What is an intuitive explanation of the Hopf fibration and the twisted Hopf fibration? As the question suggests, what is an intuitive explanation of the Hopf fibration and the twisted Hopf fibration? Many thanks in advance.
 A: I have never heard of the latter, but let me try to give some description of the former. We can identify $\Bbb R^{2n}$ with $\Bbb C^{n}$. For any element $\gamma \in \Bbb C^{n}-0$, we can associate the unique complex line between the origin and $\gamma$ which is just the $\Bbb C$ multiples of $\gamma$, $\Bbb C\gamma =\{z\gamma:z\in \Bbb C\}$. We can give the set of complex lines in $\Bbb C^n$ a topological structure (as well as a smooth structure, a complex structure, a symplectic structure etc.), and we call this space $\Bbb CP^{n-1}$. The Hopf map $H$ is simply the map $S^{2n-1}\to \Bbb CP^{n-1}$ which takes the point $\gamma$ to line $\Bbb C\gamma$. 
One can easily check that since each complex line (a real 2-plane in $\Bbb R^{2n}$) intersects $S^{2n-1}$ in a circle, so the inverse of a point is always a circle in $S^{2n-1}$. 
For $n=2$ ,this is especially interesting as $\Bbb CP^1$ is homeomorphic (diffeomorphic, biholomorphic, symplectomorphic etc) to $S^2$. Also, in $S^3$ we could possibly form interesting links from different point inverses. 
Lets consider the example where $\gamma_1=(1,0) \in \Bbb C^2$ and $\gamma_2=(0,1) \in \Bbb C^2$ and let's find the 2-component link given by $H^{-1}(\Bbb C\gamma_1) \cup H^{-1}(\Bbb C\gamma_2)$ . $H^{-1}(\Bbb C\gamma_1)=\{(x,y) \in \Bbb C^2: \ |x|+|y|=1,y=0 \}$ which bounds the disk $D(\gamma_1)=\{(x,y)\in \Bbb C^2 : |x|\leq 1, y= \sqrt {1-|x|^2}\}$ . Similarly $H^{-1}(\Bbb C\gamma_2)$ bounds the disk $D(\gamma_2)=\{(x,y)\in \Bbb C^2 : |y|\leq 1, x= \sqrt {1-|y|^2}\}$. In $S^3$, these disks intersect transversely along the arc $\{(x,y) \in \Bbb R^2 \subset \Bbb C^2: x\geq 0, y=\sqrt{1-x^2}\} $ . There is actually only one 2-component link (up to isotopy) in $S^3$ where each of the components are unlinks and disks bounding these unlinks can be arranged to intersect along a single arc transversely (say by the uniqueness of the meridian of a knot). That link is the Hopf link shown below. 

A: Circle bundles $E \mapsto S^2$ are indexed by their Euler class, an element of $H_2(S^2;\mathbb{Z}) \approx \mathbb{Z}$. One can characterize this integer using obstruction theory, as the obstruction to the existence of a section of the bundle map $E \mapsto S^2$.
"The" Hopf fibration is (up to oriented bundle equivalence) the circle bundle of Euler class $+1$. 
The circle bundle of Euler class $0$ is simply the projection map to $S^2$ of the Cartesian product $E = S^2 \times S^1$. In general, the total space of the circle bundle of Euler class $n$ is the Lens space $L(n,1)$ (thanks to @PVAL-inactive for this correction)). I believe (the OP might want to confirm that this is the intent of the question) that these are the "twisted Hopf fibrations". Perhals, just as for the Hopf fibration itself (as explained in the answer of PVAL), one might be able visualize the Euler class in each of these cases as the linking number in the total space $L(n,1)$ of the fibers over two points of the base space $S^2$.
