# Banach space geometry without bounded operators?

I understand that $B(X)$ can be think of as the collection of symmetries of a Banach space $X$, and that they provide important information concerning the geometric structure of the space. But I am just curious, how far can one go without $B(X)$?

Or in other words, what kinds of properties are so intrinsic to a space that can even be realised without thinking about the operators? I guess the notion of symmetry should be implicit behind all kinds of geometry, but how implicit can they be?

This question is rather of backwards nature but still might be interesting. Thanks!

-
Completeness, compactness, separability, closedness and complementability of subspaces. What is your definition of symmetries? –  Norbert Oct 20 '12 at 15:08
@Norbert Yes. All these are intrisic properties of the underlying space. By symmetries of a space I refer to the endomorphisms. –  Hui Yu Oct 20 '12 at 15:34

Geometric properties can often be described as properties of $B(X)$ (or subspaces of it) or as properties on $X$. (Of course we only call them 'geometric' if we can describe them in $X$). Sometimes geometric properties were first observed as properties of operators. Let me give an example:
In 1964, if I'm not mistaken, Daugavet observed that on $C([0,1])$, every compact $T\in K(C[0,1])$ (denoting by $K(X)$ the space of compact operators on $X$) has maximal distance to the identity, that is we have $\|\mathrm{Id} + T\| = 1 + \| T\|, \quad T \in K(C[0,1])$ This property seems first to be a property of operators, but intrestingly, this property, which was afterwards called the Daugavet property, can be described geometrically. A space $X$ is said to be (a) Daugavet (space), if it shares this property. Now for the geometric description: The part of the unit ball $B_X$ which lies on one side of a hyperplane is called a slice, that is a slice a set (for $x^* \in S_{X^*}$ and $\epsilon > 0$) $S(x^*,\epsilon) = \{x \in B_X \mid x^*(x) \ge 1 -\epsilon \}$ A space is now exactly then Daugavet iff these slices have 'big' diameter, that is given a slice $S(x^*, \epsilon)$ an $y \in S_X$, and a $\delta > 0$, there is an $x \in S(x^*, \epsilon)$ with $\|x-y\| \ge 2 - \delta$.
When you are talking about $S_{X^*}$ you are already using functionals (operators into $\mathbb{C}$) –  Norbert Oct 20 '12 at 15:38
It seems very interesting. But can yo give some more details about the $S_X$? How is it defined? Thanks! –  Hui Yu Oct 20 '12 at 15:39
@Norbert You are right, but I think we can get rid of the functionals by defining a Hyperplane to be a 1-codimensional closed subspace $H$. Given such an $H$, which intersects $B_X$ and a $x \in X\setminus H$, we can define $S(H, x)$ as the $\{y \in B_X \mid \operatorname{conv}\{y,x\} \cap H = \emptyset\}$. So we can descibe a slice without using functionals. –  martini Oct 20 '12 at 15:46
@HuiYu $S_X$ denotes the unit sphere $\{x \in X \mid \|x\| = 1\}$. –  martini Oct 20 '12 at 15:47