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I've just been playing around with the simplest notion of a connection on $\mathbb{R}^n$, that is

$$(\nabla_{v}X)^i=v(X^i)$$

with $X$ a vector field and $v$ a tangent vector at $p$.

From the definition of parallel transport, I know that we should regard $v$ as parallel to $X$ at $p$ iff $\nabla_vX=0$. I can't quite see how this relates to the notion of parallelism in $\mathbb{R}^n$ though!

I know that $v(X^i)$ tells me the rate of change of the component function $X^i$ in the direction of $v$ at $p$. I've convinced myself that the geometrical consequences of $v(X^i) = 0$ are the following.

In an infinitesimal neighbourhood of $p$ the vector field $X$ remains the same as you transport it along the straight line defined by $v$. Then for $v$ an arbitrary vector field we have that $X$ is a parallel vector field along the integral curves of $v$.

Are these the strongest things we can say? Many thanks!

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1 Answer 1

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Parallel in the old fashion Euclidean sense and parallel in the Riemannian geometry sense have little to do with one another.

In $\mathbb{R}^2$, consider the vector field which always points right and has unit length. That is, $v_{(x,y)} = (1,0)$ at every point $p\in \mathbb{R}^2$.

First, let $X_{(x,y)} = (e^x , 0)$. This is a vector field which always points right but as you get larger $x$ values, the arrows get longer. In the classical geometry sense, $X$ and $v$ are parallel at every point. However, if you compute, you'll see that $\nabla_v X \neq 0$, so these are not parallel in the Riemannian geometry sense.

Second, let $X_{(x,y)} = (0,1)$. This is a vector field which always points up with length $1$. In the classical geometry sense, since $v$ points right and $X$ points up, there is no way they are parallel anywhere.

Nonetheless, $\nabla_v(X) = 0$ so they are parallel in the Riemannian geometry sense.

The idea of Riemannian parallelism is that, to say $X$ is parallel along $v$ should mean that as you move along $v$, $X$ doesn't change. In the first example, $X$ does change as you move along $v$, while in the second example, it doesn't.

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Thank you - that's straightened things out for me a bit. Are my intuitions correct about the strongest things I can deduce in my (edited) question above? There by "parallel vector field" I mean this in the Riemannian sense, that is the same at every point. –  Edward Hughes Jan 6 '13 at 14:56
    
That paragraph sounds fine to me. –  Jason DeVito Jan 6 '13 at 15:00
    
Marvellous - I understand it now. I had become confused by considering the simple example $X = (x,y)$, $v = (1,1)$, $p=(1,1)$ and $(\nabla_vX)^1 = 1 \neq 0$. But of course this is not parallel in the Riemannian sense either, because of same scaling problem in your answer. I'll have to be careful to adjust my notion of parralellism depending on the context. –  Edward Hughes Jan 6 '13 at 15:35

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