I recently read about Hopf fibrations when I was looking into qubits and the Bloch sphere. The analogy is that the fibre is the phase circle $S^1$, base space the Bloch sphere $S^2$, and the total space $E \cong S^3$.
What I fail to understand is the reason this is a non-trivial bundle. I tried searching around and found homotopy based arguments, which I cannot comprehend.
Is there an intuitive explanation to this? Why is it not the simple Cartesian product $S^2\times S^1$?
Your question amounts to prove that $S^3$ and $S^2\times S^1$ are not homeomorphic. A first argument is homotopic. The fundamental group of $S^2\times S^1$ is isomorphic to $\mathbb{Z}$, generated by the loop $t\mapsto (0,e^{2i\pi t})$, whereas $S^3$ has trivial homotopy group.
As you seem to be unfamiliar with homotopy, I will give a hint on why this result is true. My hint is not a complete proof as it relies on a deep theorem (the Jordan-Brouwer theorem, of which standard proof uses jhomology, thus algebraic topology), but is quite intuitve.
Assume there is a homeomorphism $f : S^3 \rightarrow S^2 \times S^1$.
Let $a\in S^1$, we know that so that $S^1\backslash\{a\}$ is connected as it is homeomorphic to $\mathbb{R}$. Thus when removing $S^2\times \{a\}$ to $S^2\times S^1$, we get $S^2\times S^1\backslash\{a\}$, which is still connected.
Let $B = f^{-1}(S^2\times \{a\})$ which is homeomorphic to $S^2$ as $f$ is a homeomorphism. We get a homeomorphism $f : S^3\backslash B \rightarrow S^2 \times S^1\backslash\{a\}$. As the latter is connected, the first should be connected. But it is not :
Choose $c\in S^3 \backslash B$, then $S^3\backslash \{c\}$ is homeomorphic to $\mathbb{R}^3$, so $S^3\backslash (B \cup \{c\})$ is homeomorphic to $\mathbb{R}^3 \backslash B'$, where $B'\cong B \cong S^2$. This has two connected components : one for the interior of $B'$ $(\ast)$ and one for the rest.
We can then see that $S^3\backslash B$ has two connected components, one for the interior of $B$, and then one for the connected components of $c$.
Thus $f : S^3\backslash B \rightarrow S^2 \times S^1\backslash\{a\}$ is a homeomorphism between spaces that don't have the same number of connected components. This is impossible and shows the inextence of $f$.
Note : $(\ast)$ is in fact the Jordan-Brouwer theorem and is quite hard to prove. It says more generally that $\mathbb{R}^n \backslash S$ where $S$ is homeomorphic to $\mathbb{S}^{n-1}$ has two connected components, one for the interior of $S$ and one for the exterior. A proof using homology is found in Allen Hatcher, Algebraic Topology, proposition 2.B.1.
Since you seem to prefer a geometric argument over a topological one, here it goes, but keep in mind that it is more complicated and in essence is "the same" argument, and contains some hand-waving. The claim is that $S^1\times S^2$ (unlike $S^3$) does not admit a metric of positive sectional curvature (or even positive Ricci curvature). The idea is to "unwind" the manifold, i.e., replace $S^1$ by $\mathbb R$ to get a periodic metric on $M=\mathbb R\times S^2$ called the lift of the metric on $S^1\times S^2$. The manifold $M$ is the universal cover of $S^1\times S^2$ (this notion can be defined without reference to the fundamental group).
$M$ is not compact but the periodic metric makes it complete. One can show that $M$ necessarily has infinite Riemannian diameter. On the other hand, the Bonnet-Myers theorem asserts that a uniform lower bound for the curvature gives an upper bound on the diameter. The contradiction shows that $S^1\times S^2$ admits no metric of positive curvature. This argument is more complicated than the one using the fundamental group, but it does not mention the fundamental group explicitly. It should serve as motivation to learn about the fundamental group.
