# Connections and forms

1. Nov 30, 2014

### mikeeey

Hello every one .
Can someone please prove me the TORSION form and the CURVATURE form ( Maurer-Cartan form)
in details, and what is the canonical solder 1-form?!

2. Dec 2, 2014

### Drakkith

Staff Emeritus
That sounds like quite a bit of work. How far have you gotten on the problem?

3. Dec 2, 2014

### mikeeey

well i know how . but there is a extra term in both of these tensor . it appears because of non-holomonic basis . i want to know how to get that term .

4. Dec 2, 2014

### Drakkith

Staff Emeritus
Well, my math skills are far below what's needed here, but perhaps you can show exactly where you're having trouble and someone more knowledgeable than myself may be able to help you.

5. Dec 2, 2014

### mikeeey

for example . the general Torsion tensor in non-holomonic basis is .
$\T^\a_\bc,\ = \Gamma^\a_\bc,\ + \Gamma^\a_\cb,\ + \gamma^\a_\bc,\$
the term
$\gamma^\a_\bc,\$
how can I find it from the derivation ?!

6. Dec 2, 2014

### Terandol

You haven't really asked a question we can answer yet. What is it you want to see proven about the torsion tensor or the curvature tensor and where are we supposed to start from? If you want to see a derivation of the components of the tensor what is the definition of the torsion tensor you are using? The derivation will look somewhat different if for example we start out with the torsion form defined as the covariant exterior derivative of the canonical one form (by the way this is the standard example of a solder form) on the frame bundle than if we define it's action on vector fields directly so you need to specify where you want the derivation to start.

The most common definition of the torsion tensor is probably thinking of $T\in \Gamma(TM\otimes\Lambda^2 T^*M)$ and defining it's action on vector fields by $$T(X,Y)=\nabla_X(Y)-\nabla_Y(X)-[X,Y].$$ In this case, to derive the formula in your post choose some local coordinates $(U,x^i)$ so that we have a frame of the tangent bundle on $U$ given by $\frac{\partial}{\partial_{x^1}},\cdots, \frac{\partial}{\partial_{x^n}}$. Define the $\gamma^a_{bc}$ by $\left[\frac{\partial}{\partial x^b},\frac{\partial}{\partial x^c}\right]= \gamma^a_{bc}\frac{\partial}{\partial x^a}$. Now just plug the basis vectors into $T$ and essentially directly from the definition, the first two terms of $T\left(\frac{\partial}{\partial x^b},\frac{\partial}{\partial x^c} \right)$ involving the covariant derivative are $\Gamma^a_{bc} \frac{\partial}{\partial x^a}-\Gamma^a_{cb}\frac{\partial}{\partial x^a}$ and the last term involving the commutator gives $-\gamma^a_{bc} \frac{\partial}{\partial x^a}$ yielding the components $T^a_{bc}=\Gamma^a_{bc} -\Gamma^a_{cb}- \gamma^a_{bc}$ (note the minus sign in the second and third terms which which you did not have in your post.)

To get a formula for the components of the curvature tensor, do a similar thing by starting out with some invariant definition of the two form and plugging in the various basis elements to extract the components explicitly.

Last edited: Dec 2, 2014
7. Dec 2, 2014

### mikeeey

Thank You my friend Terandol for the explanation , but u did not derive me the the two tensors quite write using CALCULUS , u just wrote the tensor formula,
here is the derivation in [ (holomoonic )COORDINATE BASIS ]

NOW can someone please derive me the two tensor like this way in orthonormal ( non-holomonic coordinate basis) to get the last term in each tensor ,
Terandol u used holomonic basis so the lie brackets will give zero in fact

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8. Dec 3, 2014

### Terandol

Well yes, I chose a frame of the tangent bundle coming from a coordinate chart, which implies that the Lie derivative vanishes, since the conventions I am familiar with reserve the symbol $\Gamma^a_{bc}$ only for these types of bases. You can simply replace the special frame $\left\{\frac{\partial}{\partial x^b} \right\}$ I chose in my previous post with an arbitrary local frame $\{e_i \}$ of the tangent bundle and the identical argument works but now the symbols $\Gamma^a_{bc}$ are usually not called Christoffel symbols and I've always seen them denoted by different symbols.

In any case I believe you are asking about the definition of the torsion tensor not proving anything about it. In fact the derivation you posted is already a complete general definition of the torsion tensor. Defining a tensor in every coordinate coordinate chart of some atlas for your manifold completely defines the tensor and you don't need to even consider the existence of nonholonomic coordinate bases when giving the definition. So we have a tensor defined in local charts by $T^i_{jk}=\Gamma^i_{jk}-\Gamma^i_{kj}$ and your question is simply asking what form this tensor takes when you write it in terms of some basis which did not come from a chart.

From this definition, if we have some (possibly nonholonomic) local frame $\{ e_1,\cdots, e_n\}$, you can derive the equation $T(e_b,e_c)=\left(\Gamma^a_{bc}-\Gamma^a_{cb}-\gamma^a_{bc}\right)e_a$ by choosing a chart $(U,x^i)$ which overlaps the domain of our local frame around any given point and then expanding the basis elements as $e_b=\sum_{\alpha=1}^n K_b^\alpha \frac{\partial}{\partial x^\alpha}$. I don't have time to type up the complete computation since it is a bit tedious but there are no tricks, just direct computation.

Just to get you started, by definition $T^a_{bc}e_a =T(e_b,e_c) =T\left( K_b^j \frac{\partial}{\partial x^j}, K_c^k \frac{\partial}{\partial x^k}\right) =K^j_b K_c^kT\left( \frac{\partial}{\partial x^j},\frac{\partial}{\partial x^k}\right) =K^j_b K_c^k\left( \Gamma^i_{jk}-\Gamma^i_{kj} \right) \frac{\partial}{\partial x^i}.$

Now do a similar thing using the inverse of the matrix $K_b^\alpha$ and the definition of the $\Gamma$'s in terms of covariant derivatives to write the terms of the form $\Gamma^k_{ij}$ in terms of $\Gamma^a_{bc}$. Substituting this back into the equation for $T(e_b,e_c)$ will result in the formula you are trying to derive. The Lie derivative of vector fields appears in the computation since you will need to use the Leibniz rule to compute the covariant derivative at some point which will provide the extra terms.

9. Dec 3, 2014

### mikeeey

Thank you very much terandol for the beautiful explanation . but can you please complete the computations ,
Im sorry if i made feeling tired of this explanation .

Thank you

10. Dec 3, 2014

### Terandol

I should warn you that I almost certainly made some indexing errors in following.

Write $e_b=K^i_b\frac{\partial}{\partial x^i}$ and $\frac{\partial}{\partial x^i}=L^b_i e_b$ so that $K^j_bL_j^\beta=\delta^\beta_b$ since these are inverse
matrices. Then
\begin{align*} \Gamma^i_{jk}\frac{\partial}{\partial x^i} &=\nabla_{\frac{\partial}{\partial x^j}} \left(\frac{\partial}{\partial x^k} \right)\\ &=\nabla_{L^\beta_j e_\beta} (L^\gamma_k e_\gamma)\\ &=L^\beta_j\nabla_{e_\beta}(L^\gamma_ke_\gamma)\\ &=L^\beta_j\left( e_\beta(L^\gamma_k)e_\gamma +L^\gamma_k \nabla_{e_\beta}e_\gamma \right)\\ &=L^\beta_je_\beta(L^\gamma_k)e_\gamma +L^\beta_jL^\gamma_k \Gamma^\alpha_{\beta\gamma}e_\alpha. \end{align*}
Plug this back into the equation and use that $K$ and $L$ are inverse matrices
to get
\begin{align*} T(e_b,e_c) &=K^j_bK^k_c\left(\Gamma^i_{jk}-\Gamma^i_{kj}\right) \frac{\partial}{\partial x^i}\\ &=K^j_bK^k_c\left( L^\beta_je_\beta(L^\gamma_k)e_\gamma +L^\beta_jL^\gamma_k \Gamma^\alpha_{\beta\gamma}e_\alpha -L^\beta_ke_\beta(L^\gamma_j)e_\gamma -L^\beta_kL^\gamma_j \Gamma^\alpha_{\beta\gamma}e_\alpha \right)\\ &=\left(\Gamma^\alpha_{bc}-\Gamma^\alpha_{cb}\right)e_\alpha + \left( K^k_ce_b(L^\gamma_k)-K^j_b e_c(L^\gamma_j) \right) e_\gamma \\ &=\left(\Gamma^\alpha_{bc}-\Gamma^\alpha_{cb} +K^k_ce_b(L^\alpha_k)-K^j_b e_c(L^\alpha_j) \right) e_\alpha. \end{align*}

Finally we need to show that the last two terms really are the Lie derivative.
To do this use the fact that $0=e_b(\delta^\alpha_c)=e_b(K^k_cL^\alpha_k) =K^k_ce_b(L^\alpha_k)+L^\alpha_ke_b(K^k_c)$ so $K^k_ce_b(L^\alpha_k)=-L^\alpha_ke_b(K^k_c)$. Then

\begin{align*} \left(K^k_ce_b(L^\alpha_k)-K^j_b e_c(L^\alpha_j) \right) e_\alpha &=e_c(K^j_b)L^\alpha_je_\alpha-e_b(K^k_c)L^\alpha_ke_\alpha\\ &=e_c(K^j_b)\frac{\partial}{\partial x^j} -e_b(K^k_c)L^\alpha_ke_\alpha\frac{\partial}{\partial x^k}\\ &=\left(e_c(K^j_b)-e_b(K^j_c)\right)\frac{\partial}{\partial x^j}\\ &=[e_c,e_b]\\ &=-[e_b,e_c]\\ &=-\gamma^\alpha_{bc}e_\alpha \end{align*}

(The fourth line above follows from the third line since $e_b=K^i_b\frac{\partial}{\partial x^i}$ and $e_c=K^i_c\frac{\partial}{\partial x^i}$ so $[e_c,e_b]=(e_c(K^i_b)-e_b(K^i_c))\frac{\partial}{\partial x^i}$.)

This gives us the result since we have shown
$$T^a_{bc}e_\alpha =T(e_b,e_c) =\left(\Gamma^a_{bc}-\Gamma^\alpha_{cb}-\gamma^\alpha_{bc}\right) e_\alpha$$
which proves $T^a_{bc}=\Gamma^a_{bc}-\Gamma^\alpha_{cb}-\gamma^\alpha_{bc}$.

Last edited: Dec 3, 2014
11. Dec 3, 2014

### mikeeey

Thank you very very very much Terandol , i followed your account .

THANK YOU .

12. Dec 3, 2014

### Terandol

In case anyone reads this thread, there is an obvious minor error in my previous post.

I wrote
\begin{align*} \left(K^k_ce_b(L^\alpha_k)-K^j_b e_c(L^\alpha_j) \right) e_\alpha &=e_c(K^j_b)L^\alpha_je_\alpha-e_b(K^k_c)L^\alpha_ke_\alpha\\ &=e_c(K^j_b)\frac{\partial}{\partial x^j} -e_b(K^k_c)L^\alpha_ke_\alpha\frac{\partial}{\partial x^k}\\ \end{align*}

but of course the last term of the last line should just be $e_b(K^k_c)\frac{\partial}{\partial x^k}$ not $e_b(K^k_c)L^\alpha_ke_\alpha\frac{\partial}{\partial x^k}.$ This mistake does not affect any of the rest of the derivation though since the next line is correct in my previous post.

13. Dec 3, 2014

### mikeeey

YES , its ok Terandol , Thank You
now i'm trying to do this computation for the curvature tensor , but i think it would be a lot of items