Quadratic Forms on a (finite dimensional real) vector space with same zero set are scalar multiples?

Question

Let $V$ be a finite-dimensional vector space over $\Bbb{R}$, let $g,h:V \times V \to \Bbb{R}$ be bilinear symmetric functionals, and let $Q_g,Q_h:V \to \Bbb{R}$ be the associated quadratic forms. Suppose that $Q_g, Q_h$ have the same zero set; $Q_g^{-1}(\{0\}) = Q_h^{-1}(\{0\})$.

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Questions:

1. Is it then true that $Q_g$ and $Q_h$ are scalar multiples of each other?
2. If (1) isn't true as stated, could it perhaps be made true by making certain restrictions on the quadratic forms, such as restricting them to have Lorentz signature (i.e with the notation below, $p=1, n$ arbitrary and $k=0$). This is the case I'm mainly interested in, but I'd of course like to know it in the more general case if it's true.
3. Can this result (if true) be generalized to arbitrary multilinear, symmetric functionals on $V$ and their associated homogeneous polynomials (as opposed to bilinear functionals and their quadratic forms). If so could you outline such a proof/ provide a reference.

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My Attempt

This reminds me of a similar result in linear algebra, namely that if $\phi,\psi \in V^*$ have the same kernel then they are scalar multiples of each other. So, my attempt at "proving" (1) was to mimic that proof as much as possible. I know that every quadratic form over $\Bbb{R}$ can be "diagonalized" (Sylvester's Law?), in the sense that we can find a basis $\beta$ for $V$ such that the matrix representation of $g$ is of the type \begin{align} [g]_{\beta} &= \begin{pmatrix} I_p & & \\ & -I_n & \\ & & 0_k \end{pmatrix} \end{align} such that $p+n+k = \dim V$. But from here, I'm not sure how to proceed. Any help is appreciated.

Answer 1

(1) isn't true as stated because it is certainly possible to have inner products which are not scalar multiples; we need to make certain restrictions on the signature. In particular, everything holds for Lorentzian signature. Finally, I'm still not sure about (3).

Theorem ($2$):

Let $n,p\geq 1$ be integers, $V$ a vector space over $\Bbb{R}$ of dimension $d=n+p$, and let $h$ be a pseudo-inner product of signature $(n,p)$ ($n$ negatives , $p$ positive). If $g$ is any bilinear, symmetric functional on $V$ such that $Q_h^{-1}(\{0\})\subset Q_g^{-1}(\{0\})$, then there is a constant $C\in\Bbb{R}$ such that $g=Ch$. Furthermore, if we assume $n\neq p$, and that $g$ also has signature $(n,p)$, then $C>0$.

Before proving this let's establish some conventions: I shall use greek indices to run through $\{1,\dots, n\}$ (i.e the "time-like" indices) and latin ones for $\{n+1,\dots, n+p\}$ (the "spacelike indices").

Now, fix a basis $\{v_1,\dots v_d\}$ of $V$ such that $h$ has the matrix representation $\begin{pmatrix} -I_n & 0 \\0 & I_p\end{pmatrix}$ (this is always possible by a type of Grahm-Schmidt argument, I believe it's called Sylvester's law of inertia?). For each $x\in\Bbb{R}^n,y\in \Bbb{R}^p$ having unit norms, consider the vector $w=x^{\alpha}v_{\alpha} + y^iv_i\in V$ (summation convention implied, and for convenience I shall write $y=(y^{n+1},\dots, y^{n+p})$ just to make the indices match up). Then, $h(w,w)=-\lVert x\rVert^2+ \lVert y\rVert^2=0$, which means by assumption, $g(w,w)=0$ as well. If we write this out more explicitly, we get \begin{align*} g_{\alpha\beta}x^{\alpha}x^{\beta} + 2g_{\alpha i}x^{\alpha}y^i + g_{ij}y^iy^j &= 0. \tag{$*$} \end{align*} First replace $y$ with $-y$ and subtract the equations to get \begin{align} g_{\alpha i}x^{\alpha}y^i &= 0. \end{align} Since this is true for all $x,y$ with unit norms, it follows each $g_{\alpha i}=0$. With this, $(*)$ reduces to \begin{align*} g_{\alpha\beta}x^{\alpha}x^{\beta} &= -g_{ij}y^iy^j. \tag{$**$} \end{align*} Since we have "separated" the $x$ and $y$'s and they are arbitrary, it follows all the "off-diagonal" terms vanish. More explicitly, for each $\alpha\neq \beta$, take $x_{\pm}=\frac{1}{\sqrt{2}}(e_{\alpha}\pm e_{\beta})$. Then, since the RHS doesn't depend on $x$ we find that $g_{\alpha\beta}=-g_{\alpha\beta}$, and hence it is $0$. Similarly, if $i\neq j$, then $g_{ij}=0$. Finally, letting $x$ range over $e_1,\dots, e_n$ then letting $y$ range over $e_{n+1},\dots, e_{n+p}$, we see that

\begin{align*} [g]&=-g_{11} \begin{pmatrix} -I_p & 0\\ 0& I_q \end{pmatrix}, \end{align*} or equivalently, $g=-g_{11}\cdot h$. Therefore, we have shown that $g$ and $h$ are related by a scale factor. Finally, if we assume $n\neq p$ and that both $g,h$ have signature $(n,p)$ then it follows that $-g_{11}>0$ (otherwise the signatures wouldn't match up). This completes the proof.

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Note in particular that for Lorentzian signature $(-,+,+,+)$ (or $(+,-,-,-)$, it doesn't matter) that if $g,h$ have the same light cones then $g=Ch$ for some $C>0$; and this is exactly the statement made in Landau and Lifshitz when they say something along the lines of "since $ds^2$ and $ds'^2$ are infinitesimals of the same order they are proportional".

Answer 2

1. Is not true clearly. In fact you can choose two different scalar products $f,g$ on $V=\mathbb{R}^n$, for example, such that

$Q_f(x)=||x||$;

$Q_g(x)=||x||_1=\sum_{i} |x_i|$

Now you have $Q_f^{-1}(0)=\{0\}=Q_g^{-1}(0)$ but $Q_f$ is not a multiple of $Q_g$.

2. I don’t know a condition of this type and I think there isn’t .

Answer 3

Partial answer:

As can be seen in the positive definite case, it does not generally hold that two bilinear forms are scalar multiplies if they have the same zero set.

However, the statement becomes true for the case of $\dim V \leq 3$ if we extend the bilinear form to a $\Bbb C$-bilinear form over $\Bbb C$. Without loss of generality, take $V = \Bbb R^3$. Plugging in $v_1 = (1,x,x^2)$, we see that $$ Q_g(v_1) = g_{11} + 2g_{12} x + 2(g_{22} + g_{13}) x^2 + 2g_{23} x^3 + g_{33}x^4. $$ Plugging in $v_2 = (-1,x,x^2)$ yields $$ Q_g(v_2) = -g_{11} - 2g_{12} + 2(g_{22} - g_{13}) x^2 + 2g_{23} x^3 + g_{33} x^4. $$ If two polynomials over $\Bbb C$ have the same zero-sets, then those polynomials must be multiples. Thus, if $h$ is such that $Q_h(v_1) = Q_h(v_2) = 0$ for all $x \in \Bbb C$, then it must hold that $h_{ij}$ is a multiple of $g_{ij}$ for all $j$, and hence $h$ is a multiple of $g$.