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9

Perhaps this has already been said in some form, but the identity $$[[a,b,c],d,e] = [a, [b,d,e], [c,d,e]]$$ is exactly the $(2,2,2)$-associative law of clone theory. I think that this is the most natural way this identity arises. A clone is a closed set of operations. Starting with a collection of finitary operations on some set, close them under composition,...

7

With the risk of making a simple thing confusing, here's another point of view. Two functions $f,g$ on an interval $I$ give a map to the projective line $I \dashrightarrow \mathbb P^1$ by $x\mapsto [f(x):g(x)]$ (some indeterminacies can occur if both functions vanish at some point). You can ask: when does this map have a critical point? If you choose the ...

6

As mentioned by @zeb, the condition $(iii) \quad \forall a,b,c,d,e\in M, \quad [[a,b,c],d,e] = [a, [b,d,e], [c,d,e]]$. is known as the third median axiom. A set $M$ endowed with a ternary operation $M^3\to M$, $(a,b,c) \mapsto [a,b,c]$, which satisfies this axiom, along with $(i) \quad \forall a,b,c\in M, \quad [a,b,c] = [a,c,b]=[b,c,a]$ $(ii) \quad \forall ... 5 It's false. Take the subalgebra of$M_3(K)$generated by the matrices$\begin{bmatrix} 0 & 0&0\\ 1 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$and$\begin{bmatrix} 0 & 0 & 0\\ 0 &1&0\\ 1& 0&1\end{bmatrix}$. These two elements form a two element right zero semigroup and so the algebra they generate is just their span ... 4 Let$S$b be the set of 3 x 3 matrices whose lower left 2 x 2 block equals zero. Then$S$is an algebra satisfying the conditions, but containing no invertible matrix. 4 Not in general. For an easy example, let$C$and$D$be derived equivalent algebras. Then$A=C\times C$and$B=C\times D$are derived equivalent, and$A=C\otimes_K(K\times K)$has a tensor decomposition, but$B$will usually not. There are also connected examples. For example,$KA_2\otimes_K KA_2$is derived equivalent to$KD_4$(where$KA_2$and$KD_4$are ... 3 There is a one-to-one correspondence between the grouplike elements and the simple,$1$-dim subcoalgebras. So if the only grouplike element is$1_H$, then there is a unique$1$-dim simple subcoalgebra (which is$k\cdot 1_H$). In that case, the HA is - by definition- called: connected HA. Also, notice that a connected HA is the same thing as an irreducible HA.... 3 I believe the answer to question 1 is no. But the question is slightly ambiguous. I can give a quiver presentation with only$\pm 1$appearing as coefficients in the relations and where being Frobenius depends on the characteristic of the field. Maybe you are asking if you can always find some quiver presentation of a commutative Frobenius algebra with$\pm ...

2

As already indicated, this is a result depending only on the underlying group (rather than ring) and you can find in my earlier question, MO 105400, a comment from me that points to: Martin, G. A. (1988). A class of Abelian groups arising from an analysis of a proof. The American Mathematical Monthly, 95(5), 433-436. JSTOR; Sci-Hub. Here are the relevant ...

2

Question 1: See Clifford and Preston, volume 1. Question 2: $m$ is the period, $n$ is called the index of the element. See this Wikipedia text. Question 3: If $n=1$, the element is called a group element of finite order. If $n=0$, it is called a unit of finite order. Or you can just call it an element of index 1 (resp. 0).

1

I think that the focus is usually on finitely length modules (those that have both ACC and DCC on submodules) because of the uniqueness part of the Krull-Schmidt theorem. In general, indecomposable decompositions are quite ill-behaved without additional hypotheses. The fact that uniform modules are finite direct sums of indecomposables follows quickly from ...

1

I had a good go at this but I couldn't nail down a proof (either by searching the literature or making my own). What I believe to be true is that for any regular-nilpotent element $X\in\mathfrak{g}$ it is contained in a minimal parabolic subalgebra $\mathfrak{p}$ equal to $\mathrm{Ker}(\mathrm{ad}_X^k)$ for some $k$. Here $k-1$ is the height of the ...

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