# Generalising the parametric transversality theorem to a foliation

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.

• 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. Commented Aug 18, 2015 at 7:21
• 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. Commented Aug 18, 2015 at 8:13
• 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. Commented Aug 18, 2015 at 14:25
• 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. Commented Aug 19, 2015 at 15:31
• Is having $F(s,p)=\psi_s(p)$ locally surjective (i.e. a submersion) sufficient? This is this my guess. Commented Aug 21, 2015 at 18:20

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?

• 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. Commented Jul 28, 2019 at 10:07
• Could you please ask your original question again, making clear which manifold is foliated, the hypotheses, and the expected conclusion? Commented Jul 28, 2019 at 19:58
• Won't that get flagged as duplicated? I could just explain here? Commented Jul 29, 2019 at 12:11
• can't you edit your original question? Commented Jul 30, 2019 at 13:47

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.

• 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. Commented Aug 30, 2015 at 4:27
• 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 $F^{-1}(R)\to S$. Commented Aug 31, 2015 at 2:42
• 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. Commented Aug 31, 2015 at 13:32
• What if we further knew that $\phi_s$ was globally surjective for almost all $s \in S$, would this be sufficient? Commented Sep 3, 2015 at 18:29