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  
stats  profile views  11,500 
I'm just here for asking stupid questions.
34m

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“Nyldon words”: understanding a class of words factorizing the free monoid increasingly
OK, now this means a lot of readingup for me... once the FPSAC deadline is over. Sorry for the slowness. 
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Apocryphal Maschke theorem?
It is  but seeing why is so is part of the problem. 
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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 Endrings 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

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Apocryphal Maschke theorem?
throw away some trash 
2d

revised 
Apocryphal Maschke theorem?
something broke the latex 
Nov 23 
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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 
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Determinant of the oriented adjacency matrix of a tree
@AllenKnutson: I'm not sure how explicit you want it. Algorithmically, it is simple: Reorient 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 
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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 
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“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 
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“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 
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“Nyldon words”: understanding a class of words factorizing the free monoid increasingly
I have verified Conjecture 1 on a 3letter alphabet for length up to $11$, and Conjecture 2 for length up to $9$. 
Nov 20 
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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 
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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 
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“Nyldon words”: understanding a class of words factorizing the free monoid increasingly
@DavidHill: Are you implying that Per's statement is true on a 2letter 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 
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“Nyldon words”: understanding a class of words factorizing the free monoid increasingly
@PerAlexandersson: This would follow from either of the two conjectures. 