bio | website | |
---|---|---|
location | University of California, Berkeley | |
age | 67 | |
visits | member for | 4 years |
seen | Mar 24 at 1:18 | |
stats | profile views | 898 |
Nov 15 |
awarded | Nice Answer |
Oct 10 |
awarded | Yearling |
Sep 30 |
awarded | Nice Answer |
Jun 25 |
awarded | Excavator |
Jan 11 |
revised |
Examples of common false beliefs in mathematics.
it's -> its |
Jan 11 |
comment |
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 |
comment |
$SU(2)$ 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 |
comment |
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 |
comment |
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. |
Mar 28 |
awarded | Yearling |
Mar 29 |
awarded | Yearling |
Feb 25 |
awarded | Nice Answer |
Jan 12 |
awarded | Nice Answer |
Oct 25 |
awarded | Nice Answer |
Aug 25 |
revised |
Infinite-dimensional complex polynomial or rational Lie algebras and their pseudogroups
changed title |
Aug 25 |
revised |
Infinite-dimensional complex polynomial or rational Lie algebras and their pseudogroups
Added note about the ease of computing formulas for the flows |
Aug 25 |
revised |
Infinite-dimensional complex polynomial or rational Lie algebras and their pseudogroups
probably -> surely |
Aug 25 |
asked | Infinite-dimensional complex polynomial or rational Lie algebras and their pseudogroups |
Aug 16 |
comment |
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 |
comment |
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. |