Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 :) :) – Un Chien Andalou 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? – Un Chien Andalou Jun 20 '12 at 6:44
up vote 2 down vote accepted

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 $\nu_\ell$: \[ \nu_\ell(fg)= f(m)\nu_\ell(g) + g(m)\nu_\ell(f). \]

share|cite|improve this answer
thank you very much martini – Un Chien Andalou Jun 20 '12 at 9:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.