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Let $M$ be an Riemannian Manifold and $\bigtriangledown$ be the Riemannian Connection on it. Let we think about the domain and range of $\bigtriangledown:\Gamma(M)\times\Gamma(M)\rightarrow\Gamma(M)$ and $\Gamma(M)$ contains all smooth vector fields on $M$.

However when we read an Riemannian Geometry book, there will be actually other three different domains of $\bigtriangledown$.

(1) When we talk about the parallel transport, we use the notion $\bigtriangledown_{\dot{\gamma}}X$.

(2) When we talk about the geodesic, we use the notion $\bigtriangledown_{\dot{\gamma}}{\dot{\gamma}}$.

(3) Let $i:N\rightarrow{M}$ be the immersion and $X,Y$ be two smooth vector fields on $N$. Then we note it as $\bigtriangledown_{i_*(X)}{i_*(Y)}$.

So the thing is that I am confused of it and I find no book treating this thing strictly. Does anyone can give an exact answer about how these things happen by step and step? Thanks.

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gaoxinge
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2 Answers2

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The first two cases you are referring to are included in the general definition you gave: the tangent vectors to a curve $\gamma$ in $M$ defined a vector field - in other words $\dot\gamma\in\Gamma(M)$. The last one is a little bit more delicate as in general the push forward $f_* X$ of a vector field $X$ by a map $f$ is not a vector field. For the special case that you are considering, an immersion, $i_*X$ can be seen as a section of $i_*TM$. Informally speaking, you can see it as a vector field, but defined not on the whole of $M$ but only on the submanifold $N$. Maybe someone else can give a more rigorous answer to this point.

GFR
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  • Strictly, $\dot{\gamma}:I\rightarrow\Gamma(M)$ and $i_*(X):N\rightarrow\Gamma(M)$. It is not a smooth vector field on $M$, but very like as you say. – gaoxinge Jan 25 '14 at 12:47
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The three cases are special cases of the general case you gave.
(1) The vector field $\dot{\gamma}$ is a vector field along a curve and we can find a vector field $T$ over $M$ that agrees with $\dot{\gamma}$ along $\gamma$ and in this case $\bigtriangledown_{T}X=\bigtriangledown_{\dot{\gamma}}X$ but we usually write $\bigtriangledown_{\dot{\gamma}}X$ instead.
(2) as number 1
(3) The differential of $i$ simply locally takes vector fields on $N$ to vector fields on $M$ and we can extend the image to be a vector field over $M$.

Semsem
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