How many independent components does the Riemann curvature tensor have?

[latentcorpse]

(i) show that $R_{abcd}+R_{cdab}$
(ii) In n dimensions the Riemann tensor has $n^4$ components. However, on account of the symmetries
$R_{abc}^d=-R_{bac}^d$
$R_{[abc]}^d=0$
$R_{abcd}+-R_{abdc}$
not all of these components are independent. Show that the number of independent components is $\frac{n^2(n^2-1)}{12}$

not really sure how to go about this.
i think (i) follows from the 3 properties above but i can't prove it. also i don't understand what $R_{[abc]}^d$ means, in particular the [abc] part. is this a Lie bracket? (i haven't covered these yet) so could someone explain what this is about.

thanks.

 

[CompuChip]

You can prove (i) using the symmetries, after moving the index down using the metric tensor.

The square brackets usually stand for anti-symmetrization, e.g.
$$T_{[ab]} = \tfrac12(T_{ab} - T_{ba})$$
$$T_{[abc]} = \tfrac13(T_{abc} - T_{bac} + T_{bca} )$$
and in general
$$T_{[\mu_1 \mu_2 \cdots \mu_n]} = \frac{1}{|S_n|} \sum_{\sigma \in S_n} \left( \text{sign}(\sigma) T_{\sigma(\mu_1\mu_2\cdots\mu_n)} \right)$$
where S_n is the permutation group of n elements (|S_n| = n!) and sign(...) denotes the sign (+ or -) of the permutation.

 

[latentcorpse]

hmmm.

couple fo things:
(i) what happened with my LaTeX? e.g. $R_{abc}^{d}$. how do i get the d to come after abc?
(ii) to prove that $R_{abcd}=R_{cdab}$ how do i switch the ab with the cd?
(iii) i don't understand your anti symmertrisation thing. surely for the n=3 case, [itex]|S_n|=3!=6 and so it should be 1/6 and then there will be 6 terms in the brackets? also how do u decide the sign of a permutation?

 

[CompuChip]

(i) $$R_{abc}{}^d$$ (click to see source)

(ii) I'm not sure about the proof anymore. Probably you start from the second property, because that is the only one that can interchange a and c, for example.

(iii) My apologies for going too fast. Indeed you get 1/6( u - v + w - x + y - z ), but using the symmetry in a <-> b, they combine in pairs (e.g. u = -v, w = -x, y = -z so you get 1/6(2u + 2v + 2y) = 1/3(u + v + y) -- work it out.
Permutations are important, read up on them. But as a quick answer: the sign of a permutation is the number of interchangings of adjacent elements you need. So for example, starting from abc: the permutation that gives bac has sign -1, because you just interchange a with b. The permutation that gives bca has sign +1, because you need to interchangings (for example, first a with b then a with c).

 

[latentcorpse]

thanks. got it now. and i found a proof of (i) also.

however I'm stuck on another proof.

i'm trying to show that $R_{[abc]}{}^d=0$
the proof says:
Note for an arbitrary dual vector field $\omega_a$ and any derivative operator $\nabla_a$ we have $\nabla_{[a} \nabla_b \omega_{c]}=0$
this equation can be proven from
$\nabla_a T^{b_1 . . . b_k}{}_{c_1 . . . c_l}= \tilde{\nabla_a} T^{b_1 . . . b_k}{}{c_1 . . . c_l} + \sum_i C^{b_i}{}_{ad} T^{b_1 . . d . . b_k}{}_{c_1 . . . c_l} - \sum{j} C^{d}_{a c_j} T^{b_1 . . . b_k}{}_{c_1 . . d . . c_l}$
where the ordinary derivative $\partial_a$ has been substituted for $\tilde{\nabla_a}$ and we have made use of the commutativity of ordinary derivatives and the symmetry of $C^{c}{}_{ab}=\Gamma^{c}{}_{ab}$. In differential forms notation this statement is $d^2 \omega=0$. Thus,
$0=2 \nabla_{[a} \nabla_{b} \omega_{c]} = \nabla_{[a} \nabla_{b} \omega_{c]} - \nabla_{[b} \nabla_a \omega_{c]} = R_{[abc]}{}^d \omega_d \forall \omega_d$
$\Rightarrow R_{[abc]}{}^d=0$

now i have quite a lot of problems following this:
(i) terminology: it talks about $\omega_d$ as a dual vector field. now there are two types of dual space as far as i know - algebraic dual space and continuous dual space. if its the algebraic one then we can refer to it as a 1 form. my question here is basically which is it? am i entitled to call it a 1 form?

(ii)ok so i have no idea how to get to $\nabla_{[a} \nabla_b \omega_{c]}=0$
using the method they descirbe.

(iii) i assume that since $\nabla_{[a} \nabla_b \omega_{c]}=0$, they just double both sides to get $2 \nabla_{[a} \nabla_b \omega_{c]}=0$ but then why is
$2 \nabla_{[a} \nabla_b \omega_{c]}= \nabla_{[a} \nabla_b \omega_{c]} - \nabla_{[b} \nabla_a \omega_{c]}$?

thanks

 

[latentcorpse]

what happened to the LaTeX?

 

[cristo]

what happened to the LaTeX?

use \nabla and \tilde{}

 

[latentcorpse]

thanks. does that make any sense to you now?

 

[gabbagabbahey]

(i) terminology: it talks about $\omega_d$ as a dual vector field. now there are two types of dual space as far as i know - algebraic dual space and continuous dual space. if its the algebraic one then we can refer to it as a 1 form. my question here is basically which is it? am i entitled to call it a 1 form?

I'd assume it belongs to an algebraic dual space.

(ii)ok so i have no idea how to get to $\nabla_{[a} \nabla_b \omega_{c]}=0$
using the method they descirbe.

Start with the definition of $\nabla_{[a} \nabla_b \omega_{c]}$ and work out the covariant derivatives.

(iii) i assume that since $\nabla_{[a} \nabla_b \omega_{c]}=0$, they just double both sides to get $2 \nabla_{[a} \nabla_b \omega_{c]}=0$ but then why is
$2 \nabla_{[a} \nabla_b \omega_{c]}= \nabla_{[a} \nabla_b \omega_{c]} - \nabla_{[b} \nabla_a \omega_{c]}$?

thanks

That amounts to saying that 2 times zero equals zero minus zero...do you have a problem with that statement?

 

[latentcorpse]

$\nabla_{[a} \nabla_b \omega_{c]}= \frac{1}{3!}[\nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c + \omega_c \nabla_a \nabla_b - \omega_c \nabla_b \nabla_a + \nabla_b \omega_c \nabla_a - \nabla_a \omega_c \nabla_b]$ I am not sure how much sense the terms make where $\omega_c$
isn't the third term though?
also, could you run through how to actually evaluate one of these derivatives?
thanks.

 

[gabbagabbahey]

$\nabla_{[a} \nabla_b \omega_{c]}= \frac{1}{3!}[\nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c + \omega_c \nabla_a \nabla_b - \omega_c \nabla_b \nabla_a + \nabla_b \omega_c \nabla_a - \nabla_a \omega_c \nabla_b]$ I am not sure how much sense the terms make where $\omega_c$
isn't the third term though?

Clearly, since $\omega_c$ isn't always the last term in the product, $\nabla_{[a} \nabla_b \omega_{c]}$ is a differential operator itself...So to prove that it's zero you need to show that operating on any arbitrary tensor field by it, yields zero.

In other words, you want to show that

$$\nabla_{[a} \nabla_b \omega_{c]}T^{d_1d_2\ldots d_k}{}_{e_1e_2\ldots e_l}=0$$

for arbitrary $$T^{d_1d_2\ldots d_k}{}_{e_1e_2\ldots e_l}$$

also, could you run through how to actually evaluate one of these derivatives?
thanks.

You just use equation 3.1.14 with $$\tilde{\nabla_a}=\partial_a$$ and [itex]C^{a}{}_{bc}=\Gamma^{a}{}_{bc}[/tex].

So, for example,

$$\nabla_a T^{bc}_{def}=\partial_a T^{bc}_{def}+\Gamma^{b}{}_{ai}T^{ic}_{def}+\Gamma^{c}{}_{ai}T^{bi}_{def}-\Gamma^{i}{}_{ad}T^{bc}_{ief}-\Gamma^{i}{}_{ae}T^{bc}_{dif}-\Gamma^{i}{}_{af}T^{bc}_{dei}$$

 

[latentcorpse]

so would the first term be

$\nabla_a \nabla_b ( \omega_c T^{d_1...d_k}{}_{e_1..e_l})=\nabla_a ((\nabla_b \omega_c)T^{d_1...d_k}{}_{e_1..e_l}) + \omega_c (\nabla_b T^{d_1...d_k}{}_{e_1..e_l}))$
isn't this going to be unbelievably long expression? I am going to get 14 terms when i do the first derivative and then 7 more for each of them of the next derivative. that's like 50 terms. and i just did the first term of 6 in the full expansion...the whole thing will be like 300 terms, no?

 

[gabbagabbahey]

Well, actually the first term in $\nabla_{[a} \nabla_b \omega_{c]}$ is the scalar function $\frac{1}{6}\nabla_{a} \nabla_b \omega_{c}$ (not a differential operator); so the first term in $\nabla_{[a} \nabla_b \omega_{c]}T^{d_1d_2\ldots d_k}{}_{e_1e_2\ldots e_l}$ will just be
[tex]\left(\frac{1}{6}\nabla_{a} \nabla_b \omega_{c}\right)T^{d_1d_2\ldots d_k}{}_{e_1e_2\ldots e_l}=\frac{1}{6}T^{d_1d_2\ldots d_k}{}_{e_1e_2\ldots e_l}\nabla_{a} \nabla_b \omega_{c}[/itex]

Anyways, you'll want to use $\nabla_a \nabla_b \omega_c-\nabla_b \nabla_a \omega_c=R_{abc}{}^{d}\omega_d$ several times and the generalized version of it; equation 3.2.12.

 

[latentcorpse]

erm... why can u just move the T... tensor in front of those del operators?

 

[gabbagabbahey]

Because those derivatives aren't acting on T, they are acting on $\omega_c$, and hence $\frac{1}{6}\nabla_{a} \nabla_b \omega_{c}$ will just be some tensor.

In contrast, if you look at the term $\frac{1}{6}\nabla_{a} \omega_c \nabla_b$, only the first operator acts on $\omega_c$; $\nabla_{b}$ will act on whatever tensor comes to the right of this expression.

 

[latentcorpse]

ok. got that. still though, as in post 12, i don't really want to do an expansion that will have 300 terms in it. I am sure its not meant to be that long, is it?

 

[gabbagabbahey]

You don't have to; just be smart about how you group your terms together

$$\begin{aligned}\nabla_{[a} \nabla_b \omega_{c]}&= \frac{1}{3!}[\nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c + \omega_c \nabla_a \nabla_b - \omega_c \nabla_b \nabla_a + \nabla_b \omega_c \nabla_a - \nabla_a \omega_c \nabla_b]\\&= \frac{1}{3!}[\nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c]- \frac{1}{3!}\omega_c (\nabla_a \nabla_b - \nabla_b \nabla_a)+ \frac{1}{3!}[\nabla_b \omega_c \nabla_a - \nabla_a \omega_c \nabla_b]\end{aligned}$$

Start by applying this to an arbitrary contravariant vector field $t^d$, then to an arbitrary covariant vector field $t_e$...then use a proof by induction to generalize...