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Let $k$ be an algebraically closed field. We define $Q=V(F)$ where $F$ is an irreducible (and homogeneous) polynomial of degree 2 in $X,Y,Z,T$ and hence gives rise to a quadratic form on $k^4$ which we assume to be non-degenerate. Prove that there is a homography $h$ such that h(Q) = V(XT-YZ).

[Perrin: Algebraic Geometry, II. Exercise 3]

I tried the following: I first tried to construct the homography by finding explicit zeros of $F$: $k$ is algebraically closed so there are $y,z,t\in k$ such that $[1:x:0:0]$, $[1:0:y:0]$ and $[1:0:0:z]$ are zeroes of $F$.

By assuming that the image of a quadric by an homography is still a quadric, we can send those three lines into $[1:0:0:0]$, $[0:1:0:0]$ and $[0:0:1:0]$. We can now cancel out $x^2,y^2$ and $z^2$ from the quadratic form we obtained, however there are a lot of coefficients left to remove.

I guess we need to use the fact that it's non degenerate but I don't really see how. My ultimate guess is that we can construct more independant lines that are zeroes of F, but I don't see how. Is my approach of the problem wrong?

J. Gushway
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1 Answers1

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So my answer is mostly algebraic. Since we mention non-degenerate quadratic forms, it means we are considering $k$ to be of caracteristic different from $2$.

Fist recall that any quadratic form $q:k^n\rightarrow k$ can be diagonalized, i.e. there is a linear change of variables, or isomorphism $\varphi:k^n\rightarrow k^n$ such that $q\circ\varphi(x_1,\ldots,x_n)=a_1 x_1^2+\ldots+a_n x_n^2$ for some $a_1,\ldots,a_n\in k$. Consider $b(v,w):=\frac{1}{2}(q(x+y)-q(x)-q(y))$ the associated bilinear form of $q$, its polarization. There are several ways to prove that any quadratic form (in any field s.t. $car(k)\neq2$) can be diagonalized:

  • either proceed by "completing the square, by eliminaing cross-products (cf. here for an example);
  • either by induction on the dimension of the vector space you are in; basically find yourself a vector $v$ such that $q(v)\neq 0$, then consider $$v^\perp :=\{w\in k^n;~b(v,w)=0\}$$ Considering $q$ is non-degenerate, it means that the matrix of $b$ will be non-degenerate, and so we have that $\dim v^\perp=n-1$ so we can proceed by induction.

Consider your quadratic form $F(X,Y,Z,T)$ and diagonalize it: there is a linear change of variables $A:k^4\rightarrow k^4$ such that $F\circ A(x,y,z,t)=ax^2+by^2+cz^2+dt^2$. Since $\overline{k}=k$, we have $F(A(X,Y,Z,T))=aX^2+bY^2+cZ^2+dT^2$. Since $F$ is supposed to be nondegenerate, we have $a,b,c,d\neq 0$, so we can further pose a change of variables $$B=\begin{pmatrix}\frac{1}{\sqrt a}\\ &\frac{1}{\sqrt b}\\ &&\frac{1}{\sqrt c}\\ &&&\frac{1}{\sqrt d}\end{pmatrix}$$ and set $$F\circ AB(X,Y,Z,T)= X^2+Y^2+Z^2+T^2.$$ Now we can set $i\in k$ such that $i^2=-1$, since $k$ is algebraically closed. We can thus factorize: $$X^2+Y^2+Z^2+T^2=(X+iT)(X-iT)-(iY+Z)(iY-Z).$$ We can finally set the last linear change of variables by setting $X'=X+iT,T'=X-iT,Y'=iY+Z,Z'=iY-Z$ or by writing $$C=\frac{1}{2}\begin{pmatrix}\phantom{-}1 &&& 1 \\ & -i & -i & \\ & \phantom{-}1 & -1 & \\ -i &&& i\\ \end{pmatrix}$$ and $F\circ ABC(X,Y,Z,T)=XT-YZ$. The sought after homography is the image in $PGL(4,k)$ of $ABC\in GL(4,k)$.

For a general case, you can check this other MSE answer.

Andrei.B
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