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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?

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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.

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  • $\begingroup$ Ok, thanks for that. What would it mean for my map to be Morse? My map is a map between manifolds, not a real valued scalar field on a manifold. Does the concept of a Morse function generalize? $\endgroup$ – Benjamin Jun 4 '15 at 16:05
  • $\begingroup$ Well, not directly. I just wanted to state that even in the simplest situation things get extremely complicated. $\endgroup$ – Thomas Rot Jun 4 '15 at 23:23
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Just a partial answer.

It is know that the preimage $ϕ^{-1}(v)$ 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?

There is Transversality theorem, which is generalisation of known fact about regular values that you mentioned. Narasimhan in his book "Analysis on real and complex manifolds" states it in following manner: enter image description here where trasversal means: enter image description here This theorem puts additional condtions on $\phi,$ but I believe still it is a good one.

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  • $\begingroup$ Thanks for that. I am also aware of the "parametric transversality theorem" which is even stronger and says, very approximately, that if $\phi_s$ is a parameterised family of maps and $\phi$ is transverse for one specific value of $s$ then it is for almost all values of $s$. This theorem requires that the set from which the parameter $s$ is drawn is connected. $\endgroup$ – Benjamin May 23 '15 at 2:04
  • $\begingroup$ This all being said. We don't yet know that the singular values are a manifold or even an algebraic variety in the first place as far as I can tell. $\endgroup$ – Benjamin May 23 '15 at 2:05

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