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On pg. 71 of Guillemin and Pollack they prove the $\epsilon$-Neighborhood theorem.

Here $Y$ is a compact boundaryless manifold in $\mathbb{R}^M$. They say

Proof: Let $h:N(Y)\to\mathbb{R}^M$ be $h(y,v)=y+v$. Notice that $h$ is regular at every point of $Y\times\{0\}$ in $N(Y)$, for through $(y,0)$ there pass two natural complementary manifolds of $N(Y)$, $Y\times\{0\}$ and $\{y\}\times N_y(Y)$. The derivative of $h$ at $(y,0)$ maps the tangent space of $Y\times\{0\}$ at $(y,0)$ onto $T_y(Y)$, and maps the tangent space of $\{y\}\times N_y(Y)$ at $(y,0)$ onto $N_y(Y)$.

I don't understand the last sentence "The derivative of $h$ at $(y,0)$ maps the tangent space of $Y\times\{0\}$ at $(y,0)$ onto $T_y(Y)$, and maps the tangent space of $\{y\}\times N_y(Y)$ at $(y,0)$ onto $N_y(Y)$."

Can anyone explain why that is please? Thanks.

I tried to compute out $$ dh_{(y,0)}(y_1,v)=y_1+v $$ but how does that imply $dh_{(y,0)}(T_{(y,0)}(Y\times\{0\}))=T_y(Y)$?

Geovanna Anthony
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1 Answers1

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Note that $h$ is the restriction to $N(Y)\subset\Bbb R^M\times\Bbb R^M$ of the linear map $$H\colon\Bbb R^M\times\Bbb R^M \to \Bbb R^M, \quad H(x,w)=x+w\,,$$ so $dH_{(y,0)} = H$. Thus, $dh_{(y,0)} = H\big|_{T_{(y,0)}N(Y)}$, as you suggested. When you write $(y_1,v)\in T_{(y,0)}N(Y) = T_y(Y) \times \{0\} \oplus \{0\}\times N_y(Y)$, this means $y_1\in T_y(Y)$ and $v\in N_y(Y)$. Since $dh_{(y,0)}(y_1,0) = y_1$ for every $y_1\in T_y(Y)$, this gives the surjection you desired. And similarly, because $dh_{(y,0)}(0,v)=v$ for every $v\in N_y(Y)$, you get the other surjection.

Ted Shifrin
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  • Thanks for formalising it Ted, that helps a lot! – Geovanna Anthony Jan 05 '14 at 22:23
  • Sure thing. Since you're relatively new here, you might not know that you can accept the answer (when you're satisfied), so that it will disappear from the list of unanswered questions :) – Ted Shifrin Jan 05 '14 at 22:25
  • Ok, I'll do that. Just one thing, how do you know $T_{(y,0)}N(Y) = T_y(Y) \times {0} \oplus {0}\times N_y(Y)$? – Geovanna Anthony Jan 05 '14 at 23:22
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    Think about parametrizing the normal bundle, or think about tangent vectors to curves through $(y,0)$. You have curves in $Y$ and you have curves in $N_y(Y)$, and that gives you $2\dim(Y)$ linearly independent directions :) – Ted Shifrin Jan 05 '14 at 23:26