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Let $M$ be a smooth manifold, $U\subseteq M$ an open subset and $p\in M$. Then $C^\infty(U)$ is naturally an $\mathbb R$-algebra. Define: $$I_p(U):=\{f\in C^\infty(U):\exists W\subseteq U\textrm{ open with }p\in W\textrm{ and }f|_W=0\}.$$ It is easy to see $I_p(U)$ is an ideal of the algebra $C^\infty(U)$. Hence, one might consider the quotient $\mathbb R$-algebra $$C^\infty_p(U):=\frac{C^\infty(U)}{I_p(U)}.$$

The elements of $C^\infty_p(U)$ are called germs of differentiable functions on $U$ at $p$.

In $C^\infty_p(U)$ we're identifying those elements in $C^\infty(U)$ which coincide in an open neighbourhood of $p$ in $U$.

Furtheremore, we have an isomorphism $F:C^\infty_p(M)\longrightarrow C^\infty_p(U)$ given by $$f+I_p(M)\longmapsto f|_U+I_p(U).$$

Of course, one might do the same construction for the case of analytic functions.

However, I read one can't obtain an analogous isomorphism in that case. Can anyone explain me the reason for that?

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The surjectivity of the map $f+I_p(M)\longmapsto f|_U+I_p(U)$ expresses the fact that for every $f\in C^\infty(U)$ there exists $g\in C^\infty(M)$ such that $f=g$ in a neighborhood of $p$. (Such $g$ is easy to create by multiplying $f$ by a bump function that is identically $1$ near $p$ and has support contained in $U$.)

The analogous assertion is false for analytic function, because of their fatalism. The restriction of $f$ to a neighborhood of $p$ determines the function $f$, including its fate outside of $U$. If it is destined to blow up somewhere, this will happen.

In concrete terms: let $M=\mathbb{R}$, $U=(-1,1)$, and $f(x)=1/(1-x)$. Any analytic function that agrees with $f$ in a neighborhood of $0$ must also agree with $f$ on all of $U$ (the identity principle for analytic functions). But being unbounded on $U$, it cannot be continuous on $\mathbb{R}$.