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• CommentRowNumber1.
• CommentAuthorzskoda
• CommentTimeNov 13th 2012
• (edited Nov 13th 2012)

In a collaborative work at Institute Ruđer Bošković in Zagreb, which will come out as a preprint soon, we are working on a certain noncommutative Hopf algebroid over a noncommutative base coming from Lie algebra theory. I gave a seminar talk about it in mathematics department in Zagreb and got a remark that our approach is too much coordinate based and that it would be desirable to have this work done using coordinate free approach. I came to a one month visit at l’IHÉS and soon realized that in dual language I can tell the story geometrically, though I still do not know how to prove some of the claims without algebraic coordinate-involving stuff which we had before.

So, a Hopf algebroid is made out of a left bialgebroid, right bialgebroid and an antipode map between them.The left and right bialgebroid have the same underlying algebra, but are different as bimodules and corings. I will leave the discussion of the antipode for another post and here sketch how I geometrically arrive at a left bialgebroid.

So let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ which can be realized as $\mathfrak{g}^L$, the Lie algebra of left invariant vector fields or the Lie algebra $\mathfrak{g}^R$, the Lie algebra of right invariant vector fields; specialization of those vector fields at the unit element $e\in G$ gives the isomorphism of vector spaces with the tangent space $T_e G$. The universal enveloping algebra $U(\mathfrak{g})$ can be realized also as the algebra of left invariant differential operators $U(\mathfrak{g}^L)$ and as $U(\mathfrak{g}^R)$. Consider now the algebra $H^L = Diff^\omega$ of formal differential operators at $e$. That means that you allow finite sums of partial derivatives to any finite order with coefficients which are formal functions, i.e. function supported on an infinitesimal neighborhood of unit element. The usual algebra of (global) differential operators $Diff(G)$ (which contains $U(\mathfrak{g}^L)$ and $U(\mathfrak{g}^R)$) is naturally embedded in $Diff^\omega$, hence it also contains $U(\mathfrak{g}^L)$ and $U(\mathfrak{g}^R)$. furthermore the images of the left and right copy of the $U(\mathfrak{g})$ mutually commute, what is very important for our story.

Thus we have two maps of $U(\mathfrak{g})$ into $\mathcal{X}^\omega$, namely

$Diff^\omega \stackrel\alpha\leftarrow U(\mathfrak{g}) \stackrel\beta\rightarrow Diff^\omega$

Here $\alpha$ is defined as extension from the vector fields as a homomorphism of associative algebras, while $\beta$ as antihomomorphism of associative algebras.

If we choose a basis in $\mathfrak{g}$, i.e. we have a frame in $\mathfrak{g}$, then we can consider two images in $\Gamma \mathcal{F} T G$, that is a section of the frame bundle of the tangent bundle: every element of the basis goes into the corresponding left or right invariant vector field. Thus there are also two maps

$\Gamma \mathcal{F} T G \stackrel{\mathcal{F}\alpha}\leftarrow \mathfrak{g} \stackrel{\mathcal{F}\beta}\rightarrow \Gamma \mathcal{F} T G$

corresponding to the frames by left and right invariant vector fields. Both maps are analytic and the image is the section of a principal $GL_n$-bundle, hence the difference is a matrix of analytic function on $G$. Better to say, if $\tau$ is the translation map of the principal bundle then $\tau(\alpha(fr),\beta(fr)) =: \mathcal{O}^{-1}(fr)$ (for a frame fr) defines some invertible matrix $\mathcal{O}$ of analytic functions on $G$. Of course, by the functoriality, this translation matrix does not depend on the choice of frame. We can compute this matrix in a neighborhood of unit element, and I will explain some concrete formula in a later post.

To be continued…

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeNov 13th 2012
• (edited Nov 13th 2012)

The space of formal power series is dual to the space of polynomials. This generalizes here. The vector space $S^\omega$ of formal functions at unit element of a Lie group $G$ is a dual to $U(\mathfrak{g}^L)$; indeed act by an left invariant differential operator at unit element and one gets a pairing. This is nondegenerate, as one can easily check for abelian case, and it holds also for nonabelian $\mathfrak{g}$.

The algebra $S^\omega$ has an adic filtration, hence it is a topological algebra. It inherits a coproduct by dualizing the product from $U(\mathfrak{g}^L)$ (transpose operator). The coproduct is of course into the completed tensor product $\Delta: S^\omega\to S^\omega\hat\otimes S^\omega$. As the product in $U(\mathfrak{g})$ is noncommutative, this coproduct is also noncommutative, in general.

There is, furthermore, a black action $\blacktriangleright$ of $S^\omega$ on $U(\mathfrak{g}^L)$ given by the formula

$D \blacktriangleright \hat{u} = \sum \langle D, \hat{u}_{(2)} \rangle \hat{u}_{(1)}$

where $\Delta_{U(\mathfrak{g})} (u) = \sum u_{(1)}\otimes u_{(2)}$ is the Sweedler notation for the classical, undeformed coproduct on the enveloping algebra and $\langle,\rangle$ is the pairing explained above. We can extend the black action to an action of $Diff^\omega$ on $U(\mathfrak{g}^L)$ by defining it as usual multiplication for $U(\mathfrak{g}^L)$-factors in $Diff^\omega$. There is a similar action $\blacktriangleleft$ of $U(\mathfrak{g}^R)$ on $Diff^\omega$. Both black actions are Hopf actions, when restricted to $D$ in $S^\omega$, i.e. $D\blacktriangleright( u v) = \sum (D_{(1)}\blacktriangleright u)(D_{(2)}\blacktriangleright v)$, where the coproduct used is the topological coproduct dual to the universal algebra product (also in Sweedler notation).

Now the idea is that one would like to extend the coproduct $\Delta$ from $S^\omega$ to the whole $Diff^\omega$ in such a way that $D\blacktriangleright(u v) = \sum (D_{(1)}\blacktriangleright u)(D_{(2)}\blacktriangleright v)$ would hold for every $D\in Diff^\omega$. As the black action of $U(\mathfrak{g}^L)$ on itself is simply the multiplication then we must have $\Delta(u) = u\otimes 1$ for any $u \in U(\mathfrak{g}^L)$.

In fact this task of defining the topological coproduct of that kind on the whole $Diff^\omega$ is impossible when the completed tensor product is over the ground field. But it will be possible over $U(\mathfrak{g}^L)$. Namely, $Diff^\omega$ is a $U(\mathfrak{g}^L)$-bimodule via $u.d.v = \alpha(u)\beta(v) d$ where $\alpha$ and $\beta$ were defined before.

Then we consider the tensor product $Diff^\omega \otimes_{U(\mathfrak{g})} Diff^\omega = Diff^\omega \otimes_{\mathbf{R}} Diff^\omega/ I_R$ where $I_R$ is the smallest right ideal containing all elements of the form $\beta(u)\otimes 1 - 1\otimes\alpha(u)$ in $Diff^\omega \otimes_{\mathbf{R}} Diff^\omega$.

This tensor product is not an algebra, but only a $U(\mathfrak{g}^L)$-bimodule. It can be viewed as certain coend.

In fact there is a subalgebra inside, so called Takeuchi product which is by its definition an algebra with respect to the factorwise (i.e. induced from product on tensor factors) product, and it is morally the subset for which $I_R$ behaves as a two sided ideal. This is an end of a coend.

It appears that $\Delta$ is well defined by the above rules when quotienting by $I_R$; furthermore its image is in the Takeuchi product and that corestriction is a homomorphism of algebras! This coproduct is counital in the usual sense, which is furthermore a left character of $Diff^\omega$ as the $U(\mathfrak{g}^L)$-ring.

This way, $Diff^\omega$ is an algebra equipped with source homomorphism $\alpha$ and target antihomomorphism $\beta$ from $U(\mathfrak{g}^L)$ whose full images commute; it is a coring in the sense of a counital coproduct in the sense of bimodule structure induced by multiplication by those source and target maps from the left. Furthermore, the coproduct corestricts to the Takeuchi product as an algebra homomorphism. This all together makes a left bialgebroid structure.

Similarly one makes a right bialgebroid structure from right action and over $U(\mathfrak{g}^R)$ and with source and target maps on the right. There is an antipode map from left to right bialgebroid which has a number of properties. The square of the antipode map is not the identity! Furthermore various expressions and calculations here involve very importantly passages between $\alpha$ and $\beta$ which on generators mean passage from left to right vector fields, hence using matrix $\mathcal{O}$ from the principal bundle explained in 1. The version of Hopf algebroid axioms is due Gabi Bohm and it is equivalent to the one in the work of Day and Street.

• CommentRowNumber3.
• CommentAuthorzskoda
• CommentTimeNov 13th 2012
• (edited Nov 20th 2012)

The structure of $Diff$ is a structure of a Hopf smash product, in fact it is the Heisenberg double over $U(\mathfrak{g})$, what is close to a statement I proved in a recent paper of mine (in dual, different, algebraic formulation):

(There is another action, the white action of $U(\mathfrak{g}^L)$ on $S_f$ (basically coming from the usual Fock space action) and the usual semicompleted Weyl algebra structure which can also be used for various purposes here, including for defining a dual smash product structure, which gives however the same topological algebra. )

There is a known structure of a Hopf algebroid on a Heisenberg double of any finite-dimensional Hopf algebra. The construction basically passes to the adic completions like here and is due to Lu’s 1994 work. However her notion of a Hopf algebroid is quite noncanonical, and nonsymmetric, while Gabi Bohm’s notion is symmetric and our construction is itself manifestly symmetric (left vs, right vector fields, and left and right actions in the story). Lu has one bialgebroid and the antipode defined in terms of it. Lu’s antipode treats differently source and target maps. However her construction is superimposable to ours; but her formula for $\beta$ looks different. It is in terms two copies of a canonical element in $U(\mathfrak{g})\hat\otimes U(\mathfrak{g})^*$. One has kind of twisting with those two copies is equivalent to our passage from left to right vector field,. Fortunately, I can compute such a procedure in coordinates given by the exponential map (zoranskoda) on the Lie group. One gets a very interesting identity

$\sum_{I, J} \frac{(-1)^{|J|}}{I! J!} e(x_I) \hat{x}_a e(x_J) \otimes \partial^{J+I} = \sum_{b =1}^n \hat{x}_b \otimes (e^{-\mathcal{C}})^b_a$

where $e$ is the coexponential map (symmetrization map $Sym(\mathfrak{g})\to U(\mathfrak{g})$), and its many generalizations This particular formula (and, in particular, the notation in it) is explained to some extent in formula for beta (zoranskoda). Edit: this case of the formula can be proved directly using Hadamard formula in $U(\mathfrak{g})[[\partial]]$.

• CommentRowNumber4.
• CommentAuthorzskoda
• CommentTimeNov 14th 2012
• (edited Nov 14th 2012)

The algebra $Diff^\omega$ has a structure of a left bialgebroid $H^L$ over $U(\mathfrak{g}^L)$ and a structure of a right bialgebroid over $U(\mathfrak{g}^R)^{op}$. One of the axioms of Hopf algebroid says that if $\mathcal{S}:Diff^\omega\to Diff^\omega$ is the antipode map, then $\mathcal{S}\circ \beta = \alpha$.
This shows that the antipode returns the right invariant differential operators into their left invariant counterparts. For formal functions, the antipode is simply the transpose (via the duality coming from the pairing) of the classical antipode map on the enveloping algebra. As $Diff^\omega$ is generated by $S^\omega$ and $U(\mathfrak{g}^L)$ then we can determine it on other elements by using the rule that $\mathcal{S}$ is the antihomomorphism of algebras. In particular, we infer the value of the antipode on left vector fields. For that, one writes for (frames) $\hat{x}^R = \mathcal{O}^{-1}\hat{x}^L$ uses that $\mathcal{S}$ is antihomomorphism of algebras, hence $\mathcal{S}(\hat{x}^L) = \mathcal{S}(\hat{x}^R) \mathcal{S}(\mathcal{O}^{-1}) = \hat{x}^L \mathcal{S}(\mathcal{O}^{-1})$. Thus, it is important to compute with $\mathcal{O}$.

On the other hand, one has to prove that the antipode is well defined by these rules. For this one notices that it is well defined within $S^\omega$ and $U(\mathfrak{g}^R)$, while the whole algebra is the smash product of the two. This can be viewed as a free product modulo additional relations coming from the Leibniz rule on generators. I will show the details later but basically a commutator gives a formal function. One pairs that formal function with an arbitrary element in $U(\mathfrak{g}^R)$. Then one applies the antipode to the quadratic expression and can alternatively transfer the antipode to the other side of the pairing, which is still arbitrary because $S$ is bijective on $U(\mathfrak{g})$. Thus the relations are preserved under $S$ and hence well defined on the quotient.

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeNov 14th 2012

Looks good, but I can’t make any knowledgeable comments

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeNov 14th 2012
• (edited Nov 14th 2012)

What’s the motivation? To find the Heisenberg double of $U(\mathfrak{g})$?

The text in #1 is all review, I gather. Also the first half of #2 is, right? The project starts, it seems, where in the middle of #2 you say

Now the idea is that one would like to extend the coproduct Δ from $S^\omega$ to the whole $Diff^\omega$

Maybe you could say at that point why that is your idea now, what the goal is.

A little further below you say

It appears that Δ is well defined

What does “it appears” mean here? It sounds like “conjectured but not yet proven”, is that what you mean to say?

• CommentRowNumber7.
• CommentAuthorzskoda
• CommentTimeNov 17th 2012
• (edited Nov 18th 2012)

Thank you very much for your comments. The 1 and beginning of 2 are indeed the story which is mostly obvious once formulated this way. The setup in which this Hopf algebroid structure was discovered is completely algebraic and the matrix $\mathcal{O}$ and other characters of the story are very nonobvious in that framework and all the proofs are coordinate dependent. Once formulated geometrically the construction of much of the structure is obvious, while the properties to make it Hopf algebroid are still not, and some proofs I still do not know in geometric language but only in coordinate dependent dual picture, with calculations involving very concrete commutators.

To find the Heisenberg double of $U(\mathfrak{g})$?

No, the goal is to equip that (understood as an algebra) Heisenberg double with Hopf algebroid structure, which will be geometrically sound and will correspond to our further motivations, which include the appearance of a very interesting Drinfeld twist for that context (which is so far understood only in some examples of Lie algebras, and the general case is open: so any hint on an origin of it, would be real nice).

Now the idea is that one would like to extend the coproduct Δ from $S^\omega$ to the whole $Diff^\omega$

Maybe you could say at that point why that is your idea now, what the goal is.

This coproduct is the one which makes the $U(\mathfrak{g})$ a $Diff^\omega$ module algebra (deformed Fock space) i.e. it gives the Leibniz rule for action on the products of elements in $U(\mathfrak{g})$. This action appears in so many construction in the literature in which this is motivated (Amelino-Camelia, R. Szabo, J. Wess being some of the physics authors).

What does “it appears” mean here? It sounds like “conjectured but not yet proven”, is that what you mean to say?

I can translate the story in algebraic world where we have a rather intricate proofs for all of this, and predating this story with $Diff^\omega$, which I made up last week to simplify the story and make it coordinate free as far as possible. Still there are things which are not clear even there. For example, the expressions of the form $\beta(u)\otimes 1 - 1\otimes \alpha(u)$, where $u\in U(\mathfrak{g})$ generate a right ideal in $Doff^\omega\otimes_{\mathbf{R}}Diff^\omega$, which should be the same as the annihilator of $U(\mathfrak{g})\otimes U(\mathfrak{g})$ when acted with $Diff^\omega\otimes_{\mathbf{R}} Diff^\omega$ (as a tensor square of left deformed Fock module) and then multiplied. I do not know any proof, and we can avoid this to some extent, but it would be more naturally to develop the theory knowing this in the first place.

• CommentRowNumber8.
• CommentAuthorzskoda
• CommentTimeNov 23rd 2012
• (edited Nov 23rd 2012)

Let me now post the argument why the coproduct $\Delta^L : Diff^\omega\to Diff^\omega\otimes_{U(\mathfrak{g})} Diff^\omega$ is in fact taking place in the Takeuchi’s product. First of all, the tensor product $Diff^\omega\otimes_{U(\mathfrak{g})}Diff^\omega$ is the tensor product $Diff^\omega\otimes_{\mathbf{R}} Diff^\omega$ quotiented by the ideal generated by all expressions of the form $\sum_{i=1}^k \beta(\hat{h}) s_i\otimes_{\mathbf{R}} t_i - s_i\otimes _{\mathbf{R}} \alpha(\hat{h}) t_i$, where $s_i, t_i\in Diff^\omega$ and $\hat{h}\in U(\mathfrak{g}^L)$. The image of $\Delta^L$ is spanned by all expressions of the form $(\hat{h} \otimes_A 1)\Delta^L(P) = \sum \hat{h} P_{(1)}\otimes_A P_{(2)}$ where $\hat{h}\in U(\mathfrak{g}^L)$ and $P\in S^\omega$. Now the key relation is

$\beta(\hat{h}) \blacktriangleright \hat{f} = \hat{f}\cdot\alpha(\hat{h})$

for all $\hat{h},\hat{f}\in U(\mathfrak{g}^L)$. This relation takes some effort to prove (I know the proof by a certain induction procedure in a dual language at this point, only).

The ideal of the relations when quotienting $Diff^\omega\otimes_{\mathbf{R}}Diff^\omega$ to obtain $Diff^\omega\otimes_{U(\mathfrak{g}^L)}Diff^\omega$ above coincides with the joint annihilator of the map $a\otimes b\mapsto (a\blacktriangleright \hat{f})(b \blacktriangleright\hat{g})$ from $Diff^\omega\otimes_{\mathbf{R}} Diff^\omega$ to $U(\mathfrak{g}^L)$. So we use the above relation to calculate (in Sweedler notation)

$\array{ ((P_{(1)}\beta(\hat{h}))\blacktriangleright \hat{f}))(P_{(2)}\blacktriangleright\hat{g}) &=& (P_{(1)}\blacktriangleright (\beta(\hat{h})\blacktriangleright \hat{f})))(P_{(2)}\blacktriangleright\hat{g}) \\ &=&(P_{(1)}\blacktriangleright(\hat{f}\alpha(\hat{h})))(P_{(2)}\blacktriangleright\hat{g}) \\ &=& (P_{(1)}\blacktriangleright\hat{f})(P_{(2)}\blacktriangleright\alpha(\hat{h}))(P_{(3)}\blacktriangleright\hat{g}) \\ &=& (P_{(1)}\blacktriangleright\hat{f})(P_{(2)}\blacktriangleright(\alpha(\hat{h})\hat{g})) }$

hence $\sum P_{(1)}\beta(\hat{h})\otimes P_{(2)} - P_{(1)}\otimes P_{(2)}\alpha(\hat{h})$ is in the annihilator (hence the ideal) for all $\hat{h}\in U(\mathfrak{g}^L)$ and all $P\in S^\omega$.

The general element in the image of $\Delta^L$ is of the form $\Delta^L(\hat{u} P) = \hat{u} P_{(1)}\otimes P_{(2)}$, where $\hat{u}\in U(\mathfrak{g})$ and $P\in S^\omega$, so we need also to look at the elements $\hat{u}\otimes 1$ (then it is automatic for products as the Takeuchi product is closed under factorwise products). But there the argument is much easier: $0 = \beta(\hat{h}) \hat{u}\otimes_{U(\mathfrak{g})} 1 - \hat{u} \otimes \alpha(\hat{h}) = \beta(\hat{h}) \alpha(\hat{u})\otimes_{U(\mathfrak{g})} 1 - \alpha(\hat{u}) \otimes_{U(\mathfrak{g})} \alpha(\hat{h})$ what is equal to $\alpha(\hat{u})\beta(\hat{h}) \otimes_{U(\mathfrak{g})} 1 - \alpha(\hat{u}) \otimes_{U(\mathfrak{g})} \alpha(\hat{h}) = \hat{u}\beta(\hat{h}) \otimes_{U(\mathfrak{g})} 1 - \hat{u} \otimes_{U(\mathfrak{g})} \alpha(\hat{h})$, because the images of $\alpha$ and $\beta$ mutually commute (of course, here $\alpha(\hat{u}) = \hat{u}\sharp 1 = \hat{u}$ using our identifications for the left bialgebroid).

This means that the image of $\Delta^L$ is within the Takeuchi product $Diff^\omega\times_{U(\mathfrak{g})} Diff^\omega \subset Diff^\omega\otimes_{U(\mathfrak{g})} Diff^\omega$, as required.

Of course, the proof of the fact that $\Delta^L$ is homomorphism of algebras into the Takeuchi product is quite standard, as this property is shared with other known cases of smash product constructions of bialgebroids:

$\Delta^L((u\sharp P)(v\sharp Q)) = \Delta^L(u (P_{(1)}\blacktriangleright v) \sharp P_{(2)} Q) = (u(P_{1}\blacktriangleright v)\sharp P_{(2)} Q_{(1)}) \otimes (1\sharp P_{(3)}Q_{(2)}),$

where one used that $\Delta^L|_{1\sharp S^\omega} = \Delta_{S^\omega}$ is a homomorphism by the definition (this is used when applied to the $P_{(2)} Q$ factor).

$\array{ \Delta^L(u\sharp P)\Delta^L(v\sharp Q) &=& [(u\sharp P_{(1)})\otimes(1\sharp P_{(2)})][(v\sharp Q_{(1)})\otimes(1\sharp Q_{(2)})] \\ &=& [(u\sharp P_{(1)})(v\sharp Q_{(1)})]\otimes (1\sharp P_{(2)} Q_{(2)})\\ &=& (u (P_{(1)}\blacktriangleright v)\sharp P_{(2)} Q_{(1)})\otimes(1\sharp P_{(3)} Q_{(2)}). }$
• CommentRowNumber9.
• CommentAuthorzskoda
• CommentTimeNov 26th 2012
• (edited Nov 27th 2012)

Let me comment the counitality. The counit of a left bialgebroid is $\epsilon: Diff^\omega\to U(\mathfrak{g}^L)$ and satisfies $action \circ (id\otimes\epsilon) \circ \Delta^L = id = action \circ (\epsilon\otimes id) \circ\Delta^L$ where the actions are the actions of $U(\mathfrak{g}^L)$ on $Diff^\omega$ from the right and from the left respectively, hence are given by $\beta$ and $\alpha$. Of course, the tensor product is over $U_{\mathfrak{g}}$ so $(\epsilon\otimes_{U(\mathfrak{g})} id)\Delta^L$ in fact implies the action (in other words, one uses the identification $U(\mathfrak{g})\otimes_{U(\mathfrak{g})} DIff^\omega\cong Diff^\omega$). On generators $\hat{x}_\mu\in \mathfrak{g}\hookrightarrow U(\mathfrak{g}^L)$, we have $\Delta^L(\hat{x}_\mu) = \hat{x}_\mu\otimes 1 = \mathcal{O}^\nu_\mu \otimes\hat{x}_\nu\in Diff^\omega\otimes_{U(\mathfrak{g})}Diff^\omega$ and we calculate (where we identify $\hat{x}_\mu$ with the corresponding left invariant vector field)

$right action (id\otimes \epsilon) (\hat{x}_\mu\otimes 1) = \beta(\epsilon(1))\hat{x}_\mu = \hat{x}_\mu$ $left action(\epsilon\otimes id) (\hat{x}_\mu\otimes 1) = \alpha(\epsilon(\hat{x}_\mu))1 = \hat{x}_\mu$ $right action (id\otimes \epsilon) (\mathcal{O}^\nu_\mu\otimes\hat{x}_\nu) = \beta(\epsilon(\hat{x}_\nu))\mathcal{O}^\nu_\mu =\hat{x}_\lambda(\mathcal{O}^{-1})^\lambda_\nu\mathcal{O}^\nu_\mu = \hat{x}_\mu$ $left action(\epsilon\otimes id) (\mathcal{O}^\nu_\mu\otimes\hat{x}_\nu) =\alpha(\epsilon(\mathcal{O}^\nu_\mu))\hat{x}_\nu = \alpha(\delta^\nu_\mu 1)\hat{x}_\nu = \hat{x}_\mu$

the counitality on $S^\omega$ part is clear, as the $\Delta^L|_{S^\omega}$ is simply the composition $S^\omega\stackrel{\Delta_{S^\omega}}\longrightarrow S^\omega\otimes_{\mathbf{R}} S^\omega\hookrightarrow Diff^\omega\otimes_{\mathbf{R}}Diff^\omega\to Diff^\omega\otimes_{U(\mathfrak{g}^L)}Diff^\omega$ hence clearly counital.

• CommentRowNumber10.
• CommentAuthorzskoda
• CommentTimeAug 13th 2014
• (edited Aug 13th 2014)

I am close to finish two collaborative papers related to the picture above, with two different coauthors.

There is an interesting thing that there is also a smaller Hopf algebroid inside, of very algebraic nature and without any completions in the tensor products or elsewhere. Namely, consider the matrix Hopf algebra $Fun(GL(n))$ of regular functions on $GL(n)$, let $G^\alpha_\beta$ and $(G^{-1})^\alpha_\beta$ be the algebra generators (they play the role of $\mathcal{O}^\alpha_\beta$ above), where $G^\alpha_\beta(A) = A^\alpha_\beta$. Let $\mathfrak{g}$ be an $n$-dimensional Lie algebra. Then there is a unique Hopf action of Hopf algebra $U(\mathfrak{g})$ on $Fun(GL(n))$ extending the formula (as a map $\phi_m:U(\mathfrak{g})\to End(Fun(GL(n)))$

$\phi_m(\hat{x}_\nu)(G^\alpha_\mu) = C^\alpha_{\rho\nu} G^\rho_\nu$

To get a right action, precompose with the antipode of $U(\mathfrak{g})$. So, using the right Hopf action, we can form the smash product $H_m = U(\mathfrak{g})\sharp Fun(GL(n))$ (here subscript $m$ stands for minimal). Define $\alpha_m:U(\mathfrak{g})\to H_m$ by $\alpha(\hat{f}) = \hat{f}\sharp 1$ and extend
$\beta_m(\hat{x}_\nu) = \hat{x}_\rho (G^{-1})^\rho_\nu$ to an antihomomorphism. I claim that $H_m$ is a Hopf algebroid over $U(\mathfrak{g})$ and that $\hat{x}_\nu\mapsto \hat{x}_\rho\otimes (G^{-1})^\rho_\nu$ is a coaction which is part of a braided-commutative monoid structure on $U(\mathfrak{g})$ in the category of left-right $Fun(GL(n))$-Yetter-Drinfeld modules, the left module structure being defined more or less as before.

All this structure is strictly inside the previous construction provided the structure constants are sufficiently nontrivial; otherwise some degeneration phenomena are possible.

• CommentRowNumber11.
• CommentAuthorzskoda
• CommentTimeNov 23rd 2016
• (edited Nov 23rd 2016)

I have put one of the papers on the above topic (the one which is mostly in dual algebraic language) with today’s update at https://www.irb.hr/korisnici/zskoda/halgrev.pdf. It is accepted to Lett Math Phys and a version will soon be published online (with some typographical differences). The arxiv version has some details not quite correct, which were corrected in the above version only after arXiv posting from Oct 3, I will replace there as well once I got the final response from the proofs team so that I am sure that they do not catch additional things needing improvement.