What $X$ should be homeomorphic to? Consider $X = \{ (x,y,z) \in \mathbb C^3 \mid x^2 + y^2 + z^2 = 1 \}$.
I want to find a "known" space which is homeomorphic to $X$. A point in $X$ is two real vectors $a,b \in \mathbb R^3$ such that $\langle a,b \rangle = 0$ and $\sum a_i^2 = \sum b_i^2 + 1$. 
I don't see what can I say more about the last equation ... I know a similar space was a disk bundle over a $3$-sphere, maybe this is the same here ? 
 A: Ok, let us start analysing a simpler case: the curve $x^2+y^2=1$ in $\mathbb{C}^2$. Denote it by $X'$. 
The projection $(x,y) \mapsto x$ gives a map $X' \to \mathbb{C}$. This is a double branched cover with two singular points of multiplicity two at $\pm1$. Roughly, this is because the the equation $y^2=1-x^2$ has exactly two complex solutions except when $1-x^2=0$, i.e. $x= \pm 1$. 
Now, the  double branched cover of the plane at two points with multiplicity two is the cylinder $\mathbb{R} \times S^1$. This is basic cut and paste that if you are familiar with branched covers you already know. If you are not, read chapter 12 of Kauffman's book "On Knots". 
Using the same trick for the curve $X$ (by considering the projection $(x,y,z) \mapsto (x,y)$) we get a double cover $X \to \mathbb{C^2}$ branching at the curve $x^2+y^2=1$. Thus, we can conclude that $X$ is the double branched cover of $\mathbb{R}^4$ branching at the cylinder. Now you can have fun with your cut and paste once again and completely understand this space.  
Another possibility is to do a linear change of coordinates ($p \mapsto i \cdot p$) and consider the complex hypersurface with eqauation $x^2+y^2+z^2=-1$. Its projective closure is the curve 
$$\bar{X}: \ \ \ \ \ x^2+y^2+z^2+k^2=0\ \ \ \ \ \ \subset \mathbb{CP^3} \ , $$
and it is ismorphic to $\mathbb{CP}^1 \times\mathbb{CP}^1$. An explicit homeomorphism  $\mathbb{CP}^1 \times\mathbb{CP}^1 \to\bar{X}$ is given by:
$$f: \ ([a:b], [c:d]) \mapsto [ac:ad:bc:bd] \ . $$
Your original $X$ is the complement in $\bar{X}$ of the complex curve 
$C= \{ x^2+y^2+z^2=0 , \ k=0 \}$. Thus $X= \mathbb{CP}^1 \times\mathbb{CP}^1 - f^{-1}(C)$ and you can do the explicit computation.
A: The answer of Antonio Alfieri is really good but not explicit, and I found the answer so I'm just writing in case someone else is also looking for it : $X$ is homeomorphic to the tangent bundle of $S^3$. 
