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The torsion tensor and symmetric connections November 8, 2009

Posted by Akhil Mathew in differential geometry, MaBloWriMo.
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Today I will discuss the torsion tensor of a Koszul connection. It measures the deviation from being symmetric in a sense defined below.

Torsion

Given a Koszul connection {\nabla} on the smooth manifold {M}, define the torsion tensor {T} by

\displaystyle T(X,Y) := \nabla_X Y - \nabla_Y X - [X,Y]. 

It can be checked that {T} is a tensor, using the identities for {\nabla} and those for the Lie bracket. For instance, here is the proof that {T(fX,Y)=fT(X,Y)} for a smooth {f}.

\displaystyle T(fX,Y) = \nabla_{fX}Y - \nabla_Y(fX) - [fX,Y]. 

In view of the connection identities and {[fX,Y] = f(X \circ Y) - Y \circ (f X) = f(X \circ Y - Y \circ X) - (Yf)X}, we get that this equals

\displaystyle f \nabla_X Y - f \nabla_Y X - (Yf) X - f[X,Y] + (Yf)X = fT(X,Y) . 

When {T \equiv 0}, we call {\nabla} symmetric. There is a useful motivation for this terminology: when we are in an open subset of {\mathbb{R}^n}, the condition is equivalent to {T\left( \partial_i, \partial_j\right)=\nabla_{\partial_i} \partial_j - \nabla_{\partial_j} \partial_i = 0} for all {i,j}, or equivalently

\displaystyle \Gamma^k_{ij} = \Gamma^k_{ji}, 

where the {\Gamma_{ij}^k} are the Christoffel symbols.

Vector fields along a surface

There is a useful consequence of symmetry that I will now discuss. If {s: U \rightarrow M} where {U \subset \mathbb{R}^2} is open, then we can call {s} a surface. We can define a vector field along {s} in the same way as a vector field along a curve—it is a smooth map {V: U \rightarrow TM} with {V} projecting down to {s}.

One way to get a vector field along {s} is to consider the partial derivatives {\frac{\partial}{\partial x} s, \frac{\partial}{\partial y} s}. Equivalently, these are obtained by {s_*\left( \frac{\partial}{\partial x} \right), s_{*}\left(\frac{\partial}{\partial y} \right)}.

Given a vector field {V} along {s}, we can consider the “partial covariant derivatives” {\frac{D}{\partial x} V, \frac{D}{\partial y} V} defined respectively at {(x,y)} as the covariant derivative of the vector field along a curve {x \rightarrow V(x,y)} or {y \rightarrow V(x,y)}. These are also vector fields along {s}. It can be checked that these operations preserve smoothness.

Another way to think of this is {\frac{D}{\partial x} V = \nabla_{s_*\left( \frac{\partial}{\partial x} \right)} V}, and similarly for {y}. There is a slight abuse of notation because {V} is only a vector field along {s}, but at least when {s} is an immersion this is ok because you can locally extend vector fields along {s}.

Here is a useful consequence of symmetry.

Proposition 1

Let {s} be a surface in {M}, and let {\nabla} be a symmetric connection on {M}. Then\displaystyle \frac{D}{\partial x} \frac{\partial}{\partial y} s = \frac{D}{\partial y} \frac{\partial}{\partial x} s.\ \ \ \ \ (1) 

 

So symmetry is just a version of Clairaut’s theorem.

One way to prove this is to compute in a coordinate system.

The following argument is incorrect.   I will post a corrected proof later today.

Another, more conceptual way, to argue is as follows. Given an embedding {i: N \rightarrow M}, we can obtain a connection {\nabla'} on {M} by defining

\displaystyle \nabla'_{X} Y := \nabla_{i_* X} i_* Y. 

Of course, there is some checking to be done here. {i_* X,i_* Y} aren’t really vector fields, but one can check (as with the definition of the covariant derivative) that it doesn’t matter, because you just extend them locally to vector fields. Then if {X',Y' \in \Gamma(N)} are {i}-related to {X,Y \in \Gamma(M)}, it follows that

\displaystyle T(X',Y') \ \mathrm{is \ } i-\mathrm{related \ to} \ T(X,Y). 

Consider an immersion {s: U \rightarrow M} for {U \subset \mathbb{R}^2} open. Then

\displaystyle \frac{D}{\partial x} \frac{\partial}{\partial y} s = \nabla_{s_{*} \left( \frac{\partial}{\partial x} \right) } s_{*}\left( \frac{\partial}{\partial y} \right) = \nabla_{s_{*} \left( \frac{\partial}{\partial y} \right) } s_{*}\left( \frac{\partial}{\partial x} \right) - s_{*} T'\left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right). 

Now {\nabla'} is symmetric, so {T'=0} and this turns into

\displaystyle \nabla_{s_{*} \left( \frac{\partial}{\partial y} \right) } s_{*}\left( \frac{\partial}{\partial x} \right) = \frac{D}{\partial y} \frac{\partial}{\partial x} s. 

We’ve now handled the case of {s} an immersion. Now consider the general case. Fix an arbitrary {s: U \rightarrow M} with {p \in U}, and consider the two quantities in (1) at {p}. If we choose an immersion {t: D_r(p) \rightarrow M} where {r} is very small (possible if {\dim M \geq 2}, which we can assume), then {s + \epsilon t} is an immersion at {p} if {\epsilon} is very small. Thus

\displaystyle \frac{D}{\partial x} \frac{\partial}{\partial y} ( s + \epsilon t) = \frac{D}{\partial y} \frac{\partial}{\partial x} ( s + \epsilon t)  

at {p}, at least. Letting {\epsilon \rightarrow 0}, the two quantities are equal.

So what’s next?  First, I ought to do something like this for the curvature tensor, though it’ll go a bit more quickly since the pattern is similar.  Then, I should talk about the Levi-Civita connection associated to a Riemannian metric and how geodesics turn out to be locally distance-minimizing paths.  This will provide ample material for a couple of more posts, and then I have other differential geometry topics in mind.

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