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2

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....

0

I apologize for the slow processing. The case of Poisson ideals coming from linear Poisson structures have been systematically studied in the literature. Similarly there are the symplectic reductions. For details and examples one may consult the recent https://arxiv.org/pdf/2107.04204.pdf For polynomial Poisson structures of higher degree I cannot find any ...

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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).

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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,...

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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 $(i) \quad \forall a,b,c\in M, \quad [a,b,c] = [a,c,b]=[b,c,a]$ $(ii) \quad \forall ... 1 I had a good go at this but I couldn't nail down a proof (either by searching the literature or making my own). What I believe to be true is that for any regular-nilpotent element$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$is the height of the ... 2 As already indicated, this is a result depending only on the underlying group (rather than ring) and you can find in my earlier question, MO 105400, a comment from me that points to: Martin, G. A. (1988). A class of Abelian groups arising from an analysis of a proof. The American Mathematical Monthly, 95(5), 433-436. JSTOR; Sci-Hub. Here are the relevant ... 4 Let$S$b be the set of 3 x 3 matrices whose lower left 2 x 2 block equals zero. Then$S$is an algebra satisfying the conditions, but containing no invertible matrix. 5 It's false. Take the subalgebra of$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}$. These two elements form a two element right zero semigroup and so the algebra they generate is just their span ... 3 I believe the answer to question 1 is no. But the question is slightly ambiguous. I can give a quiver presentation with only$\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 ...

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 ...

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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 ...

2

Viewing cartesian operads as algebraic theories, your question may be rephrased as: "How can we present the (multisorted) algebraic theory whose models are (monosorted) algebraic theories?" This is given by the theory of abstract clones. You can check that the category of monosorted algebraic theories is isomorphic to the category of abstract ...

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The answer to the first question is yes: Claim. Let $a, b \in \{0, 1 \}^{\ast}$ and let $n \ge 0$ be the least integer such that $a(i) = b(i) = 0$ for every $i > n$. Then the equation $$a +_2 x = b$$ has a unique solution $x \in \{0, 1 \}^{\ast}$ which is recursively defined by $x(0) = a(0) + b(0)$, $x(i) = a(i - 1)x(i - 1) + a(i) + b(i)$ for $0 < i \... 3 No, a free Boolean algebra$R$on an infinite cardinal$\kappa$(e.g., if$\kappa = \aleph_0$,$R$is the Cantor algebra), is a commutative von Neumann regular ring which is not well complemented as an$R$-module. 4 The answer is no. Take a compact totally disconnected space$X$with no isolated points, like the Cantor set. Let$K$be any field and let$R$be the ring of locally constant functions$f\colon X\to K$with pointwise operations. This is a commutative von Neumann regular ring. The idempotents of$R$are precisely the characteristic functions$1_K\$ of ...

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