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"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.

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    Let $y\to\infty$.2012-02-03
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    @Jonas Meyer I'm not following. Would you explain that further?2012-02-03
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    Fix $x$. The limit of the right-hand side as $y\to\infty$ is $-\infty$, but the same is not true for the left-hand side.2012-02-03
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    «It *seems* daunting»?! Have you tried? I hate when my students say that kind of things to me... :/2012-02-03
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    @Mariano: Endurance and persistence seem to be rather uncommon traits with young ones these days, I'm told.2012-02-03
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    Indeed! I sometimes tell them that I won't answer the question unless they bring a stack of three or four pages of calculation...2012-02-03
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    @Mariano: Of course I have tried. The complicated result led me to believe that I was approaching the problem the wrong way, so I thought I would ask about it here. It *seems* to me that it is a reasonable question; if you believe it is somehow inappropriate then close it.2012-02-03
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    @I.J.Kennedy It was in my original post.2012-02-17

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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.

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    This has the added benefit of almost telling us what to do in the complex case: since an affine equivalence must take asymptotic directions to asymptotic directions, the matrix of the equivalence is determined.2012-02-03
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  • 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»...)

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    An enlightening question is: why does this break over the complex numbers?2012-02-03
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    Yes, but I was hoping for a way that doesn't use that fact (we haven't proved it in class yet). My apologies for not making that clear in the question.2012-02-03
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    What haven't you proved in class? In any case, my two bullet points are simply suggestions for you to actually prove them, so if you did not cover them in class it does not matter :D2012-02-03
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    We have not proved the second statement. My mistake for interpreting your two bullet points as an answer to the question.2012-02-03
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    I also did not prove Sylvester's Law of Inertia! :P2012-02-03
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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.

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    I had to double-take when you said "the other Mariano."2012-02-03
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    @anon: there are way too many Marianos in this thread... ;)2012-02-03
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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...

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    I would say that for this particular problem, expanding *is* the best way of doing it :D2012-02-03
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    Agreed. When looking at the expanded form I failed to notice the significance of the $(b^2 + d^2)y^2$ term.2012-02-03