I need to show that, given a manifold $M$ and its diagonal $\Delta\subset M\times M$, we have $T\Delta\cong\mathcal{N}_{\Delta|M\times M}$, where $T\Delta$ is the tangent bundle and $\mathcal{N}_{\Delta|M\times M}$ is the normal bundle of $\Delta$.
The definition of normal bundle my professor told me is the following. For every $p\in \Delta$, define $$ (\mathcal{N}_{\Delta|M\times M})_p:=\frac{T_p(M\times M)}{T_p\Delta} $$ Gluing all together we obtain $$ \mathcal{N}_{\Delta|M\times M}=\frac{T(M\times M)|\Delta}{T\Delta} $$ I was looking at Bott-Tu classic book, and I found out a different definition of normal bundle. They say that it's defined by the exact sequence $$ 0\rightarrow T\Delta\rightarrow T(M\times M)|\Delta\rightarrow\mathcal{N}_{\Delta|M\times M}\rightarrow 0 $$. I think that to solve my initial problem is better to use this definition. But what I can't understand is
- Why the two definitions are equivalent?
- How is exactly defined the exact sequence above, i.e. how are those maps defined?
The problem is almost tautological, which makes it a bit hard to explain on the internet.
Let's forget about bundles, but consider just vector spaces. If $V\subset W$ is a subspace there is a canonical map $V\rightarrow W$ which just maps the element $v\in V$ to $v\in W$. This map is the inclusion and is injective.
Similarly, there is a canonical map $W\rightarrow W/V$ mapping $w\in W$ to $[w]$ (the equivalence class of all vectors in $W$ that differ from $w$ by an element in $V$). This map is called the projection and is surjective.
Clearly for $v\in V$, we see that $v\in W$ and $[v]=0\in W/V$ which shows that the sequence
$$ 0\rightarrow V\rightarrow W\rightarrow W/V\rightarrow 0 $$
is exact.
In general, saying that we have an injective map $i:V\rightarrow W$ (which we can view as a way of seeing $V$ as a subspace of $W$) is the same as saying that we have a (very) short exact sequence
$$ 0\rightarrow V\rightarrow W $$
We can always complete it to a short exact sequence $$ 0\rightarrow V\rightarrow W\rightarrow W/i(V)\rightarrow 0 $$
By projecting to the quotient as before. There is nothing more to say about bundles.