Reading off connection 1-forms from Cartan's structural equation $de=-\omega\wedge e$

Question

Suppose we have a Lorentzian metric of the form \begin{align} g&=-f(r)^2\,dt^2+ h(r)^2(dr^2+r^2\,d\theta^2+r^2\sin^2\theta\,d\phi^2) \end{align} Where $f,h$ are say strictly positive functions. We use the Levi-Civita connection. I introduced the 1-forms \begin{align} e^0=f(r)\,dt,\quad e^1=h(r)\,dr,\quad e^2=rh(r)\,d\theta,\quad e^3=rh(r)\sin\theta\,d\phi \end{align} which diagonalize the metric, and now I'm trying to use these to calculate the connection 1-forms $\omega^a_{\,b}$ using Cartan's structural equation $de=-\omega\wedge e$ (since Levi-Civita connection is torsion free).

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Question 1.

The issue I'm facing is that once I calculate $de$, I'm not sure how to identify $\omega$ from those equations: initially I tried the most naive thing by just looking at the appropriate coefficient and calling that the appropriate component of $\omega$, but I think this naive approach is wrong, probably because the wedge-product of non-zero forms can still be zero (so "cancelling" terms won't work).

To be more explicit, I calculated \begin{align} \begin{cases} de^0= f'(r)\,dr\wedge dt\\ de^1= 0\\ de^2=(h(r)+rh'(r))\,dr\wedge d\theta\\ de^3= (h(r)+rh'(r))\sin\theta\,dr\wedge d\phi+ rh(r)\cos\theta\,d\theta\wedge d\phi \end{cases} \end{align} When I first did the calculation, I naively concluded that \begin{align} de^0=f'(r)\,dr\wedge dt=-\left[-\frac{f'(r)}{f(r)}\,dr\right]\wedge e^0, \end{align} and thus that $\omega^0_0=-\frac{f'(r)}{f(r)}\,dr, \omega^0_1=\omega^0_2=\omega^0_3=0$. Next, from $de^1=0$ I naively concluded that $\omega^1_{\,b}=0$ for all $b=0,1,2,3$. I did a similar thing with the other equations. But now I realize this is wrong, because for example, we can also write \begin{align} de^0=f'(r)\,dr\wedge dt= -\left[\frac{f'(r)}{h(r)}\,dt\right]\wedge e^1, \end{align} so if I were to use my above logic, I would have $\omega^0_0=0, \omega^0_1=\frac{f'(r)}{h(r)}\,dt, \omega^0_2=\omega^0_3=0$. So clearly my mistake stems from the fact that the wedge of non-zero forms can be zero. But now I'm not sure what the correct approach is.

I have read this answer by @Ted Shifrin, and it seems like the correct answer is the second approach, but I'm not sure why. Also, I can't really understand that answer because it's not clear to me why certain certain $\omega^a_b$ are equal to certain functions and why others are multiples of some $e^i$, and why some others are zero.

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Question 2.

The equation $de=-\omega\wedge e$ consists of four equations relating $2$-forms. However, $\omega$ being a $4\times 4$ matrix (in this case) of $1$-forms, consists a-priori of 16 unknowns. I believe in this case due to the Lorentzian signature and the diagonalizability of the metric, there is some relationship between $\omega^a_b$ and $\omega^b_a$, so that it can be written as \begin{align} [\omega^a_b]&= \begin{pmatrix} 0&\alpha_1&\alpha_2&\alpha_3\\ \alpha_1&0&\beta_1&\beta_2\\ \alpha_2&-\beta_1&0&\beta_3\\ \alpha_3& -\beta_2&-\beta_3&0 \end{pmatrix} \end{align} for some 1-forms $\alpha_1,\beta_i$. So, now there are only 6-unknowns, but this is still too many unknowns for the number of equations.

So my question is whether we can always use this structural equation to determine $\omega$ completely? I believe the answer is yes because for the case of Christoffel symbols $\Gamma^i_{jk}$ we have explicit formulas for it in terms of the metric, and now since $\omega$ are related to $\Gamma$ in some fashion, the same ought to hold true; but now I'm not sure how to reconcile this with the above counting argument (6 unknowns vs 4 equations).

Answer 1

The answer to your first question still boils down to the uniqueness. If one has a solution that satisfies the structure equations and (skew-) symmetry, then it must be the answer.

In the case of my solution to which you linked, remember, for starters, that $\omega_j^j=0$ for $j\ge 1$.

I really should have written that solution in a slightly different order. I agree that a priori we might conceivably have had, say, $\omega_3^1=e^2$. Then we'd have $\omega_2^1=e^3$ and so $\omega_1^2 = -e^3$. But we establish that $\omega_1^2$ is a multiple of $e^2$, and so this is impossible. Establishing the ones in red and imposing symmetry conditions completely fills out the table.

Answer 2

As mentioned by @Ted Shifrin's comments and answer, since we are working with the Levi-Civita connection (torsion-free, and metric-compatible) it follows that there exist unique connection 1-forms which satisfy the desired equations. Below, I shall go into more detail, but the gist is that it is much more efficient to make strategic guesses for the $\omega$'s rather than systematically calculating them, because that would be analogous to trying to calculate the Christoffel symbols using the formula $\Gamma^i_{jk}=\frac{1}{2}g^{ia}\left(\frac{\partial g_{aj}}{\partial x^{k}}+ \frac{\partial g_{ka}}{\partial x^j} - \frac{\partial g_{jk}}{\partial x^a}\right)$, which is painful for explicitly calculating things because many terms often turn out to be 0 after a tedious calculation. Using strategic guesses, if we find a solution for the $\omega$'s, then by the uniqueness result, that has to be it.

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(Existence and) Uniqueness Result.

We first start with a pretty straightforward lemma about differential forms.

Lemma.

Let $M$ be a smooth $n$-dimensional manifold, and $\{\epsilon^i\}_{i=1}^n$ a local co-frame, and $\{\omega_{ij}\}_{i,j=1}^n$ a collection of 1-forms skew-symmetric in $i,j$ and such that for all $1\leq j\leq n$, we have$\sum_{i=1}^n \omega_{ij}\wedge \epsilon^i=0$. Then, each $\omega_{ij}=0$.

The proof is pretty easy. Since the $\epsilon$'s form a co-frame, we can write $\omega_{ij}=\sum_k f_{ijk}\cdot \epsilon^k$ for some smooth functions $f_{ijk}$ ($1\leq i,j,k\leq n$) which are skew-symmetric in $i,j$. Now, plugging this into our hypothesis, we have for all $j$, \begin{align} 0&=\sum_{i=1}^n\omega_{ij}\wedge\epsilon^i\\ &=\sum_{i,k=1}^nf_{ijk}\epsilon^k\wedge \epsilon^i\\ &=\sum_{i<k} f_{ijk}\epsilon^k\wedge \epsilon^i + \sum_{i=k}f_{ijk}\epsilon^k\wedge \epsilon^i + \sum_{i>k}f_{ijk}\epsilon^k\wedge \epsilon^i\\ &=\sum_{i<k}(f_{ijk}-f_{kji})\epsilon^k\wedge \epsilon^i \end{align} Now, $\{\epsilon^i\}_{i=1}^n$ being a local basis of 1-forms implies that the set of 2-forms $\epsilon^k\wedge \epsilon^i$ with $1\leq i<k\leq n$ is a local basis. Hence, for all $1\leq j\leq n$ and all $1\leq i<k\leq n$, we have $f_{ijk}=f_{kij}$. But from here it clearly follows that for all $1\leq i,j,k\leq n$, we have $f_{ijk}=f_{kij}$. So, we have skew-symmetry in first two indices and symmetry in the first and last. This actually implies skew-symmetry in last two indices because $f_{ikj}=f_{jki}=-f_{kji}=-f_{ijk}$. From here, we easily see all the $f$'s vanish: \begin{align} f_{ijk}=f_{kji}=-f_{jki}=f_{jik}=-f_{ijk}, \end{align} so $f_{ijk}=0$, and hence each $\omega_{ij}=0$. There may be a slightly more efficient way to shuffle the indices to arrive at this conclusion, but anyway the proof is complete.

Now, we come to the main result

Let $(M,g)$ be an $n$-dimensional pseudo-Riemannian manifold. Then there exists a unique metric-compatible, torsion-free connection $\nabla$, such that for every smooth local 'orthonormal' coframe $\{\epsilon^1,\dots, \epsilon^n\}$, we have that the corresponding connection 1-forms satisfy $\omega_{ij}=-\omega_{ji}$ and $d\epsilon_i=-\omega_{ij}\wedge \epsilon^j$ (where $\epsilon_i=g_{ia}\epsilon^a$).

Some remarks before the proof. The condition $\omega_{ij}=-\omega_{ji}$ in an 'orthonormal' (co)frame is equivalent to metric-compatibility of the connection (simply differentiate the expression $g_{ij}=g(e_i,e_j)$ and play around with the indices, and product rule). Also, note that due to 'orthonormality', we have $g_{ij}$'s are constant and equal to $\pm\delta_{ij}$. I'm working with the lowered-indices version $\epsilon_i=g_{ia}\epsilon^a$ so that we have skew-symmetry of the $\omega$'s so that we can use the previous lemma. Next, the condition $d\epsilon=-\omega\wedge \epsilon$ with appropriate indices conveys that the connection is torsion free (with torsion, it should read $d\epsilon=-\omega\wedge \epsilon+\theta$ where $\theta$ with indices are the torsion 2-forms).

Below, I'll present two proofs for the uniqueness. The first is the slick way, invoking the lemma above. The second is more constructive/systematic/explicit and actually gives a formula which can be turned around to give a proof of existence of the Levi-Civita connection (similar to the above mentioned formula for the $\Gamma$'s).

Proof 1. of Uniqueness

Suppose $\omega_{ij}$ and $\eta_{ij}$ are two such connection 1-forms. Consider their difference $\zeta_{ij}=\omega_{ij}-\eta_{ij}$. Then, we have $\zeta_{ij}=-\zeta_{ji}$ and $\zeta_{ij}\wedge \epsilon^i=0$, so by the lemma each $\zeta_{ij}=0$, thereby proving the uniqueness.

Proof 2 of Uniqueness.

Write $\omega_{ij}=f_{ijk}\epsilon^k$ and $d\epsilon_a=\frac{1}{2}h_{ija}\epsilon^i\wedge \epsilon^j$ for some smooth functions $f,h$ which are skew-symmetric in the first two slots. Now, the structure equations $d\epsilon_a=-\omega_{ab}\wedge \epsilon^b=\omega_{ba}\wedge \epsilon^b$ say that \begin{align} \frac{1}{2}h_{ija}\epsilon^i\wedge \epsilon^j&= f_{bak}\epsilon^k\wedge \epsilon^b\\ &=f_{jai}\epsilon^i\wedge \epsilon^j\\ &=\frac{1}{2}(f_{jai}-f_{iaj})\epsilon^i\wedge \epsilon^j \end{align} so it follows that $h_{ija}=f_{jai}-f_{iaj}$, or by some index renaming, $f_{ijk}-f_{kji}=h_{kij}$. Now, write the same thing 3 times with the indices cyclically permuted: \begin{align} \begin{cases} f_{ijk}-f_{kji}&=h_{kij}\\ f_{jki}-f_{ikj}&=h_{ijk}\\ f_{kij}-f_{jik}&=h_{jki} \end{cases} \end{align} Now, take the first equation, subtract the second, add the third, and divide by 2 (and use skew-symmetry of $f$ in its first two slots) to get $f_{ijk}=\frac{1}{2}(h_{kij}-h_{ijk}+h_{jki})$. This proves that the $f$'s (and hence $\omega$'s) are uniquely determined by the $h$'s (and hence $\epsilon$'s and $d\epsilon$'s). This proves the uniqueness and gives an explicit formula.

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Once again to emphasize, while the second proof does give us an explicit and systematic approach to calculating the $\omega$'s, it is not efficient at all. It's much better to make educated guesses for the $\omega$'s especially since in many of the hands on calculations (atleast the ones I'm trying to practice calculations with) the metric always takes a special form with certain symmetries (eg spherically symmetric), so many of the connection 1-forms will end up being zero. As a result, it's best to just look for simple solutions, and then verify in the end they're the right ones.

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Calculating $\omega$'s for my Question.

We have $g_{ab}=\text{diag}(-1,1,1,1)$, and let us write $[\omega_{ij}]= \begin{pmatrix} 0&a_1&a_2&a_3\\ -a_1&0&b_1&b_2\\ -a_2& -b_1&0&b_3\\ -a_3&-b_2&-b_3&0 \end{pmatrix}$. We thus have

\begin{align} &\begin{cases} \epsilon^0&=f(r)\,dt\\ \epsilon^1&=h(r)\,dr\\ \epsilon^2&=rh(r)\,d\theta\\ \epsilon^3&=rh(r)\sin\theta\,d\phi \end{cases} \quad \text{and}\quad \begin{cases} \epsilon_0&=-f(r)\,dt\\ \epsilon_1&=h(r)\,dr\\ \epsilon_2&=rh(r)\,d\theta\\ \epsilon_3&=rh(r)\sin\theta\,d\phi \end{cases}. \end{align} Hence from the structural equation $d\epsilon_i=-\omega_{ij}\wedge \epsilon^j$, we have \begin{align} \begin{cases} d\epsilon_0= -f'(r)\,dr\wedge dt&= -(a_1\wedge \epsilon^1+a_2\wedge \epsilon^2+a_3\wedge \epsilon^3) \\ d\epsilon_1= 0&=-(-a_1\wedge \epsilon^0+b_1\wedge\epsilon^2+b_2\wedge \epsilon^3)\\ d\epsilon_2=(h(r)+rh'(r))\,dr\wedge d\theta&= -(-a_2\wedge \epsilon^0-b_1\wedge \epsilon^1+b_3\wedge \epsilon^3)\\ d\epsilon_3= (h(r)+rh'(r))\sin\theta\,dr\wedge d\phi+ rh(r)\cos\theta\,d\theta\wedge d\phi&=-(-a_3\wedge \epsilon^0-b_2\wedge \epsilon^1-b_3\wedge \epsilon^2) \end{cases} \end{align}

• Note that $\epsilon^0=f(r)\,dt$ involves a $dt$ term, while $d\epsilon^1,d\epsilon^2,d\epsilon^3$ do not have any $dt$ term. So, it makes sense to set $a_2=a_3=0$ (and by uniqueness this has to be correct). Also, we have $d\epsilon^0=-\left(-\frac{f'(r)}{h(r)}\,dr\right)\wedge \epsilon^0$, so we can set $a_1=-\frac{f'(r)}{h(r)}\,dr$. Then, all the equations will be satisfied by $a$.
• From the equation $d\epsilon_1=0=-(b_1\wedge\epsilon^2+b_2\wedge \epsilon^3)$, we see that $\begin{pmatrix} b_1\\ b_2 \end{pmatrix}= \begin{pmatrix} f_{11}& f_{12}\\ f_{12}&f_{22} \end{pmatrix}\cdot \begin{pmatrix} \epsilon^2\\ \epsilon^3 \end{pmatrix} $ for some functions $f_{11},f_{12},f_{22}$.
• Going to the next equation, we have $d\epsilon_2=-(-b_1\wedge\epsilon^1+f_{12}\epsilon^2\wedge \epsilon^3)=b_1\wedge \epsilon^1-f_{12}\epsilon^2\wedge \epsilon^3$. On the other hand, we have $d\epsilon_2=-\left(1+\frac{rh'(r)}{h(r)}\right)\,d\theta\wedge \epsilon^1$. Hence, it makes sense to set $b_1=-\left(1+\frac{rh'(r)}{h(r)}\right)\,d\theta$ (which is a pure multiple of $\epsilon^2$), and hence to also set the "cross term" $f_{12}=0$. So, $b_1$ is a multiple of $\epsilon^2$ and $b_2$ is a multiple of $\epsilon^3$. The choice of setting $f_{12}=0$ is also reinforced by the fact that $d\epsilon_2$ doesn't involve any $d\phi$ (which is proportional to $\epsilon^3$) terms.
• Finally, \begin{align} d\epsilon_3&=(h(r)+rh'(r))\sin\theta\,dr\wedge d\phi+ rh(r)\cos\theta\,d\theta\wedge d\phi\\ &=\left[-\left(1+\frac{rh'(r)}{h(r)}\right)\sin\theta\,d\phi\right]\wedge \epsilon^1+ [-\cos\theta\,d\phi]\wedge \epsilon^2 \end{align} Choose the first square bracket to be $b_2$ (which is indeed a multiple of $\epsilon^3$ as required above) and let $b_3$ be the second square bracket.

Summarizing, we have \begin{align} [\omega_{ij}]&= \begin{pmatrix} 0& -\frac{f'(r)}{h(r)}\,dt& 0 & 0\\ \frac{f'(r)}{h(r)}\,dt & 0 & -\left(1+\frac{rh'(r)}{h(r)}\right)\,d\theta& -\left(1+\frac{rh'(r)}{h(r)}\right)\sin\theta\,d\phi\\ 0& \left(1+\frac{rh'(r)}{h(r)}\right)\,d\theta & 0 & -\cos\theta\,d\phi\\ 0& \left(1+\frac{rh'(r)}{h(r)}\right)\sin\theta\,d\phi & \cos\theta\,d\phi & 0 \end{pmatrix}\\\\ \text{and}\quad [\omega^i_{\,j}]&= \begin{pmatrix} 0& \frac{f'(r)}{h(r)}\,dt& 0 & 0\\ \frac{f'(r)}{h(r)}\,dt & 0 & -\left(1+\frac{rh'(r)}{h(r)}\right)\,d\theta& -\left(1+\frac{rh'(r)}{h(r)}\right)\sin\theta\,d\phi\\ 0& \left(1+\frac{rh'(r)}{h(r)}\right)\,d\theta & 0 & -\cos\theta\,d\phi\\ 0& \left(1+\frac{rh'(r)}{h(r)}\right)\sin\theta\,d\phi & \cos\theta\,d\phi & 0 \end{pmatrix} \end{align} As a sanity check, $[\omega_{ij}]$ is indeed skew-symmetric and we can quickly verify that it satisfies the structure equations, so this is the correct answer. The matrix for $\omega^i_{\,j}$ is obtained by multiplying row 0 of $[\omega_{ij}]$ by a factor of $-1$ (due to Lorentz signature $(-1,1,1,1)$).

Having found the $\omega^i_{\,j}$ we can then go on to calculate the curvature 2-forms and so on.