User daniel asimov - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:39:46Z http://mathoverflow.net/feeds/user/5484 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/24038#24038 Answer by Daniel Asimov for Examples of common false beliefs in mathematics. Daniel Asimov 2010-05-09T18:40:39Z 2010-10-03T14:11:15Z <p><b>Complex variables</b>: "An entire function that is onto and locally one-to-one is globally one-to-one."</p> <p>Counterexample: <code>$f(z) := \int_0^z \exp(\zeta^2)\,d\zeta$</code></p> <p>I'll leave the proof that this is indeed a counterexample as a pleasant exercise.</p> <p>(I believe this example is due to Lawrence Zalcman.)</p> http://mathoverflow.net/questions/36700/infinite-dimensional-complex-polynomial-or-rational-lie-algebras-and-their-pseudo Infinite-dimensional complex polynomial or rational Lie algebras and their pseudogroups Daniel Asimov 2010-08-25T22:11:44Z 2010-08-26T00:41:15Z <p>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:</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{(a z<sup>2</sup> + b z + c) d/dz&nbsp;&nbsp;|&nbsp;&nbsp;(a,b,c) ∊ ℂ<sup>3</sup>}</p> <p>form precisely the Lie algebra whose nonzero elements generate the linear fractional transformations, i.e., PSL(2,ℂ).</p> <ul> <li>Is there some underlying reason for this? (Beyond direct calculation, which provides an easy proof.)</li> </ul> <p>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.</p> <ul> <li>Is there some easy-to-explain reason for this?</li> </ul> <p>The next "simplest" such polynomial Lie algebras of vector fields on C <em>seem</em> to be those defined by all polynomial and all rational functions:</p> <p>(*)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<strong>V</strong><sub>P</sub> := {P(z) d/dz &nbsp;|&nbsp; P(z) ∊ C[z]}&nbsp;&nbsp; and &nbsp;&nbsp;<strong>V</strong><sub>R</sub> := {R(z) d/dz &nbsp;|&nbsp; R(z) ∊ C(z)}.</p> <ul> <li><p>Contrariwise, do there exist finite-dimensional Lie algebras of vector fields on ℂ defined by rational functions -- other than the polynomial ones mentioned above?</p></li> <li><p>In case <strong>V</strong><sub>P</sub> and/or <strong>V</strong><sub>R</sub> 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 <strong>pseudogroups</strong>, but perhaps they behave like Lie groups.</p></li> </ul> <p>[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).]</p> <ul> <li><p>In any case, are there standard names for the Lie algebras <strong>V</strong><sub>P</sub> and <strong>V</strong><sub>R</sub> ? </p></li> <li><p>References to the above matters would also be appreciated.</p></li> </ul> http://mathoverflow.net/questions/35468/widely-accepted-mathematical-results-that-were-later-shown-wrong/35587#35587 Answer by Daniel Asimov for Widely accepted mathematical results that were later shown wrong? Daniel Asimov 2010-08-14T18:13:40Z 2010-08-15T06:00:14Z <p>One part of Hilbert's 16th problem is to determine whether a polynomial vector field in ℝ<sup>2</sup>: </p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;V(x,y) = (P(x,y),Q(x,y)) </p> <p>has at most a finite number of limit cycles.</p> <p>In 1923, Dulac published a paper supposedly proving this.</p> <p>Around 1980-81, Ecalle and Ilyashenko independently recognized that the proof had serious gaps.</p> <p>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.</p> <p>See Ilyashenko's paper, "A centennial history of Hilbert's 16th problem" at <a href="http://www.ams.org/journals/bull/2002-39-03/S0273-0979-02-00946-1/S0273-0979-02-00946-1.pdf" rel="nofollow">http://www.ams.org/journals/bull/2002-39-03/S0273-0979-02-00946-1/S0273-0979-02-00946-1.pdf</a>.</p> <p>(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.)</p> http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals Topological spaces that resemble the space of irrationals Daniel Asimov 2010-07-30T22:11:20Z 2010-08-09T17:53:29Z <p>(This question actually arose in some research on number theory.)</p> <p>I once learned that any countable dense subspace of any Euclidean space ℝ<sup>n</sup> is homeomorphic to the rationals ℚ.</p> <p>Now I wonder if something similar is true for the irrationals <strong>J</strong> := ℝ - ℚ (with the subspace topology from ℝ).</p> <p>Let <strong>c</strong> denote the cardinality of the continuum.</p> <blockquote> <p><strong>I</strong>. Is each cartesian power <strong>J</strong><sup>n</sup> homeomorphic to <strong>J</strong> ?</p> </blockquote> <p>Also, how far can this be pushed?</p> <blockquote> <p><strong>II</strong>. Let X be a dense totally disconnected subspace of ℝ<sup>n</sup> such that every neighborhood of each point of X contains <strong>c</strong> points. Is X homeomorphic to <strong>J</strong> ?</p> </blockquote> <p>What about for such subspaces of fairly nice subspaces of ℝ<sup>n</sup> ?</p> <blockquote> <p><strong>IIa</strong>. Let X be any subspace of ℝ<sup>n</sup> as described in <strong>II</strong>., and let B denote any subspace of ℝ<sup>n</sup> homeomorphic to [the open unit ball in ℝ<sup>n</sup> union any subset of its boundary]. Then is X ∩ B homeomorphic to <strong>J</strong> ?</p> </blockquote> <p>And what about greater generality ?</p> <blockquote> <p><strong>III</strong>. Is there a simple set of conditions that describe exactly all spaces (or subspaces of ℝ<sup>n</sup>) that are homeomorphic to <strong>J</strong> ? What about <strong>J</strong><sup>n</sup> ? (Perhaps the word <em>homogeneous</em> or <em>metric</em> needs to be included.)</p> </blockquote> <p>(I found nothing relevant via Google, in MathSciNet, or here on MathOverflow.)</p> http://mathoverflow.net/questions/17960/google-question-in-a-country-in-which-people-only-want-boys/31066#31066 Answer by Daniel Asimov for Google question: In a country in which people only want boys Daniel Asimov 2010-07-08T15:37:44Z 2010-07-08T23:27:03Z <p>A colleague, Eugene Salamin, came up with what I would consider the "Book" solution:</p> <p><em>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.</em></p> <p>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. </p> <p>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.)</p> <p>(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> <p>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.</p> http://mathoverflow.net/questions/5499/which-mathematicians-have-influenced-you-the-most/30663#30663 Answer by Daniel Asimov for Which mathematicians have influenced you the most? Daniel Asimov 2010-07-05T20:10:59Z 2010-07-06T09:42:22Z <p>(I think that for a question like this with the answers being entirely personal, the voting is of little or no significance.)</p> <p>For me there are so many that I hardly know where to begin. Initially, <strong>Martin Gardner</strong>. Among those I knew personally: my undergrad profs (espcially I.M. <strong>Singer</strong>) who taught me what math is. Then Bill <strong>Thurston</strong>, with whom I shared an office in grad school. Stephen <strong>Smale</strong>, my de facto co-thesis advisor.</p> <p>Notably <strong>Gauss</strong>, <strong>Riemann</strong>, <strong>Klein</strong>, <strong>Poincaré</strong>, <strong>Milnor</strong>.</p> <p>Above all, my thesis advisor, Morris <strong>Hirsch</strong>, with whom I've had a continuing connection since 1970.</p> http://mathoverflow.net/questions/17960/google-question-in-a-country-in-which-people-only-want-boys/30679#30679 Answer by Daniel Asimov for Google question: In a country in which people only want boys Daniel Asimov 2010-07-05T21:11:48Z 2010-07-05T21:27:44Z <p>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.</p> <p>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. </p> <p>The number of boys is obviously 1 for any outcome (of the form G<sup>n</sup>B), and so its expectation is 1.</p> <p>The expected number of girls is given by the summation of n&middot;p(n) for n = 0 to &infin;, where p(n) = the probability of the outcome G<sup>n</sup>B, which is 1/2<sup>n+1</sup>. This sum is perhaps surprisingly also 1, which is easy to verify.</p> <p>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> <p>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.</p> http://mathoverflow.net/questions/30425/which-manifolds-admit-a-diffeomorphism-of-order-n/30546#30546 Answer by Daniel Asimov for Which manifolds admit a diffeomorphism of order $n$? Daniel Asimov 2010-07-04T20:00:52Z 2010-07-04T20:00:52Z <p>For a closed orientable surface M<sub>g</sub> of genus g >= 2, there are only a finite number of possible orders of a diffeomorphism or homeomorphism h:M<sub>g</sub> -> M<sub>g</sub>. One constraint on such orders is that the map induced on first homology </p> <p>&nbsp;&nbsp; h<sub>*</sub>:H<sub>1</sub>(M<sub>2g</sub>) -> H<sub>1</sub>(M<sub>2g</sub>) </p> <p>belongs to GL(Z<sup>2g</sup>), and there is only a finite set of possible orders for elements of this group. (Note that if h<sub>*</sub> is the identity on first homology with g >= 2, then h is homotopic to the identity on M<sub>g</sub> and cannot have order > 1.)</p> <p>For this reason, e.g., on the double torus M<sub>2</sub> there can be no homeomorphism of order greater than 12. For more information see, e.g., <em>Finite groups of matrices whose entries are integers</em>, James Kuzmanovich and Andrey Pavlichenkov, American Mathematical Monthly, Vol. 109, No. 2 (Feb., 2002), pp. 173-186.</p> <p>(In the converse direction, I suspect any isomorphism of the cohomology ring H<sup>*</sup>(M<sub>g</sub>) can be realized by some self-homeomorphism of M<sub>g</sub>, but do not have a reference for this offhand.)</p> http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/27892#27892 Answer by Daniel Asimov for Examples of common false beliefs in mathematics. Daniel Asimov 2010-06-12T00:40:56Z 2010-06-12T00:40:56Z <p>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.</p> http://mathoverflow.net/questions/27881/who-is-the-last-mathematician-that-understood-all-of-mathematics/27890#27890 Answer by Daniel Asimov for Who is the last mathematician that understood all of mathematics. Daniel Asimov 2010-06-12T00:18:33Z 2010-06-12T00:18:33Z <p>Many people have mentioned Jean Dieudonn&eacute; (1906-92) in this regard.</p> http://mathoverflow.net/questions/27749/what-are-some-correct-results-discovered-with-incorrect-or-no-proofs/27873#27873 Answer by Daniel Asimov for What are some correct results discovered with incorrect (or no) proofs? Daniel Asimov 2010-06-11T21:01:33Z 2010-06-11T21:01:33Z <p>When Stephen Smale was a graduate student, he thought he had a proof of the Poincar&eacute; Conjecture as follows: Take a compact simply-connected 3-manifold M and remove the interiors of two disjoint 3-balls to get, say, M<sub>1</sub> having as boundary two copies of S<sup>2</sup>. It is easy to show that M<sub>1</sub> has a nonsingular vector field entering along one S<sup>2</sup> 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 M<sub>1</sub> must be S<sup>2</sup> x [0,1] and hence by replacing the removed 3-balls, M must have been S<sup>3</sup>. QED. </p> <p>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 Poincar&eacute; Conjecture for dimensions ≥ 5.</p> http://mathoverflow.net/questions/26474/analytic-ode-with-complex-time/26513#26513 Answer by Daniel Asimov for Analytic ODE with complex time Daniel Asimov 2010-05-30T22:38:26Z 2010-05-30T22:38:26Z <p>One thing to be careful about is that even for an analytic ODE given on ℂ via</p> <p>dz/dt = f(z)</p> <p>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) . . .</p> <p>. . . <strong>but</strong> 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 </p> <p>O<sub>0</sub> x (-K,K) and </p> <p>O<sub>1</sub> x (-K,K) </p> <p>in ℂ x ℝ such that for some t<sub>0</sub> in (-K,K) we have, e.g.,</p> <p>Φ(z<sub>1</sub>,t<sub>0</sub>) = z<sub>1</sub> for all z<sub>1</sub> ∈ O<sub>1</sub>, although</p> <p>Φ(z<sub>0</sub>,t<sub>0</sub>) ≠ z<sub>0</sub> for all z<sub>0</sub> ∈ O<sub>0</sub>.</p> <p>This seems to violate permanence, but for a subtle reason does not.</p> <p>For a concrete example: let P(z) := i(z<sup>3</sup> - z), and let O<sub>j</sub> be a small open neighborhood of j in ℂ. Then the flow given by Φ(z,t) := z(t) satisfying</p> <p>dz/dt = P(z)</p> <p>is defined for all real time t on both O<sub>0</sub> and O<sub>1</sub>.</p> <p>But setting t<sub>0</sub> = &pi;, we have</p> <p>Φ(z,&pi;) - z = 0 for all z in O<sub>1</sub>, although</p> <p>Φ(z,&pi;) - z ≠ 0 for all z in O<sub>0</sub>.</p> http://mathoverflow.net/questions/26462/noninteger-iterates-of-functions-how-to-get-ode-from-flow-at-a-given-time/26495#26495 Answer by Daniel Asimov for Noninteger iterates of functions: How to get ODE from flow at a given time? Daniel Asimov 2010-05-30T20:20:58Z 2010-05-30T21:47:27Z <p>If g is a real-analytic function defined near x<sub>0</sub> with g(x<sub>0</sub>) = x<sub>0</sub> and 0 &lt; λ &ne; 1 where λ := g'(x<sub>0</sub>), then Koenigs proved that there exists a real-analytic homeomorphism h defined near x<sub>0</sub> such that hgh<sup>-1</sup>(x) = L(x) where </p> <p>L(x) := x<sub>0</sub> + λ(x-x<sub>0</sub>)</p> <p>and h is unique up to a constant factor*. </p> <p>This allows g to be embedded in a local flow: Φ(x,t) := h<sup>-1</sup> L<sup>t</sup> h(x), where </p> <p>L<sup>t</sup>(x) := x<sub>0</sub> + λ<sup>t</sup> (x-x<sub>0</sub>),</p> <p>such that Φ(x,1) = g(x) where defined.</p> <p>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</p> <p>V(x) := ∂Φ(x,t)/∂t |<sub>t=0</sub>.</p> <p>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 </p> <p>g<sub>c</sub>(x) := c<sup>x</sup>. </p> <p>Then for 1 &lt; c &lt; e<sup>1/e</sup> the function g<sub>c</sub> has two distinct fixed points each satisfying the hypotheses of the Koenigs theorem, and these give two distinct flows into which g<sub>c</sub> embeds as the time-1 map. </p> <p>Concretely, set c = &radic;2, so that g<sub>c</sub>(x) = x for both x = 2 and x = 4, with derivatives </p> <p>g<sub>c</sub>'(2) = ln(2) and </p> <p>g<sub>c</sub>'(4) = ln(4). </p> <p>Calculating the respective real-analytic flows Φ<sub>2</sub>(x,t) and Φ<sub>4</sub>(x,t), both are defined for (x,t) = (3, 1/2). </p> <p>But Φ<sub>2</sub>(3, 1/2) and Φ<sub>4</sub>(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.</p> <hr> <p>&#42; This is actually true in greater generality; see J. Milnor's book <em>Dynamics in One Complex Variable</em>, 3rd ed., Princeton University Press, 2006.</p> http://mathoverflow.net/questions/7063/a-problem-of-an-infinite-number-of-balls-and-an-urn/25910#25910 Answer by Daniel Asimov for A problem of an infinite number of balls and an urn Daniel Asimov 2010-05-25T17:57:48Z 2010-05-25T17:57:48Z <p>The reason this problem has the generally accepted answer of 0 balls in the jug at midnight -- among <em>mathematicians</em> -- 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.</p> <p>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.</p> <p>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 <em>Mathematical Games</em> column in <em>Scientific American</em>, where "zero balls at midnight" was given as the correct solution. (See <em>A Mathematician's Miscellany</em>, J.E. Littlewood, Methuen, 1953.)</p> <p>Philosophers, on the other hand, are still debating this problem. (See, for example, <em>Paradoxes from A to Z</em> by Michael Clark, 2nd ed., Routledge, 2007.)</p> http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice/23217#23217 Answer by Daniel Asimov for Why worry about the axiom of choice? Daniel Asimov 2010-05-02T01:06:41Z 2010-05-02T01:06:41Z <p>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.</p> <p>But after I learned that AC is logically equivalent to </p> <pre><code> **The cartesian product of any non-empty collection of non-empty sets is non-empty** </code></pre> <p>I found it impossible not to believe it. My only conclusion could be that my mathematical intuition was not well-developed. </p> <p>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.</p> <p>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).</p> <hr> <p>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</p> <p>(*) &nbsp; &nbsp; &nbsp; z<sup>2</sup> + 1 = 0</p> <p>(and similar situations). There is no basis for selecting between i and -i . . . or even <em>naming</em> 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.</p> http://mathoverflow.net/questions/19619/why-do-dynamicists-worry-so-much-about-differentiability-hypotheses-in-smooth-dyn/19673#19673 Answer by Daniel Asimov for Why do dynamicists worry so much about differentiability hypotheses in smooth dynamics? Daniel Asimov 2010-03-28T22:48:29Z 2010-03-28T22:48:29Z <p>There are some fascinating phenomena in dynamical systems and related fields whose existence depends on the degree of differentiability.</p> <p>Among all of these, my favorite is the fact, proved by Haefliger, that although there exist C<sup>&infin;</sup> codimension-1 foliations of S<sup>3</sup>, there does not exist any C<sup>&omega;</sup> codimension-1 foliation of S<sup>3</sup>.</p> <p>A much more basic such result, due to Denjoy, is that a C<sup>2</sup> 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 C<sup>1</sup>.</p> http://mathoverflow.net/questions/19661/maximally-symmetric-smooth-projective-varieties-in-cp2 Maximally symmetric smooth projective varieties in CP^2 Daniel Asimov 2010-03-28T20:12:00Z 2010-03-28T22:14:56Z <p>Let P(X,Y,Z) be a homogeneous polynomial in &#8450;[X,Y,Z] whose locus M in &#8450;&#8473;<sup>2</sup> is a nonsingular curve of genus &ge; 2.</p> <p>Define M to be <em>maximally symmetric</em> if the following is <strong>not</strong> true: </p> <hr> <p>There exists a continuous family {&nbsp;P<sub>t</sub> &nbsp; |&nbsp; t &#8714; [0,1]&nbsp;} of homogeneous polynomials in &#8450;[X,Y,Z] such that 1), 2), and 3) hold:</p> <p>1) P<sub>0</sub> = P.</p> <p>2) The locus M<sub>t</sub> in &#8450;&#8473;<sup>2</sup> of each P<sub>t</sub> is nonsingular.</p> <p>3) There is a group G such that the ambient isometry groups <br/>G<sub>t</sub> := Isom<sub>A</sub>(M<sub>t</sub>) are all isomorphic to G for 0 &le; t &lt; 1, but G<sub>1</sub> contains G as a proper subgroup.</p> <p>Here the "ambient isometry group" Isom<sub>A</sub>(M<sub>t</sub>) of a projective curve M in &#8450;&#8473;<sup>2</sup> means the subgroup of Isom(&#8450;&#8473;<sup>2</sup>) = PSU(3) that carries M to itself.</p> <hr> <p>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 <em>especially</em> their ambient isometry groups.</p> <hr> http://mathoverflow.net/questions/54252/are-there-smooth-bodies-of-constant-width/54260#54260 Comment by Daniel Asimov Daniel Asimov 2013-01-11T04:50:26Z 2013-01-11T04:50:26Z I took a look at the Fillmore paper, and just before his Corollary to Theorem 2 -- which reads &quot;Corollary. There exists an analytic hypersurface of constant width in E^n having the same group of symmetries as a regular n-simplex.&quot; -- he writes &quot;If we imitate the construction of a Reuleux triangle . . .. Thus:&quot; 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. http://mathoverflow.net/questions/11821/su2-and-the-three-sphere/11823#11823 Comment by Daniel Asimov Daniel Asimov 2012-12-24T17:27:06Z 2012-12-24T17:27:06Z It's not so much that you <i>also</i> need det(x) = 1, as that this is exactly the same as saying |a|&lt;sup&gt;2&lt;/sup&gt; + |b|&lt;sup&gt;2&lt;/sup&gt; = 1. http://mathoverflow.net/questions/93879/riemann-zeta-at-even-integers/93884#93884 Comment by Daniel Asimov Daniel Asimov 2012-09-14T02:51:13Z 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). http://mathoverflow.net/questions/104451/irreducible-homology-3-spheres-that-bound-smooth-contractible-manifolds/104526#104526 Comment by Daniel Asimov Daniel Asimov 2012-08-13T20:18:16Z 2012-08-13T20:18:16Z The usual contractible 2-complex discovered by Bing is called a &quot;House With Two Rooms&quot;, and this is <i>not</i> 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. http://mathoverflow.net/questions/34334/how-well-can-we-localize-the-exoticness-in-exotic-r4/35208#35208 Comment by Daniel Asimov Daniel Asimov 2010-08-16T06:37:50Z 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 Poincar&#233; 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. http://mathoverflow.net/questions/35655/measure-on-real-grassmannians/35658#35658 Comment by Daniel Asimov Daniel Asimov 2010-08-15T15:45:40Z 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 -&gt; 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 <i>metric</i> 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. http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals/35027#35027 Comment by Daniel Asimov Daniel Asimov 2010-08-10T15:59:09Z 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. http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals/33982#33982 Comment by Daniel Asimov Daniel Asimov 2010-08-07T18:56:33Z 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. http://mathoverflow.net/questions/33972/how-many-people-fully-understand-the-proof-of-fermats-last-theorem/34080#34080 Comment by Daniel Asimov Daniel Asimov 2010-08-01T17:45:35Z 2010-08-01T17:45:35Z The first sentence of this answer (in either revision) a verb. http://mathoverflow.net/questions/32174/infinite-games-are-they-well-defined/32192#32192 Comment by Daniel Asimov Daniel Asimov 2010-07-16T17:58:48Z 2010-07-16T17:58:48Z Oops, that should have read, &quot;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. http://mathoverflow.net/questions/32174/infinite-games-are-they-well-defined/32192#32192 Comment by Daniel Asimov Daniel Asimov 2010-07-16T17:54:43Z 2010-07-16T17:54:43Z (Cont'd.) Assume &quot;whose turn it is&quot; 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.) http://mathoverflow.net/questions/32174/infinite-games-are-they-well-defined/32192#32192 Comment by Daniel Asimov Daniel Asimov 2010-07-16T17:47:04Z 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 &lt; lambda). This is easily accomplished in a &quot;fair&quot; 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.) http://mathoverflow.net/questions/17960/google-question-in-a-country-in-which-people-only-want-boys/17963#17963 Comment by Daniel Asimov Daniel Asimov 2010-07-15T09:26:12Z 2010-07-15T09:26:12Z &quot;... larger populations have more girls ...&quot; 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. http://mathoverflow.net/questions/17960/google-question-in-a-country-in-which-people-only-want-boys/31066#31066 Comment by Daniel Asimov Daniel Asimov 2010-07-08T23:43:26Z 2010-07-08T23:43:26Z 1. The words &quot;a social convention cannot override biology&quot; (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 &quot;Stop as soon as a certain consecutive string of B's and G's occurs&quot; 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? http://mathoverflow.net/questions/17960/google-question-in-a-country-in-which-people-only-want-boys/30679#30679 Comment by Daniel Asimov Daniel Asimov 2010-07-06T23:12:51Z 2010-07-06T23:12:51Z Zare falls into exactly the trap I mention in my first paragraph.