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Let $a=(a_0, a_1, ..., a_n )$, $b=(b_0, b_1, ..., b_n )$ that belong to ${\mathbb R}^{n+1}$. Define polynomials $f_a (t)=a_0 +a_1 t+ ... + a_n t^n$ and $f_b (t)=b_0 +b_1 t+ ... + b_n t^n$ and let $f_{ab}(t)=f_a (t)f_b '(t)-f_a '(t)f_b (t)=c_0 +c_1 t+ ... + c_{2n-2} t^{2n-2}$, where $'$ denotes derivative. From the above setting we may define the (bilinear) map $F$ : ${\mathbb R}^{2n+2} \to {\mathbb R}^{2n-1}$, $F(a,b)=c$.

QUESTION: Is $F$ onto?

If so, is it known what the fiber $F^{-1}(c)$ will look like?

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  • $\begingroup$ Please write the formulas in LaTeX. $\endgroup$ – Mikhail Borovoi May 21 '16 at 6:46
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    $\begingroup$ Reformulated question: let $\mathbb{R}_k[t]$ denote the space of real polynomials of degree $\le k$. Let $F:\mathbb{R}_n[t]\times\mathbb{R}_n[t]\to \mathbb{R}_{2n-2}[t]$ be the skew-symmetric bilinear map $(A,B)\mapsto AB'-A'B$ (note the collapsing of the two main coefficients, which makes this valued here rather than in $\mathbb{R}_{2n-1}[t]$). Is $F$ surjective? $\endgroup$ – YCor May 21 '16 at 6:54
  • $\begingroup$ The differential at $(1,t^n)$ is surjective, if I'm correct. So, the image has non-empty interior (and, at the complex level, the image contains a Zariski dense open subset). $\endgroup$ – YCor May 21 '16 at 9:36
  • $\begingroup$ Yes, but the problem is real surjectivity, for which just a single differential is not sufficient. $\endgroup$ – Alex Degtyarev May 21 '16 at 10:02
  • $\begingroup$ I know, that's why I didn't claim it as an answer. And even in the complex case a single surjective differential is not enough. $\endgroup$ – YCor May 21 '16 at 10:03
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No, $F$ is not onto in general. For example, it is not difficult to show that, when $n=3$, the polynomial $$ c = c_0 + c_1t + c_2 t + c_3 t^3 + c_4 t^4 $$ is in the image of $F$ if and only if $$ 12\,c_0c_4 - 3\,c_1c_3 + {c_2}^2 \ge 0. $$

To see this, note that if $c = F(a,b)$ as above, then $$ 12\,c_0c_4 - 3\,c_1c_3 + {c_2}^2 = (3a_0b_3-3a_3b_0-a_1b_2+a_2b_1)^2\ge0. $$ To prove the converse direction, note that, if the above inequality holds, then, taking the square root of the above equation, one finds that one can solve for all of the expressions $a_ib_j-a_jb_i$ for $0\le i < j \le 3$, and this determines $a$ and $b$ uniquely up to a unimodular linear combination.

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  • $\begingroup$ If $c=c_0+\dots+c_4t^4$ is as above with $12c_0c_4-3c_1c_3+c_2^2<0$ (for instance, $c=t^4-1$) do you know if it still holds that $c$ does not belong to the image for any $n\ge 3$? $\endgroup$ – YCor May 21 '16 at 12:05
  • $\begingroup$ @YCor: No, that doesn't happen. For example, if $f_a(t) = 1 + \tfrac13t^4$ and $f_b(t) = -t$, then one has $$f_{ab}(t) = t^4-1.$$ $\endgroup$ – Robert Bryant May 21 '16 at 17:34
  • $\begingroup$ @YCor: Further, taking $f_a(t) = 1$, one can solve $f'_b(t) = c(t)$ for any given polynomial $c(t)$ to get a polynomial $f_b(t)$, so not bounding the degrees makes the problem trivial. $\endgroup$ – Robert Bryant May 21 '16 at 18:56

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