Can we choose smoothly the singular vectors of a matrix? Let $X$ be the space of all real $n \times n$ matrices, with strictly negative determinant, and pairwise distinct singular values. $X$ is an open subset of the space of all real square matrices. Is there a way to choose smoothly the singular vectors over $X$?
More precisely, I want smooth maps $U:X \to \text{SO}_n,V:X \to \text{O}_n^-$ such that 
$$ A=U(A)\Sigma(A)V(A)^T,$$ holds for every $A \in X$, where
$\Sigma(A) = \text{diag}\left( \sigma_1(A),\dots\sigma_n(A) \right)$, and $\sigma_1(A)$ is the smallest singular value of $A$. (i.e. I specifically want the first diagonal element of $\Sigma(A)$ to be minimal).
Do such smooth maps $U,V$ exist?
Edit: I now see that that if such $U,V$ exist, then the map $A \to \Sigma(A)$ is also smooth. I guess that in general, as long as the singular values are distinct, they can be chosen smoothly (this should probably follow from more general theory of smooth selection of eigenvectors), but I am not sure that one can choose them in a way which keeps the "identity" of the smallest one. Anyway, any reference on that part would be appreciated (although I am really interested in the "harder" question of a possible smooth selection of the singular vectors themselves, not just the singular values).
 A: Let $\mathcal D$ be the space of $n\times n$ diagonal matrices with distinct non-zero positive entries. This has $n!$ connected components corresponding to the ways to order the elements on the diagonal. Fix a connected component $\mathcal D_0\subset \mathcal D.$ I claim that the map
\begin{align*}
\mu: SO_n\times \mathcal D_0\times O^-_n\to X\\
(U,\Sigma,V)\mapsto U\Sigma V^T
\end{align*}
is a $2^{n-1}$-fold smooth covering map from a connected space (for $n\geq 2$). So it can be inverted locally but has no global section.
$A\mapsto \Sigma(A)$ by itself is a smooth function, because $\Sigma\in\mathcal D_0$ is uniquely determined. It may be worth mentioning that while the $k$'th singular value is not smooth for arbitrary matrices - consider $\mathrm{diag}(1,t)$ as $t\in(0,2)$ - it is Lipschitz continuous in the singular (i.e. operator) norm. See Golub-van Loan, Matrix Computations, Corollary 8.6.2.
Although $\Sigma$ is uniquely determined by $U\Sigma V^T,$ the matrices $U$ and $V$ are not. If $\mu(U',\Sigma,V')=\mu(U,\Sigma,V)$ then $U^{-1}U'$ and $V^{-1}V'$ are equal and of the form $\operatorname{diag}(\pm1,\dots,\pm1)$ with an even number of $-1$'s - see for example the answer at How unique (on non-unique) are U and V in Singular Value Decomposition (SVD)?. This explains the $2^{n-1}.$ In short: let $\hat U=U^{-1}U'$ and $\hat V=V^{-1}V'^T.$ Then $\hat U\Sigma \hat V^T=\Sigma,$ so $\hat U\Sigma^2 \hat U^T=(\hat U\Sigma \hat  V^T)(\hat U\Sigma \hat V^T)^T=\Sigma^2,$ which means $\hat U$ commutes with $\Sigma.$ This forces $\hat U$ to be diagonal. Similarly $\hat V$ is diagonal. My intuition for this is that if $A=U\Sigma V^T$ then $U$ has to map standard basis vectors to the corresponding left singular vectors of $A,$ and if we ignore the condition $\det U=1$ for a moment, there are exactly two ways to do that for each vector because only the sign is ambiguous. $V$ also has to map standard basis vectors to right singular vectors of $A,$ but the sign is already determined by the choice of $U.$
To see that it's a local diffeomorphism, use the inverse function theorem. By a symmetry argument it suffices to check the derivative at $U=V=1.$ We need to check that $(u,s,v)\mapsto D\mu(u,s,v)=u\Sigma + s + \Sigma v$ is an injective linear map, where $u,v$ are skew-symmetric and $s$ is diagonal. The diagonal entries are just those of $s.$ The $i,j$ entry is $u_{ij} \sigma(j)+\sigma(i) v_{ij},$ and the $j,i$ entry is $u_{ji} \sigma(i)+\sigma(j) v_{ji}=-u_{ij} \sigma(i)-\sigma(j) v_{ij}.$ Since $\begin{pmatrix}\sigma(j)&\sigma(i)\\-\sigma(i)&-\sigma(j)\end{pmatrix}$ is invertible, we can recover $u_{ij}$ and $v_{ij}$ from these.
Finally, $\mu$ is a covering map because each fiber has the same (finite) cardinality. Indeed for any $x\in X$ by the local homeomorphism and Hausdorff properties, there are disjoint open sets $U_1,\dots,U_{2^{n-1}}$ each mapped diffeomorphically by $\mu$ to a (possibly different) open neighborhood of $x.$ This implies that $\bigcap_i\mu(U_i)$ is an evenly covered neighborhood of $x.$

The way I originally thought of this problem is to consider the map $f:SO_2\to \mathbb R^{2\times 2}$ that sends $U$ to $U\cdot \mathrm{diag}(2,1)\cdot U^T.$ This is a slightly different situation, but perhaps a bit easier to visualize. $f(U)$ is a positive definite matrix, and $x^Tf(U)x=1$ describes a certain ellipse with axes of length 1 and $1/\sqrt 2.$ The question is whether we can smoothly choose the rotation matrix $U,$ given the ellipse. This is impossible because continuously rotating by 180 degrees gives the same ellipse, but forces $U$ to end up with an extra 180 degree rotation. Actually $f:SO_2\to f(SO_2)$ is just the map $S^1\to S^1$ of degree $2.$
This argument relies on the fact that the fibers are discrete. Otherwise you'd get a more general fibration.
