In the lecture we had the immersion $f: U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$.

Now we said that a map $\lambda : U \rightarrow T^*f$ is a covector field. Okay, that's fine.

Also, we defined the covariant derivative for vector fields $v : U \rightarrow Tf$ which we denoted by $\nabla_X v$, where $X$ was either also a vector field or a vector in the tangential space $T_pU$.

For $X = \sum_{i=1}^{2} \xi_i f_i$ and $Y = \sum_{i=1}^{2} \eta_i f_i$ (both vector fields) the covariant derivative is defined by

$$\nabla_X Y := \sum_{i,k} \xi_i \left( \partial_i \eta_k + \sum_j \eta_j \Gamma_{ij}^{k} \right) f_i $$

I guess that it is pretty clear, that this is by definition also a map $\nabla_XY : U \rightarrow Tf.$ Thus, it is a vector field.

Then we tried to explain what a covariant derivative for a covector field is.

First, we noted that a covariant derivative of a covector field $\lambda$ is still taken along a vector field $X$ and is supposed to give us a map

$$\nabla_X \lambda : U \rightarrow T^*U,$$ so we get again a covector field. (Actually, we can identify $T^*f$ with $T^*U$, so although we would have expected $T^*f$, this does not really matter)

But I have troubles to understand the definition.

In our lecture we defined $$\nabla_{e_i} \lambda(e_j)= \partial_i (\lambda (e_i))- \lambda ( \nabla_{e_i}(e_j)),$$

which seems to be nonsense. I mean $e_j$ is NOT an element of $U$, so we cannt apply $\nabla_{e_i} \lambda$ to this element.

Also, the wikipedia definition does not seem to be more meaningful.

You can find it here

This was the first question: How do we properly define the covariant derivative of covector fields in this context?

The second question is: Apparently, there is also some ambiguity between vector fields that map from $X : U \rightarrow Tf$ or vector fields that map from $X : U \rightarrow TU$. (of course it does not really matter, but maybe something here is the more canonical definition). It occurs to me that our definition suggests that the covariant derivative for vector fields and vector fields itself map from $ U \rightarrow Tf$. Is this really the canonical definition? But apparently, we wanted to use the other notion for covector fields ( as maps $\lambda: U \rightarrow T^*U$), is this true?

As I am really a beginner in differential geometry, I am not familiar with Riemannian manifolds or anything harder than euclidean submanifolds. Therefore, since this seems to be a problem in differential geometry, I can only accept answers that do not use new definitions or symbols that are not explained in the answer. Thus, I would highly appreciate it, if you would keep it simple.

  • $\begingroup$ take a look juanmarqz.wordpress.com/2010/11/27/… for a version $\endgroup$ – janmarqz Dec 16 '14 at 0:48
  • $\begingroup$ 1) there are to cotangent bundles $T^*U\to U$ and $T^*fU\to fU$, then you can take sections in both bundle to give covectors fields on $U$ and on $fU$ $\endgroup$ – janmarqz Dec 18 '14 at 20:07
  • $\begingroup$ 2) to get a subbundle $\xi$ in $T^*U$ related to $f$ there is a construction called pullback to get $\xi=f^*(T^*fU)$ $\endgroup$ – janmarqz Dec 18 '14 at 20:13
  • $\begingroup$ 3) where in this two bundles do you what to do $\nabla_X\omega$? where $X$ is a vector field and $\omega$ is a covector field $\endgroup$ – janmarqz Dec 18 '14 at 20:20
  • $\begingroup$ @janmarqz never heard the word bundle, isn't there a more basic explanation? ( I am only considering a surface in $\mathbb{R}^3$, which is not fancy at all) $\endgroup$ – user159356 Dec 18 '14 at 20:50

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.

  • $\begingroup$ I mean, the problem is that I don't see what $\lambda (Y) $ should be. $Y$ maps from $U$ to $Tf$ and $\lambda : U \rightarrow Tf^*$, what is $\lambda(Y)$ then? $\endgroup$ – user159356 Dec 28 '14 at 20:32
  • $\begingroup$ At each point $p$, $Y(p)$ is a vector in the vector space $T_p \Sigma$, whilst $\lambda(p)$ is a vector in the dual space $T^\ast_p \Sigma := (T_p \Sigma)^\ast$ of $T_p \Sigma$, i.e., is a linear transformation $\lambda(p) : T_p \Sigma \to \mathbb{R}$. Thus, $\lambda(Y)(p) := \lambda(p)(Y(p))$ is scalar. $\endgroup$ – Branimir Ćaćić Dec 28 '14 at 21:38
  • $\begingroup$ so wikipedia says that we have $(\nabla_{\mathbf v}\alpha)({\mathbf u})=\nabla_{\mathbf v}(\alpha({\mathbf u}))-\alpha(\nabla_{\mathbf v}{\mathbf u}).$ and now you say that this term after the equal sign is the same as $v(\alpha(u))$ but what does this mean again? or your $X( \lambda (e_j))$? I mean $X$ takes vectors in $U$, but $\lambda(e_j)$ is a real number. $\endgroup$ – user159356 Dec 29 '14 at 18:05
  • $\begingroup$ For each $p$, $\lambda(e_j)(p) = \lambda(p)(e_j(p))$ is a scalar, so that $\lambda(e_j) : p \mapsto \lambda(p)(e_j(p))$ defines a scalar field, i.e., a real valued function, so that you can form its directional derivative $X(\lambda(e_j)) = \nabla_X(\lambda(e_j))$ along the vector field $X$. $\endgroup$ – Branimir Ćaćić Dec 29 '14 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy