bio | website | quora.com/Alon-Amit |
---|---|---|
location | Los Altos, CA | |
age | 46 | |
visits | member for | 5 years, 10 months |
seen | Jul 13 at 7:47 | |
stats | profile views | 4,095 |
Dabbler in things.
Jun 17 |
awarded | Popular Question |
Jun 9 |
awarded | Nice Answer |
Jun 4 |
awarded | Notable Question |
May 21 |
revised |
Philosophy behind Yitang Zhang's work on the Twin Primes Conjecture
Trying to fix the first two links to Wikipedia |
May 6 |
answered | Useful tricks in experimental mathematics |
Apr 15 |
revised |
What's the “best” proof of quadratic reciprocity?
(Typo correction) |
Feb 17 |
awarded | Notable Question |
Jan 1 |
revised |
Which curves cut the Hyperelliptic locus?
edited title |
Dec 2 |
revised |
Proofs without words
deleted 34 characters in body; added 49 characters in body; deleted 43 characters in body |
Sep 29 |
awarded | Yearling |
Jul 27 |
revised |
Point sets in Euclidean space with a small number of distinct distances
Fixed exponent of n to 77/141. |
Jul 24 |
awarded | Nice Answer |
Jul 7 |
awarded | Nice Answer |
Sep 30 |
awarded | Yearling |
Sep 19 |
revised |
Number of finite simple groups of given order is at most 2 - is a classification-free proof possible?
Added a few LaTeX tags. |
Sep 19 |
comment |
out-trees and least upper boundness
I'm still very confused. Yes, lattices contain semicycles, that's just the point - you were asking if every LUB graph is a (kind of) tree, and it isn't. You now changed the question to ask if every connected digraph without semicycles (which further satisfies LUB) must be a (kind of) tree. Well, it is, by definition, right? The underlying undirected graph certainly is a tree (connected and cycle free). This is just an out-tree except that we haven't chosen a specific root. |
Sep 16 |
comment |
Finite simple groups and conjugacy classes with 2p elements
Is it known that this can't occur with the alternating groups? Empirically it looks like the sizes of the conjugacy classes of $A_n$ all have at least 3 prime factors once $n>8$. |
Sep 16 |
comment |
Spanning trees in 3 regular graphs.
What do "clip" and "half edges" mean? |
Sep 15 |
answered | out-trees and least upper boundness |
Sep 11 |
comment |
Second eigenvalue of suspension of a graph
Shouldn't you be looking at the Laplacian, rather than the adjacency matrix? For non-regular graphs I'm not sure if the second largest eigenvalue is the thing that controls mixing. Also, the operation you're interested in seems more like taking the cone over $G$, rather than a suspension. |