Can we define diffeomorphisms from R^n to lower-dimensional manifolds?

[lavinia]

I understand, I am aware of the existence of a unique maximal differentiable structure for the Reals, that the two structures are diffeomorphic .I was mentioning that while the differentiable structures are diffeomorphic, I believe, this map is not a diffeomorphism. This is all I meant.

Right. I misread your statement. I thought you were saying that if one takes x^1/3 as a chart then one would obtain a different smooth version of R. My mistake. I thought you were responding to my statement about smooth manifolds that are topologically equivalent but not diffeomorphic.

Interestingly, in the case of the two different differentiable structures on R, the identity map is not a diffeomorphism but x^3 is while in your case the identity is a diffeomorphism but x^3 is not.

 

[WWGD]

No problem, I have made plenty of mistakes myself .

 

[mathwonk]

WWGD: I am just saying we were left to fill in your statement, since you used only partial words. And since you did not present a map at all, we were not inclined to realize you were thinking of a particular map.

I.e. you said:

"I guess the standard example of homeo but not diffeo is that of R and (x,x3):x∈R."

That can be filled in several ways. Since you presented examples of two manifolds, but no map between them, I thought you meant to say:

" "I guess the standard example of homeo[morphic] but not diffeo[morphic manifolds] is that of R and (x,x3):x∈R." which is false.

You however apparently meant to say:

"I guess the standard example of [a] homeo[morphism] but not [a] diffeo[morphism] is that of [the horizontal projection between the y axis, i.e.] R and [the graph] (x,x3):x∈R." which is true.

So I made the mistake of assuming I understood what you meant.

 

[WWGD]

OK, yes, I was sloppy in this post; I will be more careful from now on, sorry.

 

[zinq]

I can't help but mention a case of too-low differentiability that the graph of y = x^3 is related to:

Consider this graph and its copies after being translated by each possible amount to the right or left. In other words, the set of curves

G_c = {((x, (x-c)³ | x ∈ ℝ}

for each real number c.

It's clear that these curve represent the trajectories of some vector field. But in too many ways! That is, one trajectory through (0,0) could be G_0 (the graph of y = x³). But another trajectory containing (0,0) of the same vector field would have to be the x-axis (since all vectors on the x-axis have slope 0).

The existence and uniqueness theorem for ordinary differential equations tells us that a C¹ vector field — one that is (at least) once-continuously differentiable — always has a unique solution for any given initial conditions.

Conclusion: Although the set of curves {G_c | c ∈ ℝ} seem as smooth as can be, in fact there is no vector field tangent to them that is continuously differentiable.

 

[mathwonk]

you are probably right but just as an intuitive response, what about a vector field that equals zero at every point of the x axis? then the unique solution through a point of the x-axis would be the constant ciurve at that point. I.e, no solution could flow along the x-axis since the velocity is always zero there.

 

[zinq]

I meant — by a vector field tangent to all the curves G_c — a non-zero vector field tangent to these curves at each point. (One way to define such a field uniquely is to choose the vector of length = 1 having positive x-component that is tangent

"The" ordinary differential equation defined by the family of curves (for any c ∈ ℝ) G_c = {(x, (x-c)³) | x ∈ ℝ} would be any non-zero vector field V(x,y) tangent at every point (a,b) to the unique curve G_c that passes through (a,b). (This would be the curve G_c for the unique number c that satisfies (a-c)³ = b.)

But this vector field would be nowhere zero, so would not have zero velocity anywhere.