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The parametric transversality theorem states that, given a parameterised family of smooth maps of $C^{\infty}$ manifolds $\phi_s:M \rightarrow N$ and a submanifold $R < N$ then for almost all values of the parameter $s$, $\phi_s$ is transverse to $R$ whenever the set of values of $s$ is a connected manifold and the adjoint map $F:M \times S \rightarrow N$, defined by $F(p,m)=\phi_s(p)$, is transverse to R. Here $S$ is the manifold of possible values of $s$.

The stronger property of a smooth map $\psi$ being transverse to a foliation $\mathcal{F}$ of $M$, is that $\psi$ is transverse to each leaf of the foliation.

Can we go one further and conclude that a family of maps smooth $\phi_s$ is transverse to a foliation of $M$ for almost all values of $s$? If not, under what conditions on $M,N, \psi_s$ and a given foliation would this hold?

I'm particularly interested in the case of a family of smooth maps from a real vector space to a compact Lie group. In my application, the foliation is into level sets of a given smoth function which is known to have only a single maximum.

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    $\begingroup$ Something seems incorrect in your statement of the theorem. Are you sure you have it right? I'm thinking of plenty of counter-examples. Take any family of smooth maps that are transverse for all but one parameter family, then you can use a bump function construction to replace it by a family that's not transverse in a neighbourhood of the original non-transverse point. $\endgroup$ Commented Aug 18, 2015 at 7:21
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    $\begingroup$ If I remember correctly the statement should have the assumption that the adjoint map $M \times S \to N$ is transverse to $R$, where $S$ is the manifold parametrizing the family. ... or something like this. I'll try to look it up later if the OP doesn't do it first. $\endgroup$ Commented Aug 18, 2015 at 8:13
  • $\begingroup$ Hmm, I'm actually seeing conflicting statements around. I think @chris is correct, this is one way at least. It would be good to know the weakest requirements for the conclusion to hold. Thanks for spotting that. $\endgroup$
    – Benjamin
    Commented Aug 18, 2015 at 14:25
  • $\begingroup$ Ok, after a google-a-thon, it seems that it is now correct as stated and that this is the standard statement. The confusion was all mine. $\endgroup$
    – Benjamin
    Commented Aug 19, 2015 at 15:31
  • $\begingroup$ Is having $F(s,p)=\psi_s(p)$ locally surjective (i.e. a submersion) sufficient? This is this my guess. $\endgroup$
    – Benjamin
    Commented Aug 21, 2015 at 18:20

2 Answers 2

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One could answer if the question were asked with precision (hypotheses and expected conclusion). If I understand correctly, the source manifold M has a foliation Fol; you consider a map F:MxS->N; and for each s the map F_s:M->N; and a submanifold R in N. You assume that F|(LxS) is transverse to R in N, for each leaf L. You ask if for almost every s and each leaf L, the map F_s|L is transverse to R in N. Is that it?

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  • $\begingroup$ Do you have a good source for this? A quick google resulted in some really impenetrable stuff. What I really need to understand is the sufficient conditions for this version of the ptt to be true. $\endgroup$
    – Benjamin
    Commented Jul 28, 2019 at 10:07
  • $\begingroup$ Could you please ask your original question again, making clear which manifold is foliated, the hypotheses, and the expected conclusion? $\endgroup$ Commented Jul 28, 2019 at 19:58
  • $\begingroup$ Won't that get flagged as duplicated? I could just explain here? $\endgroup$
    – Benjamin
    Commented Jul 29, 2019 at 12:11
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    $\begingroup$ can't you edit your original question? $\endgroup$ Commented Jul 30, 2019 at 13:47
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The theorem says that if $F:M\times S\to N$ is transverse to $R$ then for almost every $s\in S$ the map $\phi_s:M\to N$ given by $m\mapsto F(m,s)$ is transverse to $R$. ($S$ being connected is irrelevant.) It can be proved by observing that $F^{-1}(R)$ is a submanifold of $M\times S$ and that the regular points for the projection $F^{-1}(R)\to S$ are precisely the points $(m,s)\in F^{-1}(R)$ such that $\phi_s$ is transverse to $R$ at $m$, and using Sard's Theorem.

Maybe your question is, if $F$ is transverse to some foliation of $N$ then does it follow that for almost all $s$ the map $\phi_s$ is transverse to the foliation? But this is not true. For example, take a foliation of $S$. The projection $M\times S\to S$ is a submersion, and therefore transverse to every immersed submanifold of $S$, in particular to every leaf of every foliation. But the corresponding map $\phi_s:M\to S$ is the constant map $s$, and is not transverse to the leaf containing the point $s$ except in the extreme case where leaves have codimension zero.

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    $\begingroup$ Are you sure that $S$ being connected is not relevant? I've seen both $S$ connected and $M$ compact stated among the premises in different sources. I'll re-check Lee's book. So, the question is, what new premises are needed in order to conclude that a family of smooth maps $\phi_s$ is transverse to a given foliation for almost all $s$? The issue of the weakest requirements for this to be true would be interesting to know. $\endgroup$
    – Benjamin
    Commented Aug 30, 2015 at 4:27
  • $\begingroup$ I sketched the proof above. There is a simple local statement: If $F$ is transverse to $R$ at the point $(m,s)\in F^{-1}(R)$, then a neighborhood of that point in $F^{-1}(R)$ is a manifold; and the question of whether that point is a regular point for the projection of that neighborhood to $S$ is easily seen to be equivalent to the question of whether the map $\phi_s$ is transverse to $R$ at the point $m$. Thus if $F$ is transverse to $R$ everywhere then the set of all $s$ such that $\phi_s$ is transverse to $R$ is the set of regular values of the projection $F^{-1}(R)\to S$. $\endgroup$ Commented Aug 31, 2015 at 2:42
  • $\begingroup$ Ok, I see that connectedness can be dropped, I'm not sure why this is in some sources. However, the real question is about what conditions on $M,N,S,F,\mathcal{F}$ are sufficient for the generalized version to hold. $\endgroup$
    – Benjamin
    Commented Aug 31, 2015 at 13:32
  • $\begingroup$ What if we further knew that $\phi_s$ was globally surjective for almost all $s \in S$, would this be sufficient? $\endgroup$
    – Benjamin
    Commented Sep 3, 2015 at 18:29

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