Is there a fibre bundle with fibre homeomorphic to $\mathbb R^k$ which cannot be given the structure of a vector bundle?

Solution 1:

This question is answered here :https://mathoverflow.net/questions/104869/examples-for-open-disc-bundle-which-is-not-vector-bundle

Quoting the mathoverflow question:

William Browder showed in "Open and closed disc bundles", Ann. of Math. (2) 83 (1966), 218-230 that there are open disc bundles over some complex which cannot be isomorphic to a vector bundle.

Theorem 1 of the above paper gives the desired counterexample.

The unique loop (quasigroup with unit) $L$ of order $5$ satisfying $x^2 = 1$ for all $x \in L$

What is the reason behind the Pythagorean relation in a hyperbola?

Is the sequence defined by the recurrence $ a _ { n + 2 } = \frac 1 { a _ { n + 1 } } + \frac 1 { a _ n } $ convergent? [duplicate]

Invertiblity of the Derivative Matrix ?! (to use Inverse function Theorem)

Prob. 8, Sec. 3.5 in Erwin Kreyszig's Introductory Functoinal Anlaysis With Applications

Proof of uniqueness of conditional expectation

Hölder continuous functions are of 1st category in $C[0,1]$

Dominated convergence theorem for complex-valued functions?

Let $x$ be an irrational number. Prove that there exist infinitely many rational numbers $\dfrac pq$ that satisfy the following

Is there a orientable surface that is topologically isomorphic to a nonorientable one?

Notation for the ith row and column of a matrix

Can all natural numbers be represented as palindromes in some base $b$?

Is there a fibre bundle with fibre homeomorphic to $\mathbb R^k$ which cannot be given the structure of a vector bundle?

Solution 1:

Related

Recent Posts