# Is this structure a Banach bundle?

Let $X$ be a Banach space. Put $Y=\{ \phi\in X^{*}\mid\;\; \parallel \phi \parallel\leq 1\;\; \&\;\; \phi \neq 0\}$ which is a locally compact Hausdorf space with the weak star topology.

Is there a well known Banach bundle structure over $Y$ with the disjoint union of $\ker \phi,\;\phi \in Y$? In particular are all $\ker \phi,\; \phi \in Y$ mutually isomorphic Banach spaces?

• Given a normed vector space, all linear subspaces of the same finite co-dimension are mutually isomorphic as normed vector spaces. In your case, $\ker(\phi)$ has co-dimension $1$ in $X$ for each $\phi \in Y$ and is closed in $X$, hence complete. – Transcendental Feb 23 '15 at 18:19

Yes: Denote by $E$ this family of subspaces. Choose $x\in X$ (with $\|x\|=1$) and $\phi_x\in Y$ with $\phi_x(x)=1$ and consider the weak star open set $U_x=\{\phi\in Y: \phi(x)\ne 0\}$. Then $$E|_{U_x} \ni (y,\phi) \mapsto (y -\phi_x(y).x,\phi)\in \ker(\phi_x)\times U_x$$ is a trivializing vector bundle chart. The chart changes look like $$\ker(\phi_x)\times (U_x\cap U_y)\ni (z,\phi)\mapsto \Big(z-\frac{\phi(z)}{\phi(x)}x,\phi\Big)\mapsto \Big(z-\frac{\phi(z)}{\phi(x)}x - \big(\phi_y(z) - \frac{\phi(z)\phi_y(x)}{\phi(x)}\big)y,\phi\Big)$$ $$\in \ker(\phi_y)\times (U_x\cap U_y).$$ So this is even vector bundle with rational transition functions.

# Edit:

Answers to the questions of the OP.

It is a sub vector bundle of the trivial bundle $X\times Y$ which fixes the topology,

The topology can also be induced by the vector bundle charts. Choosing an abstract linear isomorphism from $\ker(\phi_x)$ to to a fixed Banach space (take $\ker(\phi_{x_0})$ for some fixed $x_0$) for every $x$, you may stabilize the fiber and give it a name.

• Prof. Michor, Thank you very much for your answer. To have a Banach bundle, do not we need to fix(at first) a typical fibre? (I think your fibre depends on x). – Ali Taghavi Feb 23 '15 at 19:16
• what is the structure and topology of E? – Ali Taghavi Feb 23 '15 at 19:17
• before trivialization, do not we need to establish a family of isomorphisms $\alpha_{xy}:\ker \phi_{x}\to \ker \phi_{y}$ with cocycle condition? I think your fibers still depends on x. Am I mistaken? – Ali Taghavi Feb 23 '15 at 19:37
• However a continuous choice of $\phi_{x}$'s is possible but I do not know how we construct the above cocycles (to have a fix fiber)? – Ali Taghavi Feb 23 '15 at 19:43
• $x$ is just an index, a name for the chart. It is NOT a variable in any chart change, it is a constant! – Peter Michor Feb 23 '15 at 19:43