I'm reading a paper, which considers the space $X$ of polynomials of the form $p(ax+by)$ for $p$ a single variable quadratic function (i.e. $p(x,y) = c(ax+by)^2 + d(ax+by) + e$ for $a,b,c,d,e \in \mathbb{R}$) such that $\int\int_D p(x,y) \ dx dy = 0$ and $\int \int_D p^2(x,y) \ dxdy = 1$, where $D$ is the unit disk. The paper claims that this space is homeomorphic to the Klein bottle, however I'm struggling with some of the details.
Specifically, it says that the space $P$ of all polynomials in two variables (i.e. $f(x,y) = b_0 + b_1x+b_2y + b_3x^2 + b_4 xy + b_5 y^2$) such that $\int \int_D f = 0$ and $\int \int_D f^2 = 1$ is homeomorphic to a 4-dimensional ellipsoid. That part I proved and agree with. From there, we consider the subspace $X$ of $P$. Plugging into the constraint $\int\int_D p(x,y) \ dx dy = 0$ gives me that $e = -\frac{1}{4} c (a^2+b^2)$, so $X$ has polynomials of the form $c(ax+by)^2 + d(ax+by) - \frac{c(a^2+b^2)}{4}$. Plugging into the constraint $\int \int_D p^2(x,y) \ dxdy = 1$ gave me an algebra mess.
The text says to consider the function $g:S^1 \times S^1 \rightarrow c(ax+by)^2 + d(ax+by),$ which is onto and has the relations $g(a,b,c,d) = g(-a,-b,c,-d)$. So letting $\theta = (a,b) \in S^1$ and $\phi = (c,d) \in S^1$, we have $(a,b) \sim (-a,-b) = \theta + \pi$ and $(c,d) \sim (c,-d) = 2\pi - \phi$. Then since $g$ is surjective onto $\{c(ax+by)^2 + d(ax+by) \mid a^2+b^2 = c^2 + d^2 = 1\} \equiv Y$, we have $Y = \text{Im}(g) = S^1 \times S^1 \big/ [(\theta,\phi) \sim (\theta+\pi, 2\pi-\phi)] = K$ (the Klein bottle). However, we want $X$ (not $Y$) to be homeomorphic to $K$.
My question is: how can we get a homeomorphism from $X$ to $K$? I know that:
$Y$ is a subset of a 4-dimensional ellipsoid (which is homeomorphic to $S^4$), so I could consider $g:S^4 \rightarrow Y$ such that $g(a,b,c,d) = c(ax+by)^2 + d(ax+by) - \frac{c(a^2+b^2)}{4}$, which has the same relations $(a,b,c,d) \sim (-a,-b,c,-d)$.
So $Y$ could be homeomorphic to $S^4 \big/ [(a,b,c,d) \sim(-a,-b,c,-d)]$, but I don't see how those identifications on $S^4$ yield a Klein bottle. (Also letting $(a,b) = \theta, (c,d) = \phi$ seems wrong since $(a,b), (c,d)$ each don't necessarily lie on a copy of $S^1$. )
I also know that the space $A$ of single variable quadratic functions $q(t) = -\frac{c(a^2+b^2)}{4} + c t + dt^2$ in $X$ form an ellipse ($\frac{d^2}{4\pi}+\frac{9c^2}{72\pi}) = 1$, which is homeomorphic to $S^1$. (Plugging these functions into $\int\int_D q = 0$ and $\int\int_D q^2 = 0$ is much easier.) Also, for $g(a,b,c,d) = c(ax+by)^2 + d(ax+by) - \frac{c(a^2+b^2)}{4}$ in $Y$, $g(a,b,c,d)(x,y) = q((a,b) \cdot (x,y))$. Maybe this is helpful?