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Let $M$ be a manifold endowed with a connection $\nabla$. Let $M_0$ be a submanifold of $M$, we denote by $N$ the normal bundle of $M_0$.

The following paragraph is from the book: Heat kernels and Dirac operators (page 217)

By orthogonal projection, the Levi-Civita connection $\nabla$ gives a connection $\nabla^N$ on the normal bundle $N$ which is compatible with the induced metric. Identifying $M_0$ with the zero section of $N$, we obtain a canonical isomorphism $$TN_{|M_0} \cong TM_{|M_0}$$

How to prove that there's an isomorphism between the bundles $TN_{|M_0}$ and $TM_{|M_0}$ ?

Mira
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  • I cannot find this on page 217. – andres1 Aug 10 '21 at 13:05
  • Hint: There are canonical isomorphisms $TN|{M_0}\cong TM_0\oplus N\cong TM|{M_0}$. – Kajelad Aug 10 '21 at 13:48
  • @andres1 Maybe we have different editions of the book! I'm reading chapter 7 (Equivariant differential forms) and the paragraph in my post is in the proof of theorem 7.13 (it's written just after the proof of lemma 7.14). – Mira Aug 10 '21 at 13:49
  • @Kajelad, why is $TN|_{M_0}$ isomorphic to $TM_0 \oplus N ?$ – Mira Aug 11 '21 at 09:17
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    @asma Take the vertical lifting map $l:N\to TN|{M_0}$ defined by $l(v)(f)=\frac{d}{dt}f(tv)|{t=0}$, and the differential $dz:TM_0\to TN|{M_0}$ of the zero section $z:M_0\to N$. These combine to form a vector bundle isomorpism $\psi:TM_0\oplus N\to TN|{M_0}$ defined by $\psi(u\oplus v)=dz(u)+l(v)$. – Kajelad Aug 11 '21 at 10:51
  • Hi @Kajelad , why don't you use tubular neiborhood theorem to identify the neighborhood of normal bundle with neiborhood of $M$ directly ? – yi li Oct 13 '22 at 05:21
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    @yili That would be putting the cart before the horse, in my opinion. The decomposition $TNM_0|{M_0}\cong TM\oplus NM_0\cong TM|{M_0}$ is essentially linear algebra, and is usually the starting point of the tubular neighborhood theorem, which states that (under certain conditions) this map extends to a diffeomorphism on a neighborhood of $M_0$. – Kajelad Oct 14 '22 at 20:59
  • got it thank you. – yi li Oct 15 '22 at 00:54

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