Immersions of the Klein Bottle The famous immersion of the Klein bottle lacks symmetry. (I'm talking about this one. http://en.wikipedia.org/wiki/File:Klein_bottle.svg) One can only see one plane of reflection.
However, the "square with sides identified" has (well, I think) in my opinion at least two non-trivial symmetries ; one which is reflection along a vertical line passing through the center of the square, and one is rotation of half a turn centered in the center of the square.
My question : is there an immersion of the Klein Bottle which makes the symmetry group of the immersion visible as a subgroup of the group of rotations/reflections of Euclidean 3-space?
Attempts : I simply tried to look at the known immersions of the Klein bottle in Euclidean 3-space, but looking at the Klein bagel (http://kleinbagel.com/) or the classical bottle wasn't very enlightening. This seems like an interesting question but I don't know much about the bottle itself.
 A: I have what is almost an immersion with the required symmetry, except it is not smooth at two points.
First let us consider a slightly unusual way of constructing a torus. Begin with a tube with flared ends. Pull the two circular ends back over the tube, so part of the tube is now inside the "folded over" part (a bit like turning a sock inside out). Keep this up until we can glue the two circular ends together. Here's a picture:

For the Klein bottle, we want to reflect one of the circular ends across a diameter before gluing it to the other end. Let's do this before going through the "folding over" construction just described. So start with a tube with a self-intersection in the middle. The tube is obtained by starting with a vertical circle, moving it to the right and simultaneously compressing it into a thinner and thinner elliptical shape. Eventually the moving circle becomes a line segment and then re-expands back to a circle. Like this:

Those two X's are supposed to suggest where the tube self-intersects. At the top and the bottom of the self-intersection, we have the failure of smoothness.
Now we can do the flaring, folding over, and end-gluing as before, resulting in the Klein-bottle.
Let us convince ourselves that this is a Klein bottle. We have the standard identification diagram:

After identifying $A_1$ and $A_2$ we have a horizontal tube. Now we need to identify $B_1$ and $B_2$, which have become circles after the first identification. However, one circle has to be "flipped over" so that the orientations match up as indicated by the arrows. The usual immersion accomplishes this by rotating one circle through 180 degrees, pulling the tube around behind it. However, if we focus our attention on what the 180 degree rotation does as a mapping of the circle to itself, we see it's just reflection in the y-axis: $(x,y)\mapsto (-x,y)$. (That's if we plot the circle in the xy-plane.) The "thinner, then thicker ellipse" transformation accomplishes the same thing. Analytically: $(x,y)\mapsto (tx,y)$ where $t$ varies from 1 to $-1$.
As for smoothness except at two points, note that along the line of self-intersection, we should really think of surface as passing through itself without intersecting. (Just like with the Riemann surface of $\sqrt{z}$ -- or the circle of self-intersection on the usual immersion.) Here's a parametrization of just the self-intersecting tube (before flaring and folding over). First let's put coordinates on the regular tube. Let's use $(x,y,t)$ where $t$ is the horizontal axis. Let $x$ and $y$ be the coordinates in a plane perpendicular to the plane of the paper (or computer screen, I guess). All the points of the original tube are given by $-1\leq t\leq 1$, $x^2+y^2=1$. Now let $f(x,y,t)=(tx, y, t)$. In a neighborhood of $t=0$ and $x,y$ with $x\neq 0$ (so $-1<y<1$), $f$ is smooth, and has a smooth inverse. Just to emphasize: if $U$ is this open set containing $(x,y,t)$, then $f:U\rightarrow f(U)$ is smooth and has a smooth inverse with domain $f(U)$. (However, $f(U)$ is not open in the relative topology from $\mathbb{R}^3$.) When $x=0$ (i.e., $y=\pm 1$), the inverse mapping is not smooth.
