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Differential Topology Hirsch Chapter 2 Section 4 Problem 11: Compute the Dimension of $J^r(M, N)$

$J^r(M, N)$ is the set of all $r$-jets from $M$ to $N$. This is an equivalence class $[x, f, U]_r$ of triples $(x, f, U)$, where $U \subset M$ is an open set, $x \in U$, and $f: U \rightarrow N$ is a $C^r$ map; the equivlence relation is: $[x, f, U]_r = [x' f', U']_r$ if $x = x'$ and in some (hence any) pair of charts adapted to $f$ at $x$, $f$ and $f'$ have the same derivatives up to order $r$.

I wanted to check to see if this was right:

$J^r(M, N)$ seems to me to only distinguish among different points and functions whose derivatives differ at order $r+1$ and up. Since $x \in M$ we know that the dimension of $J^r(M, N)\geq dim M$. Now we just have to figure out the dimension of all functions that differ at order $r+1$ and up and add it to dim$M$? I can only think that this is infinite...

Math_Day
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  • Isn't the $r=1$ jet at a point just all maps from $T_p M$ to $T_f(p) N$? Then this is finite dimensional... – Jason DeVito - on hiatus Sep 09 '20 at 19:57
  • I agree. But I am still having trouble computing the dimension given the equivalence relation of the jets. It definitely isn't the entire Tangent space. – Math_Day Sep 09 '20 at 20:12
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    Consider first the case when $N$ is 1-dimensional and $M$ is $m$-dimensional. Compute the space of polynomials in $m$ variables, of degree $\le r$, with zero constant term. This will be your $J^r_x(M, N)$. – Moishe Kohan Sep 16 '20 at 23:05

1 Answers1

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One useful way of thinking of jets is as a generalization of Taylor approximations, or as a coordinate-independent way of encoding all derivatives of order $\le r$. In this respect,your last paragraph seems to have things backward: the $r$-jets at $p$ do not distinguish between functions whose derivatives at $p$ differ only at order $>r$.

Choose local coordinates $x^1,\dots,x^m$ on an open domain $V\subseteq M$ and $y^1\dots,y^n$ on $W\subseteq N$, and consider jets $[p,f,U]_r$ with $p\in V$ and $f(p)\in W$. The class $[p,f,U]_r$ is uniquely determined by $p$, $f(p)$, and the partial derivatives $$ \frac{\partial f^j}{\partial x^{i_1}\dots\partial x^{i_k}},\ \ \ 1\le k\le r,\ \ \ 1\le j\le n,\ \ \ 1\le i_1\le\dots\le i_k\le m $$ Furthermore, every such collection $(x^i,y^j,A^j_{i_1\dots i_k})$ with $A^{j}_{i_1\dots i_k}\in\mathbb{R}$ is the $r$-jet of some function $f$ (this can be seen by writing $f$ as a polynomial of order $r$ in coordinates). These are called the derivative coordinates on $J^r(M,N)$ induced by $x^i,y^i$. The number of coordinates is then the dimension of $J^r(M,N)$. $$ \dim J^r(M,N)=\dim M+\dim N+S(\dim M,r)\dim(N) $$ where $S(m,r)$ is the number of nondecreasing sequences $1\le i_1\le\dots\le i_k\le m$ with length $1\le k\le r$.

Showing that the derivative coordinates do indeed form a coordinate patch requires showing that they map smoothly into $J^r(M,N)$. This will depend on how you have chosen to define the smooth structure on $J^r(M,N)$.

Kajelad
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