[Math] Covariant derivative (or connection) of and along a curve

analysiscovariancedifferential-geometrydifferential-topologyreal-analysis

Let $c:[a,b] \rightarrow M$ be a curve parametrized by arc-length.

Then $c': [a,b] \rightarrow TM$ such that $c'(t) \in T_{c(t)}(M).$

So essentially, I want to understand how $\nabla_{c'}c'$ is defined then in coordinates.

Cause normally, we are dealing with vector fields $X : M \rightarrow TM$ when we talk about the covariant derivative or connection.

In this case, we have for $X,Y: M \rightarrow TM$ that $X= \sum_{i} \eta^i \partial_i$ and $Y = \sum_{i} \zeta^i \partial_i$

In this case,

$$\nabla_YX = \sum_{i,j,k} \zeta^i( \eta^j \Gamma_{i,j}^k \partial_k + \partial_i \eta^j \partial_j)$$

and everything is well-defined.

The problem occurs, cause $c' : [a,b] \rightarrow TM$ so there is no way we can take a partial derivative in some coordinate direction of $c',$
as it just depends on time. Is there still a way to explain how $\nabla_c'c'$ is for example defined, although $c'$ is not really a vector field?

Best Answer

HINT: Write $c'(t) = \sum\limits_i a^i(t)\partial_i\big|_{c(t)}$. Then recall that for any vector field $X$ we can compute $\nabla_{c'}X = \dfrac D{dt}(X\circ c)(t)$.

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[Math] Covariant derivative for a covector field

This is a bit too long for a comment, so let me answer your first question.

Let $(\cdot,\cdot)$ denote the pairing of covector fields with vector fields, e.g., $(\lambda,Y) := \lambda(Y)$ for $\lambda$ a covector field and $Y$ a vector field. Think of the scalar field $(\lambda,Y)$ as the product of the covector field $\lambda$ with the vector field $Y$ and think of covariant differentation as a consistent way to define the directional differentiation of objects more complicated than scalar fields, particularly on curved spaces. Then, for all these notions of directional differentiation to be truly consistent, they should satisfy the obvious Leibniz rule with respect to taking the product of a covector field and a vector field, i.e., for any given covector field $\lambda$ and vector field $Y$, $$ X(\lambda,Y) = (\nabla_X \lambda,Y) + (\lambda,\nabla_X Y) $$ for any vector field $X$. What's nice is that this is actually enough to define the covariant derivative $\nabla_X \lambda$ of a covector field $\lambda$ along the vector field $X$, by defining $\nabla_X \lambda$ to be the covector field such that $$ X(\lambda,Y) = (\nabla_X \lambda,Y) + (\lambda,\nabla_X Y), $$ or equivalently, $$ \nabla_X \lambda (Y) = X(\lambda(Y)) - \lambda(\nabla_X Y), $$ for all vector fields $Y$.

Indeed, let $\Sigma \subset \mathbb{R}^3$ denote the image of $f$, let $\{e_1,e_2\}$ be your favourite frame for $T\Sigma$, i.e., a set of vector fields such that for each $p \in U$, $\{e_1(p),e_2(p)\}$ is a basis of the tangent space $T_p \Sigma \subset \mathbb{R}^3$ to $\Sigma$ at $p$, and let $\{e^1,e^2\}$ be the corresponding coframe for $T^\ast \Sigma$, i.e., a set of covector fields such that for each $p \in U$, $\{e^1(p),e^2(p)\}$ is the dual basis of the cotangent space $T^\ast_p \Sigma = (T_p \Sigma)^\ast \subset (\mathbb{R}^3)^\ast$ to $\Sigma$ at $p$ corresponding to the basis $\{e_1(p),e_2(p)\}$ of $T_p \Sigma$. Recall that any covector field $\omega$ can be uniquely written in terms of $\{e^1,e^2\}$ as $$ \omega = (\omega,e_1)e^1 + (\omega,e_2)e^2 = \omega(e_1)e^1 + \omega(e_2)e^2. $$ Then, we can solve Leibniz rule $$ X(\lambda,e_j) = (\nabla_{X}\lambda,e_j) + (\lambda,\nabla_{X}e_j), $$ i.e., $$ X(\lambda(e_j)) = \nabla_{X}\lambda(e_j) + \lambda(\nabla_{X} e_j), $$ for $\nabla_{X}\lambda(e_j)$ to get $$ \nabla_{X} \lambda(e_j) = X(\lambda(e_j)) - \lambda(\nabla_{X} e_j), $$ which therefore forces $$ \nabla_{X} \lambda = (X(\lambda(e_1)) - \lambda(\nabla_{X} e_1))e^1 + (X(\lambda(e_2)) - \lambda(\nabla_{X} e_2))e^2. $$ In particular, we find that $$ \nabla_{e_i} \lambda = (e_i(\lambda(e_1)) - \lambda(\nabla_{e_i} e_1))e^1 + (e_i(\lambda(e_2)) - \lambda(\nabla_{e_i} e_2))e^2, $$ which is exactly what you learnt in class.

[Math] Covariant derivative and geodesic

Observe that through the map $$\left(\begin{array}{c} u^1\\ u^2 \end{array}\right) \stackrel{f}\to \left(\begin{array}{c} x^1\\ x^2\\ x^3 \end{array}\right), $$ you have a GPS for points in the surface. It is with this that you get $\frac{\partial f}{\partial u^1}$ and $\frac{\partial f}{\partial u^2}$, which are the two columns of the jacobian $Jf$, and serve to define the tangent space at any point on the surface.

So to travel along a curve in the surface you need to travel along a curve in the domain $U$. This implies that $u^1,u^2$ depends on a parameter, which we can denote with $t$.

Schematically $$t\stackrel{\gamma}\to \left(\begin{array}{c} u^1(t)\\ u^2(t) \end{array}\right) \stackrel{f}\to \left(\begin{array}{c} x^1(t)\\ x^2(t)\\ x^3(t) \end{array}\right), $$ will give the way to control the positions on curve over the surface.

Now $\dot{c}=\frac{d}{dt}(f\circ\gamma)=Jf\cdot\dot{\gamma}$ and since $\dot{\gamma}=\dot{u^1}e_1+\dot{u^2}e_2$ then $$\dot{c}=Jf(\dot{u^1}e_1+\dot{u^2}e_2)=\dot{u^1}Jf(e_1)+\dot{u^2}Jf(e_2),$$ so $$\dot{c}=\dot{u^1}\frac{\partial f}{\partial u^1}+\dot{u^2}\frac{\partial f}{\partial u^2}.$$

This way you can use it in $\nabla_{\dot{c}}\dot{c}$ to get $$\nabla_{\dot{c}}\dot{c}=\sum_s(\ddot{u^s}+\sum_{ij}\Gamma^s{}_{ij}\dot{u^i}\dot{u^j})\frac{\partial f}{\partial u_s}.$$

This last formula is get as follow: $$\nabla_{\dot{c}}\dot{c}= \nabla_{\sum_s\dot{u^s}\frac{\partial f}{\partial u_s}} \sum_{\sigma}\dot{u^{\sigma}}\frac{\partial f}{\partial u_{\sigma}},$$

$$\quad=\sum_{s{\sigma}}\dot{u^s}\nabla_{\frac{\partial f}{\partial u_s}} \dot{u^{\sigma}}\frac{\partial f}{\partial u_{\sigma}},$$

$$\qquad=\sum_{s{\sigma}}\dot{u^s} \frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}}\dot{u^s}\dot{u^{\sigma}}\nabla_{\frac{\partial f}{\partial u_s}}\frac{\partial f}{\partial u_{\sigma}},$$

$$\qquad=\sum_{s{\sigma}}\dot{u^s} \frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}r}\dot{u^s}\dot{u^{\sigma}}\Gamma^r{}_{s{\sigma}}\frac{\partial f}{\partial u_r},$$

$$\qquad=\sum_{s{\sigma}} \frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{du^s}{dt}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}r}\dot{u^s}\dot{u^{\sigma}}\Gamma^r{}_{s\sigma}\frac{\partial f}{\partial u_r},$$

$$\qquad=\sum_{{\sigma}} \frac{d^2u^s}{dt^2}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s\sigma r}\dot{u^s}\dot{u^r}\Gamma^{\sigma}{}_{sr}\frac{\partial f}{\partial u_{\sigma}},$$

$$\qquad=\sum_{{\sigma}}\left( \frac{d^2u^{\sigma}}{dt^2}+\sum_{sr}\dot{u^s}\dot{u^r}\Gamma^{\sigma}{}_{sr}\right) \frac{\partial f}{\partial u_{\sigma}}.$$

Best Answer

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[Math] Covariant derivative for a covector field

[Math] Covariant derivative and geodesic

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