Let $(M,g)$ be Lorentzian manifold of the form
$$M=I\times \Sigma,\quad\quad g=-dt^{2}+h_{t}$$
where $\Sigma$ is a spacelike hypersurface and $h_{t}$ is a Riemannian metric on $\Sigma$ for every $t\in I$. Now, I have seen the following claim:
$\alpha\in\Omega^{k}(M)$ can be written as $\alpha=(\partial_{t}\lrcorner \alpha)\wedge dt+i^{\ast}_{\Sigma}\alpha$, where $i_{\Sigma}:\Sigma\to M$ denotes the embedding.
My question is: Is $i_{\Sigma}^{\ast}\alpha$ above correct?
Lets take $\alpha\in\Omega^{k}(M)$. Then, in coordinates, we have that (indiced of $\Sigma$ denoted by $i\in \{1,\dots,d\}$ and indices of $M$ by $\mu\in\{0,\dots,d\}$):
$$\alpha=\frac{1}{k!}\alpha_{\mu_{1}\dots\mu_{k}}dx^{\mu_{1}}\wedge\dots\wedge dx^{\mu_{k}}=\underbrace{\bigg (\frac{1}{(k-1)!}\alpha_{0i_{2}\dots i_{k}}dx^{i_{2}}\wedge\dots\wedge dx^{i_{k}}\bigg)}_{=\partial_{t}\lrcorner\alpha\in\Gamma(\mathrm{pr}^{\ast}\bigwedge^{k-1}T^{\ast}\Sigma)}\wedge dt + \underbrace{\frac{1}{k!}\alpha_{i_{1}\dots i_{k}}dx^{i_{1}}\wedge\dots\wedge dx^{i_{k}}}_{\in\Gamma(\mathrm{pr}^{\ast}\bigwedge^{k}T^{\ast}\Sigma)}$$
which gives a similar decomposition as above, i.e. $\alpha=(\partial_{t}\lrcorner\alpha)\wedge dt+\alpha_{\Sigma}$ where $\alpha_{\Sigma}$ is a "time-dependent $k$-form on $\Sigma$", i.e. $\alpha_{\Sigma}\in\Gamma(\mathrm{pr}^{\ast}\bigwedge^{k}T^{\ast}\Sigma)$ with $\mathrm{pr}:\mathbb{R}\times\Sigma\to\Sigma$. However, $\alpha_{\Sigma}$ is not the same as $i_{\Sigma}^{\ast}\alpha$, as the latter is equivalent to $\alpha_{\Sigma}\vert_{\Sigma}$, i.e. its not time-dependent. So, it seems to be me that the above decomposition is wrong, i.e. it should rather be $\alpha=(\partial_{t}\lrcorner \alpha)\wedge dt+\alpha_{\Sigma}$ where $\alpha_{\Sigma}$ is the time-dependent $k$-form on $\Sigma$ as defined above.