# Zero connection => Zero torsion tensor ?

1. Mar 31, 2009

### guhan

I am reading a book which says that if the components of the connection vanishes then the torsion and curvature tensors vanish. Now, I understand how this happens with curvature tensor, but I don't see how torsion tensor would vanish.

Torsion tensor:
$$\tau (X,Y) = D_X (Y) - D_Y (X) - [X,Y]$$

If connection is zero then $$D_X (Y) = D_Y (X) = 0$$.
But this still does NOT make $$\tau$$ zero right?

2. Mar 31, 2009

### nughret

This is incorrect; write out the equation in local co-ords(in which connection vanishes) then
(DXY)i = Yi,jXj

3. Mar 31, 2009

### StatusX

The connection is not a tensor, so you can only say it vanishes with respect to certain choices of basis. If it vanishes in a coordinate basis, then since the Lie bracket of coordinate basis vectors vanish, so does the last term in your expression for torsion. On the other hand, in a non-coordinate basis, the last term may contribute.

4. Apr 1, 2009

### guhan

@ nughret
Oh yes, you are right. It was an oversight on my part.

@ StatusX
I assume you meant to show both when and when not taking the coordinate bases, the statement (torsion vanishes for zero connection) holds. Thank you.

When connection, $$\Gamma$$, is zero, $$Z^i_{;j}=Z^i_{,j}$$ and so...
$$(D_XY - D_YX)^i = X^j Y^i_{;j} - Y^j X^i_{;j} = X^j Y^i_{,j} - Y^j X^i_{,j} = [X,Y]^i$$
And thus the torsion becomes zero.

5. Apr 1, 2009

### StatusX

No, it's possible to have non-zero torsion even if the connection (that is, the connection coefficients with respect to some particular basis) vanishes. In fact, I believe the connection vanishing (in some basis, coordinate or not) is precisely equivalent to the Riemann curvature tensor vanishing, but even if this tensor vanishes there may still be non-zero torsion.

6. Apr 2, 2009

### guhan

@ StatusX
You are right that zero connection and zero curvature are equivalent. But, I don't understand how zero-connection in some (non-coord) basis can give a non-zero torsion in that same basis. Can you please point out how and why this happens, in the little derivation in my last post?

7. Apr 2, 2009

### nughret

given any point x in our manifold, given that we can find some chart around x in which the connection vanishes show that the torsion tensor is zero everywhere.

The above posts provide an outline of the proof. If you do not believe such a proof is complete then provide a counterexample

8. Apr 2, 2009

### StatusX

Yes, if you define a chart, and look at the (coordinate) basis related to that chart, then zero connection implies zero torsion. But in the more general case of a non-coordinate basis (ie, one that can't be written in the form $\partial/\partial x^\mu$ for some coordinates $x^\mu$), one can still talk about connection coefficients, and then it's possible that the connection coefficients vanish in such a basis even if the torsion does not. In fact, from your formula you can see that the torsion will be given by the Lie bracket of the various basis vectors with each other, which is zero iff the basis is coordinate. So any connection which is zero in a non-coordinate basis has torsion.

As far as the derivation in the 4th post, the mistake is at:

$$X^j Y^i_{,j} - Y^j X^i_{,j} = [X,Y]^i$$

To be explicit, by $$Y^i_{,j}[/itex] you mean $e_j(Y^i)$, ie, the derivative of the scalar function $Y^i$ in the direction of the basis vector $e_j$. But if we take, eg, $X = e_m$, $Y = e_n$, for some fixed basis vectors, then the functions $X^i,Y^i$ are constant (they are equal to 1 for i=n or m respectively, and zero otherwise), and so the above formula would suggest that $[e_n,e_m]=0$, which is only true if this is a coordinate basis. In other words, the formula for the Lie bracket needs to be modified when you're working in a non-coordinate basis. I won't work out here exactly what the new formula is, but hopefully this argument convinces you that it must be something different. 9. Apr 3, 2009 ### nughret Yes I see your point Status, and as you say the connection vanishing in some basis implies only zero curvature not zero torsion. In regards to the formula for the Lie bracket using Z = [X,Y] if Z(f) = X(Y(f)) - Y(X(f)), gives all required terms. Using a non commutative basis we then find the extra terms relate directly to the Lie bracket of our basis vectors. If you want to find the explicit formula just write out the above in local coordinates remembering to write our vector field in terms of the non-com basis vectors. 10. May 10, 2009 ### guhan In the case of non-commuting bases, I think there is no reason to believe that we can even find dual bases [tex]\{\omega^i\}$$ such that $$\omega^i(e_j) = \delta^i_j$$ and so modifying the lie bracket formula may not be as direct since we would be left with $$[X,Y]=[X^ie_i,Y^je_j]=X^ie_iY^je_j-Y^je_jX^ie_i$$ and then...?

Can we use Frobenius theorem and say that for any [X,Y] to lie in the tangent space they should be spanned by the coordinate (and hence commuting) bases - $$\{ \frac{\partial}{\partial x^i} \}$$ - and thus make a transformation $$\frac{\partial}{\partial x^i} = A^j_i e_j$$? and then, use this to find lie bracket formula?

PS: Sorry, I didnt notice this thread for long. It somehow didnt show up as updated in my subscribed thread list.

11. May 14, 2009

### nughret

For vector fields V = Viei, W

[V,W] = (VjWi,j - Vi,jWi + VjWkCijk)ei

Where Cijk = ([ej,ek]d we can find these in some neighbourhood by expanding the non-commuting basis in terms of a coordinate basis

12. May 15, 2009

### guhan

How do you get the $$V^i_{,j}$$ and $$W^i_{,j}$$ terms?
And, if you are using the structure constants $$C^i_{jk}$$, then why not use the bilinear property of lie brackets and directly say that $$[V,W]=V^jW^k[e_j,e_k]=V^jW^kC^i_{jk}e_i$$?