3

Given a smooth sub-manifold $X$ of $\mathbb{R^n}$ and define the diagonal in $X \times X$ to be

$$\triangle = \{(x,x) \mid x \in X \} \subset \mathbb{R^n}\times \mathbb{R^n}$$

and normal bundle to $\triangle$ is defined to be

$$N(\triangle)=\{(y,w) \mid y\in \triangle, w\in T_y(X\times X) \text{ such that } w \cdot v =0 \text{ for all } v \in T_y(\triangle)\}$$

with projection $\pi : N(\triangle) \to \triangle$ given by $(y,w)\mapsto y$

Then we want to show that $N(\triangle)$ is equivalent to the tangent bundle $TX$.

I know (by definition of equivalent) that two vector bundles said to be equivalent if there are diffeomorphisms $$f:TX \to N(\triangle) \text{ and }\tilde{f}:X \to \triangle$$

such that $f$ takes each fibre $TX_x$ to $N(\triangle)_{\tilde{f}(x)}$ by a vector space isomorphism.

What do I really need to do? Should I try to write down $f$ and $\tilde{f}$ and show that $f$ is an isomorphism on fibres?

Any help would be thankful.

SamC
  • 1,738

1 Answers1

2

Simply note that $TX=T\Delta=\{(y, w, w)\in T_y(X\times X)=T_yX\oplus T_yX, y\in\Delta\}$ and $N\Delta=\{(y, v, -v)\in T_y(X\times X), y\in\Delta\}$. Now the isomorphism between the two vector bundles can be easily constructed.

Alex Fok
  • 4,828
  • Why there is 3 components, $(y,w,w) \in T_y(X \times X)$? isn't $X \times X \subset \mathbb{R^n} \times \mathbb{R^n}$ so shouldn't be $(w,,w) \in T_y(X \times X)$? Also, why $TX = T\triangle$? – SamC May 11 '15 at 13:32
  • I am being pedantic. $y$ denotes a point in the base manifold. You may ignore $y$ if you understand the context. – Alex Fok May 11 '15 at 13:37