Difference between revisions of "Curvature is tensorial"
(→Tensoriality in the first variable) 
(→Facts used) 

Line 24:  Line 24:  
    
 3  [[uses::Corollary of Leibniz rule for Lie bracket]] (in turn follows from [[uses::leibniz rule for derivations]] For a function <math>f</math> and vector fields <math>X,Y</math>:   3  [[uses::Corollary of Leibniz rule for Lie bracket]] (in turn follows from [[uses::leibniz rule for derivations]] For a function <math>f</math> and vector fields <math>X,Y</math>:  
−  <br><math>\! f[X,Y] = [fX,Y] + (Yf)X</math><br><math>f[X,Y] = [X,fY]  (Xf)Y</math>  +  <br><math>\! f[X,Y] = [fX,Y] + (Yf)X</math><br><math>\! f[X,Y] = [X,fY]  (Xf)Y</math> 
}  }  
Revision as of 00:16, 20 December 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  Statement with symbols 

1  Any connection is linear in its subscript argument  for any function and vector field . 
2  The Leibnizlike axiom that is part of the definition of a connection  For a function and vector fields , and a connection , we have 
3  Corollary of Leibniz rule for Lie bracket (in turn follows from leibniz rule for derivations  For a function and vector fields :

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  Fact (1): is linear in its subscript argument.  
2  Fact (2)  . 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  Fact (1)  
4  is additive in its subscript argument  
5  Fact (3)  
6  Fact (1) 
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  Fact (1)  .  
2  Fact (2)  . To make this more concrete, imagine an input . Then, the rewrite becomes . This comes setting , , in Fact (3).  
3  Fact (1)  
4  is additive in its subscript argument.  .  
5  Fact (3)  
6  Fact (1) 
Tensoriality in the third variable
Given: A function .
To prove: . 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  Fact (2)  and  
2  Fact (2)  , etc.  
3    cancellations  
4  use , definition  cancellation 