Let $F:\mathbf{R}^n\rightarrow\mathbf{R}$ be a non-degenerate quadratic forms. Let $$ O(F):=\{g\in GL_n(\mathbf{R}):F(gv)=F(v),\forall v\in \mathbf{R}^n\} $$ be the isotropy group of $F$.

Q: So how does one prove in the simplest way possible that if $O(F)=O(G)$ then there exists $\lambda\in\mathbf{R}^{\times}$ such that $F=\lambda G$?

P.S. I would like to have a proof that could be explained to undergraduate students who take an advanced linear algebra class.


A relatively easy proof also follows from using the reflection identity: First, define the inner product associated to $F$, namely $v\ \cdot_F\ w = {\frac12}\bigl(F(v{+}w)-F(v)-F(w)\bigr)$, and then, for any $v$ with $F(v)\not=0$, define the reflection in $v$ by $$ \rho^F_v(w) = w - 2\ \frac{v\ \cdot_F\ w}{v\ \cdot_F\ v}\ v $$ This map $\rho^F_v$ belongs to $O(F)$, as is easy to verify. By your assumption, it belongs to $O(G)$ as well. This works out to imply $$ \bigl(v\ \cdot_F\ v\bigr)\bigl(v\ \cdot_G\ w\bigr) = \bigl(v\ \cdot_G\ v\bigr)\bigl(v\ \cdot_F\ w\bigr). $$ Now if you replace $v$ by $v+tw$ in this equation and compare $t$-linear terms, you'll get $$ \bigl(v\ \cdot_F\ v\bigr)\bigl(w\ \cdot_G\ w\bigr) = \bigl(v\ \cdot_G\ v\bigr)\bigl(w\ \cdot_F\ w\bigr), $$ i.e., $F(v)G(w) = F(w)G(v)$, from which the result is obvious.

  • $\begingroup$ @Hugo: Thanks, but, even though I don't remember a reference for it off the top of my head, I'm sure that this argument is classical. Note that it works over any field of characteristic not equal to $2$ and it does not assume that the vector space has finite dimension. If you look up references to the Cartan-Dieudonné Theorem, you'll probably find it written up explicitly (and more elegantly, I expect) in one of those references. $\endgroup$ – Robert Bryant Feb 7 '12 at 16:01
  • $\begingroup$ Also a key observation in your argument is that the reflection $\rho_v^F$ only depends on the $\mathbf{C}$-line spanned by $v$, namely $\mathbf{C}v$. moreover there exists a unique line $L$ such that for all $w\in L$ one has that $\rho_v^F(w)=-w$ namely the line spanned by $v$. So using this obervation with the assumption that $O(F)=O(G)$ one may deduce that $\rho_v^F=\rho_v^G$. $\endgroup$ – Hugo Chapdelaine Feb 7 '12 at 17:08
  • $\begingroup$ So in other words, give a line $L$ and a quadratic form $F$, there exists a unique linear map $s:\mathbf{R}^n\rightarrow\RR^{n}$ such that $s(w)=-w$ for all $w\in L$, $s^2=1$ and $s$ preserves $F$-length. So this uniqueness result is important in your argument. $\endgroup$ – Hugo Chapdelaine Feb 7 '12 at 17:22
  • $\begingroup$ @Hugo: Umm,...no. I didn't use uniqueness at all (and it's false anyway, since, for example, the map $v\mapsto -v $ has all of these properties, too). What I used was that, for each $v\in V$ with $F(v)\not=0$, there is an explicit formula for an element $\rho^F_v$ in $O(F)$ and that assuming that all of those particular elements also belong to $O(G)$ implies the relation $F(v)G(w)=F(w)G(v)$ for all $v,w\in V$ purely by algebra. Of course, it then follows that $\rho^F_v = \rho^G_v$. I guess you could argue directly that $\rho^F_v\in O(G)$ implies $\rho^F_v = \rho^G_v$, but I didn't do that. $\endgroup$ – Robert Bryant Feb 7 '12 at 18:27
  • $\begingroup$ @Robert, you are right, it is not unique. Then how do you get the equality $\bigl(v\ \cdot_F\ v\bigr)\bigl(v\ \cdot_G\ w\bigr) = \bigl(v\ \cdot_G\ v\bigr)\bigl(v\ \cdot_F\ w\bigr).$ $\endgroup$ – Hugo Chapdelaine Feb 7 '12 at 18:48

Robert's answer involves picking any $v$ and $w$ in ${\mathbf R}^n$ and getting an identity of polynomials of degree at most 2 in $t$ away from where $F(v+tw) = 0$, which excludes at most two values of $t$. He equates the coefficients in the identity to get the desired formula.

The same technique works for nondegenerate quadratic forms over any field with at least five elements (since a quadratic polynomial function on a field is completely determined away from two points if at least three points remain). Of course we can ignore fields of size 2 and 4, since that would be in characteristic 2, but what happens in the case of quadratic forms over a field of size 3? Admittedly the question was being asked over the real numbers for the purpose of a linear algebra class, so the whole point of checking finite fields isn't essential, but it's worth noting that the result is still true. I'll indicate the steps, but leave out computational details.

We have two nondegenerate quadratic forms $Q_1$ and $Q_2$ on a finite-dimensional vector space $V$ over a field $K$ of characteristic not 2 (EDIT: I will dicuss characteristic 2 at the end), and we assume the isotropy groups of $Q_1$ and $Q_2$ are the same. We want to show if $Q_1(v) \not= 0$ that $Q_2(v) \not= 0$ and $Q_2(w) = cQ_1(w)$ for all $w \in V$, where $c = Q_2(v)/Q_1(v)$.

a) Let $v$ in $V$ be nonzero, $L \colon V \rightarrow K$ be linear, and $B \colon V \times V \rightarrow K$ be a nondegenerate bilinear form. If $B(w,v)L(w) = 0$ for all $w \in V$, then $L(w) = 0$ for all $w \in V$. (Hint: There is a basis of $V$ in the complement of any hyperplane.)

b) Let $B_1$ and $B_2$ be the symmetric bilinear forms associated to $Q_1$ and $Q_2$. Suppose $v \in V$ satisfies $Q_1(v) \not= 0$. Then by part a and a reflection as in Robert's answer, $$Q_1(v)B_2(v,w) = Q_2(v)B_1(v,w)$$ for all $w \in V$. (The reflection $s_v(w) = w - (2B_1(v,w)/Q_1(v))v$ is in the isotropy group of $Q_1$, and therefore $Q_2(s_v(w)) = Q_2(w)$ for all $w$, which implies the above equation from nondegeneracy of $B_1$.)

c) Assume from now on that $Q_1$ and $Q_2$ have the same isotropy group. Using part b, for all $v \in V$ we have $Q_1(v) \not= 0$ if and only if $Q_2(v) \not= 0$, and when $Q_1(v) \not= 0$ we have $\{w : B_1(v,w) = 0\} = \{w : B_2(v,w) = 0\}$.`

d) If $Q_1(v) \not= 0$ and $B_1(v,w) \not= 0$ then $Q_2(w) = (Q_2(v)/Q_1(v))Q_1(w)$. (To prove this, by part c we can assume $Q_1(w) \not= 0$.)

e) If $Q_1(v) \not= 0$ and $B_1(v,w) = 0$ then $Q_2(w) = (Q_2(v)/Q_1(v))Q_1(w)$. (By part c, $B_2(v,w) = 0$. We have $B_1(v,v+w) = B_1(v,v) = Q_1(v) \not= 0$, so we can apply part d with $v+w$ in place of $w$.)

Now what happens in characteristic 2? The bijection between quadratic forms and symmetric bilinear forms breaks down, so the notion of isotropy group or nondegeneracy has to carefully distinguish between quadratic forms and symmetric bilinear forms. Let me set up the terminology that I'll use, to minimize the chance of confusion.

Let $K$ be a field of characteristic 2. A quadratic form on a finite-dimensional $K$-vector space $V$ is a function $Q \colon V \rightarrow K$ that looks like a homogeneous quadratic polynomial in one (equivalently, any) choice of basis of $V$, and this is the same thing as saying the function $B(v,w) := Q(v+w) - Q(v) - Q(w)$ from $V \times V$ to $K$ is a bilinear form, and it is obviously symmetric. (Classically we'd want to divide this $B$ by 2, so $B$ here is analogous to $2B$ outside characteristic 2.) We have $B(v,v) = Q(2v) - 2Q(v) = 2Q(v) $, and in characteristic 2 this is 0, so we can't recover values of $Q$ from values of $B$. The points is that many $Q$'s can have the same $B$.

Example: On $K^2$, let $Q(x,y) = ax^2 + bxy + cy^2$. Then you can check from the characteristic being 2 that $B((x,y),(x',y')) = b(xy' + yx')$. In particular, $a$ and $c$ don't appear in $B$, so $x^2 + xy$ and $xy$ are two different quadratic forms on $K^2$ that have the same associated symmetric bilinear form.

An isometry of $(V,Q)$ is defined to be a linear map $A \colon V \rightarrow V$ such that $Q(Av) = Q(v)$ for all $v \in V$. An isometry of $(V,B)$ is defined to be a linear map $A \colon V \rightarrow V$ such that $B(Av,Aw) = B(v,w)$ for all $v$ and $w$ in $V$. Because $B(v,w)$ can be expressed in terms of values of $Q$, any isometry of $(V,Q)$ is an isometry of $(V,B)$, but in characteristic 2 the converse is false: the isometry group (or what you call isotropy group) of a quadratic form in characteristic 2 is smaller than the isometry group of its associated bilinear form.

Example: Let $Q(x,y) = ax^2 + bxy + cy^2$ with $b \not= 0$. Up to scaling by $b$, the bilinear form attached to $Q$ is $B((x,y),(x',y')) = xy' + yx'$, which is the standard alternating pairing on $K^2$, so the isometry group of $B$ is the symplectic group of $K^2$, and ${\rm Sp}(K^2) = {\rm SL}_2(K)$. By a direct calculus, a matrix $(\begin{smallmatrix}\alpha&\beta\\ \gamma&\delta\end{smallmatrix})$ with determinant 1 is an isometry of $(V,Q)$ if and only if $Q(\alpha,\gamma) = a$ and $Q(\beta,\gamma) = c$. For instance, $(\begin{smallmatrix}0&1\\1&0\end{smallmatrix})$ is in ${\rm SL}_2(K)$ and it is an isometry of $(V,Q)$ if and only if $a = c$, so when $a \not= c$ (ex: $x^2 + xy$) this particular $2 \times 2$ matrix that is an isometry of $B$ is not an isometry of $Q$.

We can define reflection in characteristic 2 as follows: if $Q(v) \not= 0$ then we set $s_v \colon V \rightarrow V$ by $$s_v(w) = w - \frac{B(v,w)}{Q(v)}v.$$ The $2B$ appearing in the classical reflection formula is exactly the same thing as the $B$ here, so this definition for a reflection is truly independent of characteristic (but depending on whether or not we're in characteristic 2 there are different conventions of what the bilinear form associated to a quadratic form is). You can check by a calculation that $Q(s_v(w)) = Q(w)$ for all $w$, so $s_v$ is in the isometry group of $(V,Q)$, and thus also is in the isometry group of $(V,B)$, which is a larger group in characteristic 2. And you can check by a direct calculation that $s_v^2$ is the identity on $V$, which is a property reflections should have. The catch is that $s_v$ might be the identity function. This happens if $v \in V^\perp = \{u : B(u,V) = \{0\}\}$, and here you're going to say "well, we are only working in the nondegenerate case". Now we have to deal with the issue of how you actually define nondegeneracy in characteristic 2, which is a bit different from what you're used to outside characteristic 2 (at least in some references you find a difference, so let's address it).

For a symmetric bilinear form $B$ the notion of nondegeneracy is the same in all characteristics: it means if $w \mapsto B(v,w)$ is identically 0 then $v = 0$. For a quadratic form $Q$ with associated symmetric bilinear form $B$, outside characteristic 2 we call $Q$ nondegenerate when $B$ is nondegenerate (and there is no difference between $B$ or $2B$ being called the bilinear form of $Q$ when $2 \not= 0$). But in characteristic 2 we do not define $Q$ to be nondegenerate when $B$ is nondegenerate.

Definition: If $Q$ is a quadratic form (in any characteristic) with associated symmetric bilinear form $B$, we call $Q$ nondegenerate when the two conditions $Q(v) = 0$ and $B(v,V) = \{0\}$ imply $v = 0$.

Outside characteristic 2, saying $Q(v) = 0$ is the same as saying $B(v,v) = 0$, so the condition $Q(v) = 0$ would be implied by $B(v,V) = \{0\}$ and thus that condition on $Q$ can be dropped from the above definition: outside characteristic 2 the above definition is the same thing as the usual notion of nondegeneracy for a quadratic form. But in characteristic 2 we can have $Q(v) \not= 0$ while $B(v,V) = \{0\}$.

Example: On $K^3$, when $K$ has characteristic 2, let $Q(x,y,z) = x^2 + xy + y^2 + z^2$. The associated symmetric bilinear form is given by $B((x,y,z),(x',y',z')) = xy' + y'x$. Notice there is no $z$ or $z'$ in this formula. If $B(v,K^3) = \{0\}$ then $v = (0,0,z)$, so $Q(v) = z^2$. Thus if $Q(v) = 0$ and $B(v,K^3) = \{0\}$ then $v = (0,0,0)$. But we can have $B(v,K^3) = \{0\}$ with $v$ being nonzero, such as $(0,0,1)$. The bilinear form $B$ is degenerate but we say the quadratic form $Q$ is nondegenerate.

Of course if it happens to be the case that $B$ is nondegenerate then $Q$ is nondegenerate by the above definition, so a quadratic form in char. 2 whose associated symmetric bilinear form is nondegenerate is a nondegenerate quadratic form, but we can have nondegenerate quadratic forms whose associated symmetric bilinear form is degenerate, as in the above example.

Returning to the reflection formula, we can have $Q(v) \not= 0$ while $B(v,V) = \{0\}$, and the so-called reflection $s_v$ is the identity function: $s_v(w) = w$ for all $w$. That is dumb. So we only define reflections $s_v$ when $Q(v) \not= 0$ and $B(v,V) \not= \{0\}$, since we need both conditions for $s_v$ to be defined and to have order 2.

Now it's time to go back to the 5 steps a, b, c, d, e above and see how they fare in characteristic 2: if $Q_1$ and $Q_2$ are nondegenerate quadratic forms with the same isometry group (or what you call isotropy group), does $Q_2 = cQ_1$ for some $c \in K^\times$?

Part a goes through without a problem since it isn't a property of quadratic forms at all. However, the nondegeneracy condition on $B$ there is going to be felt when we apply part a to later parts in characteristic 2.

As for part b, to make this go through you need to assume $Q_1(v) \not= 0$ and $B_1(v,V) \not= \{0\}$, which are two genuinely different conditions: the first does not imply the second like it would outside char. 2, where $Q$-values are special cases of $B$-values. If you assume $Q_1$ and $Q_2$ have the same isometry group, then for any $v$ with $Q_1(v) \not= 0$ and $B_1(v,V) \not= \{0\}$ then you can derive $$ Q_1(v)B_2(v,w) = Q_2(v)B_1(v,w) $$ for all $w \in V$ using the reflection $s_v$, just like part b is done outside characteristic 2. (When I wrote in the original part b that you use nondegeneracy of $B_1$, what you need is that $w \mapsto B_1(v,w)$ is not identically 0 with the specific $v$ where $Q_1(v) \not= 0$. It is, strictly speaking, weaker than nondegeneracy to assume that about $v$.) Notice, alas, that the two conditions $Q_1(v) \not= 0$ and $B_1(v,V) \not= \{0\}$ are not implied by nondegeneracy of $Q_1$l for a nondegenerate $Q_1$ we could have a $v$ where $Q_1(v) \not= 0$ while $B_1(v,V) = \{0\}$. The 3-dimensional nondegenerate quadratic form with degenerate bilinear form that I mentioned earlier is an example of this, with $v = (0,0,1)$.

To get the conclusion of part c to work in characteristic 2, it seems to me that you need to assume $B_1$ and $B_2$ are nondegenerate (which is stronger than saying $Q_1$ and $Q_2$ are nondegenerate). With this assumption, if $Q_1$ and $Q_2$ have the same isometry group then the conclusions of part c remain true in char. 2 by the same proofs as one uses outside char. 2.

Part d goes through in characteristic 2 if you assume $B_1$ and $B_2$ are nondegenerate (stronger than assuming $Q_1$ and $Q_2$ are nondegenerate).

Now we're at the last step, part e. Unfortunately, here there appears to be a serious problem related to $Q$-values not being $B$-values. We assume $B_1$ and $B_2$ are nondegenerate, $Q_1$ and $Q_2$ have the same isometry group, $Q_1(v) \not= 0$, and $B_1(v,w) = 0$ (so also $B_2(v,w) = 0$). We want to derive $Q_2(w) = (Q_2(v)/Q_1(v))Q_1(w)$, as in part d. But the hint I gave in part e breaks down: $B_1(v,v+w) = B_1(v,v)$, which is 0, not $Q_1(v)$, since the bilinear form associated to a quadratic form in char. 2 is alternating. In fact, for any $a$ and $b$ in $K$ we have $B_1(v,av+bw) = aB_1(v,v) + bB_1(v,w) = aB_1(v,v) = 0$. Thus there is no $u$ in the subspace spanned by $v$ and $w$ such that $B_1(v,u) = 0$.

If you try to use Robert's method, assuming $K$ has characteristic 2 and at least 5 elements (so we give up on $K = {\mathbf F}_2$ or $K = {\mathbf F}_4$), things break down because when $B_1(v,w) = 0$ we have $B_1(x,y) = 0$ for all $x$ and $y$ in the $K$-span of $v$ and $w$.

If we could get the conclusion of part e to work by some other method then we'd have the result you want, but at this point I don't know what that other method would be. I suggest posting a question on MO for the experts on quadratic forms: in characteristic 2 does the isotropy group of a nondegenerate quadratic form determine the quadratic form up to a nonzero scaling factor? I would not be surprised if there might be weird counterexamples when $K$ is small and $V$ has small dimension.

By the way, you don't gain much by switching to the viewpoint of the isometry group of the symmetric bilinear forms, instead of quadratic forms, in characteristic 2: such bilinear forms are automatically alternating, so if they are nondegenerate then their isometry group is the symplectic group for that dimension, hence the group is determined up to isomorphism by the dimension. I don't think it's that interesting to ask about two symplectic isometry groups being literally equal rather than isomorphic.

  • $\begingroup$ Thanks Keith, it is good to have it for future reference! So what should one do in characteristic $2$ in order to save the statement? The naive definition for the inner product fails since we divide by $2$, moreover one should also replace the isotropy group by something a bit larger but what would be such a replacement... $\endgroup$ – Hugo Chapdelaine May 24 '12 at 13:22
  • $\begingroup$ Hugo: I edited my answer to discuss characteristic 2, but unfortunately the final result is inconclusive. I don't see how to make part e of my outline work in characteristic 2, but I don't have a specific counterexample to offer in characteristic 2 that would contradict what you are asking about. I suggest you post a new MO question that I indicate in the second to last paragraph of the updated answer for the people who are more familiar than I am with quadratic forms in characteristic 2. $\endgroup$ – KConrad May 24 '12 at 22:43
  • $\begingroup$ So just to let you know that I posted a continuation of your "quite generous" post on quadratic forms in characteristic 2! see mathoverflow.net/questions/98003/… $\endgroup$ – Hugo Chapdelaine May 26 '12 at 2:14

Without changing anything important, we can orthogonally diagonalize the Gram matrix of the first form, call that $A.$ Call the other matrix $C.$ Your hypotheses are that $A$ has nonzero elements on the diagonal, and for any invertible matrix $P$, we know $$ P^T A P = A \; \Leftrightarrow \; P^T C P = P. $$ Take two indices $i < j,$ with the square block at entries $ii, ij, ji,jj$ being

$$ A_2 \; = \; \left( \begin{array}{cc} a & 0 \\\ 0 & b \end{array} \right) , $$ where the first case we do is $a,b > 0.$ Then we may take $P$ to have all 0 entries except at the same four positions, $$ P_2 \; = \; \left( \begin{array}{cc} \cos t & \sqrt{\frac{b}{a}} \sin t \\\ - \sqrt{\frac{a}{b}} \sin t & \cos t \end{array} \right) , $$ giving us $P^T A P = A. $ You need to show the students that $P^T C P $ also alters in exactly the way we expect in those four positions. Then, with $$ C_2 \; = \; \left( \begin{array}{cc} r & s \\\ s & t \end{array} \right) , $$ and $P^T C P = C, $ we write out the three equations, resulting in $t \left( \frac{a}{b} \right) = r$ and $s=0.$ Similar with both $a,b < 0.$

Now suppose we chose two indices with diagonal elements of opposite sign, $$ A_2 \; = \; \left( \begin{array}{cc} a & 0 \\\ 0 & -b \end{array} \right) , $$ then take $$ P_2 \; = \; \left( \begin{array}{cc} \cosh t & \sqrt{\frac{b}{a}} \sinh t \\\ \sqrt{\frac{a}{b}} \sinh t & \cosh t \end{array} \right) , $$ and again $P^T C P = C $ tells us that the 2 by 2 block of $C$ is daigonal with the same ratio of diagonal entries. Do this for all the $(n^2 - n)/2$ pairs of indices. We have shown that $C$ is diagonal and all ratios of diagonal elements match the ratios for $A.$ Since $C$ is also nondegenerate, all the diagonal entries of $C$ are nonzero and a constant times the same element in $A.$

  • $\begingroup$ Very nice computational proof @Will, thanks a lot! $\endgroup$ – Hugo Chapdelaine Feb 7 '12 at 13:11

If $v=0$, then we have $F(v)= G(v)=0$ Hence obviously we have $F(v)= \lambda. G(v)$ for some non zero real $\lambda$.

Now if $v\neq 0$, then for any $g\in GL_n(\mathbb R)$, we have $g(v)\neq 0$ and $g^{-1}(v)\neq 0$ and hence $F(g^{-1}(v))\neq 0$ and $G(g^{-1}(v))\neq 0$, as $F$ and $G$ are non-degenerate.

If $g\in O(F)$ we have $g\in O(G)$. We have $\forall v\in \mathbb R^n, v\neq 0$,

$$ F(g(v))= F(v)\text{ and } G(g(v))= g(v)$$ $$F(g(v)). G(v)= F(v). G(g(v))$$ $$F(g^{-1}(g(v))). G(g^{-1}(v))= F(g^{-1}(v)). G(g.g^{-1}(v))$$ $$F(v)= \frac{F(g^{-1}(v))}{G(g^{-1}(v))}.G(v)$$

Edited: But it doesn't says that $\frac{F(g^{-1}(v))}{G(g^{-1}(v))}$ is independent of $v$. So proof is incomplete.

  • 1
    $\begingroup$ Why is $\frac{F(g^{-1}(v))}{G(g^{-1}(v))}$ independent from $v$, i.e. constant ? $\endgroup$ – Ralph Feb 7 '12 at 13:42
  • $\begingroup$ @Ralph, You are right,... I cnt see reason for $\frac{F(g^{-1}(v))}{G(g^{-1}(v))}$ to be independent of $v$. $\endgroup$ – zapkm Feb 8 '12 at 5:21

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