# Equivalence generated by Jacobian minors

Let $$f,g:\mathbb{R}^m \to \mathbb{R}^n$$ be two smooth functions and let $$k$$ be a strictly positive integer. Write $$f \sim_k g$$ if at each point in the domain, the determinants of all $$k \times k$$ minors of the Jacobians of $$f$$ and $$g$$ coincide. This is clearly an equivalence relation; and more generally, one could let $$f$$ and $$g$$ be smooth maps of smooth manifolds $$X \to Y$$, and ask that for each $$k$$-form $$\omega$$ on $$Y$$ the pullbacks $$f^*\omega$$ and $$g^*\omega$$ agree as $$k$$-forms on $$X$$.

Have such equivalences been studied and given an alternate characterization? Here's the sort of thing I'm looking for: if $$k=1$$ and we are mapping between Euclidean spaces, $$f$$ and $$g$$ must lie in the same orbit of the translation group acting on the codomain. Perhaps there is a similar Lie-theoretic reformulation of $$\sim_k$$ for maps $$X \to Y$$ even when $$k > 1$$?

• If $k$-forms have the same pullbacks, then we can make $k$-forms with very small support, take a limit, and find that the maps $f$ and $g$ agree. Jul 6 '20 at 17:44

## 1 Answer

First, let me point out that the OP's suggested generalization to arbitrary target manifolds $$Y$$ of asking that $$f^*\omega = g^*\omega$$ for all $$k$$-forms on $$Y$$, is not equivalent to the question about equality of $$k$$-by-$$k$$ minors.

To see this, consider the simplest case, $$X = Y=\mathbb{R}^1$$ and $$k=1$$. The general $$1$$-form is of the form $$\omega = h(x)\,\mathrm{d}x$$ (where $$x:\mathbb{R}\to\mathbb{R}$$ is the identity function), and $$f^*\omega = g^*\omega$$ for all $$1$$-forms $$\omega$$ is equivalent to $$h\bigl(f(x)\bigr)f'(x) = h\bigl(g(x)\bigr)g'(x)$$ for all functions $$h:\mathbb{R}\to\mathbb{R}$$, which clearly implies that $$f(x) = g(x)$$ for all $$x\in\mathbb{R}$$. Meanwhile, $$f^*\mathrm{d}x = g^*\mathrm{d}x$$ only implies $$f'(x) = g'(x)$$, which is the same as requiring that all the $$1$$-by-$$1$$ Jacobian minors of $$f$$ and $$g$$ are equal, which does imply that they differ by an additive constant.

Instead, I think that the natural generalization is that one endows the $$n$$-manifold $$Y$$ with a coframing, i.e., a basis $$\omega = (\omega^1,\ldots,\omega^n)$$ of the $$1$$-forms on $$Y$$ (which, of course, requires that $$Y$$ be parallelizable, in fact, $$\omega:TY\to\mathbb{R}^n$$ defines a linear isomorphism $$\omega_y:T_yY\to\mathbb{R}^n$$ for each $$y\in Y$$) and then require that $$f^*(\omega^{i_1}\wedge\cdots\wedge\omega^{i_k}) = g^*(\omega^{i_1}\wedge\cdots\wedge\omega^{i_k})$$ for all $$i_1.

Even this will generally force some 'finite' equations on $$f$$ and $$g$$ if $$\omega$$ is chosen generally (and the dimension of $$X$$ is greater than $$k$$). For a multi-index $$I = (i_1,\ldots,i_p)$$ with $$1\le i_1, write $$|I|=p$$ and $$\omega^I$$ for $$\omega^{i_1}\wedge\cdots\wedge\omega^{i_p}$$. Then there will be functions $$h^I_J$$ on $$Y$$ such that $$\mathrm{d}\omega^I = \sum_{|J|=|I|{+}1} h^I_J\,\omega^J.$$ If the functions $$h^I_J$$ are not constant, and $$f$$ and $$g$$ both have differential rank at least $$k{+}1$$, then applying the exterior derivative to the equations $$f^*\omega^I = g^*\omega^I$$ for $$|I|=k$$ will generally force some relations on the functions $$f^*h^I_J$$ and $$g^*h^I_J$$. One can avoid this 'problem' by assuming that the functions $$h^I_J$$ be constants. For example, when one takes the standard coordinate coframing on $$Y = \mathbb{R}^n$$, one has $$h^I_J=0$$. More generally, if $$Y$$ is a Lie group and the $$\omega^i$$ are a basis for the left-invariant forms on $$Y$$, then the $$h^I_J$$ are constants. In this latter case, when $$X$$ is connected, one will have $$f^*\omega^i=g^*\omega^i$$ for all $$i$$ if and only if $$g = \lambda_y\circ f$$ where $$\lambda_y:Y\to Y$$ is left multiplication by $$y\in Y$$ (regarded as a Lie group). So I think that this is the natural generalization of the OP's case of $$\mathbb{R}^n$$.

Second, let me point out that, if the differential ranks of $$f$$ and $$g$$ are both less than $$k$$ at every point, then, of course, $$f^*\omega = g^*\omega=0$$ for all all $$k$$-forms $$\omega$$ on $$Y$$, so there is no further condition implied than the rank condition. Thus, to get an interesting problem, one must assume that the differential ranks of $$f$$ and $$g$$ are at least $$k$$ in order to get an interesting theory.

Once one assumes this, there are some reasonable things to say. For example, if one assumes that the differential ranks of $$f$$ and $$g$$ are both at least $$k$$, then the condition $$f^*\omega^I = g^*\omega^I$$ for all $$|I|=k$$ implies that $$\mathrm{ker}(f'(x)) = \mathrm{ker}(g'(x))\subset T_xX$$ and $$\omega_x\bigl(f'(x)(T_x)\bigr) = \omega_x\bigl(g'(x)(T_x)\bigr) \subset\mathbb{R}^n$$ for all $$x\in X$$. Moreover, if one sets $$K_x = \mathrm{ker}(f'(x)) = \mathrm{ker}(g'(x))\subset T_xX$$ and $$Q_x = \omega_x\bigl(f'(x)(T_x)\bigr) = \omega_x\bigl(g'(x)(T_x)\bigr) \subset\mathbb{R}^n$$, then the induced isomorphisms $$[f'(x)],[g'(x)]:T_xX/K_x\to Q_x$$ satisfy, for all $$x\in X$$, $$E_k\bigl([f'(x)]\bigr) = E_k\bigl([g'(x)]\bigr):E_k(T_xX/K_x)\to E_k(Q_x),$$ where $$E_k$$ is the '$$k$$th-exterior power' functor on the category of vector spaces and linear maps.

When the differential ranks of $$f$$ and $$g$$ are equal to $$k$$, this is not a very strong condition, so $$f$$ and $$g$$ need not be closely related. For example when $$m=n=k$$, and $$X = Y = \mathbb{R}^k$$, then the OP's condition on $$f$$ and $$g$$ reduces to the assumption that the $$f$$- and $$g$$- pullbacks of the standard volume form on $$Y$$ are equal, and, of course, there are many such pairs $$f$$ and $$g$$ besides the translations.

When the differential ranks of $$f$$ and $$g$$ are both greater than $$k$$, though, this is a much stronger condition. In fact, one finds that, when $$k$$ is odd, this is equivalent to $$[f'(x)] = [g'(x)]$$ while, when $$k$$ is even, this is equivalent to $$[f'(x)] = \pm[g'(x)]$$. Once one is in this situation, at least when one assumes that the differential ranks of $$f$$ and $$g$$ are constant, the Cartan equivalence method can be applied. For example, one has the following result:

Proposition: Suppose that $$f,g:\mathbb{R}^m\to\mathbb{R}^n$$ are smooth maps of constant differential rank greater than $$k\ge1$$ and suppose that all their corresponding $$k$$-by-$$k$$ Jacobian minors are equal. If $$k$$ is odd, then $$g = c + f$$ where $$c\in\mathbb{R}^n$$ is a constant. If $$k$$ is even, then $$g = c \pm f$$ where $$c\in\mathbb{R}^n$$ is a constant.

• Every time I read this answer, I learn something more. Thank you for taking the time to set everything down so carefully, and moreover, for answering the question that I should have asked! Aug 22 '20 at 20:19