"standard" Square is not a Smooth Submanifold of R^2


by Bacle
Tags: smooth, square, standard, submanifold
Bacle
Bacle is offline
#1
Aug25-10, 01:45 AM
P: 662
Hi, everyone:
I am trying to show the standard square in R^2, i.e., the figure made of the line
segments joining the vertices {(0,0),(0,1),(1,0),(1,1)} is not a submanifold of R^2.

Only idea I think would work here is using the fact that we can immerse (using inclusion)
the tangent space of a submerged manifold S into that of the ambient manifold M
, so that, at every p in S, T_pS is a subspace of T_pM .

Then the problem would be clearly at the vertices. I think we can choose a tangent
vectorX_p at, say, T_(0,0) S, and show that X_p cannot be identified with
a tangent vector in T_(0,0) R^2.

Seems promising, but it has not yet been rigorized for your protection.

Any ideas for making this statement more rigorous.?

Thanks.
Phys.Org News Partner Science news on Phys.org
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
quasar987
quasar987 is offline
#2
Aug25-10, 03:16 AM
Sci Advisor
HW Helper
PF Gold
quasar987's Avatar
P: 4,768
Immagine the square S is embedded. Then at a corner point of the square, there is a smooth chart of Rē of the form D-->Rē where D is a epsilon-open disk centered at the corner that maps the corner to 0 and D cap S to R\subset Rē; the line (x,0).

Show that the inverse of this map is not differentiable at 0, thus spanning a contradiction. (the inverse of this map is actually the transition function btw this chart and the standard chart (Rē,id) or Rē, which is supposed to be smooth by hypothesis. That'S why this is a contradiction.)
Hurkyl
Hurkyl is offline
#3
Aug25-10, 04:01 AM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101
Quote Quote by Bacle View Post
Then the problem would be clearly at the vertices. I think we can choose a tangent
vectorX_p at, say, T_(0,0) S, and show that X_p cannot be identified with
a tangent vector in T_(0,0) R^2.
This sounds like a workable proof idea. It probably helps name elements of the tangent space at (0,0) via limits of tangent vectors defined on the two adjacent sides. Why? We understand the sides, but the corners are mysterious, so we use what we know to help us understand what we don't know.

Hurkyl
Hurkyl is offline
#4
Aug25-10, 04:03 AM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101

"standard" Square is not a Smooth Submanifold of R^2


Quote Quote by quasar987 View Post
Immagine the square S is embedded. Then at a corner point of the square, there is a smooth chart of Rē of the form D-->Rē where D is a epsilon-open disk centered at the corner that maps the corner to 0 and D cap S to R\subset Rē; the line (x,0).
Maybe it's just too early in the morning, but that's not obvious to me.
quasar987
quasar987 is offline
#5
Aug25-10, 04:16 AM
Sci Advisor
HW Helper
PF Gold
quasar987's Avatar
P: 4,768
Oh? Well, one well-known definition\characterisation of an embedded submanifold N of a manifold M is that at every point q of N, there exists a chart (U,f) of M around q such that f(U cap N)=R^k x {0}.

Here, N=S and M=R^2, and from such a chart (U,f) around a corner point q, I think it is clear how we can construct a chart of domain D an epsilon-disk like in my post(?)


Register to reply

Related Discussions
How the Berkeley edition Calculus book by Stewart differs from the "standard" one? Calculus 0
What is a "smooth" function? Calculus 1
"Normalizing" standard deviation between two data sets Set Theory, Logic, Probability, Statistics 5
"standard perturbation theory" - what exactly is meant? Quantum Physics 0
Can "smooth" and "analytic" be used interchangeably? Calculus 3