User daniel asimov - MathOverflow

Answer by Daniel Asimov for Examples of common false beliefs in mathematics.

2010-05-09T18:40:39Z

Complex variables: "An entire function that is onto and locally one-to-one is globally one-to-one."

Counterexample: $f(z) := \int_0^z \exp(\zeta^2)\,d\zeta$

I'll leave the proof that this is indeed a counterexample as a pleasant exercise.

(I believe this example is due to Lawrence Zalcman.)

Infinite-dimensional complex polynomial or rational Lie algebras and their pseudogroups

2010-08-25T22:11:44Z

In studying the transformation groups generated by holomorphic vector fields V(z) d/dz on ℂ, I've noticed the (surely well-known) fact that the complex quadratic vector fields:

{(a z² + b z + c) d/dz | (a,b,c) ∊ ℂ³}

form precisely the Lie algebra whose nonzero elements generate the linear fractional transformations, i.e., PSL(2,ℂ).

Is there some underlying reason for this? (Beyond direct calculation, which provides an easy proof.)

Other than the further Lie algebras {0}, {a d/dz}, and {(a z + b) d/dz} over ℂ of trivial, constant, and linear (affine) vector fields, there seem to be no other polynomial finite-dimensional Lie algebras of vector fields on ℂ.

Is there some easy-to-explain reason for this?

The next "simplest" such polynomial Lie algebras of vector fields on C seem to be those defined by all polynomial and all rational functions:

(*) V_P := {P(z) d/dz | P(z) ∊ C[z]} and V_R := {R(z) d/dz | R(z) ∊ C(z)}.

Contrariwise, do there exist finite-dimensional Lie algebras of vector fields on ℂ defined by rational functions -- other than the polynomial ones mentioned above?
In case V_P and/or V_R generate well-studied (infinite-dimensional) Lie "groups" of transformations from open sets of ℂ into open sets, then what are these groups? Properly, these are pseudogroups, but perhaps they behave like Lie groups.

[Note: It's not hard to compute formulas for such transformations -- the flows -- directly from an expression for the vector field in terms of its zeroes (and poles, if any).]

In any case, are there standard names for the Lie algebras V_P and V_R ?
References to the above matters would also be appreciated.

Answer by Daniel Asimov for Widely accepted mathematical results that were later shown wrong?

2010-08-14T18:13:40Z

One part of Hilbert's 16th problem is to determine whether a polynomial vector field in ℝ²:

V(x,y) = (P(x,y),Q(x,y))

has at most a finite number of limit cycles.

In 1923, Dulac published a paper supposedly proving this.

Around 1980-81, Ecalle and Ilyashenko independently recognized that the proof had serious gaps.

In 1991-92, Ilyashenko and Ecalle independently published (quite different) proofs that a polynomial vector field in the plane does indeed have at most a finite number of limit cycles.

See Ilyashenko's paper, "A centennial history of Hilbert's 16th problem" at http://www.ams.org/journals/bull/2002-39-03/S0273-0979-02-00946-1/S0273-0979-02-00946-1.pdf.

(Many related questions remain unsolved, such as finding sharp or even good upper bounds for the maximum number of limit cycles in terms of the degrees of the polynomials P and Q.)

Topological spaces that resemble the space of irrationals

2010-07-30T22:11:20Z

(This question actually arose in some research on number theory.)

I once learned that any countable dense subspace of any Euclidean space ℝⁿ is homeomorphic to the rationals ℚ.

Now I wonder if something similar is true for the irrationals J := ℝ - ℚ (with the subspace topology from ℝ).

Let c denote the cardinality of the continuum.

I. Is each cartesian power Jⁿ homeomorphic to J ?

Also, how far can this be pushed?

II. Let X be a dense totally disconnected subspace of ℝⁿ such that every neighborhood of each point of X contains c points. Is X homeomorphic to J ?

What about for such subspaces of fairly nice subspaces of ℝⁿ ?

IIa. Let X be any subspace of ℝⁿ as described in II., and let B denote any subspace of ℝⁿ homeomorphic to [the open unit ball in ℝⁿ union any subset of its boundary]. Then is X ∩ B homeomorphic to J ?

And what about greater generality ?

III. Is there a simple set of conditions that describe exactly all spaces (or subspaces of ℝⁿ) that are homeomorphic to J ? What about Jⁿ ? (Perhaps the word homogeneous or metric needs to be included.)

(I found nothing relevant via Google, in MathSciNet, or here on MathOverflow.)

Answer by Daniel Asimov for Google question: In a country in which people only want boys

2010-07-08T15:37:44Z

A colleague, Eugene Salamin, came up with what I would consider the "Book" solution:

Phooey, this isn't at all a mathematical puzzle. A social convention cannot override biology, so the proportion of boys and girls is the biologically determined one, nominally 1/2, 1/2.

I didn't immediately understand his reasoning. But if all families are enumerated 1,2,3,... and you imagine each family's sequence of children placed in numerical order to make one infinite (or very long) sequence, then the resulting sequence of B's and G's is statistically identical to one you would get by repeatedly flipping a fair coin.

Viewed this way, the rule for stopping when the first B is reached is clearly a red herring! And clearly the proportion of boys and girls will be equal. (At least asymptotically, with probability 1, by the Strong Law of Large Numbers.)

(Likewise, if the original question is varied so that Prob(B) = p and Prob(G) = q, p+q=1, then by the same reasoning the ultimate proportions of boys and girls are p and q, respectively.)

P.S. On the other hand, this does not work for each possible stopping rule. Say we're back to the usual assumption of each birth having an equal chance of being a boy or girl. In an imaginary world, suppose each family stopped having children when the proportion of the girls in their family first exceeded 2/3. Then the ratio of girls to boys in the population will clearly be greater than 2.

Answer by Daniel Asimov for Which mathematicians have influenced you the most?

2010-07-05T20:10:59Z

(I think that for a question like this with the answers being entirely personal, the voting is of little or no significance.)

For me there are so many that I hardly know where to begin. Initially, Martin Gardner. Among those I knew personally: my undergrad profs (espcially I.M. Singer) who taught me what math is. Then Bill Thurston, with whom I shared an office in grad school. Stephen Smale, my de facto co-thesis advisor.

Notably Gauss, Riemann, Klein, Poincaré, Milnor.

Above all, my thesis advisor, Morris Hirsch, with whom I've had a continuing connection since 1970.

Answer by Daniel Asimov for Google question: In a country in which people only want boys

2010-07-05T21:11:48Z

The correct answer has nothing to do with the number of families. This is a very tricky problem, and many people fall into the trap of trying to average each possible fraction of girls, weighted only by the probability of that outcome. But in fact they would need to be weighted also by the size of the population, if that strategy is used to find the answer.

Google's reasoning is perfectly correct, but here is another route to the same result. We just find the expected number of boys and the expected number of girls for one family.

The number of boys is obviously 1 for any outcome (of the form GⁿB), and so its expectation is 1.

The expected number of girls is given by the summation of n·p(n) for n = 0 to ∞, where p(n) = the probability of the outcome GⁿB, which is 1/2ⁿ⁺¹. This sum is perhaps surprisingly also 1, which is easy to verify.

Thus each family's expected number of children is 1+1 = 2, and for N families, this just becomes N+N = 2N. And so on average, the population will have an equal number of children of each sex.

P.S. I will agree that Google's phrasing could have been more precise. But that is the case with virtually any math problem that is phrased as a problem in the real world, and I believe Google's intended problem is sufficiently clear that there is no real value in debating all its possible meanings.

Answer by Daniel Asimov for Which manifolds admit a diffeomorphism of order $n$?

2010-07-04T20:00:52Z

For a closed orientable surface M_g of genus g >= 2, there are only a finite number of possible orders of a diffeomorphism or homeomorphism h:M_g -> M_g. One constraint on such orders is that the map induced on first homology

h_*:H₁(M_2g) -> H₁(M_2g)

belongs to GL(Z^2g), and there is only a finite set of possible orders for elements of this group. (Note that if h_* is the identity on first homology with g >= 2, then h is homotopic to the identity on M_g and cannot have order > 1.)

For this reason, e.g., on the double torus M₂ there can be no homeomorphism of order greater than 12. For more information see, e.g., Finite groups of matrices whose entries are integers, James Kuzmanovich and Andrey Pavlichenkov, American Mathematical Monthly, Vol. 109, No. 2 (Feb., 2002), pp. 173-186.

(In the converse direction, I suspect any isomorphism of the cohomology ring H^*(M_g) can be realized by some self-homeomorphism of M_g, but do not have a reference for this offhand.)

Answer by Daniel Asimov for Examples of common false beliefs in mathematics.

2010-06-12T00:40:56Z

Perhaps the most prevalent false belief in math, starting with calculus class, is that the general antiderivative of f(x) = 1/x is F(x) = ln|x| + C. This can be found in innumerable calculus textbooks and is ubiquitous on the Web.

Answer by Daniel Asimov for Who is the last mathematician that understood all of mathematics.

2010-06-12T00:18:33Z

Many people have mentioned Jean Dieudonné (1906-92) in this regard.

Answer by Daniel Asimov for What are some correct results discovered with incorrect (or no) proofs?

2010-06-11T21:01:33Z

When Stephen Smale was a graduate student, he thought he had a proof of the Poincaré Conjecture as follows: Take a compact simply-connected 3-manifold M and remove the interiors of two disjoint 3-balls to get, say, M₁ having as boundary two copies of S². It is easy to show that M₁ has a nonsingular vector field entering along one S² and exiting along the other. Clearly by the simply-connectedness of M, each orbit entering on one boundary component must exit on the other one. Thus M₁ must be S² x [0,1] and hence by replacing the removed 3-balls, M must have been S³. QED.

I'm not sure who first pointed out the error, but undoubtedly understanding examples like this helped him appreciate the subtlety of the problem and ultimately prove the Generalized Poincaré Conjecture for dimensions ≥ 5.

Answer by Daniel Asimov for Analytic ODE with complex time

2010-05-30T22:38:26Z

One thing to be careful about is that even for an analytic ODE given on ℂ via

dz/dt = f(z)

where f is an entire function, the solutions Φ(z,t) always exist for all (z,t) in some open neighborhood of ℂ x {0} in ℂ x ℂ or just ℂ x ℝ (if we just consider real time) . . .

. . . but even for f(z) = a mere polynomial P(z), a paradoxical phenomenon can occur, even just considering real time: The flow Φ(z,t) can be defined, for some K > 0, on two disjoint open sets

O₀ x (-K,K) and

O₁ x (-K,K)

in ℂ x ℝ such that for some t₀ in (-K,K) we have, e.g.,

Φ(z₁,t₀) = z₁ for all z₁ ∈ O₁, although

Φ(z₀,t₀) ≠ z₀ for all z₀ ∈ O₀.

This seems to violate permanence, but for a subtle reason does not.

For a concrete example: let P(z) := i(z³ - z), and let O_j be a small open neighborhood of j in ℂ. Then the flow given by Φ(z,t) := z(t) satisfying

dz/dt = P(z)

is defined for all real time t on both O₀ and O₁.

But setting t₀ = π, we have

Φ(z,π) - z = 0 for all z in O₁, although

Φ(z,π) - z ≠ 0 for all z in O₀.

Answer by Daniel Asimov for Noninteger iterates of functions: How to get ODE from flow at a given time?

2010-05-30T20:20:58Z

If g is a real-analytic function defined near x₀ with g(x₀) = x₀ and 0 < λ ≠ 1 where λ := g'(x₀), then Koenigs proved that there exists a real-analytic homeomorphism h defined near x₀ such that hgh^-1(x) = L(x) where

L(x) := x₀ + λ(x-x₀)

and h is unique up to a constant factor*.

This allows g to be embedded in a local flow: Φ(x,t) := h^-1 L^t h(x), where

L^t(x) := x₀ + λ^t (x-x₀),

such that Φ(x,1) = g(x) where defined.

Then Φ satisfies the differential equation dx(t)/dt = V(x(t)) where the velocity V (denoted by f in the Question) is given by

V(x) := ∂Φ(x,t)/∂t |_t=0.

Note, however, that the flow Φ(x,t) into which the original function g embeds as the time-1 map need not be unique when there exist more than one fixed point of g. As an example, for x > 0 consider the function

g_c(x) := c^x.

Then for 1 < c < e^1/e the function g_c has two distinct fixed points each satisfying the hypotheses of the Koenigs theorem, and these give two distinct flows into which g_c embeds as the time-1 map.

Concretely, set c = √2, so that g_c(x) = x for both x = 2 and x = 4, with derivatives

g_c'(2) = ln(2) and

g_c'(4) = ln(4).

Calculating the respective real-analytic flows Φ₂(x,t) and Φ₄(x,t), both are defined for (x,t) = (3, 1/2).

But Φ₂(3, 1/2) and Φ₄(3, 1/2) first differ in the 25th decimal place. Hence they are solutions of distinct differential equations. I.e., they have different velocity functions.

* This is actually true in greater generality; see J. Milnor's book Dynamics in One Complex Variable, 3rd ed., Princeton University Press, 2006.

Answer by Daniel Asimov for A problem of an infinite number of balls and an urn

2010-05-25T17:57:48Z

The reason this problem has the generally accepted answer of 0 balls in the jug at midnight -- among mathematicians -- is that for any given ball, one may follow its itinerary: at some point it goes into the jug, and then at some point no earlier it leaves the jug, never to be moved again.

Thus it would be easy to formalize a generalized class of problems so that one may prove rigorously that as long as each ball n is moved only finitely many times before midnight, it is at midnight where its last motion took it to.

The infinitely-many-marbles problem was probably created by the mathematician J.E. Littlewood in the early 1950s, and was popularized in Martin Gardner's Mathematical Games column in Scientific American, where "zero balls at midnight" was given as the correct solution. (See A Mathematician's Miscellany, J.E. Littlewood, Methuen, 1953.)

Philosophers, on the other hand, are still debating this problem. (See, for example, Paradoxes from A to Z by Michael Clark, 2nd ed., Routledge, 2007.)

Answer by Daniel Asimov for Why worry about the axiom of choice?

2010-05-02T01:06:41Z

When I first encountered AC as an undergraduate student, like most math students I thought it was seriously questionable since it led to weird and counter-intuitive things like non-measurable sets and, worst of all, the Banach-Tarski Paradox.

But after I learned that AC is logically equivalent to

 **The cartesian product of any non-empty collection of non-empty sets is non-empty**

I found it impossible not to believe it. My only conclusion could be that my mathematical intuition was not well-developed.

I came to accept the consequences of AC as natural aspects of mathematics, and they no longer seem nearly as weird or counter-intuitive as they did at first.

And I would be fairly unhappy if there existed a vector space without a basis. Or if there existed two sets A, B without an injection, bijection, or surjection A -> B (i.e., a failure of Trichotomy).

On the other hand, I have begun to be philosophically troubled by the common and probably harmless assumption that mathematicians can choose between the two complex roots of

(*) z² + 1 = 0

(and similar situations). There is no basis for selecting between i and -i . . . or even naming them i and -i ! So although I intend to continue referring to i for convenience, it feels to me that technically one has no right to do so; instead a technically correct discussion should always refer to the set of roots of (*), but never just one of them alone.

Answer by Daniel Asimov for Why do dynamicists worry so much about differentiability hypotheses in smooth dynamics?

2010-03-28T22:48:29Z

There are some fascinating phenomena in dynamical systems and related fields whose existence depends on the degree of differentiability.

Among all of these, my favorite is the fact, proved by Haefliger, that although there exist C^∞ codimension-1 foliations of S³, there does not exist any C^ω codimension-1 foliation of S³.

A much more basic such result, due to Denjoy, is that a C² diffeomorphism of the circle having irrational rotation number is topologically conjugate to the rotation of the circle having that same rotation number. But there are counterexamples for diffeomorphisms that are merely C¹.

Maximally symmetric smooth projective varieties in CP^2

2010-03-28T20:12:00Z

Let P(X,Y,Z) be a homogeneous polynomial in ℂ[X,Y,Z] whose locus M in ℂℙ² is a nonsingular curve of genus ≥ 2.

Define M to be maximally symmetric if the following is not true:

There exists a continuous family { P_t | t ∊ [0,1] } of homogeneous polynomials in ℂ[X,Y,Z] such that 1), 2), and 3) hold:

1) P₀ = P.

2) The locus M_t in ℂℙ² of each P_t is nonsingular.

3) There is a group G such that the ambient isometry groups
G_t := Isom_A(M_t) are all isomorphic to G for 0 ≤ t < 1, but G₁ contains G as a proper subgroup.

Here the "ambient isometry group" Isom_A(M_t) of a projective curve M in ℂℙ² means the subgroup of Isom(ℂℙ²) = PSU(3) that carries M to itself.

Question: I'd like pointers to the literature regarding what may be known about a classification of such "maximally symmetric" projective curves up to ambient isometry, their defining polynomials, and especially their ambient isometry groups.

Comment by Daniel Asimov

2013-01-11T04:50:26Z

I took a look at the Fillmore paper, and just before his Corollary to Theorem 2 -- which reads "Corollary. There exists an analytic hypersurface of constant width in E^n having the same group of symmetries as a regular n-simplex." -- he writes "If we imitate the construction of a Reuleux triangle . . .. Thus:" This seems to imply that he is assuming that [the intersection of four balls in 3-space, centered at the vertices of a regular tetrahedron and each with radius = the side-length of the tetrahedron] is a body of constant width. But this is known to be false.

Comment by Daniel Asimov

2012-12-24T17:27:06Z

It's not so much that you also need det(x) = 1, as that this is exactly the same as saying |a|<sup>2</sup> + |b|<sup>2</sup> = 1.

Comment by Daniel Asimov

2012-09-14T02:51:13Z

In the functional equation for ζ(s), the term Γ(s) should be Γ(s/2), and the term Γ(1-s) should be Γ((1-s)/2).

Comment by Daniel Asimov

2012-08-13T20:18:16Z

The usual contractible 2-complex discovered by Bing is called a "House With Two Rooms", and this is not what is depicted in the image above this answer. The 2-complex depicted is not contractible (since a loop around either inner cylinder is a nontrivial 1-cycle, as is easily verified). To obtain the House With Two Rooms, one needs to add two disjoint rectangles IxI to the image, each one intersecting one inner cylinder in an interval, the outer cylinder in an interval, and the 2-complex depicted in its entire (rectangular) boundary circle.

Comment by Daniel Asimov

2010-08-16T06:37:50Z

Removing a standard $D^4$ from a potentially-exotic smooth $S^4$ yields a potentially-exotic $\mathbb{R}^4$ that's standard at infinity. So if it were known that the latter must be globally standard, then replacing the $D^4$ would imply the original $S^4$ is standard, and hence the 4-dimensional smooth Poincaré conjecture. And there's essentially only one way to replace the $D^4$, since Gamma_4 = 0 (Cerf) implies oriented diffeos of $S^3$ are smoothly isotopic.

Comment by Daniel Asimov

2010-08-15T15:45:40Z

For one application, an explicit curve {C(t) : t in [0,oo)}, dense in a Grassmannian of 2-planes in n-space, is the basis for the animation technique in statistical computer graphics known as the Grand Tour. It's important to ensure that as t -> oo, the curve C spends time in any open set U proportional to the invariant measure* of U. * Though the invariant measure on a Grassmannian is unique up to a scalar multiple, the invariant metric is not in the sole case of 2-planes in 4-space. This oriented Grassmannian's metric is the product of two round 2-spheres whose radii may be in any ratio.

Comment by Daniel Asimov

2010-08-10T15:59:09Z

Hello, Ethan. Yes, indeed -- as you may recall, I proved that (as well as an n-dimensional version) in a class of yours on PL topology around 1970.

Comment by Daniel Asimov

2010-08-07T18:56:33Z

It's also amusing that given n, the complement of any countable dense subset of R^n is homeomorphic to the complement of any other such.

Comment by Daniel Asimov

2010-08-01T17:45:35Z

The first sentence of this answer (in either revision) a verb.

Comment by Daniel Asimov

2010-07-16T17:58:48Z

Oops, that should have read, "There must be a first countable ordinal beta at which there are no unclaimed members of J; the player whose turn it is at beta is defined as the winner.

Comment by Daniel Asimov

2010-07-16T17:54:43Z

(Cont'd.) Assume "whose turn it is" has been defined for all countable limit ordinals, and hence by alternation for all countable ordinals. Then it's not even necessary to predefine the length of the game. A simple example is this: let J be any countably infinite set. Each player on their turn claims some member of J. There must be a first countable ordinal at which the last member of J is claimed: the player who did that is defined as the winner. (I hope to expand on this soon in a short article.)

Comment by Daniel Asimov

2010-07-16T17:47:04Z

Incidentally, there's no reason that the plays of an infinite ordinal game must be indexed by omega. Any infinite ordinal lambda will do, though let's just be concerned about countable ordinals here. Special care must be taken to know whose turn it is at each play (whose index is some ordinal < lambda). This is easily accomplished in a "fair" way as long as whose turn it is is defined for limit ordinals; this is not difficult. (Successor ordinals just follow the alternating turn rule, as for the case lambda = omega.)

Comment by Daniel Asimov

2010-07-15T09:26:12Z

"... larger populations have more girls ..." No, since the population size is meaningful only if something is known about the number of families. In order to use the Strong Law of Large Numbers we assume infinitely many families #1,#2,#3,... with each one's sequence of births (G^n)B being concatenated in order of family #. Then the stochastic process generating this infinite sequence of G's and B's is isomorphic (up to measure 0) with repeated flips of a fair coin. Hence the SLLN implies the asymptotic fraction of B's or G's is 1/2, each with probability = 1. This is, to me, conclusive.

Comment by Daniel Asimov

2010-07-08T23:43:26Z

1. The words "a social convention cannot override biology" (not mine) mean just that the ultimate proportions of boys and girls are the same as the proportions in which boys and girls are born. 2. The stopping rule (in question) is a red herring because any stopping rule of the form "Stop as soon as a certain consecutive string of B's and G's occurs" will result in the same ratio of 1:1 (or more generally p:q) as the probabilities of B vs. G (or H vs. T) are in. 3. You are right that the 2:1 stopping rule is not almost certain to occur. Oops. 4. What modification are you thinking of?

Comment by Daniel Asimov

2010-07-06T23:12:51Z

Zare falls into exactly the trap I mention in my first paragraph.