# Can we perturb a map $\mathbb{R}^n \to \mathbb{R}^n$ to have distinct singular values?

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

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.

• 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. – 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 $$W^{1,2}(\mathbb D^2,\mathbb R^2).$$

• @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. – Dap Nov 11 '19 at 6:37
• 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. – Dap Nov 11 '19 at 7:03
• 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 $\phi$? – Asaf Shachar Nov 11 '19 at 12:44
• @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 – Dap Nov 18 '19 at 10:39
• Thank you. So the idea is to replace $\zeta$ with $\zeta+N\psi$, right? – Asaf Shachar Nov 18 '19 at 14:06