3
$\begingroup$

This (which is a follow-up to Lifting a determinant map) must be standard, and yet I am failing to find a reference. Consider a map $f:\mathbb{R}^n \to M^{n\times n}.$ You can pick its degree of regularity to maximize your happiness, but the question is whether (and how) you can find a map $g: \mathbb{R}^n \to \mathbb{R}$ such that its Hessian is given by $f.$ Now, if one does it in steps: first find a function $h: \mathbb{R}^n \to \mathbb{R}^n$ of which $f$ is the Jacobian, and then find $f$ of which $h$ is the gradient, you get a debauche of integrability conditions, which seems hard to make sense of, but there must be a nice way to express it.

A related (and, I am guessing, easier) question is as above with $\mathbb{R}^n$ replaced by $\mathbb{C}P^n,$ matrices now being complex (if, instead of matrices, you would prefer your favorite algebraic group, that is of interest too), and the map should be a rational map.

Improve this question
$\endgroup$
5
  • $\begingroup$ But you don’t get a debauche of integrability conditions. If you write down all of the equations that $g$ and $h$ need to satisfy, the only integrability condition needed is that third partials commute. The proof is essentially the same as for the Poincaré Lemma. $\endgroup$
    – Deane Yang
    Commented Feb 13, 2018 at 5:10
  • $\begingroup$ @DeaneYang third partials of the mysterious $g?$ So, what does this translate to for the data ($f$)? That $\frac{\partial f_{ij}}{x_k} = \frac{\partial f_{ik}}{x_j}?,$ plus the symmetry of the matrix? (the latter needed to go from $h$ to $g$). This is somehow unsatisfying, because it is not very geometric to me - if I have a disk sitting in $S^{n\times n}$ (symmetric matrices), how do I tell a disk which comes from such a construction from one which does not?! $\endgroup$
    – Igor Rivin
    Commented Feb 13, 2018 at 5:35
  • $\begingroup$ the map $f$ matters, not just its image. $\endgroup$
    – Deane Yang
    Commented Feb 13, 2018 at 10:03
  • $\begingroup$ @DeaneYang Yes, of course. Does it mean that every image can be reparametrized? $\endgroup$
    – Igor Rivin
    Commented Feb 13, 2018 at 13:02
  • $\begingroup$ good question. I don’t know $\endgroup$
    – Deane Yang
    Commented Feb 13, 2018 at 18:32

1 Answer 1

Reset to default
4
$\begingroup$

This is essentially the Poincaré lemma. You can write it as a first order system of equations: $$ \partial_ig = p_i $$ $$ \partial_jp_i = f_{ij} $$ The obvious necessary conditions are: $$ \partial_if_{jk} = \partial_jf_{ik} $$ They can be shown to be sufficient by integrating the equations one coordinate at a time as follows:

Assume that the equations above hold for $1 \le i,j,k \le n-1$ along the hyperplane $x^n = 0$. You can extend $g, p_i$ to all of $\mathbb{R}^n$ by integrating the equations $$ \partial_n g = p_n $$ $$ \partial_n p_i = f_{ni},\ 1 \le i \le n $$ You then check that the equations for $1 \le i,j,k \le n-1$ must also continue to hold on all of $\mathbb{R}^n$, because they hold, by inductive assumption, on $x^n = 0$ and, by the equations above, $$ \partial_n(\partial_ig - p_i) = \partial_i(\partial_nu - p_n) = 0 $$ $$ \partial_n(\partial_jp_i - f_{ij}) = \partial_j(\partial_np_i - f_{ni}) = 0 $$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .