- #1
Bacle
- 662
- 1
"standard" Square is not a Smooth Submanifold of R^2
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.
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.