f[X,Y]</math>.  
}  }  
===Tensoriality in the second variable===  ===Tensoriality in the second variable===  
−  +  '''Given''': <math>f:M \to \R</math> is a <math>C^\infty</math>function.  
−  <math>\! R(X,fY) = f R(X,Y)</math>  +  '''To prove''': <math>\! R(X,fY) = f R(X,Y)</math>, or more explicitly, <math>\nabla_X\nabla_{fY}  \nabla_{fY}\nabla_X  \nabla_{[X,fY]} = f (\nabla_X\nabla_Y  \nabla_Y\nabla_X  \nabla_{[X,Y]}</math>. 
We start out with the left side:  We start out with the left side:  
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<math>\nabla_X\nabla_{fY}  \nabla_{fY}\nabla_X  \nabla_{[X,fY]}</math>  <math>\nabla_X\nabla_{fY}  \nabla_{fY}\nabla_X  \nabla_{[X,fY]}</math>  
−  +  Each step below is obtained from the previous one via some manipulation explained along side.  
−  <math>(Xf)\nabla_Y + f(\nabla_X\nabla_Y  \nabla_Y\nabla_X  +  { class="sortable" border="1" 
−  +  ! Step no. !! Current status of left side !! Facts/properties used !! Specific rewrites  
−  +    
−  +   1  <math>\nabla_X(f\nabla_Y)  f\nabla_Y\nabla_X  \nabla_{[X,fY]}</math>  By definition of a connection, <math>\nabla</math> is <math>C^\infty</math>linear in its subscript argument  <math>\nabla_{fY} \to f\nabla_Y</math>.  
−  <math>f(\nabla_X\  +   
−  +   2  <math>(Xf)\nabla_Y + f(\nabla_X\nabla_Y)  f\nabla_Y\nabla_X  \nabla_{[X,fY]}</math>  Fact (3), the Leibnizlike axiom for connection  <math>\nabla_X(f\nabla_Y) \to (Xf)\nabla_Y + f(\nabla_X\nabla_Y)</math>. To make this more concrete, imagine an input <math>Z</math>. Then, the rewrite becomes <math>\nabla_X(f\nabla_Y(Z)) \to (Xf)\nabla_Y(X) + f(\nabla_X\nabla_Y(Z))</math>. This comes setting <math>A = X</math>, <math>f = f</math>, <math>B = \nabla_YZ</math> in Fact (3).  
−  +    
−  +   3  <math>f(\nabla_X\nabla_Y  \nabla_Y\nabla_X)  \nabla_{[X,fY]} + \nabla_{(Xf)Y}</math>  <math>\nabla</math> is <math>C^\infty</math>linear in its subscript argument.  <math>(Xf)\nabla_Y \to \nabla_{(Xf)Y}</math>  
−  <math>\  +   
−  +   4  <math>f(\nabla_X\nabla_Y  \nabla_Y\nabla_X)  \nabla_{[X,fY]  (Xf)Y}</math>  <math>\nabla</math> is additive in its subscript argument.  <math>\nabla_{[X,fY]}  \nabla_{(Xf)Y} \to \nabla_{[X,fY]  (Xf)Y}</math>.  
−  +    
−  +   5  <math>f(\nabla_X\nabla_Y  \nabla_Y\nabla_X  \nabla_{[X,Y]}</math>  Fact (2)  <math>[X,fY]  (Xf)Y \to f[X,Y]</math>  
−  <math>f (\nabla_X\nabla_Y  \nabla_Y\nabla_X  \nabla_{[X,Y]}</math>  +  } 
−  
−  
===Tensoriality in the third variable===  ===Tensoriality in the third variable=== 
Revision as of 21:00, 25 February 2011
This article gives the statement, and possibly proof, that a map constructed in a certain way is tensorial
View other such statements
Contents
Statement
Let be a connection on a vector bundle over a differential manifold . The Riemann curvature tensor of is given as a map defined by:
We claim that is a tensorial map in each of the variables .
Related facts
 Curvature is antisymmetric in first two variables
 Curvature is antisymmetric in last two variables
 Curvature is symmetric in the pairs of first and last two variables
Facts used
Fact no.  Name  Full statement 

1  Leibniz rule for derivations  For a vector field and functions , we have: 
2  Corollary of Leibniz rule for Lie bracket  For a function and vector fields :

3  The Leibnizlike axiom that is part of the definition of a connection  For a function and vector fields , and a connection , we have 
Proof
To prove tensoriality in a variable, it suffices to show linearity in that variable. This is because linearity in functions guarantees linearity in a function that is 1 at exactly one point, and zero at others.
The proofs for and are analogous, and rely on manipulation of the Lie bracket and the property of a connection being in the subscript vector. These proofs do not involve any explicit use of . The proof for relies simply on repeated application of the product rule, and the fact that .
Tensoriality in the first variable
Given: is a function.
To prove: , or more explicitly,
We start out with the left side:
Each step below is obtained from the previous one via some manipulation explained along side.
Step no.  Current status of left side  Facts/properties used  Specific rewrites 

1  By definition of a connection, is linear in its subscript argument.  
2  Fact (3), the Leibnizlike axiom for connection.  . To understand this more clearly imagine an input to the whole expression, so that the rewrite becomes . In the notation of fact (3), , , and .  
3  is linear in its subscript argument  
4  is additive in its subscript argument  Failed to parse (syntax error): \nabla_{(Yf)}X} + \nabla_{[fX,Y]} = \nabla_{(Yf)X + [fX,Y]}  
5  Fact (2)  . 
Tensoriality in the second variable
Given: is a function.
To prove: , or more explicitly, .
We start out with the left side:
Each step below is obtained from the previous one via some manipulation explained along side.
Step no.  Current status of left side  Facts/properties used  Specific rewrites 

1  By definition of a connection, is linear in its subscript argument  .  
2  Fact (3), the Leibnizlike axiom for connection  . To make this more concrete, imagine an input . Then, the rewrite becomes . This comes setting , , in Fact (3).  
3  is linear in its subscript argument.  
4  is additive in its subscript argument.  .  
5  Fact (2) 
Tensoriality in the third variable
Let be a scalar function. We will show that:
We start out with the left side:
Now we apply the Leibniz rule for connnections on each term:
We again apply the Leibniz rule to the first two term groups:
After cancellations we are left with the following six terms:
But since , the last three terms vanish, and we are left with: