Compute connection 1-forms of warped product manifold using method of moving frames

Question

Let $(\bar{M},\bar{g})$ and $(\dot{M},\dot{g})$ be two Riemannian manifolds and let $f\in C^{\infty}(\bar{M})$ be nowhere zero. Let $(M^n,g)$ be the warped product of the two manifolds with warping function $f$; that is, $M=\bar{M}\times\dot{M}$ and \begin{equation} g=\bar{g}\times_f\dot{g}:=\bar{\pi}^*\bar{g}+(f\circ\bar{\pi})^2\dot{\pi}^*\dot{g} \end{equation} where $\bar{\pi}:\bar{M}\times\dot{M}\to\bar{M}$ and $\dot{\pi}:\bar{M}\times\dot{M}\to\dot{M}$ are the natural projections. (If you like, we can simply write $g=\bar{g}+f^2\dot{g}$).

Convention on indices: $1\leq a,b,c,\cdots\leq q$ for $(\bar{M},\bar{g})$, $q+1\leq\alpha,\beta,\gamma,\cdots\leq n$ for $(\dot{M},\dot{g})$ and $1\leq i,j,k,\cdots\leq n$ for $(M,g)$. Einstein summation convention is assumed.

Let $\left\{\bar{\omega}^a\right\}_{a=1}^q$ and $\left\{\dot{\omega}^{\alpha}\right\}_{\alpha=q+1}^n$ be local orthonormal coframes on $(\bar{M},\bar{g})$ and $(\dot{M},\dot{g})$ respectively. Then a local orthonormal coframe on $(M,g)$ can be given by \begin{align} \omega^i:=\left\{ \begin{array}{ccl} \bar{\pi}^*\bar{\omega}^i & \mbox{if} & 1\leq i\leq q \\ (f\circ\bar{\pi})\dot{\pi}^*\dot{\omega}^i & \mbox{if} & q+1\leq i\leq n \end{array} \right. \end{align} (Again, if you like, we can simply write $\bar{\omega}^i$ and $f\dot{\omega}^i$ respectively). Moreover, under these two coframes, denote the connection 1-forms by $\bar{\omega}^b_a$ and $\dot{\omega}^{\beta}_{\alpha}$ respectively.

My goal is to compute the connection 1-forms of $(M,g)$. By taking exterior derivative of the definition of $\omega^i$, applying the Cartan's 1st structural equation and gather the terms to the LHS, I arrive at \begin{gather} \omega^b\wedge\big(\omega^a_b-\bar{\pi}^*\bar{\omega}^a_b\big)+\omega^{\beta}\wedge\omega^a_{\beta}=0 \\ \omega^b\wedge\left(\omega^{\alpha}_b-\frac{(f\circ\bar{\pi})_b}{f\circ\bar{\pi}}\omega^{\alpha}\right)+\omega^{\beta}\wedge\big(\omega^{\alpha}_{\beta}-\dot{\pi}^*\dot{\omega}^{\alpha}_{\beta}\big)=0 \end{gather} where $(f\circ\bar{\pi})_i$ is defined via $d(f\circ\bar{\pi})=(f\circ\bar{\pi})_i\omega^i$.

Now it is tempting to conclude directly from above that \begin{align} \omega^a_b&=\bar{\pi}^*\bar{\omega}^a_b \\ \omega^{\alpha}_{\beta}&=\dot{\pi}^*\dot{\omega}^{\alpha}_{\beta} \\ \omega^{\alpha}_b&=\frac{(f\circ\bar{\pi})_b}{f\circ\bar{\pi}}\omega^{\alpha} \end{align} but I don't think this is valid in general for sum of wedge products. The Cartan's lemma tells us that at most we only can conclude that the expressions in the parentheses can be written as a linear combination of the $\omega^i$'s (this is trivial here) with coefficients satisfying some symmetry on the indices.

Hence, I would like to ask for the way to proceed. How do we continue to compute the connection 1-forms? Any comment, hint and answer is welcomed and appreciated.

P.S. This is not a homework problem. I was just trying to explore and play around the things on my own. Thus, there is a chance that what I have written contains some errors. Feel free to correct me if I have written something wrong.

Answer 1

Your instincts are right. What you're missing is the usual symmetry/skew-symmetry argument combining the Cartan lemma with skew-symmetry of the Levi-Civita connection forms to prove uniqueness. Recall that if $\sum \omega^i\wedge \omega^j_i = 0$, then $\omega^j_i = \sum a^j_{ik}\omega^k$, with $a^j_{ik}=a^j_{ki}$ by the Cartan lemma. But the fact that these are connection forms for the Levi-Civita connection tells you that $a^j_{ik} = -a^i_{jk}$. Thus, $$a^j_{ik} = -a^i_{jk} = -a^i_{kj} = a^k_{ij} = a^k_{ji} = -a^j_{ki} = -a^j_{ik},$$ so $\omega^j_i = 0$, as you wish.

Answer 2

Let me try to note down the proof (as complete as possible) here by using Ted Shifrin's hint (thanks for the answer).

By Cartan's lemma, it follows that \begin{align} \omega^a_b-\bar{\pi}^*\bar{\omega}^a_b&=A^a_{bi}\omega^i & & (1) \\ \omega^a_{\beta}&=A^a_{\beta i}\omega^i & & (2) \\ \omega^{\alpha}_b-\frac{f_b}{f}\omega^{\alpha}&=B^{\alpha}_{bi}\omega^i & & (3) \\ \omega^{\alpha}_{\beta}-\dot{\pi}^*\dot{\omega}^{\alpha}_{\beta}&=B^{\alpha}_{\beta i}\omega^i & & (4) \end{align} where $A^a_{ij},B^{\alpha}_{ij}$ satisfy $A^a_{ij}=A^a_{ji}$ and $B^{\alpha}_{ij}=B^{\alpha}_{ji}$ (symmetric on the two lower indices). Note that (1) and (2) share the same '$A$' since they come from the same equation. Similarly for (3) and (4). At this point we use a different '$B$' for (3) and (4) to keep the generality, but soon it can be observed, by (2) and (3), that '$B$' and '$A$' are somehow related.

Now, using the usual symmetry/skew-symmetry argument (as suggested by Ted), we have \begin{align} A^a_{bc}=-A^b_{ac}=-A^b_{ca}=A^c_{ba}=A^c_{ab}=-A^a_{cb}=-A^a_{bc} \end{align} and so $A^a_{bc}=0$. Similarly $B^{\alpha}_{\beta\gamma}=0$. Thus we may reduce (1) and (4) to \begin{align} \omega^a_b-\bar{\pi}^*\bar{\omega}^a_b&=A^a_{b\beta}\omega^{\beta} & & (5) \\ \omega^{\alpha}_{\beta}-\dot{\pi}^*\dot{\omega}^{\alpha}_{\beta}&=B^{\alpha}_{\beta b}\omega^b & & (6) \end{align} We would now like to show that the R.H.S. of (5) is indeed zero as desired. For this purpose we observed from (2) that $A^a_{\beta i}=-A^{\beta}_{ai}$. Thus \begin{align} A^a_{b\beta}=A^a_{\beta b}=-A^{\beta}_{ab}=-A^{\beta}_{ba}=A^{b}_{\beta a}=A^{b}_{a\beta}=-A^a_{b\beta} \end{align} and so $A^a_{b\beta}=0$. This proves, by (5), that \begin{align} \omega^a_b=\bar{\pi}^*\bar{\omega}^a_b \end{align} Moreover, (2) can now also be reduced to \begin{align} \omega^a_{\beta}&=A^a_{\beta\alpha}\omega^{\alpha} & & (7) \end{align}

Next, we would also like to show that the R.H.S. of (6) is zero as desired. First we observe from (6) that $B^{\alpha}_{\beta b}$ is antisymmetric in $\alpha$ and $\beta$. On the other hand, as we mentioned at the beginning, '$B$' and '$A$' are related via (2) (or the reduced (7)) and (3) as follow: \begin{align} B^{\alpha}_{\beta b}&=B^{\alpha}_{b\beta} \\ &=\left(\omega^{\alpha}_b-\frac{f_b}{f}\omega^{\alpha}\right)(e_{\beta}) & & e_i\text{ is dual to }\omega^i \\ &=-\omega^b_{\alpha}(e_{\beta})-\frac{f_b}{f}\delta_{\alpha\beta} \\ &=-A^b_{\alpha\beta}-\frac{f_b}{f}\delta_{\alpha\beta} & & \text{By (7)} \end{align} Since both of $A^b_{\alpha\beta}$ and $\delta_{\alpha\beta}$ are symmetric in $\alpha$ and $\beta$, thus so does $B^{\alpha}_{\beta b}$. This makes $B^{\alpha}_{\beta b}$ to be both antisymmetric and symmetric in $\alpha$ and $\beta$, hence it must vanish. This proves, by (6), that \begin{align} \omega^{\alpha}_{\beta}=\dot{\pi}^*\dot{\omega}^{\alpha}_{\beta} \end{align} Finally, $B^{\alpha}_{\beta b}=0$ implies $A^b_{\alpha\beta}=-\frac{f_b}{f}\delta_{\alpha\beta}$ and so \begin{align} \omega^{\beta}_a&=-\omega^a_{\beta} \\ &=-A^a_{\beta\alpha}\omega^{\alpha} & & \text{By (7)} \\ &=\frac{f_a}{f}\delta_{\beta\alpha}\omega^{\alpha} \\ &=\frac{f_a}{f}\omega^{\beta} \end{align} This completes the proof of connection $1$-forms of warped product manifold.