bio | website | |
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location | University of California, Berkeley | |
age | 67 | |
visits | member for | 4 years, 8 months |
seen | Mar 24 at 1:18 | |
stats | profile views | 971 |
Jan 11 |
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Are there smooth bodies of constant width?
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. |
Dec 24 |
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$SU(2)$ and the three sphere
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. |
Sep 14 |
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Riemann zeta at even integers
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). |
Aug 13 |
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Irreducible homology 3-spheres that bound smooth contractible manifolds
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. |
Aug 16 |
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How well can we localize the “exoticness” in exotic R^4?
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. |
Aug 15 |
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Measure on real Grassmannians
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. |
Aug 10 |
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Topological spaces that resemble the space of irrationals
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. |
Aug 7 |
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Topological spaces that resemble the space of irrationals
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. |
Jul 16 |
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Infinite games: are they well defined?
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. |
Jul 16 |
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Infinite games: are they well defined?
(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.) |
Jul 16 |
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Infinite games: are they well defined?
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.) |
Jul 15 |
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Google question: In a country in which people only want boys
"... 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. |
Jul 8 |
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Google question: In a country in which people only want boys
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? |
Jul 6 |
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Google question: In a country in which people only want boys
Zare falls into exactly the trap I mention in my first paragraph. |
Jul 5 |
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Which manifolds admit a diffeomorphism of order $n$?
Wow, that's an interesting example. Thank you. |
Jul 4 |
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How is it that you can guess if one of a pair of random numbers is larger with probability > 1/2?
I think you mean Randall Munroe. |
Jun 17 |
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Examples of common false beliefs in mathematics
There are many situations where one needs to speak of a set of two numbers that may or may not be equal. E.g.: "Let x<sub>1</sub>, x<sub>2</sub> ∈ ℝ. Then among all the open intervals containing the set {x<sub>1</sub>, x<sub>2</sub>}, none of them is contained in all the others." If one is addressing mathematicians, there is no need to specify that x<sub>1</sub> might be equal to x<sub>2</sub>. |
Jun 12 |
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Examples of common false beliefs in mathematics
In case that wasn't clear: F(x) = ln(x) + C_1 for x > 0, and F(x) = ln(-x) + C_2 for x < 0, where C_1 and C_2 are arbitrary real constants. |
Jun 12 |
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Examples of common false beliefs in mathematics
Really? What about the function F(x) given by ln(x) + C_1, x > 0 F(x) = ln(-x) + C_2, x < 0 for arbitrary reals C_1, C_2 ? (The appropriate technical condition is that an antiderivative be differentiable on the same domain as the function it's the antiderivative of is defined on.) |
Jun 12 |
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What are some correct results discovered with incorrect (or no) proofs?
Indeed Heawood proved the 5-color theorem. But I'm not aware that he gave an incorrect proof of the 4-color theorem. What he is known for doing is finding a flaw in an 1879 supposed proof, by Kempe, that had stood for 11 years. Perhaps at least as impressive, he determined the "Heawood number" -- an upper bound for the chromatic number -- for every compact surface, and conjectured it was the actual chromatic number. This number turned out to be the actual chromatic number of every compact surface except the Klein bottle, as shown by Ringel & Youngs (except for the sphere) in 1968. |