2
$\begingroup$

Given the tetrad basis $\{(e_{\mu})\}$, i.e. smooth vector fields for which:

$$ (e_{\mu})^a (e_{\nu})_a = \eta_{\mu \nu} $$

we wish to find the components of the Ricci tensor in this basis, in terms of the Riemann tensor components (in the same basis), given by:

$$ R_{\rho \sigma \mu \nu} = R_{abcd} (e_{\rho})^a (e_{\sigma})^b (e_{\mu})^c (e_{\nu})^d $$

(For full clarity, abstract index notation is used for Latin indices. Also, I see the index of the basis vector fields as an indication of which basis vector is referred to, NOT as a component.) The book states, without derivation, that the components are given by:

$$ R_{\rho \mu} = \eta^{\sigma \nu} R_{\rho \sigma \mu \nu} $$

I tried deriving this out of curiosity but didn’t succeed:

$$ R_{abcd} = R_{\rho \sigma \mu \nu} (e_{\rho})_a (e_{\sigma})_b (e_{\mu})_c (e_{\nu})_d $$ $$ R_{ac} = R_{abc}^b = R_{\rho \sigma \mu \nu} (e_{\rho})_a (e_{\sigma})_b (e_{\mu})_c (e_{\nu})^b = \eta^{\sigma \nu} R_{\rho \sigma \mu \nu} (e_{\rho})_a (e_{\mu})_c $$

Of course, repeated Greek indices (referring to components) are summed over. The issue is, that when I try to insert $(e_{\alpha})^a$ and $(e_{\beta})^b$ into the tensor to decide the components, I run into an issue:

$$ R_{\alpha \beta} = R_{ac} (e_{\alpha})^a (e_{\beta})^c = R_{\rho \sigma \mu \nu} \eta^{\sigma \nu} \eta^{\rho}_{\alpha} \eta^{\mu}_{\beta} $$

The problem here is that $\eta$ can be both $\pm 1$, meaning that the Ricci component might be the negative of the equation for certain indices. I’m struggling to find where my thinking goes wrong. It seems to me that since the basis doesn’t fulfill the demand $(e_{\mu})_a (e_{\nu})^a = \delta_{\mu \nu}$, this problem always arises when inserting basis vectors to get components?

Any help is appreciated! This has been bothering me for a while.

$\endgroup$
1
  • 1
    $\begingroup$ It is by definition. $\endgroup$
    – kobe
    Jul 5, 2022 at 2:10

2 Answers 2

1
$\begingroup$

Write $e_\rho{}^a$ for the basis vectors and $e^\rho{}_a$ for the basis covectors. Note that $R_{abcd}$ is four times covariant, so it should be a linear combination of (tensor producs of) the basis covectors: \begin{align} R_{abcd} &= R_{\rho\sigma\mu\nu}e^\rho{}_ae^\sigma{}_be^\mu{}_ce^\nu{}_d. \end{align} Meanwhile, $R^d{}_{abc}$ has expansion \begin{align} R^d{}_{abc} &= g^{xd}R_{xabc} \\ &= g^{\alpha\beta} R_{\rho\sigma\mu\nu} e_\alpha{}^x e_\beta{}^d e^\rho{}_x e^\sigma{}_a e^\mu{}_b e^\nu{}_c \\ &= g^{\alpha\beta} R_{\rho\sigma\mu\nu} \delta_\alpha^\rho e_\beta{}^d e^\sigma{}_a e^\mu{}_b e^\nu{}_c \\ &= g^{\rho\beta} R_{\rho\sigma\mu\nu} e_\beta{}^d e^\sigma{}_a e^\mu{}_b e^\nu{}_c. \end{align} (In particular its components are $R^\beta{}_{\sigma\mu\nu}=g^{\rho\beta} R_{\rho\sigma\mu\nu}$, but we don't need this). Now to contract, set $d=b$ so that \begin{align} R_{ac}=R^b{}_{abc} &= g^{\rho\beta} R_{\rho\sigma\mu\nu} e_\beta{}^b e^\sigma{}_a e^\mu{}_b e^\nu{}_c \\ &= g^{\rho\beta} R_{\rho\sigma\mu\nu} \delta_\beta^\mu e^\sigma{}_a e^\nu{}_c \\ &= g^{\rho\mu} R_{\rho\sigma\mu\nu} e^\sigma{}_a e^\nu{}_c. \end{align} This works for any basis. In particular, for a tetrad you get $g^{\rho\mu}=\eta^{\rho\mu}$.

$\endgroup$
9
  • $\begingroup$ Thank you for your answer! The only thing that confuses me is the assumption $e_{\mu}^a e^{\nu}_a = \delta_{\mu}^{\nu}$ . What bothers me is how the tetrad basis assumes $e_{\mu}^a e^{\nu}_a = \eta_{\mu}^{\nu}$, which isn’t necessarily the identity map. Will your derivation really work for the tetrad basis in that case, and if so, why exactly? $\endgroup$
    – Max
    Jul 5, 2022 at 8:19
  • $\begingroup$ I did some thinking. Since $g^{\mu \nu} = g^{ab} e^{\mu}_a e^{\nu}_b = e_{\mu}^a e^{\nu}_a = \eta^{\mu \nu}$, we should have $\eta_{\alpha}^{\rho} = g_{\alpha}^{\rho}$ and $\eta_{\beta}^{\nu} = g_{\beta}^{\nu}$, and then I can see why it works. Is this correct? $\endgroup$
    – Max
    Jul 5, 2022 at 11:56
  • 1
    $\begingroup$ You are thinking of the coefficients of the metric, which are (in case of a tetrad) the diagonal matrix $\eta_{\mu\nu}=g_{ab}e^a_\mu e^b_\nu$. This matrix may have some plus or minus $1$ on the diagonal. On the other hand, once you choose any basis, $e_\mu^a$ (which are vectors) the dual basis $e^\nu_b$ is defined to be the unique covectors such that $e^\mu_a e_\nu^a$ is the identity matrix $\delta^\mu_\nu$. $\endgroup$ Jul 6, 2022 at 3:52
  • 1
    $\begingroup$ @Max Yeah, exactly. The object $(e_\mu)_a$ is just (in my opinion, bad, so I won't use it) notation for the covector $g_{ab}e_\mu^b$, while $e^\nu_a$ is the dual basis that satisfies $e_\mu^a e^\nu_a=\delta^\nu_\mu$. In general they are related by $g_{ab}e_\mu^a=g_{\alpha\beta}e^\alpha_a e^\beta_b e_\mu^a =g_{\mu\beta}e^\beta_b$. For a tetrad this is $g_{ab}e_\mu^a = \eta_{\mu\beta}e^\beta_b$. If for example the metric has signature $(+,-,-,-)$, then this means that $g_{ab}e_0^a = e^0_b$ but $g_{ab}e_i^a = -e^i_b$ for $i=1,2,3$. $\endgroup$ Jul 6, 2022 at 12:41
  • 1
    $\begingroup$ @Max That's good to hear. I'll probably add some details of this discussion in the comments to my answer. $\endgroup$ Jul 7, 2022 at 8:07
1
$\begingroup$

As commented, this is all by definition. The summation of the inverse Minkowski metric with the fourth lower index of Riemann over sigma results in a raised index: $$\eta^{\sigma \nu} R_{\rho \sigma \mu \nu} \equiv R_{\rho \sigma \mu}^{\sigma}$$ No need to mess with inputing vectors and covectors into the tensors; summations with the metric and inverse metric are compacted into this notational process of raising and lowering indices.

Then, the Ricci tensors components are defined to be the Riemann tensor with a summation over the top and middle lower index (it is like a trace of the Riemann tensor): $$R_{ab} \equiv R_{a c b}^c$$ Hence $$R^{\sigma}_{\rho \sigma \mu} = R_{\rho \mu}$$ In fact, the refrence to any specific tetrad basis is somewhat misleading. Anytime you write down a tensor as a symbol with some upper and/or lower indices, the use of a basis is implicit. In fact, this is the power of tensors, the algebra looks exactly the same no matter the basis. So we usually leave it implicit unless converting between coordinate systems (applying the tensor transformation laws) and simply work purely in components.

$\endgroup$
3
  • $\begingroup$ Thanks for the answer! Two things that still bother me are: (1) When using a tetrad basis, the geometry isn’t necessarily flat right? In other words, $\eta$ isn’t the actual metric. How can we then assume that it can be used to upper/lower indices? (2) I am still curious to know where my derivation goes wrong. I completely agree that it’s a bit redundant to do it that way, but I just have to know why it doesn’t work the way I do it. Do you have any ideas of this? $\endgroup$
    – Max
    Jul 5, 2022 at 8:25
  • 1
    $\begingroup$ Okay I actually do see that the answer to (1) is that $\eta_{\mu \nu}$ is the metric components in the tetrad basis, so the only problem remaining is (2) $\endgroup$
    – Max
    Jul 5, 2022 at 9:51
  • 1
    $\begingroup$ I think you are missing the point. Indeed, many authors define these tensors by their components, but not everyone. This question assumes that the Riemann tensor was defined without reference to coordinates, and hence the equality of components is not a definition but a theorem (not too hard to show, but still). $\endgroup$ Jul 8, 2022 at 0:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .