Let $E \rightarrow S^1$ be a smooth fiber bundle whose fiber $F = \Sigma_g$ is an oriented surface of genus $g$. By Whitney's embedding theorem, we can embed the total space $E$ into $\mathbb{R}^6$. If the bundle is trivial, then it can be embedded into $\mathbb{R}^5$ (since the surface can be embedded into $\mathbb{R}^3$ and $S^1$ can be embedded into $\mathbb{R}^2$). I have two questions:
- Is there a surface bundle over $S^1$ that can be embedded into $\mathbb{R}^4$?
- What's the minimal $n \in \mathbb{N}$ such that every surface bundle over $S^1$ can be embedded to $\mathbb{R}^n$?
First of all, since $n$ is not a power of two, $E$ can always be embedded in $\mathbb{R}^5$ (see https://en.wikipedia.org/wiki/Whitney_embedding_theorem, paragraph : Sharper results). However, you can't do better in general. Consider $f$ a homeomorphism of $\Sigma_g$ that reverses orientation. Let, $$ E = (\Sigma_g \times [0,1])/((x,0) \sim (f(x),1)), \qquad \pi : \left\{\begin{array}{rcl} E & \rightarrow & \mathbb{S}^1 \\ [x,t] & \mapsto & e^{2i\pi t}\end{array}\right. $$ Then $E$ is a smooth manifold and $\pi$ is a fiber bundle of fiber $\Sigma_g$. The fact that $f$ reverses orientation implies that $E$ is non orientable (see Let $M=(\mathbb{S}^2\times [0,1])/\sim$, where $(p,0)\sim (-p, 1)$. Prove that M is a compact, differentiable manifold. Is it orientable? but replace $\mathbb{S}^2$ by $\Sigma_g$, the proof is the same). However, any compact manifold $K$ of dimension $n$ that embeds in $\mathbb{R}^{n + 1}$ is orientable so $E$ doesn't embed in $\mathbb{R}^4$.
And as said @Michael Albanese in your comments, the trivial bundle embeds in $\mathbb{R}^4$.