Riemannian properties of Clifford torus as a boundary. As far as I know, the Clifford torus
$$C \mathbb{T}^2 = \left\{ (z,w) \in \mathbb{C}^2 : |z|^2 = |w|^2 = \frac{1}{2} \right\}$$
is a closed minimal ($H_g=0$) hypersurface in the unit sphere $\mathbb{S}^{3}$,
$$\mathbb{S}^3 = \left\{ (z,w) \in \mathbb{C}^2 : |z|^2 + |w|^2 = 1 \right\},$$
it has no umbilic points (its principal curvatures are differente $-1 \neq 1$ everywhere) and it is not totally geodesic ($II_g \neq 0$).
I also know that we can consider the Clifford torus as the boundary of "half" $\mathbb{S}^{3}$. But I heard that "the Clifford torus divides $\mathbb{S}^{3}$ in two solid torus". So, I'm confused.
What exactly does it mean "divide" here? I mean if I consider "half" of $\mathbb{S}^{3}$ which boundary is the Clifford torus, then am I considering a (flat) solid torus which boundary is the Clifford torus? Then it has all curvatures equal zero in the interior instead of positive, as I would expect for a hemisphere? What are its Riemannian properties (secctional, Ricci, scalar curvatures)?
And, please, correct me if I gave any wrong information above.
Thank you in advance.
 A: Perhaps try think of $S^3$ as the boundary of $D^4$. Then this is the same as the boundary of $D^2\times D^2$, which is $S^1\times D^2 \cup D^2\times S^1$. The $D$ are then both solid tori and one then glues along their common boundary. So 'divide' here just means the opposite to gluing.
A: Consider the following two subset of $\mathbb S^3$: 
\begin{align}
S_1& = \{ (z, w)\in \mathbb S^3: |z| \ge |w|\}  \\
S_2& = \{ (z, w)\in \mathbb S^3: |z| \le |w|\}  
\end{align}
Then the intersection 
$$S_1\cap S_2 = \{ (z, w)\in \mathbb S^3: |z| = |w|\}$$
is exactly the Clifford torus. Each $S_i$ is diffeomorphic to the solid torus $\mathbb S^1 \times D^2$, where $D^2$ is the disc of radius $1/\sqrt 2$. An explicit diffeomorphism is given by
$$ f_1 : \mathbb S^1 \times D^2 \to S_1 , \ \ \ f(e^{i\theta}, z) = \left(\sqrt{1-|z|^2} e^{i\theta}, z\right)$$
and similar for $S_2$. 
You have to be precise when you talk about curvature. The instrinsic curvature of the Clifford torus is flat, but the extrinsic curvature (i.e. second fundamental form) is non-trivial. Indeed it is not totally geodesic. I would the analogy to hemisphere and equator is misleading, in that case the equator is totally geodesic. 
