Show that there is no affine transformation that takes a circle to a hyperbola in $\mathbb{R}^2$

Question

"Show that the standard circle (defined by $f(x,y) = x^2 + y^2 - 1$) is not equivalent to the standard hyperbola (defined by $g(x,y) = x^2 - y^2 - 1$). That is, show that there is no $[A,\overline{s}] \in \text{Aff}(\mathbb{R}^2)$ such that $[A,\overline{s}] \cdot f(x,y) = g(x,y)$. Check that there is such an $[A,\overline{s}]$ if we allow $A \in \text{GL}_2(\mathbb{C}).$"

I've reduced this to showing that there are no $a,b,c,d,s,t \in \mathbb{R}$ such that $$f(ax+as+by+bt,\: cx + cs + dy+dt) = g(x,y).$$ $$\Rightarrow(ax+as+by+bt)^2+(cx + cs + dy+dt)^2 - 1=x^2-y^2-1$$ How should I proceed? Expanding that expression probably isn't the best way to do it.

Answer 1

Why think when you can simply compute?

So $f(x,y)=x^2+y^2-1$. Consider the affine transformation given by $$\left\{\begin{array}{l}x\leftarrow ax+by+s\\y\leftarrow cx+dy+t\end{array}\right.$$ Applying it to $f$ we get $$f(ax+by+s, cx+dy+t)=(a^2+c^2) x^2+2(a b + c d)xy+(b^2 + d^2)y^2+\text{terms of lower degree}.$$ If this is to be equal to $g(x,y)=x^2-y^2-1$, then, in particular, looking at the coefficient of $y^2$ we see that we must have $$b^2+d^2=-1.$$ Of course, this is not going to work...

Answer 2

• In $\mathbb R^2$, a circle is a bounded set and a hyperbola is not.

• An invertible affine transformation of the plane maps bounded sets to bounded sets.

(One can replace «bounded» by «compact» or «connected»...)

Answer 3

The other Mariano is of course cheating: this is algebraic geometry after all...

The gentlemanly way of doing this is to observe how the coefficients of the homogeneous part of top degree changes when you do an affine transformation, and then invoke Sylvester's Law of Inertia.

In more detail... Let $f\in \mathbb R[x,y]$ be a polynomial of degree two, which we can write uniquely as $$f(x,y) = v^tav+b^tv+c$$ with $v$ the vector $\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)$, $a$ a symmetric $2\times 2$ matrix, $b\in\mathbb R^2$ and $c\in\mathbb R$.

If $[A,\bar s]$ is an affine transformation, then $$[A,\bar s]\cdot f=v^ta'v+b'^tv+c'$$ for some new $a'$, $b'$ and $c'$ which we can compute explicitly. Most interestingly, we have $$a'=A^taA.$$ Sylvester's Law of Inertia then implies that $a$ and $a'$ have the same number of positive eigenvalues, the same number of negative eigenvalues, and the same number of zero eigenvalues.

Now, the matrix $a$ corresponding to your $f$ is $\begin{pmatrix}1&0\\0&1\end{pmatrix}$, while that of $g$ is $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. Since they have different numbers of negative eigenvalues, there is no affine transformation mapping one to the other.

Answer 4

Let us call $f_0$, $f_1$ and $f_2$ the parts of $f$ which are homogeneous of degree $0$, $1$ and $2$, respectively, so that in particular $f=f_2+f_1+f_0$, and similarly for $g$.

If $[A,s]$ is an invertible affine transformation and $[A,s]\cdot f=g$, then one checks easily that $[A,s]\cdot f_i=g_i$ for each $i\in\{0,1,2\}$. In particular, $$[A,s]\cdot(x^2+y^2)=x^2-y^2.$$ But two polynomials which are affinely equivalent are either both irreducible or both reducible, yet $x^2+y^2$ is irreducible and $x^2-y^2$ is not.

This argument is «geometric»: we are using the fact that a circle has no asymptotic directions while the hyperbola has two, and that the property of having asymtotic directions is preserved under affine equivalence.