(adsbygoogle = window.adsbygoogle || []).push({}); "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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Standard Square is not a Smooth Submanifold of R^2

**Physics Forums | Science Articles, Homework Help, Discussion**