bio | website | people.virginia.edu/~deh4n |
---|---|---|
location | ||
age | ||
visits | member for | 5 years, 1 month |
seen | 2 days ago | |
stats | profile views | 788 |
Apr 8 |
comment |
An application of Maschke's theorem
@KConrad $7\cdot 7 = 49 = (5\sqrt{2}-1)(5\sqrt{2}+1)$? |
Mar 3 |
awarded | Yearling |
Nov 21 |
comment |
“Nyldon words”: understanding a class of words factorizing the free monoid increasingly
Also, can you define you ordering more explicitly. Do you read words left-to-right or right-to-left? |
Nov 21 |
comment |
“Nyldon words”: understanding a class of words factorizing the free monoid increasingly
I don't think your comment above is helpful. I think it was appropriate to delete it. |
Nov 21 |
comment |
“Nyldon words”: understanding a class of words factorizing the free monoid increasingly
Isn't (3) already false for $w=101$, $u=10$ and $v=1$? |
Nov 20 |
comment |
“Nyldon words”: understanding a class of words factorizing the free monoid increasingly
@TheMaskedAvenger This is definitely not what is going on. The modifications you are proposing would yield Lyndon words for the associated ordering. Nyldon words behave very differently. For example, the Nyldon words $101$ and $1011$ defy this kind of description. |
Nov 18 |
comment |
“Nyldon words”: understanding a class of words factorizing the free monoid increasingly
@darijgrinberg: sorry, no. I was not saying that. I was just pointing out that things that seem to happen in a 2 letter alphabet are unlikely to be true for bigger alphabets. For example, certain palindromes are Nyldon, but cycling the first letter to the end will not make them Lyndon Honestly, these Nyldon words are baffling. |
Nov 18 |
comment |
“Nyldon words”: understanding a class of words factorizing the free monoid increasingly
@PerAlexandersson, I don't think this coincidence in a 2 letter alphabet generalizes. The thing about Lyndon words is that they are filled with patterns. For example, every Lyndon work looks like $w=w_1^kw_1'i$, where $w_1$ is Lyndon, $w_1'$ is a (possibly empty) left factor of $w_1$ and $i$ is a letter such that $w_1'i>w_1$ (this was proved by Leclerc). In contrast, Nyldon words seem to be pattern avoiding. |
Nov 7 |
comment |
Irreducible representations of Weyl group of F$_4$ on zero weight spaces?
I don't have Bourbaki with me. Is $\varpi_1$ short or long? |
Nov 3 |
answered | Given a locally nilpotent derivation over a field of characteristic 0 and a local slice, how is the ring homomorphism below defined? |
Nov 2 |
comment |
When exactly and why matrix multiplication became a part of undergraduate curriculum?
Just to focus in, is your question about when US universities adopted linear algebra in their core curriculum? And, if this can be established, who were the advocates of this that made it happen? |
Nov 1 |
answered | The formula for a perhaps basic identity (move from stackexchange) |
Oct 31 |
awarded | Yearling |
Oct 22 |
answered | Does Schur's Lemma hold in this case? Regular representations of $S_n$ over $\mathbb R$ |
Oct 18 |
revised |
Homomorphisms from irreducible spaces to reducible spaces
added 25 characters in body |
Oct 18 |
answered | Homomorphisms from irreducible spaces to reducible spaces |
Oct 17 |
comment |
Shift-invariant symmetric functions in representation theory?
The permutation representation of $S_n$ is an $n$-dimensional vector space $V$ with basis $x_1,\ldots,x_n$. The symmetric algebra is the ring of symmetric functions in $n$-variables. The standard $n-1$ dimensional representation (call it $W$) is the subspace of $V$ with basis $x_i-x_{i+1}$. Now, $S(W)\subset S(V)$ consists of shift invariant symmetric functions. |
Oct 9 |
comment |
Generators of invariant polynomials of semisimple Lie algebra
Wouldn't `Schur functions' be the canonical choice? |
Oct 8 |
comment |
A class of matrix determinants between Wronskians and Vandermondes
are you specifically interested in equation (4.2) in the link? This seems much more tractable than the general problem. |
Oct 8 |
comment |
A class of matrix determinants between Wronskians and Vandermondes
@Alex R. I seem to have added the assumption that $d_1=n-1$ in my answer. As you point out, it is much more complicated if this is not the case. I think that if $d_1=n+k-1$, then something like I wrote holds if $d_1-d_2=k_1+k$ (with $G=0$ for $>k_1+k)$. |