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As stated in the title, suppose $f,g$ are bilinear forms $V \times V \rightarrow T$. Suppose $K: V \times V \rightarrow T$, and $f(a,b) = K(a,b) g(a,b)$.

Does this imply that $K$ is constant? I don't seem to be able to prove this, though intuitively it makes sense - as soon as we multiply a bilinear form by anything non constant, it is no longer a bilinear form.

If it true for the case of e.g. $T = \mathbb{R}$, $V=\mathbb{R}^2$, please provide a rigorous proof.

This question is motivated by a similar step that is used in a cross product identity proof - Coordinate free proof for $a\times (b\times c) = b(a\cdot c) - c(a\cdot b)$ and Do the BAC-CAB identity for triple vector product have some intepretation?

Where it's shown that $a \times (b \times c) = B'(a \cdot c)b + C'(a \cdot b)c$ and we wish to show that $B'$ is constant, i.e. not dependent on either $b$ or $c$. I'd be interested in whether it's a valid step in this case in particular.

RobPratt
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Jake1234
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    Although the constraint is on $f$ and not $K$ at the points where $g(x, y) = 0$, so without further assumptions on $K$ the best you should expect is that $K$ is constant on ${g \ne 0}$. – ronno May 27 '23 at 14:37
  • @ronno I've added a particular case to the question in the case of a cross product related identity. – Jake1234 May 27 '23 at 14:53

3 Answers3

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Since the condition on $K$ is vacuous at the points where $g(x, y) = 0$, without further assumptions on $K$ the best you should expect is that $K$ is a constant on $U = \{(x, y)\in V \times V \mid g(x, y) \ne 0\}$. Of course if $K$ is a polynomial (or more generally continuous, if say $T = \mathbb{R}$) and $g$ is not the $0$ bilinear form so that $U$ is dense then it follows that $K$ is a constant everywhere.

Consider $x$ such that $y \mapsto g(x, y)$ is not the $0$ function, ie $(x \times V) \cap U \ne \varnothing$. Now define $f_x(y) = f(x, y)$ and $g_x(y) = g(x, y)$. By assumption, $f_x$ vanishes on $\ker g_x \subsetneq V$, so descends to an element of $(V/\ker g_x)^*$. But this vector space is generated by $g_x$, so $f_x = K_x g_x$ for some $K_x \in T$. Comparing, we get that $K(x, y) = K_x$ for each $y$ such that $(x, y) \in U$, so $K$ is constant on $(x \times V) \cap U$. This holds for each $x$ and by a symmetric argument $K$ is also constant on each $(V \times y) \cap U$.

Now it suffices to prove that for any $(x_1, y_1), (x_2, y_2) \in U$, we can always find $y \in V$ such that $(x_1, y), (x_2, y) \in U$, by applying the above to $x_1 \times V$, $V \times y$ and $x_2 \times V$. Consider the linear functionals $g_1 = g_{x_1}$ and $g_2 = g_{x_2}$. By assumption, each is non-zero, and we want to find a point where they both do not vanish, ie we want to show that $W_1 \cup W_2 \ne V$, with $W_i = \ker g_i \subsetneq V$.

This is proved eg here in more generality but the argument applied to this case is as follows. Note that $y_1 \notin W_1$ and $y_2 \notin W_2$. If $y_2 \notin W_1$ then we are done. Otherwise $y_1 + \lambda y_2$ is never in $W_1$ and in $W_2$ for at most one $\lambda \in T$, and since $T$ must have more than one element, we can take $y = y_1 + \lambda y_2$ for any other $\lambda$.

ronno
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  • This still feels more circuitous than it needs to be. Suggestions for simplifications are welcome. – ronno May 27 '23 at 15:20
  • Thanks for the detailed argument. A more basic argument than what's in the second paragraph could be to argue roughly as $f(x,cy) = cf(x,y) = cK(x,y)g(x,y)$ and also $f(x,cy) = K(x,cy)g(x,cy) = cK(x,cy)g(x,cy)$ so $K(x,y)=K(x,cy)$. But don't modify the answer please - I'm just not knowledgeable enough to understand the "But this vector space is generated by g_x" or how those functionals would get mapped to $(V/\ker g_x)*$ in a rigorous manner, could you please provide a linear algebra reference that would cover something like this? – Jake1234 May 27 '23 at 16:14
  • Also, I'm a bit surprised it's this complicated. Is there any simpler way to show this in the case of the cross product identity concerning $a \times (b\times c)$ that I've linked? Non of the answers there ever justify this claim, so I assumed I was missing some very simple solution... maybe this problem is slightly different in some sense, I don't know. – Jake1234 May 27 '23 at 16:22
  • I don’t understand why $g_x$ generates the space. If $T$ isn’t one dimensional, then this seems to fail. – Eric May 27 '23 at 16:22
  • @ronno I posted an answer -- is it too simplistic? I think I might be missing something. – Elchanan Solomon May 27 '23 at 16:45
  • @Eric Did you mean to say "If $(V / \ker g_x)^*$ isn't one dimensional"? $T$ is the field. – Jake1234 May 27 '23 at 17:25
  • @ElchananSolomon I think your answer is fine but I don't assume continuity, whereas I don't see how to avoid the continuity at $0$ in your argument. $K$ will be continuous on ${g \ne 0}$ automatically but it doesn't need to be at $0$. – ronno May 27 '23 at 18:54
  • @Jake1234 I don't see how the multiplication by $c$ simplifies that step, note that I directly show that $T$ is continuous on all of $(x \times V) \cap U$, not just $x \times Ty$. By the first isomorphism theorem, $V/\ker g_x \xrightarrow{g_x} T$ is an isomorphism, in particular $V/\ker g_x$ is $1$-dimensional. – ronno May 27 '23 at 18:55
  • @Eric I was assuming $T$ is the base field. If $f$ and $g$ are just bilinear maps, not bilinear forms, you can apply this to "each coordinate" of $f$ and $g$. – ronno May 27 '23 at 18:58
  • @Jake1234 As in my first comment, I too think this argument is too complicated. Feel free to unaccept and wait longer for a simpler answer before accepting this one. – ronno May 27 '23 at 18:59
  • If you apply it to each coordinate, how do you know that the $K$’s match? – Eric May 27 '23 at 20:24
  • @Eric $K(x, y)$ should still be in the base field, so it doesn't depend on which coordinate you look at. Maybe I don't understand your objection. – ronno May 27 '23 at 20:30
  • The constant value that you get might be different for different coordinates. You’d need to show that either $g(x,y)$ could give nonzero coordinates in multiple coordinates at the same time which feels true not nonobvious to me. – Eric May 27 '23 at 20:46
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    @Eric Note that if some coordinate of $g$ is identically zero then that coordinate of $f$ is also identically zero. For all the other coordinates, the points where they don't vanish is a dense open set, and intersection of finitely many dense open sets is dense open. – ronno May 27 '23 at 20:54
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The following shows that if $K$ is continuous then it is constant.

We calculate:

$$\lambda f(a,b) = f(\lambda a,b) = K(\lambda a,b) g(\lambda a,b) = \lambda K(\lambda a,b)g(a,b)$$

which implies that $K(a,b)g(a,b) = f(a,b) = K(\lambda a,b)g(a,b)$, and hence $K(\lambda a,b) = K(a,b)$ for all $\lambda$ and $(a,b)$ with $g(a,b) \neq 0$. If $g$ is not the zero bilinear form (in which case $f$ is also zero and the question is trivial), the set of $(a,b)$ with $g(a,b) \neq 0$ is dense in $V \times V$, so, again by continuity, $K(\lambda a,b) = K(a,b)$ for all $\lambda$ and all $(a,b)$.

By a similar argument, $K(a,\mu b) = K(a,b)$ for all $\mu$ and all $(a,b)$.

Then taking limits as $\lambda, \mu \to 0$, $K(a,b) = K(\lambda a, \mu b) = K(0,0)$, so $K$ is constant

Elchanan Solomon
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  • "The set of (a,b) with $g(a,b) \neq 0$ is dense in $V\times V$" - that doesn't seem right for e.g. $g(a,b) = a_1 * b_1$. – Jake1234 May 27 '23 at 16:52
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    The ratio of continuous function don’t need to be continuous. For example, $|x|=sign(x)x$, but $sign$ isn’t continuous. I think this is uncorrectable since $g(0,0)=0$ is explicitly not “invertible” where your proof requires it. – Eric May 27 '23 at 16:57
  • @Jake1234: It is true for that example. The set of $(a,b)$ for which $ab \neq 0$ is all of $\mathbb{R}^2$ with the coordinate axis removed, which is dense in $\mathbb{R}^2$. – Elchanan Solomon May 27 '23 at 18:14
  • What I meant was $g:\mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}$, and taking $g((a_1, a_1), (b_1, b_2)) = a_1 b_1$. Then $g((0,x),(y,z))=0$ for all $x,y,z$. Though thinking about it, it's probably still dense. – Jake1234 May 27 '23 at 18:35
  • @Jake1234 ${g \ne 0}$ is dense in that example and in fact whenever $g \ne 0$. More generally, if a polynomial vanishes on any open ball then it must be identically $0$. – ronno May 27 '23 at 20:32
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The basic idea is that we first hold $a$ constant and show it holds, then extend that to everything.

Let $f$ and $g$ be linear maps such that $f(v)=K(v)g(v)$.

If $g$ is zero everywhere, then so must $f$, so for any fixed C, $f(v)=C g(v)$.

Otherwise choose $a$ such that $g(a)\neq 0$. Let $C=K(a)$, so $f(a)=C g(a)$

Let $b\neq a$ be any other vector.

If $g(b)=0$, then $f(b)=0=C g(b)

If $g(a)$ and $g(b)$ are linearly independent then $K(a)g(a)+K(b)g(b)=f(a)+f(b)=f(a+b)=K(a+b)(g(a)+g(b))$, so $K(a)=K(a+b)=K(b)$.

Otherwise, there’s some $r \neq 0$ such that $r g(a)=g(b)$, so $0=g(b-r a)=0$, so $0=K(b-ra)g(b-r a)=f(b-r a)=f(b)-r f(a)=K(b)g(b)-r K(a)g(a)=(K(b)-K(a))g(b)$. Thus, $K(b)=K(a)=C$.

Thus, there’s some $C$ such that $f(v)=Cg(v)$ everywhere

Applying this to the original problem gives that there are functions $K1$ and $K2$ so that $f(x,y)=K1(x)g(x,y)=K2(y)g(x,y)$ everywhere.

Thus, whenever $g(x,y)$ is nonzero, $K1(x)=K2(y)$

For any $x1,y1$ and $x2,y2$ where $g(x1,y1)$ and $g(x2,y2)$ are nonzero, consider $g(xi,y1),g(xi,y2),g(xi,y1+y2)$. The third is the sun of the other two and $g(xi,yi)$ is nonzero, so at least 2 out of the 3 are nonzero for a given $i$. Thus, at least one of $y1,y2,y1+y2$ must have nonzero values for both $x1$ and $x2$, for that $y3$, $K1(x1)=K2(y3)=K1(x2)$, so where ever $g$ is nonzero $K1$ gives the same value, so $K$ is constant.

Eric
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