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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?

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    $\begingroup$ 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. $\endgroup$ – Transcendental Feb 23 '15 at 18:19
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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.

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  • $\begingroup$ 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). $\endgroup$ – Ali Taghavi Feb 23 '15 at 19:16
  • $\begingroup$ what is the structure and topology of E? $\endgroup$ – Ali Taghavi Feb 23 '15 at 19:17
  • $\begingroup$ 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? $\endgroup$ – Ali Taghavi Feb 23 '15 at 19:37
  • $\begingroup$ 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)? $\endgroup$ – Ali Taghavi Feb 23 '15 at 19:43
  • $\begingroup$ $x$ is just an index, a name for the chart. It is NOT a variable in any chart change, it is a constant! $\endgroup$ – Peter Michor Feb 23 '15 at 19:43

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