# Question about definition of normal bundle from the quotient space

According to Wikipedia, the definition of normal bundle is defined as,

Defintion $$1$$. [Normal bundle]

Let $$(M,g)$$ be a Riemannian manifold, and $$S\subset M$$ a Riemannian submanifold.

• For a given $$p \in S$$ , a vector $$n \in T_pM$$ to be normal to $$S$$ if whenever $$g(n,v)=0$$ for all $$v \in T_pS$$. Then the set $$N_pS$$ of all such $$n$$ is called the normal space to $$S$$ at $$p$$.
• $$NS:= \coprod _{s \in S}N_pS$$ is called the total space of normal bundle to $$S$$ at $$p$$. ............. $$(*)$$

From the above definition, I understand

$$N_p S := \left\{ n \in T_pM : g(n,v)=0~ for~ all~~ v ~\in T_pS \right\}$$

For example, let $$M=\mathbb{R}^2$$ and $$S=S^1$$ and pick a point $$p\in S$$. Then, since $$S$$ is embedded in $$M$$, naturally $$p\in M$$. and $$T_pM$$ is also well defined. then since the vector $$n$$ is easily constructed ( clearly the every vector $$v$$ who lives in $$T_pM$$ is perpendicular to $$n$$, literally, $$g(n,v)=0$$. ) and like the second bullet, $$NS$$ is also well defined for any point $$p \in M$$. (The following image is just visualizing my description. the pink vector is one of element of $$N_pS$$ ) Meanwhile, how about the formal definition of normal bundle? Even though such definition is slightly different from each source, essentially, the basic idea seems to use a quotient space, However, I think that the formal definition contradicts the first definition, $$(*)$$. To begin with, based on wikipedia description,

one can define a normal bundle of $$N$$ in $$M$$, by at each point of $$N$$, taking the quotient space of the tangent space on $$M$$ by the tangent space on $$N$$. For a Riemannian manifold one can identify this quotient with the orthogonal complement, but in general one cannot...

I will define for convenience,

Defintion2. [General definition of normal bundle] Let $$(M,g)$$ be a Riemannian manifold, and $$S\subset M$$ a Riemannian submanifold. Then the quotient space $$TM/TS$$ is called a normal bundle to $$S$$ at $$p$$.

For example, also consider $$M=\mathbb{R}^2$$ and $$S=S^1$$ and metric tensor still is Riemannian metric, $$g$$. if I pick two vector bundle $$v,w \in TM$$, the equivalence relation ~ is given,

$$v\sim w ~if~and~only~if~ v-w \in TS ...... (**)$$

Then we can viusalize both vector and $$v$$ and $$w$$ like the below picture, and since $$\sim$$ is equivalent relation, we consider a representation $$\nu \in TM/TS$$. Then the representation $$\nu$$ would be a normal bundle. Of course, when considering $$(**)$$, the representation is descirbed

$$\nu=\left\{ w + \alpha u : w \in TM , \alpha \in \mathbb{R}, u \in TS \right\}$$

However, when comparing to the first definition, $$\nu$$ clearly does not represent normal vector. Obviously, for any point $$p \in S$$ , then $$w+u$$ is not perpendicular to the tangential vector $$T_pS$$

Therefore, I cannot understand why the formal definition of normal bundle , Defintion2, is a reasonable statement when considering Definition1.

• Definition $2$ doesn't require a metric (in particular, you don't need to talk about Riemannian manifolds and Riemannian submanifolds, just smooth manifolds and smooth submanifolds). Your expression for $\nu$ is not the normal bundle, it is all of $TM$. I'm not sure what you intended it to be. Jun 12, 2022 at 22:15

Indeed the Riemannian Manifold structure of $$M$$ ties $$TM/TS$$ with the idea of normality, but, as you have realized, it isn't that direct. Let me draw a picture for you. Let $$M = \mathbb{R}^3$$ be endowed the standard Riemannian manifold structure and take $$S := \mathbb{S}^2 \subseteq M$$ to be the unit sphere with a Riemannian submanifold structure. Fix an arbitrary point $$p \in S$$ and draw the tangent plane $$T_p S.$$

What does $$T_p M / T_p S$$ look like? Visually, it corresponds to those tangent vectors at $$p \in M$$ whose difference resides in a plane that is a translation of the tangent plane $$T_p S.$$ As a result, one can view $$T_p M / T_p S$$ as a collection of planes where the "zero" element is precisely the plane $$T_p S$$ based at $$p.$$ Okay. So what does this have to do with the (more intuitive) definition of the normal space at $$p$$ which you denote $$N_p S$$? I'll use your notation for the metric $$g_p: T_p M \times T_p M \to \mathbb{R}.$$ Fix an element of the quotient space $$V_p \in T_p M / T_p S.$$ Note that $$V_p$$ is not a vector space, but can be taken to be an affine plane in the vector space $$T_p M.$$ I claim that, in this subspace, there exists a unique vector $$q \in V$$ so that

$$g_p(q, v) = 0,\qquad \forall v \in V_p,$$

where I'm using the natural identification of elements in $$V_p$$ to vectors in $$T_p M.$$ By this formula and the uniqueness of the solution, every hyperplane $$V_p \in T_p M / T_p S$$ can be tied to a single vector $$q \in T_p M.$$ Visually, This identification can be done for every element of the quotient space producing, Here $$q$$ is a vector that is normal to $$S$$ at $$p$$ in the sense derived from the metric. This is precisely what is meant by the comment:

For a Riemannian manifold one can identify this quotient with the orthogonal complement, but in general one cannot...

Without the inner product, I can collapse each and ever plane $$V \in T_p M / T_p S$$ to some vector in $$V$$ but it would be arbitrary. For example, below is one such (arbitrary) choice of collapsing the planes $$V$$ to a point in $$T_p M.$$ Worse still, it can be done in a way that is not smooth. Again, the metric is doing the heavy lifting here and ensures the smoothness of the subbundle $$NS \subseteq TM$$ that is produced.

Why do some texts use the quotient instead of the extrinsic definition? Just as the commenter points out, it is because even without the metric one can still formally construct $$E := \sqcup_{p\in S} T_p M / T_p S$$ as a vector bundle. It just won't be a subbundle of $$TM$$ and so one cannot think of it a smooth assignment of vector spaces normal to $$S$$ (ask yourself what the ambient vector space is and what "normal" would mean). It still has value, however, as its rank still indicates the dimension of the directions "transverse" to $$S.$$