Torsion is tensorial: Difference between revisions

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{{tensoriality fact}}
==Statement==
==Statement==


===Symbolic statement===
Let <math>M</math> be a [[differential manifold]] and <math>\nabla</math> be a [[fact about::linear connection]] on <math>M</math> (viz., <math>\nabla</math> is a [[connection]] on the [[tangent bundle]] <math>TM</math> of <math>M</math>).


Let <math>M</math> be a [[differential manifold]] and <math>\nabla</math> be a [[linear connection]] on <math>M</math> (viz., <math>\nabla</math> is a [[connection]] on the [[tangent bundle]] <math>TM</math> of <math>M</math>).
Consider the [[fact about::torsion of a linear connection|torsion]] of <math>\nabla</math>, namely:
 
Consider the [[torsion of a linear connection|torsion]] of <math>\nabla</math>, namely:


<math>\tau(\nabla): \Gamma(TM) \times \Gamma(TM) \to \Gamma(TM)</math>
<math>\tau(\nabla): \Gamma(TM) \times \Gamma(TM) \to \Gamma(TM)</math>
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given by:
given by:


<math>\tau(\nabla)(X,Y) = \nabla_Y X - \nabla_X Y - [X,Y]</math>
<math>\tau(\nabla)(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]</math>


Then, <math>\tau(\nabla)</math> is a [[tensorial map]] in both coordinates.
Then, <math>\tau(\nabla)</math> is a [[fact about::tensorial map]] in both coordinates. In other words, the value of <math>\tau(\nabla)</math> at a point <math>p \in M</math> depends only on <math>\nabla, X(p), Y(p)</math> and does not depend on the values of the vectors fields <math>X,Y</math> at points other than <math>p</math>.


==Proof==
More explicitly, for any point <math>p \in M</math>, <math>\tau(\nabla)</math> defines a bilinear map:
 
<math>\! \tau(\nabla): T_p(M) \times T_p(M) \to T_p(M)</math>
 
Further, since in fact [[torsion is antisymmetric]], this is an alternating bilinear map.
 
==Related facts==


===Tensoriality in the first coordinate===
* [[Curvature is tensorial]]
* [[Torsion is antisymmetric]]
* [[Curvature is antisymmetric in first two variables]]


We'll use the fact that tensoriality is equivalent to <math>C^\infty</math>-linearity.
==Facts used==


''To prove'': <math>\tau(\nabla)(fX,Y) = f\tau(\nabla)(X,Y)</math>
{| class="sortable" border="1"
! Fact no. !! Name !! Statement with symbols
|-
| 1 || Any connection is <math>C^\infty</math>-linear in its subscript argument || <math>\nabla_{fA} = f\nabla_A</math> for any <math>C^\infty</math>-function <math>f</math> and vector field <math>A</math>.
|-
| 2 || The Leibniz-like axiom that is part of the definition of a connection || For a function <math>f</math> and vector fields <math>A,B</math>, and a connection <math>\nabla</math>, we have <math>\nabla_A(fB) = (Af)(B) + f\nabla_A(B)</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>
|}


''Proof'': We prove this by expanding everything out:
==Proof==


<math>\tau(\nabla)(fX,Y) = \nabla_Y(fX) - \nabla_{fX}(Y) - [fX,Y] = f \nabla_YX  - f \nabla_Y X - (Yf)(X) - [fX,Y]</math>
To prove tensoriality in a variable, it suffices to show <math>C^\infty</math>-linearity in that variable. This is because linearity in <math>C^\infty</math>-functions guarantees linearity in a function that is 1 at exactly one point, and zero at others.


To prove the equality with <math>f \tau(\nabla)(X,Y)</math>, we need to show:
The proofs for <math>X</math> and <math>Y</math> are analogous, and rely on manipulation of the Lie bracket <math>[fX,Y]</math> and the property of a connection being <math>C^\infty</math> in the subscript vector.
===Tensoriality in the first coordinate===


<math>(Yf)(X) = f[X,Y] - [fX,Y]</math>
'''Given''': <math>f:M \to \R</math> is  <math>C^\infty</math>-function


To prove this, we need to show that both sides evaluate to the same expression for any function <math>g</math>. Plugging a function <math>g</math>, we see that the right side becomes:
'''To prove''': <math>\tau(\nabla)(fX,Y) = f\tau(\nabla)(X,Y)</math>


<math>f(X(Yg) - Y(Xg)) - f(X(Yg)) + Y((fX)g) = Y(f(Xg)) - f(Y(Xg)) </math>
'''Proof''': We start out with the left side:


Applying the Leibniz rule for <math>Y</math> on the product of functions <math>f</math> and <math>Xg</math>, this simplifies to:
<math>\tau(\nabla)(fX,Y)</math>


<math>(Yf)(Xg) + Y(Xg)(f) - f(Y(Xg)) = (Yf)(Xg)</math>
Each step below is obtained from the previous one via some manipulation explained along side.


which is precisely equal to the left side.
{| class="sortable" border="1"
! Step no. !! Current status of left side !! Facts/properties used !! Specific rewrites
|-
| 1 || <math>\nabla_{fX}(Y) - \nabla_Y(fX) - [fX,Y]</math> || Definition of torsion || whole thing
|-
| 2 || <math>f \nabla_X Y  - \nabla_Y(fX) - [fX,Y]</math> || Fact (1): <math>C^\infty</math>-linearity of connection in subscript argument || <math>\nabla_{fX} \mapsto f\nabla_X</math>
|-
| 3 || <math>f \nabla_X Y  - (f \nabla_Y X + (Yf)(X)) - [fX,Y]</math> || Fact (2): The Leibniz-like axiom that's part of the definition of a connection || <math>\nabla_Y(fX) \mapsto f\nabla_YX + (Yf)(X)</math>
|-
| 4 || <math>f \nabla_X Y  - f \nabla_Y X - ((Yf)(X) + [fX,Y])</math> || parenthesis rearrangement || --
|-
| 5 || <math>f \nabla_X Y  - f \nabla_Y X - f[X,Y]</math> || Fact (3) || <math>(Yf)(X) + [fX,Y] \mapsto f[X,Y]</math>
|-
| 6 || <math>f(\nabla_X Y - \nabla_Y X - [X,Y])</math> || factor out || --
|-
| 7 || <math>f\tau(\nabla)(X,Y)</math> || use definition of torsion || <math>\nabla_X Y - \nabla_Y X - [X,Y] \mapsto \tau(\nabla)(X,Y)</math>
|}


===Tensoriality in the second coordinate===
===Tensoriality in the second coordinate===
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The proof is analogous to that for the first coordinate.
The proof is analogous to that for the first coordinate.


{{fillin}}
''To prove'' <math>\tau(\nabla)(X,fY) = f \tau(\nabla)(X,Y)</math>
 
''Proof'': This is similar to tensoriality in the first coordinate.

Latest revision as of 17:56, 6 January 2012

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

Statement

Let M be a differential manifold and be a linear connection on M (viz., is a connection on the tangent bundle TM of M).

Consider the torsion of , namely:

τ():Γ(TM)×Γ(TM)Γ(TM)

given by:

τ()(X,Y)=XYYX[X,Y]

Then, τ() is a tensorial map in both coordinates. In other words, the value of τ() at a point pM depends only on ,X(p),Y(p) and does not depend on the values of the vectors fields X,Y at points other than p.

More explicitly, for any point pM, τ() defines a bilinear map:

τ():Tp(M)×Tp(M)Tp(M)

Further, since in fact torsion is antisymmetric, this is an alternating bilinear map.

Related facts

Facts used

Fact no. Name Statement with symbols
1 Any connection is C-linear in its subscript argument fA=fA for any C-function f and vector field A.
2 The Leibniz-like axiom that is part of the definition of a connection For a function f and vector fields A,B, and a connection , we have A(fB)=(Af)(B)+fA(B)
3 Corollary of Leibniz rule for Lie bracket (in turn follows from Leibniz rule for derivations For a function f and vector fields X,Y:


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

Proof

To prove tensoriality in a variable, it suffices to show C-linearity in that variable. This is because linearity in C-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 in the subscript vector.

Tensoriality in the first coordinate

Given: f:MR is C-function

To prove: τ()(fX,Y)=fτ()(X,Y)

Proof: We start out with the left side:

τ()(fX,Y)

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 fX(Y)Y(fX)[fX,Y] Definition of torsion whole thing
2 fXYY(fX)[fX,Y] Fact (1): C-linearity of connection in subscript argument fXfX
3 fXY(fYX+(Yf)(X))[fX,Y] Fact (2): The Leibniz-like axiom that's part of the definition of a connection Y(fX)fYX+(Yf)(X)
4 fXYfYX((Yf)(X)+[fX,Y]) parenthesis rearrangement --
5 fXYfYXf[X,Y] Fact (3) (Yf)(X)+[fX,Y]f[X,Y]
6 f(XYYX[X,Y]) factor out --
7 fτ()(X,Y) use definition of torsion XYYX[X,Y]τ()(X,Y)

Tensoriality in the second coordinate

The proof is analogous to that for the first coordinate.

To prove τ()(X,fY)=fτ()(X,Y)

Proof: This is similar to tensoriality in the first coordinate.