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2009 October 6th - questions/answers

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1 Sune

Is it possible to explicitly define (without AC) a total ordering on a set with cardinality greater than \mathbb{R}?

FIRST ANSWER by Eric Wofsey.

Yes. Here's a kind of cheap example; there's probably a less ad hoc one you could find. Let X be any well-ordered set which does not inject into \mathbb{R} (such a set can be proved to exist without AC). Then \mathbb{R} \sqcup X with the ordering that all of X is greater than all of \mathbb{R} gives an example.

SECOND ANSWER by Todd Trimble.

The first answer might be question-begging, since the follow-up might be, "How do you construct such a well-ordered set X?" But this isn't hard: let X be the set whose elements are isomorphism classes of well-ordered subsets, i.e., subsets of \mathbb{R} which have been equipped with a well-ordering. Any two such classes are comparable (one is isomorphic to an initial segment of the other), and this doesn't require AC: just match up their first elements, second elements, etc. and see which one runs out first. In fact, X is well-ordered by this comparability relation. But \mathbb{R} can't be in bijection with X; if it were, we could transfer the well-ordering of X to a well-ordering of \mathbb{R} along the bijection, and this latter well-ordering gives an element of X, i.e., X is isomorphic to one of its initial segments, which is a contradiction. Notice this is just a variant of the argument behind the Burali-Forti paradox.

THIRD ANSWER by George Lowther.

Thanks Todd, I was just wondering that, but after a quick search discovered the Hartogs number of X, which you are describing.

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2 Yuhao

Let K be the function field of an algebraic curve C. Then by the birational nature of the genus, it determines the genus g of the curve. How can we see g directly from the field?

(One way is to PICK a function in K and we get a ramified morphism C
\rightarrow P^1, then use Riemann-Hurwitz to compute the genus; However, I think there should be a more "intrinsic" way to see that, i.e. without picking a function in an arbitrary manner. The genus is a well-defined invariant for any extension of the ground field of tr.deg. 1, right?)


FIRST ANSWER by David Speyer

Here is one answer, although you might not find it very satisfying: let k be the ground field. The Kahler differentials \Omega_{K/k} are the K vector space generated by formal symbols dg subject to d(f+g)=df+dg, d(fg) = f dg + g df and da=0 for a \in k. This is a one dimensional K vector space.

Let \omega be a differential. For any valuation v on K, let t be such that v(t)=1. We say that \omega is regular at v if v(\omega/dt) \geq 0. We say that \omega is regular if it is regular at every valuation. Then the space of regular differentials is a k vector space of dimension g.

FURTHER THOUGHTS by David Speyer

Both Ben's answer and mine used the set of valuations of K/k. This essentially means that we used the ground field k. A valuation of K/k is defined as a valuation of K which is trivial on k; conversely, the ground field can be recovered from the valuations that respect it by the formula k = \bigcap_{v} v^{-1}(\mathbb{R}_{\geq 0}).

Here is a cautionary example to show that there can not be any solution which only uses properties of the field K, without reference to k. Let C and D be two irreducible curve of different genuses. Let K be the field of meromorphic function on C \times D, let k and \ell be the fields of functions on C and D. Then K is a transcendence degree 1 extension of both k and \ell, but has different genuses when considered in these two ways.


Remark by Ian Agol

Danny Calegari has a blog post about this for an affine curve. I suppose to get this from the function field, one would have to choose two generic elements of the field, and determine a minimal polynomial relation between them, then compute the number of points in the interior of the Newton polygon of this polynomial.


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3 Anon

Determine explicitly a partition of the plane into two sets A and B such that neither A nor B contains (the image of) a non-constant continuous curve.


FIRST ANSWER by George Lowther

Let S be a subset of the reals such that S\cap[a,b] and S^c\cap[a,b] cannot be written as a countable union of closed sets for any a<b. This can be done (eg, explicit examples of non-Borel sets achieve this). Let Q be the rationals. Then, A=(S\times Q)\cup(S^c\times Q^c) and B=(S\times Q^c)\cup (S^c\times Q) should do it. Btw, I also posted this answer, with a proof, on mathoverflow.net

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4 An H

Let K denote the cyclotomic field obtained by adjoining pth roots of unity to Q, where p is prime. Let q be a prime element in the ring of integers of K which is coprime to p. Let L denote the Kummer extension of K by adjoining a pth root of q to K.

Question: is it always true that p divides the exponent of the prime ideal (1-\zeta_p) in the relative discriminant D_{L/K}? (as usual, \zeta_p denotes a primitive pth root of unity)

FIRST ANSWER by An Huang.

No, not always. Counterexample: p=3, q=5, then the exponent equals 4. (One can calculate the relative discriminant by using local and global conductors, and solving some simple congruence equations explicitly.)

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5 Critch

In the categories Set and AbGrp (and I believe any topos), finte limits (e.g. kernels, products, equalizers...) commute with directed colimits (aka direct limits). Give an example category where this fails.

FIRST ANSWER by Peter LeFanu Lumsdaine.

A straightforward, if slightly unsatisfying, answer: Set^{op}.

For an example of how finite colimits fail to commute with filtered (i.e. co-directed) limits in Sets: let A_n = \{n,n+1,\ldots\}, and consider the two constant maps c^n_0,c^n_1 : A_n \to 2. These pairs stack into a nice ladder, with 2 sitting constant on the right, and the obvious injections A_{n+1} \to A_n marching down on the left.

Now the (filtered) limit over n of the coequalisers of c^n_0,c^n_1 is 1, since each coequaliser is 1; but the coequaliser of the limits is 2, since the limits give the diagram 0 \stackrel{\to}{\to} 2.

There are more satisfying answers too—i.e. nice concrete "right-way-round" categories where it fails—but they elude me now, I'm afraid (it's late).

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6 Anton

Can a coequalizer in the category of Schemes fail to be surjective? (Note: it must hit all closed points in the target, because otherwise the closed point could be removed to make the coequalizer smaller.)


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7 Darsh

Let f: \mathbb{C}\cup\infty \to \mathbb{C}\cup\infty be rational.

In what generalized notions of convergence does the sequence x,f(x),f^2(x),\ldots converge?


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8 Harold

In the category of smooth manifolds, when does the fibre product of two maps exist?

A) Necessary conditions

B) Sufficient conditions (e.g. that each map is a fibration).


FIRST ANSWER by Todd Trimble.

Let f: E \to M and f': E' \to M be two smooth maps. A sufficient condition for their pullback to exist in the category of manifolds is that they be transverse to each other, meaning that whenever f(e) = x = f'(e'), the images of the tangent spaces at e, e' under the induced maps Df: TE \to TM, Df': TE' \to TM together span the tangent space at x. (The condition is equivalent to f \times f': E \times E' \to M \times M meeting the diagonal submanifold \Delta: M \to M \times M transversally, so that the inverse image of the diagonal becomes a submanifold of E \times E'.)

FIRST ANSWER by Dmitri Pavlov.

If I remember everything correctly, the following two statements are true: A sufficient condition is that at least one of the maps is a submersion. If for all smooth maps f: X → Y the fiber product of f and g: Z → Y exists, then g must be a submersion.


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9 James T

Which bounded linear maps on a Hilbert space are the exponentials of other maps? I.e., what is the image of the map exp: B(H)\to B(H)?


FIRST ANSWER by Dmitri Pavlov.

The image of exp contains an open neighborhood of 1. Hence it contains the connected component of 1 in the group of invertible elements of B(H), which is an open subset of B(H). The image of a connected space is connected, hence the image of exp coincides with the connected component of 1.

Correction: Of course, this argument only proves that the set of all products of elements in the image of exp coincides with the connected component of 1, which is the set of all invertible elements.

SECOND ANSWER by Todd Trimble:

Dmitri, I don't quite follow. Why does the second sentence follow from the first? (Warning: we don't always have exp(X) \cdot exp(Y) = exp(X + Y) since two exponentials might not commute.)

THIRD ANSWER by George Lowther

Actually, by the polar decomposition, every invertible operator can be written as exp(X)exp(Y), so the invertible elements is already connected and Dmitri's answer says every invertible operator is an exponential (which seems wrong). It is the case that every finite dimensional invertible matrix has a logarithm (consider Jordan normal form), but seems unlikely in the infinite dimensional case.

FOURTH ANSWER by Anonymous

As far as I know this is open and likely to have no nice answer. This is a special case of the known-to-be-essentially-impossible general problem of characterizing the range of the exponential map on a Banach algebra. What is easy to see is that anything close enough to the identity must be in the range of the exponential map, and using Dmitri's argument it follows that the set of finite products of things in the range of the exponential map is the connected component of the identity in the group of invertible elements. This much is true for a general Banach algebra. When the algebra is commutative, of course, this means that anything invertible is in the range of the exponential map.

[passing edit from Yemon Choi: the last statement is not quite correct, it should be "anything in the connected component of the identity is in the range of the exponential map". Consider C(T), the algebra of continuous functions on the circle with pointwise multiplication, and look at the function sending exp(i\theta) to exp(i\theta). This is invertible, but has winding number 1 and so can't be in the same component of the invertible group as the identity of the algebra, which has winding number 0. More generally one has the Arens-Royden theorem for commutative Banach algebras, saying that the invertibles modulo their connected component form a group isomorphic to 1st Cech cohomology of the maximal ideal space (integral coefficients, I think).]

To continue George's remark, one way to see that there are invertible bounded operators on Hilbert space with no logarithms is to note that there are invertible bounded operators on Hilbert space with no square roots (there is a paper by Halmos and maybe some more people from the 1950s, something like "operators with square roots," containing examples).

There is a paper by Conway and Morrel from the 80s, "Roots and logarithms of bounded operators" which identifies the closure and interior of the set of logarithm-able operators. This is probably the best that functional analysis can do. Not that that means the problem has been worked over; I just don't think functional analysts are inclined to ask or work on questions like this. (Or maybe the ones that are, are more interested in the invariant subspace problem.)

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10 Pablo

Give a simple example of a (necessarily infinte dimensional) Lie algebra that is not the lie algebra of any (necessarily infinite dimensional) lie group.

FIRST ANSWER by Andrew Stacey

(Source: "Loop Groups" by Pressley and Segal)

The polynomial loop Lie algebra, Lpol𝔤, is "associated" to the polynomial loop group, LpolG, but there is no exponential map between them.

SECOND ANSWER by Allen Knutson

Of course it depends on what "is the Lie algebra of" means. If you feel that an exponential map is required, then the example above works. I personally start from the Lie group and induce a Lie algebra on the tangent space to the identity, in which case the example above is not interesting.

Again from Pressley and Segal: let diff(S^1) be the Lie algebra of vector fields on the circle. This is the Lie algebra of a real group, Diff(S^1), the diffeomorphisms of the circle. But its complexification diff(S^1) tensor C is not the Lie algebra of any group. (It wants to be the Lie algebra of the Neretin semigroup of contractions of the unit disc, but that's as close as you can get.)

THIRD ANSWER by Alan Weinstein

Let G be the unitary group of an infinite dimensional Hilbert space. Its center is a circle. In the product G x G, the center is a torus. Let R be a dense 1-dimensional subgroup of this torus. It is normal in G x G, so its Lie algebra r is an ideal in g x g. Now the quotient g x g/r is a Lie algebra which is not the Lie algebra of any Lie group. The proof uses the fact that an infinite-dimensional unitary group is contractible (Kuiper's theorem.) Otherwise, you could get a group by passing to universal coverings, which works in the finite dimensional case.

I believe that this example is due to Douady.

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11 Theo

See mathoverflow.net.


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