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In topology and analysis we define the support of a continuous real function $f:X\rightarrow \mathbb R$ to be $ \left\{ x\in X:f(x)\neq 0\right\}$. This is the complement of the fiber $f^{-1} \left\{0 \right\}$. So it looks like the support is always an open set. Why then do we take its closure?

In algebraic geometry, if we look at elements of a ring as regular functions, then it's tempting to define their support the same way, which yields $\operatorname{supp}f= \left\{\mathfrak p\in \operatorname{Spec}R:f\notin \mathfrak p \right\}$. But these are exactly the basic open sets of the Zariski topology. I'm just trying to understand whether this is not a healthy way to see things because I've been told "supports should be closed".

  • In differential geometry, there is the notion of (smooth) functions (and more generally differential forms) with compact support. It's an important subset of all (smooth) functions, and since we're talking about manifolds here, the support cannot be compact if it's not closed. That's one reason for supports being closed, at least. – Arthur Dec 18 '15 at 11:23
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    We could also defined the support using germs : say a function is zero around $x$ if it is zero in a neighborhood of $x$. Then the support of a function $f$ is the set of $x$ where $f$ is not zero around $x$. Using this definition, it is automatically closed. – Roland Dec 18 '15 at 11:37
  • I think it is a matter of technical convenience. For instance, when constructing partitions of unity, one often wants supports to be closed, not open. – Zhen Lin Dec 18 '15 at 11:40
  • @Roland so is it kind of unhealthy to think of supports as a basis for the Zariski topology? –  Dec 18 '15 at 11:41
  • @Exterior I think that support should be closed because this is the usual definition, because the support of a section of a sheaf (the definition I gave work for every sheaf) is closed, because the support of a distribution, measure, $L^p$ function... are all closed (and there is no other definition). – Roland Dec 18 '15 at 12:00
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    Your first section is confusing: in topology and analysis the support of a continuous real function is defined as the closure of $ \left{ x\in X:f(x)\neq 0\right}$. – Georges Elencwajg Dec 18 '15 at 13:21
  • I'm trying to understand why we take the closure, i.e why it's defined this way. –  Dec 18 '15 at 14:41

1 Answers1

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Given a scheme $X$ there are two notions of support $f\in \mathcal O(X)$:
1) The first definition is the set of of points $$\operatorname {supp }(f)= \left\{ x\in X:f_x\neq 0_x\in \mathcal O_{X,x}\right\}$$ where the germ of $f$ at $x$ is not zero.
This support is automatically closed: no need to take a closure.
2) The second definition is the good old zero set of $f$ defined by $$V(f)=\{ x\in X:f[x]=\operatorname {class}(f_x)\neq 0\in \kappa (x)=\mathcal O_{X,x}/ \mathfrak m_x\}$$ It is also automatically closed.
3) The relation between these closed subsets is$$ V(f)\subset \operatorname {supp }(f)$$with strict inclusion in general:
For a simple example, take $X=\mathbb A^1_\mathbb C=\operatorname {Spec}\mathbb C[T],\: f=T-17$ .
Then for $a\in \mathbb C$ and $x_a=(T-a)$ we have $f[a]=a-17\in \kappa(x_a)=\mathbb C$ and for the generic point $\eta=(0)$ we have $f[\eta]=T-a\in \kappa(\eta)=\operatorname {Frac}(\frac {\mathbb C[T]}{(0)})=\mathbb C(T)$.
Thus $f[x_{17}]=0$ and $f[P]\neq 0$ for all other $P\in \mathbb A^1_\mathbb C$ , so that $$V(f)=\{x_{17}\}\subsetneq \operatorname {supp }(f)=\mathbb A^1_\mathbb C$$

Georges Elencwajg
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