Embedding Klein bottles in 4-space A question about topology from an ignorant logician, so please be kind if this is obvious!
We all know that the Klein bottle, unlike the torus, cannot be embedded in 3-space. And we all know (because we were told) that it can be embedded in 4-space.  I can see how to embed the torus in 3-space, using only quadratic stuff (I can do that with stuff I learnt at school) but how does one embed the Klein bottle in 4-space?  I am asking because I think if I had an equation to pour over I might start to get a feel for how the injection works and what the bottle looks like.  In particular, I would like to know if the isometry group of the embedded bottle looks like the isometry group of the embedded torus in 3-space.  I'm hoping it does, because that would put flesh on the assertion that there isn't really any self-intersection.
I asked this of a visiting topologist but she didn't know off the top of her head so I'm trusting that this isn't a stupid question!
 A: The parametrization for the Klein bottle provided by Will Brian is
\begin{align}
x &= (a+b\cos v)\cos u\\
y &= (a+b\cos v)\sin u\\
z &= (b\sin v)\cos(u/2)\\
t &= (b\sin v)\sin(u/2)\\
\end{align}
where $a>b>0$.
This leads to the defining conditions:
\begin{align}
4a^2(x^2+y^2)&=(a^2-b^2+t^2+x^2+y^2+z^2)^2\\
y(z^2-t^2)&=2txz\\
tyz&>0
\end{align}
The first equation comes from expressing $b \cos v$ both in $x,y$-terms and in $z,t$-terms.
The second equation comes from expressing $\tan u$ both in $x,y$-terms and in $z,t$-terms.
The inequality comes from $(a+b \cos v)(b \sin u \sin v)^2 > 0$.
There may be one other independent inequality also.
A: If embedding means smooth embedding, the non-orientability of the Klein bottle $K$ implies that its stable normal bundle will also not be trivial.  This implies that $K$ cannot be smoothly described as the solutions to an equation $f(\bf x)=\bf b$, with $\bf b \in \mathbb R^2$ a regular value of a smooth function $f: \mathbb R^4 \rightarrow \mathbb R^2$, i.e. solutions to two equations in four unknowns, which seems to be your goal.
(Guilliman and Pollack's differential topology book is a friendly place to read about the mathematics here.)
A: Step by step. First, we have the unit circle in the complex plane:
$$ S\ :=\ \{s\in\Bbb C:\ |s|=1 \} $$
We may even have a larger circle:
$$ C\ :=\ \{5\!\cdot\! s:\ s\in S\} $$
Then, a torus $\ T\ $ is a surface around circle
  $\ C\times\{0\}\subseteq \Bbb C\times\Bbb R:$
$$ T\ :=\ \{ ((5+\Re s)\cdot c,\, \Im s)\,:\,\ (c\ s)\in S\times S\}
   \quad\subseteq\quad\Bbb C\times\Bbb R $$
Of course:
$$ T\ =\ \{ ((5+\Re s)\cdot c^2,\, \Im s)\,:\,\ (c\ s)\in S\times S\}
   \quad\subseteq\quad\Bbb C\times\Bbb R $$
Finally, let's obtain Klein bottle $\ K\subseteq\Bbb C\times\Bbb C\ $
by twisting $\ T\ $ half as fast as points of $\ T\ $ rotate around
$\ (\mathbf 0\ 0)\in\Bbb C\times\Bbb R:$
$$ K\ :=\ \{ ((5+\Re s)\cdot c^2,\, \Im s\cdot c)\,:
             \,\ (c\ s)\in S\times S\}
   \quad\subseteq\quad\Bbb C\times\Bbb C $$
A: $\bigg\{ (z_1,z_2)\in \mathbb{C}^2 | |z_1|^2+|z_2|^2=1\bigg\}$ is
3-dimensional sphere $S$. When $ \Sigma =\bigg\{ (z_1,z_2)\in
\mathbb{C}^2 | |z_1| = |z_2| = \frac{1}{\sqrt{2}} \bigg\}$ is a torus in
$S$, then we know that $S$ is a union of two solid torus. 
Here $((R+a\cos\ t)\cos\ \theta,(R+a\cos\ t)\sin\ \theta,a\sin\ t)$
is a parametrization for torus $\Sigma$ in $\mathbb{R}^3$. Further,
a circle $$\bigg((R+a\cos\ \frac{\theta }{2} )\cos\ \theta, (R+a\cos\
\frac{\theta }{2} )\sin\ \theta,a\sin\ \frac{\theta
 }{2} \bigg) $$ in $\Sigma$ can be a boundary of Mobius band, which is in
 solid torus. Hence since Klein bottle $K$ is a union of two Mobius
 bands, then $K$ is in a $3$-dimensional sphere. 
