My Lecturer mentioned at the beginning of my differential geometry course that you need at least $4$-D to embed a $2$-D shape on an ambient space ( not too sure what ambient space means ....)
My question is why is this so ? We're only a few weeks in but hopefully the explanation wont require any concepts which are too advanced.
A guess...
A famous theorem of Hassler Whitney--the Whitney embedding theorem--states that any smooth manifold of dimension $n$ may be embedded smoothly in $\mathbb{R}^{2n}$. So, roughly, an $n$-dimensional smooth manifold may be considered as a submanifold of an ambient Euclidean space of twice the dimension. For you, this means that every surface may be embedded in $\mathbb{R}^4$.
Now, the $2n$ in the theorem is best possible in general, meaning that there are examples of an $n$-dim manifold failing to embed in $\mathbb{R}^{2n-1}$. So, if you want a globally true theorem, $\mathbb{R}^{2n}$ has to be the "answer" for all $n$-manifolds. But, it does NOT say that some $n$-manifolds can't "do better" than $2n$. This might be your confusion. For a very simple example, an open $2$-disk is already a submanifold of $\mathbb{R}^3$, which is of dimension strictly less than $4$.
An smoothly embedded closed surface in $\mathbb{R}^3$ will have a well-defined "outward" normal and so will be orientable. Hence if you try to embed a non-orientable closed surface, such as the Klein bottle, into $\mathbb{R}^3$ you will necessarily get a self-intersection. By going one dimension up to $\mathbb{R}^4$ and pull the self-intersection out you get an embedding.
Of course, for orientable surface (such as a sphere) you could do it in $\mathbb{R}^3$, but your lecturer is referring to the lowest dimension to embed all surfaces.
