## Representation Theory – Explicit Examples of Finite-Dimensional, Involving Hopf Algebras with a Traceless Antipode?

$$required {AMScd}$$

In the newspaper [1]It is shown that there are finally three-dimensional Hopf algebras $$H$$ where the antipode $$S: H to H$$ is without a trace.

Unfortunately, the proof method is in [1] seems to me rather inexplicable. I understand that the authors use the fact that every fusion category
$$mathcal {C}$$ Fulfilling some conditions is the representation category of some Hopf algebra. Then you convert the trace condition $$text {Tr} (S) = 0$$ in a condition of the fusion category $$mathcal {C}$$ and show that they can easily create categories that meet this new condition, in addition to the required conditions $$mathcal {C}$$ be a representation category $$H$$,

The paper [1] is now eight years old, so I hope that some more explicit examples in the literature could be at this point. My question is:

Question: What are some explicit examples of a finite, inclusive Hopf algebra (across a field)? $$k$$ of characteristic $$0$$) where the antipode is without a trace? By explicit I mean given by a generator / relationship representation or as some structure tensors on a certain basis.

Here I weakened the hypothesis to assume it $$H$$ is only indicative, i. $$S ^ 2 = text {Id}$$, A sentence by Larson and Radford [2] says that semisimple Hopf algebras are involved, though $$text {char} (k) = 0$$,

Alternatively, I would appreciate some advice that I need to know to get an explicit example [1]if necessary.

## \$ E_ infty \$ algebras and Tor Unital rings

Remember that this is a non-unitalial ring $$R$$ is called Tor-unital, though $$Tor ^ 1_ {R _ +} ( mathbb Z, mathbb Z) cong 0$$ from where $$R _ +$$ is the unitalization of $$R$$, See, e.g. https://arxiv.org/pdf/1610.04998.pdf. If $$R$$ is Tor-unital, does it $$bigsqcup_n BGL_n (R)$$ have a natral $$E_ infty$$-Algebra structure?

## Braid Groups – Does the diversity of algebras match the Yang-Baxter equation of their finite members?

Suppose that $$f, g: X ^ {2} rightarrow X$$, and $$T: X ^ {2} rightarrow X ^ {2}$$ is the function where $$T (x, y) = (f (x, y), g (x, y))$$, Then $$(X, f, g)$$ should satisfy the Yang-Baxter equation, though $$(T times 1_ {X}) circ (1_ {X} times T) circ (T times 1_ {X}) = (1_ {X} times T) circ (T times 1_ { X}) circ (1_ {X} times T)$$, In other words, $$(X, f, g)$$ meets the Yang-Baxter equation if and only if it satisfies the following identities:

1. $$f (f (x, y), f (g (x, y), z)) = f (x, f (y, z))$$
2. $$g (f (x, y), f (g (x, y), z)) = f (g (x, f (y, z)), g (y, z))$$
3. $$g (g (x, y), z) = g (g (x, f (y, z)), g (y, z))$$

The collection of all algebras that meet the Yang Baxter equations is diverse. Is this diversity created by its finite members?

Is there an inverse system? $$((X_ {n}) _ {n}, ( phi_ {m, n}) _ {m, n})$$ of finite algebras that fulfill the Yang-Baxter identity along with it
a sequence $$(e_ {n}) _ {n} in varprojlim X_ {n}$$ where everyone $$X_ {n}$$ is generated by $$e_ {n}$$ and where $$(e_ {n}) _ {n}$$ creates a free subalgebra of $$varprojlim X_ {n}$$?

The motivation of this question is based on the fact that the Yang-Baxter equation is a generalization of self-distribution ability, and if $$T (x, y) = (x * y, x)$$, then $$T$$ meets the Yang-Baxter equation if and only if $$T$$ is self-distributing. The multitude of self-distributing algebras is generated by their finite members (in particular the multigeneric Laver tables) and the free self-distributing algebra of a generator, which is embedded in an inverse boundary of the finite self-distributing algebras called classical load tables.

## Finite generation of flat deformations of algebras

To let $$R = mathbb C[q^{pm 1}]$$ and let it go $$A$$ graded (possibly not commutative) $$R$$-Algebra, $$A = oplus_ {n = 0} ^ infty A_n,$$ from where $$A_0 = R$$ and everything $$A_n$$are free $$R$$Modules.
Then $$A & # 39; = A / (q-1)$$ is a graduated algebra over $$R / (q-1) = mathbb C$$ and so
$$A$$ can be thought of as deformation of $$A & # 39 ;.$$ I could not find a definition of a flat algebra deformation, but I imagine that's an example of that, right?
(Assume that $$A & # 39;$$ is commutative if it helps.)

Main question: is that possible? $$A & # 39;$$ is finally generated $$A$$ is not (as algebras over $$mathbb$$ and $$R$$ respectively)?

## at.algebraic topology – Effect of quaternion algebras on real vector spaces

I am confused by the following example.

To let $$mathbf {H}$$ Be the departmental ring of Hamilton quaternions. As a topological space $$X$$, It is a $$4$$Euclidean space and its Betti cohomology with real coefficients is trivial in the positive area, and we have:

$$H ^ 0 (X, mathbf {R}) = mathbf {R}$$

since $$X$$ Is connected.

Now $$X$$ carries an obvious action from $$mathbf {H}$$: for all $$z in mathbf {H}$$, $$z$$ defines a continuous endomorphism $$mu_z: X to X$$ send a point $$x$$ to $$zx$$,

Because the bundle inverse picture $$mu_z ^ {- 1} mathbf {R}$$ on $$X$$ is still the constant bundle $$mathbf {R}$$We should have an action from $$mathbf {H}$$ on $$H ^ 0 (X, mathbf {R})$$However, this is impossible because the smallest real vector space is on it $$mathbf {H}$$ looks unrivaled and has a dimension $$4$$,

The action of $$mathbf {H}$$ as defined on $$H ^ 0 (X, mathbf {R})$$ can not be trivial, because everyone $$mu_z$$ has an inverse that is supposed to invert the effect of $$mu_z$$ on cohomology.

Where is the mistake?

My expectation is that to say$$mu_z ^ {- 1} mathbf {R}$$ is the constant bundle $$mathbf {R}$$"One is too sloppy: The isomorphism between the two disks may not be natural and may involve decisions that interfere with any attempt to define an action $$mathbf {H}$$ on $$H ^ 0 (X, mathbf {R})$$and that should be the salient point to avoid inconsistencies. The functoriality of Betti cohomology only indicates maps $$H ^ 0 (X, mathbf {R}) to H ^ 0 (X, mu_z ^ {- 1} mathbf {R})$$ and say "$$mu_z ^ {- 1} mathbf {R} simeq mathbf {R}$$ and therefore $$H ^ 0 (X, mathbf {R})$$ carries an action of $$mathbf {H}$$"Is clearly wrong.

## Linear map connecting two left-invariant single forms that are evaluated in different Lie algebras

As can be seen, a left-invariant unit in a Lie group evaluated in another Lie algebra can be factored by the canonical left-invariant Maurer-Cartan form of this Lie group, followed by a linear one Map between the two Lie algebras? How can one see the existence and possibly uniqueness of this linear map? Many Thanks.

## at.algebraic topology – exact reference for the equivalence of \$ E_n \$ -algebras and locally constant factorization algebra?

The result in question is a consequence of Theorem 5.4.5.15 in the higher algebra

Proposal 5.4.5.15. To let $$M$$ to be and to be a variety $$C ^ ⊗$$ Bean $$∞$$-operad. Then composition with map $$mathrm {disk} (M) ^ ⊗ → mathbb {E} ^ ⊗_M$$ Remark 5.4.5.8 effects a completely faithful embedding
$$θ: mathrm {Alg} ^ {nu} _ { mathbb {E} _M} (C) → mathrm {Alg} ^ {nu} _ { mathrm {Disk} (M)} (C) .$$
The essential picture of $$θ$$ is the full subcategory of $$mathrm {Alg} ^ {nu} _ { mathrm {disk} (M)} (C)$$ spanned by the locally constant
Objects.

When $$M = mathbb {R} ^ n$$, the $$infty$$-operad $$mathbb {E} _M$$ That's just that $$mathbb {E} _n$$-operad (the small $$nu$$ Above it only says that they are "non-unitalial" algebras.

Make sure the book does not use the language of the factorization algebras, but if you compare the definitions, you will find that there is a conservative range of functions of locally constant factorization algebras $$M$$ too constant locally $$mathrm {disk} (M)$$-algebras, this is a left-hand inverse to the functor of Theorem 5.4.5.15.

## at.algebraic topology – Is the category of rational Lie algebras monoidal?

I hate to ask such a naïve question, but this is it. Accept $$A$$ and $$B$$ are rational Lie algebras, d. H. rational vector spaces together with a bracket. Then, $$A otimes _ { mathbb {Q}} B$$ is a rational vector space. Can it be given a Lie bracket so that the category of rational Lie algebras is a monoide category? This question came to me when I read Quillen's Rational Homotopy Theory and "Equivalence of the Monoidal Model Categories" by Schwede-Shipley, in which Quilling's work is mentioned, but without claiming any monoide properties.

## at.algebraic topology – cohomology of small plates and dg algebras via \$ mathbb {F} _p \$

This is an alternative form of the question that I asked a while ago.

We, the people who do not know the topology, learn that the characteristic of DG algebras in characteristic p is not completely sufficient for applications that one must work with spectra. However, it seems that in some cases you can still work at the DG algebra level, eg. when the algebra rises $$mod p ^ 2$$ and some cohomological disappearance happens gradually $$geq 2p$$ (as in Kaledins
Proposition 5.10 of https://arxiv.org/pdf/math/0611623.pdf or the result of Deligne's illusory on the de Rham complex in characteristic p).

I wonder if there is a standard result somewhere that says, "The cohomology of small memory locations (or configuration spaces) with coefficients in $$mathbb {F} _p$$ are so and so, so DG algebras would be clearly wrong in the general public, but if something disappears, it's a good idea to work with a DG algebra that has some steenrod operations. "

I apologize for the sloppy question and am grateful to propose a better version of the same question.