Given a smooth ($C^{\infty}$) map $\phi: V \rightarrow SU(n)$ where $V$ is a (finite dim, real) vector space (of potentially very large dimension) and $SU(n)$ is the special unitary Lie group, what can be said about the of singularities of this map?
It is know that the preimage $\phi^{-1}(v)$ of a regular value $v$ is a submanifold of $V$. What is known about the preimage of a singular value? Is this also a manifold in this case? What about the preimage of all singular points?
Do the singular values form a manifold? I known they are a null set by Sard's theorem but this does not use anything specific to $SU(n)$. Does the target space being $SU(n)$ help us to say anything more about the singular points?
If we further know that the map $\phi$ is onto does this affect things?
where trasversal means:
This theorem puts additional condtions on
I don't think the target space being $SU(n)$ gives you much information:
Preimages of singular values can be extremely nasty. Any closed subset of $\mathbb{R}$ is the set of singular points of a function $\mathbb{R}\rightarrow \mathbb{R}$. Embedding $\mathbb{R}$ into $SU(n)$ gives you nasty subsets of $\mathbb{R}$. This idea generalizes to higher dimensional vector spaces (although I don't believe any closed subset is the critical set of a smooth function).
Consider the function $e^{1/x} \sin(1/x)$. The critical values are a null set (by sard), but do have an accumulation point $0$, so they do not form a manifold.
More information about $\phi$ will of course give you more information about the critical set: e.g. if $\phi:\mathbb{R}\rightarrow \mathbb{R}$ is Morse, we know the that the critical set is discrete, and we have a nice local model of the function around its critical points.