$\newcommand{\SO}[1]{\text{SO}(#1)}$ $\newcommand{\dist}{\operatorname{dist}}$

Let $\mathbb{D}^n$ be the closed $n$-dimensional unit ball, and let $f:\mathbb{D}^n \to \mathbb{R}^n$ be smooth.

Set $$X=\text{GL}^+_n \cup \{ A \in M_n \, | \text{ the singular values of } \, A \text{ are distinct }\}$$ Here $M_n$ is the space of real $n \times n$ matrices.

Do there there exist $f_n \in C^{\infty}(\mathbb{D}^n, \mathbb{R}^n)$ such that $f_n \to f$ in $W^{1,2}(\mathbb{D}^n, \mathbb{R}^n)$ and $df_n \in X$ everywhere on $ \text{int}(\mathbb{D}^n) $?

Can we at least perturb $f$ to make the points where the are recurring singular values isolated? We need to understand what happens to the zeroes of the discriminant of the characteristic polynomial of $df^Tdf$ under perturbation.

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    $\begingroup$ I'm not really sure but I feel the answer should be ''no'', because consider the following map from 3-dimensional ball: $f: (x; y; z) \rightarrow (- 1000 x(1 + x^2 + y^2); - 1000 y (1 + x^2 + y^2); -z)$ consider the map $df^t df$ (from the ball to the space of symmetric matrices). Matrices with $\sigma_1 = \sigma_2 < 0$ form a family of codimension two, and I believe that this thing intersects it transversely. Hence, it won't be possible to smoothly perturb it to remove the intersection. $\endgroup$ – Lev Soukhanov Nov 9 '19 at 16:37

Here is a positive answer for $n=2.$ The argument doesn't seem to generalize easily to higher $n.$ The idea is to write $df\in X$ as $\star df_1 + df_2\neq 0$ and make use of Hodge decomposition.

The set $X$ consists of the matrices not of the form $(\begin{smallmatrix}a&b\\b&-a\end{smallmatrix}),$ in the $(dx,dy)$ basis (for matrices with $\sigma_1=\sigma_2,$ the middle part of an SVD is scalar, so the matrix has to be a scalar multiple of an orthogonal matrix). So we want to find an approximating sequence $g^{(n)}$ with

$$\star dg^{(n)}_1 + dg^{(n)}_2\neq 0$$ on the unit ball, with the usual Hodge star operator $\star(a\;dx + b\;dy)=-b\;dx+a\;dy.$.

For the Hodge decomposition I want to replace $\mathbb D^2$ by a more convenient space. I'll use $\mathbb R^2,$ but using a non-compact manifold is not essential an the argument could use a compactification $\mathbb R^2/\Lambda\mathbb Z^2.$

We can assume $f$ extends to a function $\bar f$ in $C^\infty_c(\mathbb R^2,\mathbb R^2),$ for example using the Sobolev extension theorem plus a standard result on density of compactly-supported smooth functions in Sobolev spaces. The combination $\zeta=\star d\bar f_1+d\bar f_2$ is a compactly-supported smooth $L^2$ vector field.

By a perturbation we can arrange that, on a neighborhood of $\mathbb D^2,$ $\zeta$ is non-zero except at isolated points Specifically...

  • pick a bounded open neighborhood $U$ of $\mathbb D^2$
  • pick a $\psi$ in $C^\infty_c(\mathbb R^2,\mathbb R^2)$ that is strictly positive on $U$
  • define $\phi:U\times \mathbb R^2\to\mathbb R^2$ by $\phi(x,M)=(M-\zeta(x))/\psi(x)$
  • and consider a regular value $N\approx (0,0)$ for the restriction $\phi|_{U\times\{(0,0)\}}$

The preimage $\phi^{-1}(\{N\})$ is the graph $\{(x,\zeta(x)+N\psi(x))\}.$ The preimage $\phi|_{U\times\{(0,0)\}}^{-1}(\{N\})$ consists of isolated points $(x,(0,0))$ such that $\zeta(x)+N\psi(x)=(0,0).$ Projecting from the graph $\{(x,\zeta(x)+N\psi(x))\}\subset U\times \mathbb R^2$ to $U$ is a diffeo, so projecting a set of isolated points gives a set of isolated points. So the points $x\in\mathbb D^2$ with $\zeta(x)+N\psi(x)=(0,0)$ are isolated.

By pushing these out of the unit ball - we can approximate $\zeta$ in $L^2$ by a sequence of smooth $L^2$ vector fields $\gamma_n$ such that $\gamma_n\neq 0$ everywhere in the unit ball.

Each $\gamma_n$ has an orthogonal Hodge decomposition which we can write as $\gamma_n=\star dg^{(n)}_1 +dg^{(n)}_2$ where $g^{(n)}_1,g^{(n)}_2$ are determined up to additive constants. The components $\star dg^{(n)}_1$ and $dg^{(n)}_2$ are "longitudinal and transverse" fields defined by pointwise projections in Fourier space, and since $\gamma_n$ has bounded Sobolev norms $(\int(1+|\xi|^2)^k|\hat\gamma_n(\xi)|^2d\xi)^{1/2}$ (where $\hat \cdot$ is Fourier transform), so do $\star dg^{(n)}_1$ and $dg^{(n)}_2.$ So they're smooth.

The functions $g^{(n)}_i$ are unique if we add the requirement $\int_{\mathbb D^2} g^{(n)}_i=\int_{\mathbb D^2} f_i$ for $i=1,2.$ Because Hodge decomposition is an orthogonal decomposition, $\star dg^{(n)}_1+dg^{(n)}_2\to \star d\bar f_1+d\bar f_2$ in $L^2$ implies $dg^{(n)}\to d\bar f$ in $L^2.$ The Poincaré–Wirtinger inequality then gives $g^{(n)}|_{\mathbb D^2}\to f$ in $W^{1,2}(\mathbb D^2,\mathbb R^2).$

  • $\begingroup$ @AsafShachar: I've added some more detail. I only know of Hodge decomposition for compact manifolds and for $\mathbb R^n,$ and I thought the latter would be easier. I am using a smoothness result for the components of the Hodge decomposition, which I've added an argument for. (There was a mistake before, where I only used $\bar f$ to be $C^\infty\cap L^2,$ where I really want all derivatives to be $L^2.$) I don't use the rank assumption on $df$ - it's easy to ensure by perturbation anyway. $\endgroup$ – Dap Nov 11 '19 at 6:37
  • $\begingroup$ I'm using Hodge decomposition on $\mathbb R^n$ for $L^2$ forms. This is in "Geometric function theory and non-linear analysis" 10.6, but that is proving something stronger ($L^p$ decomposition for all $1<p<\infty$) - the $L^2$ decomposition is easy in Fourier space but I don't have a more direct reference. $\endgroup$ – Dap Nov 11 '19 at 7:03
  • $\begingroup$ Thank you, really. I am sorry to trouble you again, but I have two more questions: (1) In the definition of $\phi:U\times X^c\to \mathbb R^{2\times 2}$, should $\zeta(x)$ be replaced by $df(x)$? (otherwise this does not compile, and I think that the replacement gives you what you want). (2) Can you please elaborate on the "pushing the isolated points out of the unit ball" - via composing with a diffeomorphism $\phi:\mathbb{R}^2 \to \mathbb{R}^2$? What is the exact action you are doing? replacing $\zeta$ with $\zeta \circ \phi$, or taking the pullback or pushforward of $\zeta$ using $\phi$? $\endgroup$ – Asaf Shachar Nov 11 '19 at 12:44
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    $\begingroup$ @AsafShachar: I've fixed the perturbation argument - I had got confused about what lives where. It's no trouble to fix mistakes of course; sorry for rushing $\endgroup$ – Dap Nov 18 '19 at 10:39
  • $\begingroup$ Thank you. So the idea is to replace $\zeta$ with $\zeta+N\psi$, right? $\endgroup$ – Asaf Shachar Nov 18 '19 at 14:06

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