2

The Jordan-Brouwer separation theorem is a celebrated result of algebraic topology, which generalizes Jordan curve Theorem (it was proved independently by Lebesgue and Brouwer in 1911: see Dieudonné, A History of Algebraic Topology and Differential Topology for a detailed discussion of Lebesgue's and Brouwer's contributions).

The theorem has also a remarkable version for smooth (that is $C^{\infty}$) hypersurfaces in $\mathbb{R}^n$. This is the result as stated in Guillemin and Pollack, Differential Topology, Chapter 2, Sec.5:

The complement of the compact, connected, smooth hypersurface $X$ in $\mathbb{R}^n$ consists of two connected open sets, the "outside" $D_0$ and the "inside" $D_1$. Moreover the closure of $D_1$ is a smooth manifold $M$, with boundary $\delta M = X$.

My question is whether there is a $C^k$ version of this theorem (with $k \geq 1$), that is whether we can replace smooth by $C^k$ in the above statement. I do not know very much about differential topology, so I apologize if my question should be a trivial one.

Maurizio Barbato
  • 3,638
  • Yes, you can. The proof is exactly the same - all of the machinery in G&P works for $C^k$ things. –  Mar 25 '16 at 12:45
  • @Mike Miller: Thank you very much Mike. As I said, I am new to differential topology, so I have some big doubts sometimes! – Maurizio Barbato Mar 25 '16 at 16:13
  • No worries at all. One nice thing to do to increase your understanding is to look through the things they do and figure out where they use smoothness, and figure out how much you can weaken it. (The definition of a tangent space in other settings is subtle in the $C^k$ case, but IIRC in G&P's setting the definition doesn't need to be changed.) This approach - start smooth, then see how much you can weaken - is actually how some analysts do their research :) –  Mar 25 '16 at 16:16

0 Answers0