# A particular confusion in showing $M_m$ is naturally isomorphic with $(F_m/F_m^2)^*$

I am working through the proof of a lemma in Frank Warner's Foundations of Differentiable Manifolds. I am trying to show that the maps he defines in the proof of this lemma are in fact inverses.

Here $$F_m^k$$ is defined to be the set of k-fold products of germs at m, $$M_m$$ is the set of tangent vectors at $$m$$.

Attempt: Let $$w\in M_m$$, this is mapped to the functional in $$(F_m/F_m^2)^*$$ that sends $$f$$ to $$m(f)=m^i\frac{\partial f}{\partial x^i}$$. We want to show that $$v_w(f)=w$$. $$v_w(f)=l(\{w(f)-w(f)(m)\})=l(\{w(f)\})-l(\{w(f)(m)\})$$ But I am not sure how to continue.

• This may be helpful: math.stackexchange.com/questions/160651/… Commented Jul 29, 2019 at 18:46
• Thank you, I have read through the posts on that question before posting on question but unfortunately it does not have any useful information related to my confusion. Commented Jul 29, 2019 at 19:35

A heuristic picture to keep in mind

It's a bit more natural to think about the isomorphism $$T^*_mM\simeq F_m/F_m^2$$. Using a local coordinate system this says that a covector $$df$$ is "the same" as a function $$f(x)=df\cdot x$$ what is "the same" as having a Taylor expansion $$f(x)=f(0) + df\cdot x + O(x^2)$$ for $$f$$ such that $$f(0)=0$$ (translates to "take $$F_m$$") and dropping $$O(x^2)$$ term (translates to "divide by $$F_m^2$$").

Lenghty explanation

Let's start with a vector $$v\in T_mM$$, i. e. an $$\mathbb R$$-linear map from $$C^\infty(M)\to \mathbb R$$ satisfying: $$v(fg) = v(f)\cdot g(m) + f(m)\cdot v(g)$$

Lemma: If $$f|_O=g|_O$$ for some open neighborhood $$O\ni m$$, then $$v(f)=v(g)$$.

Proof: Using linearity this translates to $$v(h)=0$$ if $$h|_O=0$$. Using a bump function we can construct a smooth function $$\phi$$ such that $$\phi(m)=1$$ and $$\mathrm{supp}~\phi \subseteq O$$. Then the smooth function $$h\cdot \phi=0$$ (i.e. the zero function) on the whole of $$M$$. Hence, $$0 = v(0) = v(h)\cdot \phi(m) + h(m) \cdot v(\phi) = v(h)$$

The above lemma basically says that if $$(f, U)$$ and $$(g, V)$$ represent the same germ $$[f, U] = [g, V] \in C^\infty_m(M)$$ (i. e. $$f|_O=g|_O$$ for some open $$O\subseteq U\cap V$$ containing $$m$$), then $$v(f)=v(g)$$. Hence, it makes sense to speak about the value $$v([f, U]) := v(f)$$.

It's just one step further from observing that we can define the value of a germ at point $$m$$ by: $$[f, U](m) := f(m)$$

Making the stalk $$C^\infty_m(M)$$ (the set of germs) into an algebra via operation on representives we discover that $$v$$ is a linear map and moreover it's a derivation: $$v( [f, U]\cdot [g, V] ) = v([f, U])\cdot {[g, V]}(m) + v([g, V])\cdot {[f, U]}(m)$$

So far so good. Now define $$F_m\subseteq C^\infty_m(M)$$ to be the set of germs vanishing at $$m$$, i.e. $${[f, U]}\in F_m$$ only if $$f(m)=0$$. $$F_m$$ has a structure of a vector space. It also makes sense to define $$F_m^2$$ as the set of sums of the form:

$${[f_1, U_1]}\cdot {[g_1, V_1]} + \dots + {[f_n, U_n]}\cdot {[g_n, V_n]}$$

where $$f_i(m)=g_i(m)=0$$ represent the elements of $$F_m$$. This also happens to be a real vector space. In fact this is a subspace of $$F_m$$ as such expressions vanish at $$m$$. So it makes sense to speak about the coset space $$F_m/F_m^2$$ and its dual.

Observe that if $$f(m)=g(m)=0$$, then $$v(fg) = v(f) \cdot g(m) + v(g)\cdot f(m)=0$$, what translates to the statement that $$v$$ is the zero map on the $$F_m^2$$ (this is a just space of (germs corresponding to) sums of such products of functions).

We know that $$v$$ is a linear map on $$F_m$$ (as a linear function on $$C^\infty_m(M)$$) and that $$v(F_m^2)=0$$. Hence, it induces a linear map on the quotient space: $$\nu_v\colon F_m/F_m^2\to \mathbb R$$ given by $$\nu_v([f, U]) = v(f)$$.

Note that the map $$v\mapsto \nu_v$$ is linear.

Take any linear functional $$\nu\in (F_m/F_m^2)^*$$. We need to produce a vector from it -- something that takes a function living on $$M$$ and returns a real number. So take a function $$f\in C^\infty(M)$$. Subtract $$f(m)$$ giving us a function $$\tilde f\in C^\infty(M)$$. Pass to the germ $${[\tilde f, M]}\in C^\infty_k(M)$$ that happens to be 0 at $$M$$, i. e $${[\tilde f, M]}\in F_m$$. We can use the projection $$F_m\to F_m/F_m^2$$ to map this into a coset $${[\tilde f, M]}\in F_m/F_m^2$$. And then we can apply our functional $$\nu$$ to produce a real number. To improve readability (and reduce mathematical precision) we will write:

$$v_\nu(f) = \nu([f-f(m)])$$

It's not hard to prove that $$v_\nu$$ is a linear map (using linearity of the operations described above) and the proof that this is a derivation is given in the posted question. This means that we can get a vector $$v_\nu$$ out of the linear functional $$\nu$$.

Observe also that the map $$\nu\mapsto v_\nu$$ is linear.

We have related the tangent space $$T_mM$$ and the space of functionals $$(F_m/F_m^2)^*$$ using two linear maps. Modulo clutter (what makes things precise but less readable) we see that their compositions evaluate to identites:

$$v\mapsto \nu_v \mapsto v_{\nu_v} = v$$ $$\nu\mapsto v_\nu \mapsto \nu_{v_\nu} = \nu$$

what ends this logical detour.

A short excursion into the realm of commutative algebra

(This is just a sketch. For better explanation see section 2.1 in the excellent lecture notes The Rising Sea of Ravi Vakil).

$$F_m$$ happens to be an ideal in the commutative ring $$C^\infty_m(M)$$. (Basically if $$f(m)=g(m)=0$$, then $$(f+g)(m)=0$$ and for every $$h$$ we have $$(h\cdot f)(m)=0$$). So we know what $$F_m^2$$ is and that $$F_m/F_m^2$$ is a module over $$C^\infty_m(M)/F_m$$. By writing down the exact sequence: $$0\rightarrow F_m\rightarrow C^\infty_m(M) \rightarrow \mathbb R\rightarrow 0$$ one can conclude $$C^\infty_m(M)/F_m\simeq \mathbb R$$, i. e. $$F_m/F_m^2$$ is a real vector space.