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.
from Hot Weekly Questions - Mathematics Stack Exchange
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