# On Tangent vectors as jets & submanifolds

Here is my second question on understanding jets better:

For a smooth manifold $M$ the set of jets $J^1_0(\mathbb{R},M)$ is the same as the Tangent bundle $TM$. This implies that any equivalence class $j_0^1f$ of curves $f:\mathbb{R} \rightarrow M$ is a tangent vector at $f(0)$.

Now suppose $N \subset M$ is a submanifold of $M$. Then we can look on it from two different angles:

First $N$ is a manifold in its own right. In that case the jet class $j_0^1f \in J^1_0(\mathbb{R},N)$ contains smooth maps $f: \mathbb{R} \rightarrow N$ only.

But if we look on $N$ as a submanifold of $M$, then a jet class $j_0^1g \in J^1_0(\mathbb{R},M)$, that is tangent to $N$, defining the same tangent vector as $j_0^1f$, is not 'the same' as $j_0^1f$ as a set of curves.

Although $j_0^1g$ is tangential to $N$, it contains curves that intersect $N$ only in $g(0)$ just having the same first order derivative in $g(0)$.

So seen this way, although $j_0^1g$ and $j^0_1f$ define the same tangent vector in $f(0)$ they are not the same as sets of curves.

Is this right?

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Now to show that a jet $j_0^1g \in J^1_0(\mathbb{R},M)$ is actually tangent to a submanifold $N$, is it sufficient to show that there is a representative $h \in j^1_0g$ and an open neighbourhood $U$ of zero in $\mathbb{R}$ such that $h$ is locally a curve in $N$, that is $h : U \rightarrow N$ ?

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It may (though I don't guarantee it) be helpful to think about the class of all smooth functions $\pi:M\to N$ such that denoting the submanifold inclusion $i:N\to M$ we have $\pi\circ i = Id$. – Willie Wong Jan 16 '12 at 11:57

The magic concept here is functoriality.

1) Jets and tangent spaces are essentially useless if you do not say how they react to differentiable morphisms between manifolds.
More precisely, consider a $\mathcal C^1$- map $\phi: N\to M$ between two manifolds $N, M$, sending $n$ to $m$ : $\phi (n)=m$. Then it induces induces a linear map
$$T_n(\phi):T_n(N)\to T_m(M): j^1_n(f)\mapsto j^1_m(\phi\circ f)$$
(Of course there is something to prove: namely that if $f$ and $g$ have the same jet at $n$, so have $\phi \circ f$ and $\phi \circ g$ at $m$. Also, I have slightly changed your notation to emphasize the point of the manifold where you take the jet.)

2) This applies of course in your case, when $\phi$ is the inclusion $i:N\hookrightarrow M$ of the submanifold $N$ into $M$.
Suppose you have a curve $f:\mathbb R\to N$. Its jet consists of a family $j^1_{N,n}(f)$ of many brothers and sisters $g:\mathbb R\to N$.
Now the effect of $T_n(i)$ on $j^1_{N,n}(f)$ is to send it to $j^1_{M,n}(f)$.
Is it the same family?
No!
The old brothers and sisters remain brothers and sisters but the family is enlarged:
there are new brothers and sisters for $f$, namely the curves $h:\mathbb R\to M$ which do not necessarily remain in $N$ after their visit at $n$ but nevertheless are tangent to $f$.
An easy example: $N=\mathbb R\times 0\subset M=\mathbb R^2$, $f(t)=(t,0) ,\; h(t)=(t,t^2)$ .

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ok. But what about the last part of my question? – Mark Neuhaus Jan 16 '12 at 13:17
The answer to the last part of your question is "yes" because in the notation of your question we have $j^1_0(g)=T_n(i)(j^1_0(f))$. To be extremely precise, we identify $T_{N,n}$ with the subspace $T_n(i)(T_{N,n})$ of $T_{M,n}$. – Georges Elencwajg Jan 16 '12 at 13:52
Ok. Thanks for your help! – Mark Neuhaus Jan 16 '12 at 13:57