# $M_m$ is naturally isomorphic to $(F_m/F_m^2)^{*}$

Let us denote $M_m$ be the set of tangent vectors to a manifold $M$ at point $m$ and is called tangent space to $M$ at point $m$ we denote $\bar{F_m}$ be the set of all germs at point $m$ and $F_m$ be the set of germs vanishes at $m$ In warner book there is a lemma: $M_m$ is naturally isomorphic to $(F_m/F_m^2)^{*}$: In proof he says if $v\in M_m$, then $v$ is a linear function on $F_m$ vanishing on $F_m^2$ because of the derivation property ,but I do not get why is that so?Could any one explain me a explicitly why? derivation property says $v(f.g)=f(m)v(g)+g(m)v(f)$, but I do not connect this with the above line of my confusion. Thank you.

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If $\nu$ is a derivation and $f$, $g \in F_m$, then $\nu(fg) = f(m)\nu(g) + g(m)\nu(f) = 0$ (as $f(m) = g(m) = 0$), and $F_m^2$ is generated by $\{fg\mid f,g \in F_m\}$. –  martini Jun 20 '12 at 6:27
Thank you :) :) –  La Belle Noiseuse Jun 20 '12 at 6:32
Ok I want to ask about another confusion related to this question: If $l\in (F_m/F_m^2)^{*}$, he defines $v_l$ at $m$ by setting $v_l(f)=l(\{f-f(m)\})$ for $f\in\bar{F_m}$ where $\{\}$ denote the cosets in $F_m/F_m^2$, could you please explicitly show me the action or activity of $l$ and defintion of $v_l$? and how the cosets looks like here? –  La Belle Noiseuse Jun 20 '12 at 6:44

I'll write an answer now. Regarding the question in the OP: For a derivation on $\bar F_m$ and we have $f,g \in F_m$ then $\nu(fg) = f(m)\nu(g) + g(m)\nu(f) = 0$ as $f(m) = g(m) = 0$. These elements generate $F_m^2$, so we have $\nu|_{F_m^2} = 0$ and so a well-defined element $\bar\nu\in (F_m/F_m^2)^*$ by $\bar \nu(f) = \nu(f)$.
Regarding the question in your comment: If $\ell \in (F_m/F_m^2)^*$, then $\ell\colon F_m/F_m^2 \to \mathbb R$ is linear. For $f \in \bar F_m$, we have $\hat f := f - f(m) \in F_m$. The coset $\{\hat f\}$ is by definition $\{ \hat f\} = f - f(m) + F_m^2 = \{f-f(m)+h \mid h \in F_m^2\}$ As $\ell$ is a linear map on the cosets we may define $\nu_\ell(f) = \ell(\{\hat f\})$. Let us show that $\nu_\ell$ is has the derivation property (as the linearity follows directly from the linearity of $\ell$). So for $f,g \in \bar F_m$ we have \begin{align*} \hat f \hat g &= \bigl(f - f(m)\bigr)\bigl(g - g(m)\bigr)\\ &= \bigl(f- f(m)\bigr)g - \hat f g(m)\\ &= fg - f(m)\bigl( g - g(m)\bigr) - f(m)g(m) - \hat fg(m)\\ &= \widehat{fg} -f(m)\hat g - g(m)\hat f\\ \iff \widehat{fg} &= f(m)\hat g + g(m)\hat f + \hat f \hat g \end{align*} Now $\hat f \hat g \in F_m^2$, so taking cosets we have $\{\widehat{fg}\} = f(m)\{\hat g\} + g(m)\{\hat f\}$ which gives by definition of $\nu_\ell$: $\nu_\ell(fg)= f(m)\nu_\ell(g) + g(m)\nu_\ell(f).$