Thumbing through "Foundations of Differentiable Manifolds and Lie Groups" by Frank Warner today, I saw the following
Lemma: $\;$ $T_{m}M$ is naturally isomorphic to $(F_m/{F_m}^2)^\ast$.
Here $M$ is a smooth manifold, $T_mM$ is the tangent space at $m \in M$ and $F_m$ is the ideal of all germs of smooth functions from $M$ into $\mathbb R$ which vanish at $m$. The isomorphism is given by $T_mM \to (F_m/{F_m}^2)^\ast, \quad X \mapsto \left( [f] \mapsto X(f) \right)$ where $f \in C^\infty(U)$ represents a germ on some neighbourhood $U$ of $m$.
I observed a corollary of this: Since both vector spaces are finite dimensional, we obtain an isomorphism between the dual spaces $T_m^\ast M \simeq (F_m/{F_m}^2)^{\ast \ast} = F_m/{F_m}^2$, given by $[g] \mapsto {\mathrm dg}_m$.
But this then implies the following:
If $f\in C^\infty(\mathbb R^n)$ such that $f(0) = 0$ (i.e. $[f]\in F_m$ with $m=0$) and ${\mathrm df}_0 = 0$, then, there exists a neighbourhood $U$ of $0$ in $\mathbb R^n$ and functions $g,h \in C^\infty(U)$, such that $g(0) = 0, h(0) = 0$ and $f(x) = g(x) h(x) \quad \forall x\in U$
I was wondering whether one could obtain the above result directly?
For the one-dimensional case this is not too difficult: Just use Taylorexpansion to get $f(x) = f(0) + df(0)x + r(x)x^2 = r(x)x^2$
And define $h(x) = x\cdot r(x),\; g(x) = x$.
But already in dimension $2$, I don't know how one would factor something like $x^2 + y^2$. So here are my
Questions:
- Do you see any flaws my derivation of the corollary?
- Do you see a way of factoring $x^2 + y^2$ in $F_0$ (as described above)?.
- Is there a general way of doing this / a direct proof of this "factorization-property"?
I'm looking forward to your thoughts!