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

by Bacle
Tags: smooth, square, standard, submanifold
 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.
 Sci Advisor HW Helper PF Gold P: 4,758 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.)
Emeritus
PF Gold
P: 16,101
 Quote by Bacle 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.

Emeritus