11,450 reputation
234105
bio website mit.edu/~darij/www
location Karlsruhe (home), Munich (until summer 2011), Cambridge/MA (2011-)
age 26
visits member for 4 years, 11 months
seen 34 mins ago
I'm just here for asking stupid questions.

34m
comment “Nyldon words”: understanding a class of words factorizing the free monoid increasingly
OK, now this means a lot of reading-up for me... once the FPSAC deadline is over. Sorry for the slowness.
2d
comment Apocryphal Maschke theorem?
It is -- but seeing why is so is part of the problem.
2d
comment Apocryphal Maschke theorem?
@WillSawin: I want an iso of k[G]-bimodules, not of bimodules over different algebras (whatever that would be). The k[G]-bimoudle structure on the sum of the End-rings doesn't a priori look canonical. Anyway the answers spread given are simple enough -- I have bumped the question only to fix an error in my post and bad latex in an answer.
2d
revised Apocryphal Maschke theorem?
throw away some trash
2d
revised Apocryphal Maschke theorem?
something broke the latex
Nov
23
comment Determinant of the oriented adjacency matrix of a tree
Forget the sentence where I said to label the vertices increasingly. This is not necessary. Whatever way they are labelled, as long as the edges are labelled accordingly, the matrix will still be unitriangular for an appropriate choice of labels for both rows and columns (and the choice for rows is the same as the choice for columns, so different choices do not force different signs), and so the determinant will still be $1$.
Nov
23
comment Determinant of the oriented adjacency matrix of a tree
@AllenKnutson: I'm not sure how explicit you want it. Algorithmically, it is simple: Re-orient all edges away from the vertex $v$, thus making $v$ the root of the tree. Remove $v$, thus obtaining a forest and some dangling edges with only a target but no source. Label the vertices increasingly (i.e., for every edge $a \to b$, we must have $a < b$). Label the edges such that the label of every edge is that of its target (this is possibly because every vertex is now the target of exactly one edge). Then, the determinant is $1$ since the matrix is unitriangular.
Nov
23
comment Determinant of the oriented adjacency matrix of a tree
@ChrisGodsil: I don't think it will be $0$.
Nov
22
awarded  Good Question
Nov
22
comment “Nyldon words”: understanding a class of words factorizing the free monoid increasingly
Thanks a lot, but this feels like it's going to take me a while to grok. First and foremost, what is $\pi$ ? And does this all add up to a (conjectural, yet) intrinsic definition of a Nyldon word?
Nov
21
comment “Nyldon words”: understanding a class of words factorizing the free monoid increasingly
@GjergjiZaimi: This is an interesting idea, but I have no experience with Hall sets so far (besides knowing that they somehow generalize Lyndon words, or rather a lift of them to the free magma), and the definition is daunting. Do you see a way to machinally check the existence of a Hall set lifting a given subset of $\mathfrak A^\ast$ (in given degrees)?
Nov
21
revised degree of polynomials in nullstellensatz
there is no ansatz in nullstellensatz
Nov
20
awarded  Nice Question
Nov
20
revised “Nyldon words”: understanding a class of words factorizing the free monoid increasingly
added 272 characters in body
Nov
20
comment “Nyldon words”: understanding a class of words factorizing the free monoid increasingly
I have verified Conjecture 1 on a 3-letter alphabet for length up to $11$, and Conjecture 2 for length up to $9$.
Nov
20
comment Irreducible representations of $Sp(4,\mathbb{F}_2)$
I assume "coming from permutation representations" means "permutation representation of $S_6$ modulo trivial subrepresentation". In this case, it is easy to see that $S_6$ has two such representations of dimension $5$. One of them is obtained by having $S_6$ act on $\left\{1,2,3,4,5,6\right\}$ in the usual way, and the other by composing this representation with the infamous outer automorphism of $S_6$. It might be a bit of work to check that these are nonequivalent. See §1.5 in math.stanford.edu/~vakil/files/sixjan2308.pdf .
Nov
19
comment Breaking up the free Lie algebra into Gl irreps
s.lie(n) can also be accessed by calling s.gessel_reutenauer(n) once ticket #17125 is merged into Sage (which is slated to happen in the next beta).
Nov
18
comment “Nyldon words”: understanding a class of words factorizing the free monoid increasingly
@DavidHill: Are you implying that Per's statement is true on a 2-letter alphabet? This would be very interestign!
Nov
18
revised “Nyldon words”: understanding a class of words factorizing the free monoid increasingly
added 447 characters in body
Nov
18
comment “Nyldon words”: understanding a class of words factorizing the free monoid increasingly
@PerAlexandersson: This would follow from either of the two conjectures.