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Standard Square is not a Smooth Submanifold of R^2 
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#1
Aug2510, 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. 


#2
Aug2510, 03:16 AM

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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 epsilonopen 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.) 


#3
Aug2510, 04:01 AM

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#4
Aug2510, 04:03 AM

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Standard Square is not a Smooth Submanifold of R^2



#5
Aug2510, 04:16 AM

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Oh? Well, one wellknown 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 epsilondisk like in my post(?) 


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