My interpretation of your question is that you would like to see a stratification of the space of smooth maps $S^1\to M^n$. It suffices to consider maps $\newcommand{\r}{\mathbb R} \r \to \r^n$, and for simplicity I will take $n=3$.

I think this is a very interesting question. I suspect the answer is no, there is not a nice stratification "all the way down", but that you have to go to fairly high codimension before you run into problems. (See the answers to this question and the references therein.)

In particular, I don't think Ryan's example presents a problem. We just have to recognize that the set of maps with a single double point (and no other singularities) consists of several strata: a codimension 1 stratum where the double point is transverse, a codimension 3 stratum where the tangents at the double point are colinear but the higher derivatives are in general position, and higher codimension strata where the two curves are tangent to higher order.

More specifically, a generic colinear double point is (after change of coordinates in the domain and range) of the form $[t \mapsto (t, 0, 0); t \mapsto (t, t^2, t^3)]$. I claim that singularities of this form are a codimension 3 submanifold of the space of all smooth maps. As in Ryan's answer, a 3-dimensional space of perturbations transverse to this stratum has the form $t\mapsto (t, t^2+a, t^3+bt + c)$.

The link of this stratum is a 2-sphere with an embedded "figure 8". The nonsingular parts of the figure 8 correspond to single transverse double points. The central (singular) point of the figure 8 corresponds to a pair of transverse double points. Thus there is no reasonable way to make the subset of maps with a single not-necessarily-transverse double point into a nice stratified set, much less into a manifold. I think this was the point of Ryan's answer. The point of my answer is that the space of all smooth maps nevertheless has a perfectly nice stratification near this sort of singularity.

The list of indecomposable strata for maps $\r\to \r^3$ starts out like this:

• codimension 1: single transverse double point
• codim 2: zero derivative with 2nd, 3rd and 4th derivatives linearly independent, i.e. conjugate to $t\mapsto (t^2, t^3, t^4)$
• codim 3: (a) triple point with tangents linearly independent; (b) double point with tangents colinear and higher derivatives independent (as above)
• codim 4: (a) triple point with tangents coplanar; (b) double point with one of the tangents zero (and higher derivatives generic); (c) colinear double point with some higher derivatives non-generic (like $[t\mapsto (t, 0, 0); t\mapsto (t, t^2, t^4)]$; not sure about this one)

Items from the above list can of course be combined. For example, two generic double points and four generic triple points has codimension $2\cdot 1+ 4\cdot3 = 14$.

It would be nice if one could continue the above list so that the complement of all the strata had infinite codimension. I'm not sure whether that's possible.

My interpretation of your question is that you would like to see a stratification of the space of smooth maps $S^1\to M^n$. It suffices to consider maps $\newcommand{\r}{\mathbb R} \r \to \r^n$, and for simplicity I will take $n=3$.

I think this is a very interesting question. I suspect the answer is no, there is not a nice stratification "all the way down", but that you have to go to fairly high codimension before you run into problems. (See the answers to this question and the references therein.)

In particular, I don't think Ryan's example presents a problem. We just have to recognize that the set of maps with a single double point (and no other singularities) consists of several strata: a codimension 1 stratum where the double point is transverse, a codimension 3 stratum where the tangents at the double point are colinear but the higher derivatives are in general position, and higher codimension strata where the two curves are tangent to higher order.

More specifically, a generic colinear double point is (after change of coordinates in the domain and range) of the form $[t \mapsto (t, 0, 0); t \mapsto (t, t^2, t^3)]$. I claim that singularities of this form are a codimension 3 submanifold of the space of all smooth maps. As in Ryan's answer, a 3-dimensional space of perturbations transverse to this stratum has the form $t\mapsto (t, t^2+a, t^3+bt + c)$.

The link of this stratum is a 2-sphere with an embedded "figure 8". The nonsingular parts of the figure 8 correspond to single transverse double points. The central (singular) point of the figure 8 corresponds to a pair of transverse double points. Thus there is no reasonable way to make the subset of maps with a single not-necessarily-transverse double point into a nice stratified set, much less into a manifold. I think this was the point of Ryan's answer. The point of my answer is that the space of all smooth maps nevertheless has a perfectly nice stratification near this sort of singularity.

The list of indecomposable strata for maps $\r\to \r^3$ starts out like this:

• codimension 1: single transverse double point
• codim 2: zero derivative with 2nd, 3rd and 4th derivatives linearly independent, i.e. conjugate to $t\mapsto (t^2, t^3, t^4)$
• codim 3: (a) triple point with tangents linearly independent; (b) double point with tangents colinear and higher derivatives independent (as above)
• codim 4: (a) triple point with tangents coplanar; (b) double point with one of the tangents zero (and higher derivatives generic); (c) colinear double point with some higher derivatives non-generic (like $[t\mapsto (t, 0, 0); t\mapsto (t, t^2, t^4)]$; not sure about this one)

Items from the above list can of course be combined. For example, two generic double points and four generic triple points has codimension $2\cdot 1+ 4\cdot3 = 14$.

It would be nice if one could continue the above list so that the complement of all the strata had infinite codimension. I'm not sure whether that's possible.