Difference between revisions of "Curvature is tensorial"

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(Related facts)
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* [[Curvature is antisymmetric in first two variables]]
 
* [[Curvature is antisymmetric in first two variables]]
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* [[Curvature is antisymmetric in last two variables]]
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* [[Curvature is symmetric in the pairs of first and last two variables]]
  
 
==Facts used==
 
==Facts used==

Revision as of 01:54, 24 July 2009

This article gives the statement, and possibly proof, that a map constructed in a certain way is tensorial
View other such statements

Statement

Let \nabla be a connection on a vector bundle E over a differential manifold M. The Riemann curvature tensor of \nabla is given as a map \Gamma(TM) \otimes \Gamma(TM) \otimes \Gamma(E) \to \Gamma(E) defined by:

R(X,Y)Z = \nabla_X\nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z

We claim that R is a tensorial map in each of the variables X,Y,Z.

Related facts

Facts used

X(fg) = (Xf)(g) + f(Xg)

f[X,Y] = [fX,Y] + (Yf)X

f[X,Y] = [X,fY] - (Xf)Y

  • The Leibniz rule axiom that's part of the definition of a connection, namely:

\nabla_X(fZ) = (Xf)(Z) + f\nabla_X(Z)

Proof

To prove tensoriality in a variable, it suffices to show C^\infty-linearity in that variable. This is because linearity in C^\infty-functions guarantees linearity in a function that is 1 at exactly one point, and zero at others.

The proofs for X and Y are analogous, and rely on manipulation of the Lie bracket [fX,Y] and the property of a connection being C^\infty in the subscript vector. These proofs do not involve any explicit use of Z. The proof for Z relies simply on repeated application of the product rule, and the fact that XY - YX = [X,Y].

Tensoriality in the first variable

Let f:M \to \R be a scalar function. We will show that:

\! R(fX,Y) = f R(X,Y)

We start out with the left side:

\nabla_{fX}\nabla_Y - \nabla_Y \nabla_{fX} - \nabla_{[fX,Y]}

Now by the definition of a connection, \nabla is C^\infty-linear in its subscript argument. Thus, the above expression can be written as:

f\nabla_X\nabla_Y - \nabla_Y (f \nabla_X) - \nabla_{[fX,Y]}

Now applying the Leibniz rule for connections, we get:

f\nabla_X\nabla_Y - (Yf)\nabla_X - f \nabla_Y\nabla_X - \nabla_{[fX,Y]}

We can rewrite (Yf)\nabla_X = \nabla_{(Yf)X} and we then get:

f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{(Yf)X + [fX,Y]}

By the corollary stated above, we have:

\! (Yf)X + [fX,Y] = f[X,Y]

which, substituted back, gives:

f(\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})

Tensoriality in the second variable

Let f:M \to \R be a scalar function. We will show that:

\! R(X,fY) = f R(X,Y)

We start out with the left side:

\nabla_X\nabla_{fY} - \nabla_{fY}\nabla_X - \nabla_{[X,fY]}

Applying the Leibniz rule and the property of a connection being C^\infty in its subscript variable yields:

(Xf)\nabla_Y + f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{[X,fY]}

which simplifies to:

f(\nabla_X\nabla_y - \nabla_Y\nabla_X) - \nabla_{[X,fY] - (Xf)Y}

We now use the corollary stated above:

\! f[X,Y] = [X,fY] - (Xf)Y

substituting this gives:

f (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]}

which is f R(X,Y)

Tensoriality in the third variable

Let f:  M \to \R be a scalar function. We will show that:

\! R(X,Y) (fZ) = f R(X,Y) Z

We start out with the left side:

\nabla_X\nabla_Y(fZ) - \nabla_Y\nabla_X(fZ) - \nabla_{[X,Y]}(fZ)

Now we apply the Leibniz rule for connnections on each term:

\nabla_X( (Yf)(Z) + f \nabla_YZ) - \nabla_Y ((Xf)Z + f \nabla_XZ) - f \nabla_{[X,Y]}Z - ([X,Y]f) Z

We again apply the Leibniz rule to the first two term groups:

(XYf)(Z) + (Yf) \nabla_XZ + (Xf) \nabla_YZ + f \nabla_X\nabla_YZ - (YXf)Z - (Xf) \nabla_YZ - (Yf) \nabla_XZ -f \nabla_Y\nabla_XZ - f \nabla_{[X,Y]}Z - ([X,Y]f) Z

After cancellations we are left with the following six terms:

f (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})Z + ((XY - YX - [X,Y])f)Z

But since [X,Y] = XY - YX, the last three terms vanish, and we are left with:

\! f R(X,Y)Z