Geodesics in $mathbb{R}^n$ with the trivial connection












1














Define geodesic as follows:




Given a tangent bundle $TMrightarrow M$ with connection $nabla$, a geodesic is a curve $gamma:Irightarrow M$ such that $(gamma^*nabla)dot gamma = 0$.




(Notice that $(gamma^*nabla)dot gamma in Gamma(gamma^*TM)$)



I want to check that according to this definition the geodesics in $mathbb R^n$ with the trivial connection $nabla(f_1frac{partial}{partial x_1}+...+f_nfrac{partial}{partial x_n})=(df_1frac{partial}{partial x_1}+...+df_nfrac{partial}{partial x_n}) $ are straight lines $x+tv$.

Take a path $gamma(t)=(x_1(t),...,x_n(t))$. We have
$$
(gamma^*nabla)dot gamma = (gamma^*nabla)( dot x_1frac{partial}{partial x_1}vert_{gamma(-)} +...+dot x_nfrac{partial}{partial x_n}vert_{gamma(-)} )
$$
But
$$dot x_ifrac{partial}{partial x_i}vert_{gamma(-)} = dot x_igamma^*frac{partial}{partial x_i}vert_{(-)} = gamma^*(dot x_ifrac{partial}{partial x_i}vert_{(-)})
$$
So that, using the definition of pullback connection:
$$
(gamma^*nabla)( sum_i gamma^*(dot x_ifrac{partial}{partial x_i}vert_{(-)})) = gamma^*(sum_i nabla(dot x_ifrac{partial}{partial x_i})) = gamma^*(sum_i frac{ddot x_i}{dt}dtfrac{partial}{partial x_i})
$$
Now I would like to say that this is zero if and only if the $frac{ddot x_i}{dt}$ are zero. If I didn't have the $gamma^*$ that would be clear. But I have no reason to assume that $gamma^*$ is injective. If it were, I would be done.






riemannian-geometry vector-bundles geodesic connections tangent-bundle







share|cite|improve this question






















  • Locally, the curve $gamma$ is injective. I don't know if that helps but...
    – Dog_69
    Dec 26 '18 at 19:09










  • @Dog_69 I am not imposing that on my curve
    – Soap
    Dec 26 '18 at 19:25










  • Hummm, if you don't impose that then your curve is just a point.
    – Dog_69
    Dec 26 '18 at 20:47










  • @Dog_69 It just means the curve has self intersections
    – Soap
    Dec 27 '18 at 16:41






  • 1




    Yes, but I said locally. Between two self-intersections, say $gamma(t_0)$ and $gamma(t_1)$ you can find a neighbourhood of $gamma(t)$ where $gamma$ is injective. Do yo know what I mean? Maybe it is totally useless but I thought it was worth mentioning.
    – Dog_69
    Dec 27 '18 at 16:53
















1














Define geodesic as follows:




Given a tangent bundle $TMrightarrow M$ with connection $nabla$, a geodesic is a curve $gamma:Irightarrow M$ such that $(gamma^*nabla)dot gamma = 0$.




(Notice that $(gamma^*nabla)dot gamma in Gamma(gamma^*TM)$)



I want to check that according to this definition the geodesics in $mathbb R^n$ with the trivial connection $nabla(f_1frac{partial}{partial x_1}+...+f_nfrac{partial}{partial x_n})=(df_1frac{partial}{partial x_1}+...+df_nfrac{partial}{partial x_n}) $ are straight lines $x+tv$.

Take a path $gamma(t)=(x_1(t),...,x_n(t))$. We have
$$
(gamma^*nabla)dot gamma = (gamma^*nabla)( dot x_1frac{partial}{partial x_1}vert_{gamma(-)} +...+dot x_nfrac{partial}{partial x_n}vert_{gamma(-)} )
$$
But
$$dot x_ifrac{partial}{partial x_i}vert_{gamma(-)} = dot x_igamma^*frac{partial}{partial x_i}vert_{(-)} = gamma^*(dot x_ifrac{partial}{partial x_i}vert_{(-)})
$$
So that, using the definition of pullback connection:
$$
(gamma^*nabla)( sum_i gamma^*(dot x_ifrac{partial}{partial x_i}vert_{(-)})) = gamma^*(sum_i nabla(dot x_ifrac{partial}{partial x_i})) = gamma^*(sum_i frac{ddot x_i}{dt}dtfrac{partial}{partial x_i})
$$
Now I would like to say that this is zero if and only if the $frac{ddot x_i}{dt}$ are zero. If I didn't have the $gamma^*$ that would be clear. But I have no reason to assume that $gamma^*$ is injective. If it were, I would be done.






riemannian-geometry vector-bundles geodesic connections tangent-bundle







share|cite|improve this question






















  • Locally, the curve $gamma$ is injective. I don't know if that helps but...
    – Dog_69
    Dec 26 '18 at 19:09










  • @Dog_69 I am not imposing that on my curve
    – Soap
    Dec 26 '18 at 19:25










  • Hummm, if you don't impose that then your curve is just a point.
    – Dog_69
    Dec 26 '18 at 20:47










  • @Dog_69 It just means the curve has self intersections
    – Soap
    Dec 27 '18 at 16:41






  • 1




    Yes, but I said locally. Between two self-intersections, say $gamma(t_0)$ and $gamma(t_1)$ you can find a neighbourhood of $gamma(t)$ where $gamma$ is injective. Do yo know what I mean? Maybe it is totally useless but I thought it was worth mentioning.
    – Dog_69
    Dec 27 '18 at 16:53














1












1








1


1





Define geodesic as follows:




Given a tangent bundle $TMrightarrow M$ with connection $nabla$, a geodesic is a curve $gamma:Irightarrow M$ such that $(gamma^*nabla)dot gamma = 0$.




(Notice that $(gamma^*nabla)dot gamma in Gamma(gamma^*TM)$)



I want to check that according to this definition the geodesics in $mathbb R^n$ with the trivial connection $nabla(f_1frac{partial}{partial x_1}+...+f_nfrac{partial}{partial x_n})=(df_1frac{partial}{partial x_1}+...+df_nfrac{partial}{partial x_n}) $ are straight lines $x+tv$.

Take a path $gamma(t)=(x_1(t),...,x_n(t))$. We have
$$
(gamma^*nabla)dot gamma = (gamma^*nabla)( dot x_1frac{partial}{partial x_1}vert_{gamma(-)} +...+dot x_nfrac{partial}{partial x_n}vert_{gamma(-)} )
$$
But
$$dot x_ifrac{partial}{partial x_i}vert_{gamma(-)} = dot x_igamma^*frac{partial}{partial x_i}vert_{(-)} = gamma^*(dot x_ifrac{partial}{partial x_i}vert_{(-)})
$$
So that, using the definition of pullback connection:
$$
(gamma^*nabla)( sum_i gamma^*(dot x_ifrac{partial}{partial x_i}vert_{(-)})) = gamma^*(sum_i nabla(dot x_ifrac{partial}{partial x_i})) = gamma^*(sum_i frac{ddot x_i}{dt}dtfrac{partial}{partial x_i})
$$
Now I would like to say that this is zero if and only if the $frac{ddot x_i}{dt}$ are zero. If I didn't have the $gamma^*$ that would be clear. But I have no reason to assume that $gamma^*$ is injective. If it were, I would be done.






riemannian-geometry vector-bundles geodesic connections tangent-bundle







share|cite|improve this question













Define geodesic as follows:




Given a tangent bundle $TMrightarrow M$ with connection $nabla$, a geodesic is a curve $gamma:Irightarrow M$ such that $(gamma^*nabla)dot gamma = 0$.




(Notice that $(gamma^*nabla)dot gamma in Gamma(gamma^*TM)$)



I want to check that according to this definition the geodesics in $mathbb R^n$ with the trivial connection $nabla(f_1frac{partial}{partial x_1}+...+f_nfrac{partial}{partial x_n})=(df_1frac{partial}{partial x_1}+...+df_nfrac{partial}{partial x_n}) $ are straight lines $x+tv$.

Take a path $gamma(t)=(x_1(t),...,x_n(t))$. We have
$$
(gamma^*nabla)dot gamma = (gamma^*nabla)( dot x_1frac{partial}{partial x_1}vert_{gamma(-)} +...+dot x_nfrac{partial}{partial x_n}vert_{gamma(-)} )
$$
But
$$dot x_ifrac{partial}{partial x_i}vert_{gamma(-)} = dot x_igamma^*frac{partial}{partial x_i}vert_{(-)} = gamma^*(dot x_ifrac{partial}{partial x_i}vert_{(-)})
$$
So that, using the definition of pullback connection:
$$
(gamma^*nabla)( sum_i gamma^*(dot x_ifrac{partial}{partial x_i}vert_{(-)})) = gamma^*(sum_i nabla(dot x_ifrac{partial}{partial x_i})) = gamma^*(sum_i frac{ddot x_i}{dt}dtfrac{partial}{partial x_i})
$$
Now I would like to say that this is zero if and only if the $frac{ddot x_i}{dt}$ are zero. If I didn't have the $gamma^*$ that would be clear. But I have no reason to assume that $gamma^*$ is injective. If it were, I would be done.






riemannian-geometry vector-bundles geodesic connections tangent-bundle




riemannian-geometry vector-bundles geodesic connections tangent-bundle






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 26 '18 at 17:08









Soap

1,027615




1,027615












  • Locally, the curve $gamma$ is injective. I don't know if that helps but...
    – Dog_69
    Dec 26 '18 at 19:09










  • @Dog_69 I am not imposing that on my curve
    – Soap
    Dec 26 '18 at 19:25










  • Hummm, if you don't impose that then your curve is just a point.
    – Dog_69
    Dec 26 '18 at 20:47










  • @Dog_69 It just means the curve has self intersections
    – Soap
    Dec 27 '18 at 16:41






  • 1




    Yes, but I said locally. Between two self-intersections, say $gamma(t_0)$ and $gamma(t_1)$ you can find a neighbourhood of $gamma(t)$ where $gamma$ is injective. Do yo know what I mean? Maybe it is totally useless but I thought it was worth mentioning.
    – Dog_69
    Dec 27 '18 at 16:53


















  • Locally, the curve $gamma$ is injective. I don't know if that helps but...
    – Dog_69
    Dec 26 '18 at 19:09










  • @Dog_69 I am not imposing that on my curve
    – Soap
    Dec 26 '18 at 19:25










  • Hummm, if you don't impose that then your curve is just a point.
    – Dog_69
    Dec 26 '18 at 20:47










  • @Dog_69 It just means the curve has self intersections
    – Soap
    Dec 27 '18 at 16:41






  • 1




    Yes, but I said locally. Between two self-intersections, say $gamma(t_0)$ and $gamma(t_1)$ you can find a neighbourhood of $gamma(t)$ where $gamma$ is injective. Do yo know what I mean? Maybe it is totally useless but I thought it was worth mentioning.
    – Dog_69
    Dec 27 '18 at 16:53
















Locally, the curve $gamma$ is injective. I don't know if that helps but...
– Dog_69
Dec 26 '18 at 19:09




Locally, the curve $gamma$ is injective. I don't know if that helps but...
– Dog_69
Dec 26 '18 at 19:09












@Dog_69 I am not imposing that on my curve
– Soap
Dec 26 '18 at 19:25




@Dog_69 I am not imposing that on my curve
– Soap
Dec 26 '18 at 19:25












Hummm, if you don't impose that then your curve is just a point.
– Dog_69
Dec 26 '18 at 20:47




Hummm, if you don't impose that then your curve is just a point.
– Dog_69
Dec 26 '18 at 20:47












@Dog_69 It just means the curve has self intersections
– Soap
Dec 27 '18 at 16:41




@Dog_69 It just means the curve has self intersections
– Soap
Dec 27 '18 at 16:41




1




1




Yes, but I said locally. Between two self-intersections, say $gamma(t_0)$ and $gamma(t_1)$ you can find a neighbourhood of $gamma(t)$ where $gamma$ is injective. Do yo know what I mean? Maybe it is totally useless but I thought it was worth mentioning.
– Dog_69
Dec 27 '18 at 16:53




Yes, but I said locally. Between two self-intersections, say $gamma(t_0)$ and $gamma(t_1)$ you can find a neighbourhood of $gamma(t)$ where $gamma$ is injective. Do yo know what I mean? Maybe it is totally useless but I thought it was worth mentioning.
– Dog_69
Dec 27 '18 at 16:53















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