# The distributional gradient of the closest isometry to the differential of a smooth map

The setting-a "linear algebra" fact:

Let $$A$$ be a real $$n \times n$$ matrix, and suppose that $$\det A<0$$ and that the singular values of $$A$$ are distinct. Then, there exist a unique matrix $$Q(A) \in \text{SO}_n$$ which is closest to $$A$$.

(Closest w.r.t the Frobenius norm, i.e. the standard Euclidean distance). Furthermore, if we define $$V=\{ A \in \mathbb{R}^{n^2} \, | \,\, A \in \text{GL}_n^- \, \, \text{and has distinct singular values}\},$$ then the map $$A \to Q(A)$$, considered as a map $$V \to \text{SO}_n$$, is smooth. (There is explanation at the end, if you are interested).

Differential geometry:

Let $$\mathbb{D}^n$$ be the closed $$n$$-dimensional unit ball ,and let $$f:\mathbb{D}^n \to \mathbb{R}^n$$ be real-analytic . Set $$U= \{ p \in \mathbb{D}^n \, | \, df_p \in \text{GL}_n^- \, \, \text{and has distinct singular values.}\}$$

Assume that $$U$$ has full measure in $$\mathbb{D}^n$$.

For every $$p \in U$$, let $$q_p:=Q(df_p) \in \text{SO}_n$$ be the unique special orthogonal matrix which is closest to $$df_p$$. By our assumption, $$q$$ is well-defined a.e. on $$\mathbb{D}^n$$.

Question: Can we say something non-trivial about the distributional gradient of $$q$$? What can we say about the difference between the distributional gradient and the classical gradient, which is defined everywhere on $$U$$? e.g. does it have a definite sign?

As I will now explain, $$q$$ does not always lie in a Sobolev space, even though $$q|_U$$ is smooth; when we approach points where $$df$$ has repeating singular values, bad "jumps" can occur: Indeed, set

$$A=\begin{pmatrix} -\sigma_1 & 0 \\\ 0 & \sigma_2 \end{pmatrix}$$. Then $$Q(A) = \begin{cases} \text{Id}, & \text{if } \, 0<\sigma_1 < \sigma_2 \\ -\text{Id}, & \text{if } \, 0<\sigma_2 < \sigma_1 \end{cases},$$ thus when $$\sigma_1,\sigma_2$$ "cross" each other, we get a jump in $$Q$$.

Here is a concrete example: Consider $$f(x,y)=(-x-\frac{x^2}{2},y+\frac{y^2}{2})$$ as a function on $$B_{\epsilon}(0)$$ (the ball with radius $$\epsilon$$ centered at the origin): Then $$df(x,y)=\begin{pmatrix} -(1+x) & 0 \\\ 0 & 1+y \end{pmatrix},$$

hence $$q=Q(df)=\begin{cases} \text{Id}, & \text{if } \, x < y \\ -\text{Id}, & \text{if } \, y < x \end{cases}$$

does not belong to $$W^{1,p}(B_{\epsilon} (0),\mathbb{R}^{4})$$ for any $$p \ge 1$$.

Explanation on why $$q$$ is smooth on $$U$$:

For $$p \in U$$, $$q_p$$ can be expressed as follows: Let $$df=U\Sigma V^T$$ be the SVD of $$df$$. Then $$q=UD V^T$$, where $$D$$ is a diagonal matrix obtained from $$\Sigma$$, by replacing the smallest singular value with $$-1$$, and all other singular values by $$1$$. Since locally, we can choose the matrices $$U,V$$ smoothly, this implies that $$q$$ is smooth on $$U$$.

• Have you tried use Lojasiewicz's inequality? – Piotr Hajlasz Apr 15 at 14:09
• Setting $n=1$ and $f(x)=x^2$ gives $q(x)=\operatorname{sign}(x)$ which is not in $W^{1,1}(\mathbb R,\mathbb R).$ The restriction to harmonic functions rules out this example at least. – Dap Apr 15 at 17:58
• @Dap Thank you! This is a nice observation. Actually, I was more interested in the behaviour of the minimizer in $\text{SO}_n$ (rather then the orthogonal polar factor itself; they coincide only for matrices with positive determinant). I thought it would be simpler to start with looking at the polar factor, as it is given by a single "expression" on the entire domain. Your comment made me understand this was hopeless, due to the possible "jumps". I have now edited the question, so the object of study is the projection on $\text{SO}_n$. Thank you for your insight, again. – Asaf Shachar Apr 16 at 9:47