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# 2009 September 29th - questions/answers

Navigation: 20 Questions home | view only questions | last week | next week

#### 1 Harold W

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

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

In other words, if , 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 is a smooth fiber bundle with fibers diffeomorphic to .
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 be any global section of , and let denote the "vertical tangent bundle of along ", the *vector* bundle of those tangent vectors at points which project to in via pushforward along the map (so they are "tangent to the fibres" of ). This definition only involves the smooth structure of , so and are identified via the isomorphism of *fiber* bundles.

On the other hand, each fiber of 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* (now invoking the linear structure of ), and similarly , 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* (via a fiber-preserving homotopy), so it can be seen as a "linearization" of that map.

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#### 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 over all primes . If there was a multiplicative identity , it would have some finite additive order , but there would exist elements of additive order greater than , a contradiction (since plus itself times, multiplied by anything, must be ).

**SECOND ANSWER** by Dustin C

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

(denoting plus itself times) equals .

Now add to itself times:

(*p* times)

- ,

and the only rational with this property is , which is not in , a contradiction.

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#### 3 Jason F

**A)** Does ZF (without the axiom of choice) imply that a well-ordering of exists? Why/why not?

**B)** If we assume a well ordering exists on what other sets can be proven well-orderable (without using choice)?

**FIRST ANSWER** by Amit

**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.

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#### 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 is the result of removing an 3-ball from each of and 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.)

**FIRST ANSWER** by Greg Kuperberg

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.)

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#### 6 Zack

This question is about "approximating" an open set by an open set .

Say is a metric space, connected opens such that (the closed symmetric difference of and is contained in a radius-r neighborhood of the boundary of ... basically, U and V only differ near the boundary of U).

Does there exist a continuous surjection which is "r-close to the identity", meaning that . ?

**FIRST ANSWER** by Eric Wofsey

No: X could be , U could be , 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.

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#### 7 Pablo

The cardinality of {A, B, C,...., Z} can be written in base 10 as 26, in base 2 as 11010. Expressing natural numbers N,M in a certain base has the advantage that there are simple rules for how to express the numbers NM and N+M in the same base. Is there a way to express number using unique prime factorization that still allows you to do arithmetic? Or more generally is there any naming scheme for the natural numbers that doesn't use any sort of base that still allows you to do arithmetic?

For example, one might try something like saying 1 = 1, 2 = (1), 2*3 = (1)(1), 2*5 = (1)()(1), etc. But its not clear how to do arithmetic with this notation.

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