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I think this question is quite trivial. We know that the classical tubular neighborhood theorem asserts the follows:

There exists an open neighborhood of $M$ in $A$ which is diffeomorphic to the total space of the normal bundle under a diffeomorphism which maps each point $x$ in $M$ to the zero normal vector at $x$

The reference is Milnor&Stasheff Chapter11. In the book the authors further proved the following statement:

"The normal bundle $v^{n}$ associated with the diagonal embedding of $M$ in $M\times M$ is canonically isomorphic to the tangent bundle of $M$"

Now in reading Greg's paper "K-Theory and Elliptic Operators"(http://arxiv.org/abs/math/0504555), in page 16 he argued that considering a vector bundle $E$ over $X$, since $E$ is locally homeomorphic to $U\times \mathbb{R}^{m}$, $TE$ locally is isomorphic to $TX_{x}\oplus E_{x}$. Hence the tangent bundle $TE$ over $E$ admits the decomposition $$TE=\pi^{*}TX\oplus \pi^{*}E$$ with $\pi:E\rightarrow X$ be the projection map.

He applied this formula to $E=TX$ and $E=TY|_{X}$ with $X\subset Y$, the results are:

$$T(TX)\cong \pi^{*}TX\oplus \pi^{*}TX$$ $$T(TY|_{TX})\cong \pi^{*}(TY|_{x})\oplus \pi^{*}(TY|X)$$

I am not sure how he goes from $T(TY|_{X})$ to $T(TY|_{TX})$, perhaps this is a typo. Nevertheless this formula still makes sense. He explained that $T(TX)$ and $T(TY)|_{TX}$ are both bundles over $TX$ and $\pi:TX\rightarrow X$ is the projection. Assuming $N$ and $TN$ are normal bundles of $X$ and $TX$ in $Y$ and $TY$, Greg contend we have should have $N\oplus TX\cong TY|_{X}$ and $TN\oplus T(TX)\cong T(TY)|_{TX}$. Then he concluded that

$$TN\cong \pi^{*}N\oplus \pi^{*}N$$

by above reasoning. I am very perplexed with this result for the following reasons:

  1. Why we have $TN\cong N(TX)$? Greg proposed that "..extending to tangent bundles,we see that $TN$ is also a tubular neighborhood of $TX$ in $TY$ diffeomorphic to the normal bundle of $TX$ in $TY$. For a proof of the tubular neighborhood theorem, see [16,11]". But I could not see how the tubular neighborhood theorem could lead to this result. I think the statement $TN\cong N(TX)$ does make sense, but I could not make an easy proof based on the reference he cited.

  2. By the result $$TE=\pi^{*}TX\oplus \pi^{*}E$$ and assuming $\pi:TN\rightarrow X$ is the projection map, we should have isomorphism $TN\cong \pi^{*}TX\oplus \pi^{*}N$ But instead we have $TN\cong \pi^{*}N\oplus \pi^{*}N$. I do not understand why we have $\pi^{*}(TX)\cong \pi^{*}N$. I also do not see why $N\oplus TX\cong TY|_{X}$ and $TN\oplus T(TX)\cong T(TY)|_{TX}$ could be helpful to derive this. I think this statement is important for it later used directly into the construction of the topological index in the Atiyah-Singer index theorem.

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1 Answer 1

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  1. Let $\pi: N \to X$. Using what you said above, we have $TN = \pi^* TX \oplus \pi^* N$. Think of an element in $N$ as $x+n$ where $x \in X$ and $n$ is normal to $T_xX$. Then we see that $$TN = \{(x, n, v, n') : x \in X, n, n' \perp T_x, v \in T_x X\}$$. On the other hand, we have that $$N(TX) = \{(x,u,t) : x\in X, u \in T_x X, t \perp T_{(x,u)} TX\}$$. But thinking locally $T_{(x,u)} TX$ breaks into directions tangent to $T_x$ and the tangent to the tangent directions (locally $TX = U\times \mathbb R^k$ so $T_{(x,u)} TX = T_x U \times T_u \mathbb R^k$). So we can break $T_{(x,u)} TX$ up into a direct sum of normal directions to $X$, giving the desired isomorphism to $TN$.
  2. It is not the case that $\pi^* TX = \pi^* N$. But as before $$TN = \{(v_x, n,n') : n,n' \perp T_x X\}$$. If $p: TX \to X$ is the projection then $$p^* N \oplus N = \{(v,n_1,n_2) : n_i \perp T_{p(v)} X\}$$ which is just $TN$.
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Thank you, Eric. I really learned things from your response. –  Kerry May 27 '11 at 1:59

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