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Let me define the space $C^\infty_p:=\{f:U\rightarrow \Bbb{R}:\text{ U is a neighbourhood of p}\}$. Then we define a derivation to be the linear map $X:C^\infty_p\rightarrow \Bbb{R}$ s.t. it satisfies the chane rule, i.e. for all $f,g\in C^\infty_p$: $X(fg)=X(f)g(p)+f(p)X(g)$.

Now my question is, why don't we have a $p$ in $X(f)$? Because somehow $X(f)$ needs to be an element in $\Bbb{R}$ but why don't we have any dependence on the point $p$?

user123234
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    It's not entirely clear to me what the question is, but the point $p$ is part of the definition of $X$: As you write, its domain is $C^\infty(p)$, which depends on $p$. – Travis Willse Jun 08 '23 at 07:05
  • @TravisWillse My question is why don't we have something like $X(f(p))$ or something like that, i.e. why doesn't $X(f)$ depend on $p$ – user123234 Jun 08 '23 at 07:06
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    I wouldn't call it chain rule, but rather Leibniz rule (it is slightly different) and it should look like: $$X(fg)=X(f)g+fX(g)$$ so that actually $X$ maps smooth functions around p to smooth functions around p: in particular, $X(fg)$, $X(f)$, $X(g)$ are smooth functions around p, and $$X(fg)(p)=X(f)(p)g(p)+f(p)X(g)(p)$$ – ashsan98 Jun 08 '23 at 07:08
  • @ashsan98 but I mean $X$ only takes a function as an input why do I have $X(f)(p)$? – user123234 Jun 08 '23 at 07:10
  • @user123234 it takes a function as input and gives a function of the same class as output; you can then evaluate this function at a point: that is what $X(f)(p)$ means: evaluate the function $X(f)$ at the point $p$ – ashsan98 Jun 08 '23 at 07:13
  • @ashsan98 but then wouldn't my definition be wrong wouldn't it then be $X:C^\infty_p\rightarrow C^\infty_p$? – user123234 Jun 08 '23 at 07:14
  • @user123234 yes, exactly! – ashsan98 Jun 08 '23 at 07:14
  • ahh okey so $X:C^\infty_p\rightarrow C^\infty_p$ and then $X(f): M\rightarrow \Bbb{R}; p\mapsto X(f)(p)$ where $M$ is a manifold. and then I get that in general $X(fg)=X(f)g+fX(g)$ is a map $M\rightarrow \Bbb{R}$ and we have $X(fg)(p)=X(f)(p)\cdot g(p)+f(p)\cdot X(g)(p)\in \Bbb{R}$. Is this all correct? – user123234 Jun 08 '23 at 07:18
  • @user123234 yes! Does it help or did you want to understand something further? – ashsan98 Jun 08 '23 at 07:19
  • @ashsan98 I have only one remaining question to not get confused later. $X(f)$ can take any value in $M$ so unfortunately I wrote above $p\in M$ but I have fixed $p$ in the definition of $C^\infty_p$. So I can have $X(f)(q)$ for $q\in M$ right, i.e. this $p$ is only used to define $f$? – user123234 Jun 08 '23 at 07:24
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    @user123234 ah good question, I did not actually correct you properly; it is defined for a neighbourhood of the point $p$, not for the whole manifold; although one can probably extend it (by zero or so). You should then change that on your comment from $M$ to $U$, but this $U$ might be different for $f$, for $X(f)$ and so on – ashsan98 Jun 08 '23 at 07:27
  • @ashsan98 ah so somehow I get $X(f):U_1\rightarrow \Bbb{R}$ where $U_1$ is a neighbourhood of $p$ (here $f:U_1\rightarrow \Bbb{R}$) but I can have $X(f):U_2\rightarrow \Bbb{R}$ with another neighbourhood $U_2$ of $p$ (here $f:U_2\rightarrow \Bbb{R}$). So there is somehow one parameter of freedom here – user123234 Jun 08 '23 at 07:31
  • @user123234 I meant rather that while $f$ might be defined on $U$, maybe $X(f)$ is defined a priori on $V$, both neighbourhoods of $p$ but not necessarily equal: we do not care about that, we only care about having it in a neighborhood of $p$ so that we have kind of info over the value of the function at $p$ and of all the derivatives at $p$ (these are all smooth functions), I would say – ashsan98 Jun 08 '23 at 07:36
  • @ashsan98 ah but the neighbourhood on which we define $X(f)$ does not need to be the same as the one on which $f$ is defined. – user123234 Jun 08 '23 at 07:37
  • @user123234 at least not from the definition, and it is not important for our purposes; maybe you can try to find a counterexample – ashsan98 Jun 08 '23 at 07:42
  • @ashsan98 so you mean $f:(-\epsilon,\epsilon)\rightarrow \Bbb{R}$ and we have $X(f)=\frac{1}{2}f$. Then $X(fg)=\frac{1}{2}fg$ and $X(f)g+fX(g)=\frac{1}{2}fg+f\frac{1}{2}g=fg$. But this isn't a derivation is it? – user123234 Jun 08 '23 at 07:43
  • @user123234 yeah that does not work; maybe it is true, but you would have to show it, it does not come from your definition and I would say it is not relevant for the construction you do with derivations, but do not trust me on that – ashsan98 Jun 08 '23 at 07:48
  • @ashsan98 okey thanks a lot! – user123234 Jun 08 '23 at 07:49

1 Answers1

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First note that a derivation has to satisfy the product rule (aka Leibniz rule) $$X(fg) = X(f)g(p) + f(p)X(g)$$ and not the chain rule. [Recall that the product rule of elementary calculus says that $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$.]

Why don't we have any dependence of $X(f)$ on the point $p$?

It seems to me that you misunderstood the concept of derivation. There is nothing like a derivation on $M$ in an absolute sense, we only have derivations at each point $p \in M$. In fact, the domain of a derivation at $p$ is the set $C_p^\infty$ which explicitly depends on $p$ because the elements of $C_p^\infty$ are precisely the smooth function living on some open neigborhood of $p$.

If we take two distinct points $p, q \in M$, then $C_p^\infty \ne C_q^\infty$, thus no derivation $X$ at $p$ can be derivation at $q$ simply because $C_q^\infty$ contains functions $g : V \to \mathbb R$ which do not belong to $C_p^\infty$. To see this, take any $\tilde g : \tilde V \to \mathbb R$ in $C_q^\infty$. Then $V = \tilde V \setminus \{p\}$ is also an open neigborhoods of $q$ so that the restriction $g = \tilde g \mid_V : V \to \mathbb R$ is in $C_q^\infty$. But clearly $g$ is not in $C_p^\infty$.

Certainly we have $C_p^\infty \cap C_q^\infty \ne \emptyset$, but this irrelevant here.

So perhaps it would be better to write $X_p$ for a derivation at $p$ to make the dependeny on $p$ more transparent notationally. But be aware that this is really just a notational issue and a matter of "taste".

There is another problem in your definition. You say that a derivation is a linear map $X:C^\infty_p\rightarrow \Bbb{R}$ which satisfy the Leibniz rule. Unfortunately $C_p^\infty$ is not a vector space, thus there are no linear maps on it. This problem can easily be handled by introducing the concept of a germ of smooth map at $p$. A germ at $p$ is an equivalence class of element of $C_p^\infty$, and the set of germs forms a vector space. See my answer to Can germs be defined as a quotient of vector spaces?

Paul Frost
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