5
$\begingroup$

My question is about the $C^j$ topology used by Greene and Krantz in their paper "Deformations of Complex Structures, Estimates for the $\bar{\partial}$-equation, and stability of the Bergman kernel". As it is not clear to me that this topology is the same as the usual Whitney strong (or weak) $C^j$ topology and it is not explicitly defined for functions on manifolds.

  1. First they describe the $C^j$ topology for maps $f:U \to \mathbb{C}$ for any open set $U \subset \mathbb{C}^n$. This is done in a quite standard way, for instance:

$$ ||f||_{C^j(U)}:= \sum_{|\alpha|+|\beta| \leq j} \left|\left|\left(\frac{\partial}{\partial z}\right)^{\alpha}\left(\frac{\partial}{\partial \bar{z}}\right)^{\beta}f\right|\right|_{\infty} $$

Where $\alpha$ and $\beta$ are taken as multidices and $||\cdot||_{\infty}$ denotes the supremum norm.

  1. Inmediately after, they define another $C^\infty$-norm. And say that it extends to a smooth manifold "via a fixed coordinate atlas". How is this extension performed? I guess you have to take a locally finite coordinate atlas and sum over all charts the previously defined norm (?). Moreover, they make a remark saying that two functions defined on $U$ are $C^\infty$ close if they are $C^k$ close for $k$ big enough and they say that this remark extends trivially to the manifold case. So implicitly they are considering a $C^j$ norm on the space of $C^\infty$ functions defined on a manifold. What is this norm?

  2. A very similar problem arises later in page 35 when they define a topology in the space of almost-complex structures of a smooth manifold. And they claim that there are neighborhoods of the form $$S_j(\prod_{1,0},\epsilon):=\{\prod_{1,0}': \text{where }\prod_{1,0} - \prod_{1,0}' \text{ is less than } \epsilon \text{ with respect to a } C^j \text{norm}\}.$$ So again, it seems that they are considering a norm on the space of $(1,1)$ tensors on manifolds (rather than open sets) and they are taking the topology induced by this norm.

My question is, what is the precise definition of this norm that induces the $C^j$ topology for complex valued smooth functions on manifolds and how does it relate to usual Whitney topologies? (references appreciated). It looks like this topology can't be the same as the Whitney topology (otherwise the Whitney topology would be usually defined using this norm rather than the more intricate usual definition). But of course this is just a moral argument.

$\endgroup$

1 Answer 1

1
$\begingroup$

It is pretty much as you guessed in 1.) Instead of summing, you take the maximum. Although maybe in some settings both describe equivalent norms (I am no expert). The key point here is that the target space is $\mathbb{C}$ and not a general manifold. In general the Whitney topology does not come from a norm, although it does come from a metric when $M$ is compact (The distance Greene and Krantz talk about does generalize to manifolds).

In general one can define a norm on the space of sections of a vector bundle. In this particular case you are dealing with sections of the trivial vector bundle $M \times \mathbb{C}$ where $M$ is a compact manifold. For the general case of sections of any vector bundle, look at Section 3 of "The Banach manifold $C^k(M,N)$ by Johannes Wittmann.

Fix a finite coordinate atlas $\{(U_i, \phi_i)\}_{1, \ldots, \ell}$ such that $\bar{U}_i$ is compact and is still contained in a coordinate atlas. Then define:

$$||f||_{C^j(M)}:=\max_{1\leq i \leq \ell} ||f||_{C^j(U_i)}$$

The $||\cdot||_{C^j(U)}$ norm defined in Wittmann's work is slightly different but I think they are equivalent. (Witmann does not sum the norms for all partial derivatives but rather takes the maximum of them).

$\endgroup$
1
  • $\begingroup$ Thank you. I guess the only piece missing is that the norm resulting from summing the sup of all partial derivatives up to order $k$ is equivalent to the norm taking the maximum of those supremums. Do you know a reference for that? $\endgroup$
    – Pita
    Commented Aug 13, 2020 at 12:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .