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Summary: Let $M$ be a smooth manifold and $p\in M$. I know of two notions of germs of functions at $p$, the more restrictive of which can be written as a quotient vector space. I am curious whether the more general notion can also be written as a quotient vector space.


Let $U$ be a neighborhood of $p$, let $\mathscr C^\infty(U)$ denote the set of smooth functions from $U$ to $\mathbb R$, and let $I_p(U)\subseteq\mathscr C^\infty(U)$ denote the subspace of functions that vanish in a neighborhood of $p$. Then, as is noted here, the space of germs of functions on $U$ at $p$ can be defined as

$$C^\infty_p(U)=\frac{\mathscr C^\infty(U)}{I_p(U)}.$$

This is a nice definition because the resulting space is automatically a vector space. However, the restriction to $U$ is undesirable. If $M$ were an analytic manifold, we might want smooth functions on some neighborhood of $p$ to have an associated germ, regardless of whether the function extends to the whole manifold.

Let $\mathscr C^\infty_p$ denote the set of pairs $(f,U)$, where $U$ is a neighborhood of $p$ and $f:U\to\mathbb R$ is smooth. Define the equivalence relation $(f,U)\sim_p(g,V)$ if $f$ and $g$ are identical on a common neighborhood of $p$. Then we can define

$$C^\infty_p=\mathscr C^\infty_p\big/\sim_p.$$

It turns out that $C^\infty_p$ is a vector space when we define addition and scalar multiplication in a natural way: $[(f,U)]+[(g,V)]=[f|_{U\cap V}+g|_{U\cap V},U\cap V]$ and $a[(f,U)]=[(af,U)]$. But it takes some extra work to show that these operations are well defined and that the vector space axioms are satisfied. Moreover, this seems like a coincidence, because $\mathscr C^\infty_p$ does not have an obvious vector space structure. Can we define $C^\infty_p$ as a quotient of two vector spaces, in a similar way to $C^\infty_p(U)$?

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  • $\begingroup$ I don't think it's a coincidence. We're taking the equivalence relation precisely to ensure that we have well-defined addition and scalar multiplication regardless of the domain issues. $\endgroup$
    – peek-a-boo
    Commented May 9, 2022 at 20:51
  • $\begingroup$ @peek-a-boo I agree. I just think it would be nice if this were reflected in the definition of $C^\infty_p$ by taking the quotient of two vector spaces, since this is possible in the other case. $\endgroup$
    – WillG
    Commented May 9, 2022 at 20:53
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    $\begingroup$ How about this: let $V_p$ be the set of all functions with domain the entire manifold, and have whatever regularity you want in a neighborhood of $p$ (eg $C^1,C^{\infty}$, real/complex analytic). And let $Z_p$ be all functions on the manifold which vanish in a neighborhood of $p$. These are easily verified to be vector spaces (subspaces of the space of all functions from the manifold into Reals/complex), so you can consider the quotient $V_p/Z_p$? $\endgroup$
    – peek-a-boo
    Commented May 9, 2022 at 20:59
  • $\begingroup$ Of course set theoretically this is different from what you defined, but it amounts to the same thing. Also since we only care about what happens near the point p, that's what's reflected in the definitions (so this works even for the analytic case). $\endgroup$
    – peek-a-boo
    Commented May 9, 2022 at 21:00
  • $\begingroup$ @peek-a-boo I think that does the trick! This is just as good, since we only care about what happens near $p$. $\endgroup$
    – WillG
    Commented May 9, 2022 at 21:07

1 Answer 1

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Although $\mathscr C^\infty_p$ does not have a vector space structure, it comes close to a vector space. Let us write the elements of $\mathscr C^\infty(U)$ in the form $(f,U)$ to emphasize that $f$ lives on $U$. Then $$\mathscr C^\infty_p = \bigcup_U \mathscr C^\infty(U) .$$ An addition on $\mathscr C^\infty_p$ is defined by $$(f,U)+(g,V)= (f|_{U\cap V}+g|_{U\cap V},U\cap V) .$$ This gives us a natural structure of an abelian monoid. The two-sided identity element is $(0_M,M) \in \mathscr C^\infty(M)$, where $0_X: X \to \mathbb R, 0_X(x) = 0$. The only elements which have an inverse are those in $\mathscr C^\infty(M)$.

Clearly the scalar multiplication on the $\mathscr C^\infty(U)$ gives us a natural scalar multiplication on their union. Explicitly it is given by $$a(f,U)= (af,U) .$$

Thus we get an algebraic structure which comes close to a vector space; all vector space axioms except the existence of additive inverses are satisfied.

A special feature of $\mathscr C^\infty_p$ is that each $(f,U) \in \mathscr C^\infty(U)$ has a lot of "individual two-sided identity elements": These are the elements $(\zeta,V) \in \mathscr C^\infty(V)$ with $U \subset V$ and $\zeta \mid _U = 0_U$ which have the property $(f,U) + (\zeta,V) = (\zeta,V) + (f,U) = (f,U)$. In particular $(0_V,V)$ is a "local identity" on $\bigcup_{U \subset V}\mathscr C^\infty(U)$. Also note that all $(f,U) \in \mathcal C^\infty_p$ have a "near inverse"; this is the map $-(f,U) = (-f,U)$. In fact, $(f,U) + (-f,U) = (0_U,U)$.

It is now an obvious idea to make all elements of $\mathscr C^\infty_p$ invertible by declaring all local identities $(0_U,U)$ to be equivalent. What does this mean?

The set $I^\infty_p \subset \mathscr C^\infty_p$ of all local identities forms a "linear subspace" of $\mathcal C^\infty_p$ in the sense that it closed under addition and scalar multiplication, and also contains to global identity $(0_M,M)$. In fact we can write $I^\infty_p = 0 \cdot \mathcal C^\infty_p$. A sort of quotient $$\mathscr C^\infty_p / I^\infty_p$$ can be defined by declaring $(f,U)$ and $(g,V)$ to be equivalent, $(f,U) \sim (g,V)$, if there exists $\omicron \in I^\infty_p$ such that $(f,U) + \omicron = (g,V) + \omicron$. It is easy to verify that this in fact an equivalence relation on $\mathscr C^\infty_p$ and that $(f,U) \sim (g,V)$ if and only if $f \mid_W = g \mid_W$ for some $W \subset U \cap V$, i.e. $(f,U)\sim_p(g,V)$ as defined in your question. Also observe that $I^\infty_p$ is contained in the equivalence class of $(0_M,M)$. Actually $[0_M,M] =\{ (f,U) \mid f \mid_W = 0_W \text{ for some } W \subset U\}$.

Therefore we get the desired description

$$C^\infty_p = \mathscr C^\infty_p / I^\infty_p .$$ This is not a quotient of two vector spaces, but it comes very close to it. Moreover, it has the benefit of explaining the origin of the equivalence relation $(f,U)\sim_p(g,V)$.

Remark: Here is a more sophisticated approach.

Let $\mathcal U_p$ denote the set of all open neigborhoods of $p$ in $M$. On $\mathcal U_p$ define $U \le V$ if $V \subset U$. Then $\mathcal U_p$ becomes a partially ordered directed set. If $U \le V$, then the inclusion map $i_{UV} : V \to U$ induces a natural linear map $i_{UV}^* : \mathcal C^\infty(U) \to \mathcal C^\infty(V), i^*(f,U) = (f \circ i,V) = (f \mid_{V},V)$. The collection $(\mathscr C^\infty(U), i_{UV}^*)$ is a directed system of vector spaces and linear maps. It has a direct limit $$\mathscr C^\infty_p = \varinjlim (\mathscr C^\infty(U), i_{UV}^*) $$ which is again a vector space. An explicit construction of $\varinjlim (\mathscr C^\infty(U), i_{UV}^*)$ is this:

Take the disjoint union of all $\mathscr C^\infty(U)$. Since they are pairwise disjoint, this disjoint union is nothing else than $\mathscr C^\infty_p = \bigcup_U \mathscr C^\infty(U)$. On this set introduce an equivalence relation via $(f,U) \sim_p (g,V)$ if $i_{UW}^*(f,U) = i_{VW}^*(g,V)$ for some $W \ge U,V$. This equivalence relation is precisely that considered in your question.

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  • $\begingroup$ (+1), but should the \bigcup in the first display be \bigcap? $\endgroup$ Commented May 10, 2022 at 12:14
  • $\begingroup$ @AndrewD.Hwang No, $\mathscr C^\infty_p$ is the union of all $\mathscr C^\infty(U)$. The intersection is empty since the $\mathscr C^\infty(U)$ are pairwise disjoint. $\endgroup$
    – Paul Frost
    Commented May 10, 2022 at 12:44
  • $\begingroup$ Thanks; I'd (mis-)understood something like a direct limit over open neighborhoods of $p$. $\endgroup$ Commented May 10, 2022 at 13:58

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