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1 Harold W

Two questions about the forgetful functor from smooth (C-infinity) vector bundles to smooth fiber bundles:

A) Suppose that $E\to B$ is a smooth vector bundle (E,B smooth manifolds). Does $E\to B$ as a smooth fiber bundle uniquely determine $E\to B$ as a vector bundle, up to isomorphism?

In other words, if $E_1\to B_1$, $E_2\to B_2$ are vector bundles which are isomorphic as smooth fiber bundles (via some fiber-preserving diffeomorphism, not necessarily linear on fibers), are they necessarily isomorphic as smooth vector bundles (via some fiber-preserving diffeomorphism, linear on fibers)?

B) Suppose that $E\to B$ is a smooth fiber bundle with fibers diffeomorphic to $\mathbb{R}^n$. Can it be given the linear structure of a vector bundle?

FIRST ANSWER for part A by Alvaro P

Yes. Consider the second interpretation. Let $s$ be any global section of $E_i$, and let $T_s(E_i)$ denote the "vertical tangent bundle of $E_i$ along $s$", the vector bundle of those tangent vectors at points $s(b)$ which project to $0$ in $T_b$ via pushforward along the map $E_i\to B_i$ (so they are "tangent to the fibres" of $E_i$). This definition only involves the smooth structure of $E_i\to B_i$, so $T_s(E_1)$ and $T_s(E_2)$ are identified via the isomorphism $E_1\to E_2$ of fiber bundles.

On the other hand, each fiber of $E_1$ as a vector space is canonically identified with its tangent space at any point, and these glue together (this can be shown via the implicit function theorem) to give a canonical isomorphism of vector bundles $T_s(E_1)\to E_1$ (now invoking the linear structure of $E_1$), and similarly $T_s(E_2)\to E_2$, completing the isomorphism.

SECOND ANSWER for part A by Critch:

It may be worth noting that the vector bundle isomorphism in the above solution is (smoothly) homotopic to the given isomorphism of fiber bundles $E_1\to E_2$ (via a fiber-preserving homotopy), so it can be seen as a "linearization" of that map.

2 Critch

Every finitey generated abelian group can be made into a ring, by the structure theorem. What is an abelian group that cannot be made into a ring?

FIRST ANSWER by Critch & Zack

It would still be nice to see a torsion-free example, but here's a torsion one:

Take the infinite direct sum of $\mathbb{Z}/p$ over all primes $p$. If there was a multiplicative identity $e$, it would have some finite additive order $n$, but there would exist elements of additive order greater than $n$, a contradiction (since $e$ plus itself $n$ times, multiplied by anything, must be $0$).

A torsion free example: Let $R$ be the additive group of those rational numbers with squarefree denominator. Suppose it had a ring structure with multiplication $*$ and identity $e$. Let $p$ be a prime which is not a factor of the numerator or denominator of $e$. Then the rational $f=e/p$ is in $R$, and

$p.f$ (denoting $f$ plus itself $p$ times) equals $e$.

Now add $f*f$ to itself $p^2$ times:

$p^2.(f*f) = f*f+\ldots+f*f = f*(p.f)+\ldots+f*(p.f)$

(p times)

$= p.(f*e) = p.f = e$,

and the only rational with this property is $e/p^2$, which is not in $R$, a contradiction.

3 Jason F

A) Does ZF (without the axiom of choice) imply that a well-ordering of $\mathbb{R}$ exists? Why/why not?

B) If we assume a well ordering exists on $\mathbb{R}$ what other sets can be proven well-orderable (without using choice)?

A) No, cf. Thm 14.36 in Jech's "Set Theory" for a proof. The result is due to Cohen, and uses a forcing argument.

B) My guess, probably just the obvious ones.

4 Darsh

The set S of (isomorphism classes of) compact connected 3-manifolds, with the operation of connected sum, is a monoid. The connected sum $X \# Y$ is the result of removing an 3-ball from each of $X$ and $Y$ and gluing them along the boundary of this ball (this operation is well-defined up to homeomorphism).

What is a minimal set of generators for this monoid? (In the 2D case, the torus and the projective plane suffice.)

In the orientable case, Milnor's theorem says that compact, connected 3-manifolds have unique factorization with respect to connected sum. So there is a unique minimal set of generators, the irreducible 3-manifolds.

In the non-orientable case, unique factorization essentially still holds, but not literally. You simply have to modify uniqueness to take into account flipping over orientable summands. But then, you also have to modify the original question, because there are two ways to glue together two chiral 3-manifolds.

5 Scott M

Is there a good version of Artin-Wedderburn for semisimple algebra objects? (click for elaboration.)

6 Zack

This question is about "approximating" an open set $U$ by an open set $V$.

Say $X$ is a metric space, $U,V\subset X$ connected opens such that $\overline{U\Delta V} \subset B(r,\partial U)$ (the closed symmetric difference of $U$ and $V$ is contained in a radius-r neighborhood of the boundary of $U$ ... basically, U and V only differ near the boundary of U).

Does there exist a continuous surjection $f: U\to V$ which is "r-close to the identity", meaning that $\sup(d(f,id)) < r$. ?

No: X could be $S^n\vee I$, U could be $S^n\vee[0,r)$, and V could be the sphere without the basepoint. Considering the basepoint to be the north pole, at the equator a map near the identity must preserve the degree of the equator about the north pole (the winding number, if n=2). But such a map cannot be extended to the northern hemisphere without using one of the two poles, and the north pole can't be used since it's not in V and the south pole can't be used since it must be close to the identity.