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Let $f:U \to \mathbb{R}^3$ be a surface with local coordinates $f_i=\frac{\partial f}{\partial u^i}$. Let $\omega$ be a one-form. I want to express $\nabla \omega$ in terms of local coordinates and Christoffel symboles. Where $\nabla$ is the Levi-Civita connection (thus it coincides with the covariant derivative).

Let $X=X^if_i,Y=Y^if_i$ be two tangent vector fields on $f$, by definition one has: $$(\nabla_X \omega)(Y)=X^i \frac{\partial (\omega(Y))}{\partial u^i}-\omega (\nabla_XY)=X^i \frac{\partial (\omega(Y^if_i))}{\partial u^i}-\omega(\nabla_{X^if_i} Y^if_i)$$

One has $$\omega(\nabla_{X^if_i} Y^if_i)=\omega (X^i[Y^{j}(\nabla_{f_i}f_j)+Y^{j}_{,i}f_j])=\omega (X^i[Y^{j}(\Gamma^{k}_{ij}f_{k})+Y^{j}_{,i}f_j])=X^{i}Y^{j}\Gamma^{k}_{ij} \omega (f_k)+X^iY^i_{,i} \omega_{j}$$

now to compute $\frac{\partial (\omega(Y^jf_j))}{\partial u^i}=Y^{j}_{,i} \omega_{j}+Y^{j} \omega_{j,i}$,where $\omega_{i}=\omega (f_i)$.

Therefore $(\nabla_X \omega)(Y)=X^iY^j(\omega_{j,i}-\Gamma^{k}_{ij} \omega_k)$

Is there a way to simplify further?

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"Since $\omega$ is linear, the derivative of $\omega$ is itself". That is absolutely incorrect. A one form field $\omega$ can be characterised by functions $\omega_i(u)$ where $\omega_i = \omega(f_i)$ the action of the one form on the coordinate vector fields. The derivative of a one-form will in general depend on the coordinate derivatives of the coordinate components of the one-form! –  Willie Wong Sep 25 '12 at 9:25
    
Also, the convariant derivative being tensorial, means that $(\nabla_X\omega)(Y)$ should not depend on the coordinate derivatives of $Y$, but only on the value of $Y$ itself. This gives a further quick check that the result of your computations are incorrect. –  Willie Wong Sep 25 '12 at 9:27
    
$\omega(Y) = \sum \omega_i Y^i$ where $\omega_i = \omega(f_i)$. Right there is where the linearity of the one-form is used. So $\partial(\omega(Y)) = \sum_{i} Y^i (\partial\omega_i ) + \omega_i(\partial Y^i)$ etc. –  Willie Wong Sep 25 '12 at 10:41
    
Why are there four $i$s in the first term? Check your summation indices again! (Hint: the first term cancels with the last term.) –  Willie Wong Sep 25 '12 at 15:27
    
Thanks! Corrected now. Just wonder for a covariant 2-tensor $S(-,-)$ can one differentiate it similar to that of a one-form? namely $\partial (S(X,Y))$ depends on $\partial (S(f_i,f_j))=\partial (S(i,j))$? –  user31899 Sep 25 '12 at 19:50

1 Answer 1

up vote 3 down vote accepted

To answer your stated questions: Yes, you've now done the computations correctly, and no, that's about as simple as you can have the expression. One may take advantage of the tensorial nature of the covariant derivative and write $$ \nabla_i \omega_j = \partial_i \omega_j - \Gamma^k_{ij}\omega_k $$ using a bit of a sloppy mixture of abstract index notation with coordinate index notations, but that is about it.


To answer your question in the comments, we introduce the abstract notion of a derivation on the tangent bundle of a (smooth) manifold $M$. Let $\mathfrak{T}^{r,s}(M)$ denote the $\mathbb{R}$-vector space of smooth $(r,s)$ tensors over $M$. A tensor derivation is defined to be a $\mathbb{R}$-linear map $\mathscr{D}:\mathfrak{T}^{r,s}(M)\to \mathfrak{T}^{r,s}(M)$ defined for all pairs $(r,s)\in \mathbb{Z}_{\geq 0}^2$ satisfying the following conditions:

  1. If $A$ and $B$ are smooth $(r,s)$ and $(t,u)$ tensor fields respectively, we have that $\mathscr{D}(A\otimes B) = \mathscr{D}A\otimes B + A\otimes \mathscr{D}B$ (in other words, the Leibniz rule holds for tensor products).
  2. If $A$ is a smooth $(r,s)$ tensor field, and $\mathfrak{C}:\mathfrak{T}^{r,s}(M) \to \mathfrak{T}^{r-1,s-1}(M)$ is a tensor contraction, then we have $\mathscr{D}(\mathfrak{C}A) = \mathfrak{C}(\mathscr{D}A)$. (It commutes with tensor contractions.)

It is a theorem that for $r = s = 0$, any $\mathbb{R}$-linear map on $\mathfrak{T}^{0,0}(M)$ (the space of smooth functions) that satisfies the Leibniz rule (the second condition does not apply since there are no tensor contractions applicable to pure functions) can be represented as the directional derivative relative to a vector field. So we can add the unessential third condition

  • There exists a smooth vector field $V$ such that for all smooth functions $f$ over $M$, $V(f) = \mathscr{D}f$.

Remark: This natural one-to-one correspondence between derivations and smooth vector fields allow us to also interpret a derivation as a map from $\mathfrak{T}^{1,0}(M)$, the space of smooth vector fields, into the space of $\mathbb{R}$-linear maps on smooth tensor fields that satisfies conditions 1 and 2. This manifests in the notation $\nabla_X Y$ for covariant differentiation and $\mathcal{L}_X Y$ for Lie differentiation.

In any case, just given the above three conditions are not enough to specify the derivation. But we just need one more ingredient: the action of $\mathscr{D}$ on either $\mathfrak{T}^{1,0}$ or $\mathfrak{T}^{0,1}$. I'll sketch the case of $\mathfrak{T}^{1,0}(M)$.

Suppose we know what $\mathscr{D}X$ is for all smooth vector fields. Then for one-forms $\omega$, we consider the contraction $\mathfrak{C}(\omega\otimes X)$ which we probably more commonly write as $\omega(X)$ and is a function. So using the axioms of the derivation, we have

$$ \mathscr{D}[\omega(X)] = (\mathscr{D}\omega)(X) + \omega(\mathscr{D}X) $$

this is an algebraic equation in which all objects are known except for one: we know what $\omega(X)$ is given $\omega$ and $X$, we know what $\mathscr{D}X$ is by assumption, and so also $\omega(\mathscr{D}X)$. We know that $\mathscr{D}[\omega(X)]$ is since $\omega(X)$ is a function. Hence we can solve the algebraic equation to get a formula for $\mathscr{D}\omega$, using that $\mathfrak{T}^{1,0}(M)$ and $\mathfrak{T}^{0,1}$ are duals so that to specify $\mathscr{D}\omega$ it suffices to specify $(\mathscr{D}\omega)(X)$ for all $X$.

Similarly, now given an arbitrary tensor field $\Xi\in \mathfrak{T}^{r,s}(M)$, we can do the same procedure and consider

$$ \mathscr{D}\left[ \mathfrak{C}_1\mathfrak{C}_2\cdots\mathfrak{C}_{r+s} \Xi \otimes X_1\otimes X_2\otimes\cdots\otimes X_s\otimes \omega_1\otimes\cdots\otimes \omega_r\right] $$

the full contraction of $\Xi$ with $s$ arbitrarily chosen vector fields and $r$ arbitrarily chosen one forms. Expanding the above expression using the axioms of the derivation, we are left with only one unknown: $\mathscr{D}\Xi$. Everything else $\mathscr{D}X_n$, $\mathscr{D}\omega_n$ etc are all computable from the axioms, the vector field $V$ (of the third axiom), and the assumed knowledge of $\mathscr{D}X$ for all vector fields $X$.


For illustration, we can verify that the above construction is indeed "tensorial". Given that

$$ \mathscr{D}\omega(X) = \mathscr{D}(\omega\cdot X) - \omega(\mathscr{D}X) $$

one may wonder whether $\mathscr{D}\omega$ is indeed a one-form: that is, whether $\mathscr{D}\omega(fX) = f\mathscr{D}\omega(X)$ for $f$ a smooth function. We can directly compute:

$$ \mathscr{D}\omega(fX) = \mathscr{D}(\omega\cdot fX) - \omega(\mathscr{D}(fX)) = \mathscr{D}[f(\omega\cdot X)] - \omega\left( \mathscr{D}f X + f \mathscr{D}X\right) = \mathscr{D}f (\omega\cdot X) + f \mathscr{D}(\omega\cdot X) - f \omega(\mathscr{D}X) - \omega(\mathscr{D}f X) $$

Noting that $\mathscr{D}f(\omega\cdot X) = \omega(\mathscr{D}f X)$ by the tensorial property of $\omega$ as a one form, we see that indeed the Leibniz-rule + contraction rule allows us to make sure that $\mathscr{D}\omega$ (and hence $\mathscr{D}\Xi$ for an arbitrary tensor field $\Xi$) is tensorial.


Lastly, what does this have to do with connection coefficients? All this mucking about with the Christoffel symbols is just a way of saying: we know what $\mathscr{D}X$ is. That is, consider the derivation labeled $\nabla_Y$, which acts on scalar fields as the vector field $Y$ with coordinate expansion $Y^i\partial_i$. For an arbitrary vector field $X$ given in local coordinates $X^i \partial_i$, we demand that the local coordinate expression of $(\nabla_Y X)^i \partial_i$ be given by

$$ (\nabla_Y X)^i = Y^j\partial_j X^i + \Gamma^i_{jk}X^k Y^j $$

and we carry on from there.

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Thank you very much for clarifications! Our lecturer briefly introduced us about definitions and basic properties of tensor fields yesterday (He apologized in the class yesterday for giving such problem which requires superior knoledge on tensor fields). I'll incorporate your explainations to the appendix of my lecture notes. –  user31899 Sep 26 '12 at 22:55

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