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I'm having trouble understanding the proof in Lee of the above claim. More precisely, that if $M$ is a smooth manifold and $S \subset M$ is an embedded $k$-submanifold, then $S$ satisfies the local $k$-slice condition, i.e. for all $p \in S$, there exists a smooth chart of $M$, $(U,\varphi)$, such that $S \cap U$ is a $k$-slice in $U$.

The proof goes as follows: Since the inclusion map is an immersion, there are smooth charts $(U,\phi)$ and $(V,\psi)$ of $S$ and $M$ respectively such that $$ i(x^1,\dots,x^k) = (x^1,\dots,x^k,0,\dots,0) $$ This follows from the rank theorem for maps of constant rank. Then we can choose $\epsilon > 0$ such that $U$ and $V$ contains coordinate balls $U_0$ and $V_0$, respectively. Then Lee claims that $U_0 = i(U_0)$ is a $k$-slice in $V_0$. Intuitively, I believe it to be true, but would like some explanation as to why that's the case, and the significance of this step in the overall proof.

staedtlerr
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1 Answers1

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$S\cap V_0=i(U_0)$ by construction: if $p\in S\cap V_0$ then, in coordinates, because $p\in S,\ p=(x^1(p),\cdots, x^k(p))$ and because $p\in V_0\subseteq V,\ p=(x^1(p),\cdots, x^k(p),0,\cdots, 0)$, so $S\cap V_0\subseteq i(U_0).$ The other inclusion is clear.

Matematleta
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  • I am not sure $S \cap V_0 = i(U_0)$. The rank theorem for either immersion or embedding tells us the inclusion maps an open set $U_0$ in $S$ to a k-slice in an open set $V_0$ in $M$, but the k-slice is not necessarily the whole intersection $S \cap V_0$, consider irrational wrapping of the torus for an example. If it's embedding, then by subspace topology $U_0 = S \cap W$ for some open set $W$ in $M$. We now have $S \cap W \cap V_0= i(U_0)$. – liyiontheway Mar 20 '22 at 03:37