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Vladimir Dotsenko's user avatar
Vladimir Dotsenko's user avatar
Vladimir Dotsenko's user avatar
Vladimir Dotsenko
  • Member for 15 years, 1 month
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Motivation behind spectral sequences
Try to understand the case of a first quadrant double complex, that is a nonnegatively bigraded vector space with two anticommuting differentials $d_1$ and $d_2$ of bidegrees $(1,0)$ and $(0,1)$. Imagine that you want to compute the homology of $d_1+d_2$, and you know the homology of $d_1$ (all horizontal rows). What would you do? What to do to figure out which homology classes of $d_1$ extend to homology classes of $d_1+d_2$? Once you do some diagram chasing in this case, the general case will be obvious.
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Is there a $3$-commutative algebra?
@DenisSerre - yes, indeed, I completely agree. Still, it is perhaps useful to have this perspective also, since studying weak identities of various kinds has been at heart of several important breakthroughs in the PI theory in the past. In particular, this observation about $W_1$ came about in a current research project, and I find this instance of 3-commutativity one of the most natural I have seen so far :)
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Published AI-generated nonsense math papers
It is perhaps worth remarking that the author seems to have been active for many decades, much longer than the AI has been around, and that one of the previous recent papers (ieeexplore.ieee.org/document/5780078) comes with the following note on the publisher's website: "Notice of Retraction: After careful and considered review of the content of this paper by a duly constituted expert committee, this paper has been found to be in violation of IEEE's Publication Principles. We hereby retract the content of this paper. Reasonable effort should be made to remove references to this paper."
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A new (?) way of composing monads
This is very intriguing. The only immediate comment is that the multiplication on $M\times L(M)$ that you obtain is something I have already seen in a linearized context: it measures to what extent the differential of the bar complex $B(A)$ of an associative algebra $A$ is not a derivation of the concatenation product imposed on the $B(A)$ (which is naturally a coalgebra, not an algebra, hence "imposed"). It leads to a non-commutative analogue of Batalin-Vilkovisky algebras, see arxiv.org/abs/1510.03261 and references therein.
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Realize as homology a given polynomial ring
Could it possibly be that what you want are Tate-Józefiak resolutions, see, e.g. eudml.org/doc/214408 (though there is an enormous body of literature on the topic)?
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Realize as homology a given polynomial ring
"I ask for something like a free resolution that is not exact" - this sentence seem to manifestly contradict itself.
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Is there a $3$-commutative algebra?
@SalvatoreSiciliano thanks! This remark clarifies some things I heard about quandles (in the context of knots, so nontrivial quandles) before...
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Is there a $3$-commutative algebra?
If I remember the definition of the trivial quandle correctly, the quandle ring of a trivial quandle satisfies the stronger identity $(xy-yx)z=0$, right?
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Is there a $3$-commutative algebra?
@DenisT why would there be a relation? $E_n$ relaxes commutativity in a homotopical context, this relaxes commutativity in a strict context...
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Is it ethical to submit a paper to journal then to Research Square? And what is the difference between that research square and ArXiv?
Research Square seems to be a business. I do not think you have any way to assess the implications arising from using it that will arrive in a year. So why deal with this?
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So many types of subwords! How are they called?
P.S. You might also find useful that in studying a similar story for permutations rather than words, there is its own terminology: type 1 is referred to as "consecutive pattern", while type 2 is simply "pattern".
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