Confusion on notion of connection & covariant derivative

[andrewkirk]

What definition of connection are you using?

I had in mind the definition used by John Lee in his 'Riemannian Manifolds'. That is that a connection is a function $\nabla:TM\times\mathscr{T}(M)\to TM$ for which $\nabla(X,V)\equiv \nabla_XV$ is in the same tangent space as $X$, and that satisfies the three linearity conditions:

(1) $\nabla_{fX_1+gX_2}Y=f\nabla_{X_1}Y+g\nabla_{X_2}Y$ for $f,g\in C^\infty(M)$.
(2) $\nabla_X(aY_1+bY_2)=a\nabla_XY_1+b\nabla_XY_2$ for $a,b\in\mathbb{R}$
(3) $\nabla_X(fY)=f\nabla_XY+(Xf)Y$ for $f\in C^\infty(M)$

In (3) $Xf$ denotes is the derivative of scalar function $f$ in direction $X$ which I think is OK to use because - as $f$ is a scalar function - it can be defined without using the connection.

Which definition of parallel transport are you using?

Given a curve $\gamma$ that passes through $p\in M$, the parallel transport $U_{(V,\gamma)}$ of a vector $V\in T_pM$ along $\gamma$ is a (I think, unique) vector field on the image of $\gamma$ that satisfies:

(a) $\nabla_{\dot{\gamma}(s)}(U_{(V,\gamma)}(\gamma(s))=0$ for all $s$ in the domain of $\gamma$
(b) $U_{(V,\gamma)}(p)=V$

There's one thing that bothers me here, which is that the second argument to the map $\nabla$ in (a) is a vector field on image $\gamma$, whereas $\nabla$ requires it to be a vector field on an open subset of $M$. I have a construction in mind that I think could probably make that robust in a small neighbourhood of $p$, but I haven't had time to think it through fully yet, let alone to code it up here.

 

[lavinia]

There's one thing that bothers me here, which is that the second argument to the map $\nabla$ in (a) is a vector field on image $\gamma$, whereas $\nabla$ requires it to be a vector field on an open subset of $M$. I have a construction in mind that I think could probably make that robust in a small neighbourhood of $p$, but I haven't had time to think it through fully yet, let alone to code it up here.

You are right to catch this. One can separately define what a vector field along a curve is and then the covariant derivative of a vector field along a curve.

A vector field along a curve is an assignment of a tangent vector,$V_{c(t)}$, to each point $c(t)$ on the curve such that for any smooth function $f$, the function $V_{c(t)}.f$ is a smooth function on the domain of definition of $c$.

- It seems to me that this could be extended to an arbitrary vector bundle by saying that the assignment $t \rightarrow c(t) \rightarrow V_{c(t)}$ is a smooth map from the reals into the vector bundle.

Then given an affine connection on the manifold, one defines a covariant derivative of a vector field along a curve by demanding linearity and the Leibniz rule plus if $V_{c(t)}$ is the restriction of a vector field to the curve then the covariant derivative along the curve is the same as the covariant derivative of the affine connection. One then shows existence and uniqueness.

 

[Twigg]

A caveat: the curve must be a regular submanifold to inherit a vector field and a covariant derivative operator from the base manifold's tangent bundle. For example, you can't project the directional derivative on $\mathbb{R^{2}}$ onto a figure eight curve. You can get the directional derivative by parametrizing the figure eight path in polar, but from $\mathbb{R^{2}}$'s point of view, the directional derivative of any field is undefined at the point of intersection (it's a singular point).

 

[lavinia]

A caveat: the curve must be a regular submanifold to inherit a vector field and a covariant derivative operator from the base manifold's tangent bundle. For example, you can't project the directional derivative on $\mathbb{R^{2}}$ onto a figure eight curve. You can get the directional derivative by parametrizing the figure eight path in polar, but from $\mathbb{R^{2}}$'s point of view, the directional derivative of any field is undefined at the point of intersection (it's a singular point).

True but if the vector field lines up with itself as the curve reaches an intersection point you would be OK. The definition only requires that for any smooth function $f$ the map $t \rightarrow V_{c(t)}.f(c(t)$ is a smooth function from an interval of real numbers into the real numbers. So the curve does not need to be an embedded submanifold.

 

[lavinia]

Thinking about connections more generallyGiven two vector fields, $X$ and $Y$ the covariant derivative gives a new vector field $∇_{X}Y$ . For fixed $Y$, the map $X \rightarrow ∇_{X}Y$ is linear in $X$. That is: $∇_{X+Z}Y = ∇_{X}Y + ∇_{Z}Y$ and for any smooth function $f$, $∇_{fX}Y = f∇_{X}Y$. This means that one can think of $∇$ as a linear map from sections of the tangent bundle (vector fields) into sections of the tensor bundle $TM^{*}⊗TM$ where $TM^{*}$ is the dual of the tangent bundle i.e. the bundle whose sections are smooth 1 forms on the manifold $M$. So $∇$ is an operator on vector fields, $s \rightarrow ∇s$ where $∇s$ is a section of $TM^{*}⊗TM$.

In this context the Leibniz rule is $∇fs \rightarrow df⊗s + f∇s$.

If one thinks about it, $s$ does not need to be a section of the tangent bundle. It could be a section of any smooth vector bundle. The definition works in exactly the same way. So now one has the idea of a connection on an arbitrary smooth vector bundle. $∇$ maps sections of the vector bundle $ζ$ into sections of $TM^{*}⊗ζ$. It is linear over the base field and satisfies the Leibniz rule. BTW: the base field is the complex numbers when $ζ$ is a complex vector bundle.

If $s_{i}$ is a basis of vector fields on an open domain then $∇s_{i} = ∑_{j}ω_{ij}⊗s_{j}$ and the $ω_{ij}$ are the connection 1 forms with respect to this basis.

 

[Ben Niehoff]

In this context the Leibniz rule is $∇fs \rightarrow df⊗s + f∇s$.

If one thinks about it, $s$ does not need to be a section of the tangent bundle. It could be a section of any smooth vector bundle. The definition works in exactly the same way. So now one has the idea of a connection on an arbitrary smooth vector bundle. $∇$ maps sections of the vector bundle $ζ$ into sections of $TM^{*}⊗ζ$. It is linear over the base field and satisfies the Leibniz rule. BTW: the base field is the complex numbers when $ζ$ is a complex vector bundle.

This is something that took me a while to appreciate. It is tempting to generalize the above Leibniz rule to tensor products:

$$\nabla (X \otimes Y) = (\nabla X) \otimes Y + X \otimes (\nabla Y),$$
but this is not quite correct. One must instead take $TM \otimes TM$ to be the vector bundle $\zeta$ in question, and the connection to act on this bundle in the appropriate tensor product representation. In effect, the "new" factor of $T^*M$ must always appear in front: $T^*M \otimes \zeta$; rather than in the middle as it might appear from the second term in my attempted "tensor product rule".

One can write down product rules like mine above, if one also makes the mental note that the new factor of $T^*M$ always goes at the front. Alternatively, one can always work with a vector field already stuck into this slot:

$$\nabla_Z (X \otimes Y) = (\nabla_Z X) \otimes Y + X \otimes (\nabla_Z Y),$$
which is always true.

If $s_{i}$ is a basis of vector fields on an open domain then $∇s_{i} = ∑_{j}ω_{ij}⊗s_{j}$ and the $ω_{ij}$ are the connection 1 forms with respect to this basis.

In light of your previous paragraph, of course $s_i$ should be a basis of the vector bundle $\zeta$, which may not be "vector fields", per se.

(I think at this point we may have thoroughly confused the OP, though!)

 

["Don't panic!"]

(I think at this point we may have thoroughly confused the OP, though!)

Yes, a little bit! Although it has been interesting to read the conversation and try and understand the information.
I have to say, I thought I understood the meaning of the covariant derivative and connection before I started studying differential geometry in more depth, but I now feel that I don't really at all :/

 

[lavinia]

Yes, a little bit! Although it has been interesting to read the conversation and try and understand the information.
I have to say, I thought I understood the meaning of the covariant derivative and connection before I started studying differential geometry in more depth, but I now feel that I don't really at all :/

A lot if this came from your questions about the Leibniz rule. I hope it helps to see how far these ideas can go. IMO It is good to be swamped sometimes.

BTW: This general definition of connection can also be used to define curvature. Curvature makes sense for connections on arbitrary smooth vector bundles.