3
$\begingroup$

Let $V$ be a vector space and $V^*$ the dual vector space. Let $T(V)$ be the tensor algebra of $V$.

The algebras $T(V)$ and $T(V^* \otimes V)$ are bialgebras. I am trying to find some bialgebra structure on the crossed product $T(V) \rtimes T(V^* \otimes V)$. There are natural multiplication and comultiplication in $T(V) \rtimes T(V^* \otimes V)$ defined as follows.

The multiplication in $T(V) \rtimes T(V^* \otimes V)$ is defined as: for $a, b, v, w, v', w' \in V$, \begin{align} & (a \sharp (v^* \otimes w))(b \sharp (v'^* \otimes w')) \\ & = a (v^* \otimes w)_{(1)}(b) \sharp (v^* \otimes w)_{(2)} (v'^* \otimes w') \\ & = a (v^* \otimes I_{(1)})(b) \sharp (I_{(2)} \otimes w) (v'^* \otimes w'), \end{align} where $I = I_{(1)} \otimes I_{(2)} = \sum_j v_j \otimes v_j^*$, $v_1, \ldots, v_n$ is a basis of $V$.

The comultiplication on $T(V) \rtimes T(V^* \otimes V)$ is given by: for $u, v, w \in V$, \begin{align} & \Delta( u \sharp (v^* \otimes w) ) \\ & = (u_{(1)} \sharp (u_{(2)})_{(1)} (v^* \otimes w)_{(1)} ) \otimes ( (u_{(2)})_{(0)} (v^* \otimes w)_{(2)} ) \\ & = (u_{(1)} \sharp (I_{(2)} \otimes u_{(2)})(v^* \otimes I_{(1)})) \otimes (I_{(1)} \sharp (I_{(2)} \otimes w)) \\ & = (u_{(1)} \sharp (I_{(2)} v^* \otimes u_{(2)} I_{(1)}) ) \otimes (I_{(1)} \sharp (I_{(2)} \otimes w)). \end{align} Here we use the coaction of $T(V)$ on $T(V^* \otimes V)$ given by \begin{align} T(V) & \to T(V) \otimes T(V^* \otimes V) \\ v & \mapsto I_{(1)} \otimes (I_{(2)} \otimes v) = v_{(0)} \otimes v_{(1)}. \end{align}

If we want $T(V) \rtimes T(V^* \otimes V)$ to be a bialgebra, then the comultiplication is a homomoprhism of algebras and we should have \begin{align} \Delta( (1 \sharp (v^* \otimes w)) (u \sharp 1) ) = \Delta( 1 \sharp (v^* \otimes w) ) \Delta( u \sharp 1). \end{align} The left hand side is \begin{align} & \Delta( (1 \sharp (v^* \otimes w)) (u \sharp 1) ) \\ & = \Delta( (v^* \otimes I_{(1)})(u) \sharp (I_{(2)} \otimes w) ) \\ & = ( ((v^* \otimes I_{(1)})(u))_{(1)} \sharp ( I_{(2)} I_{(2)} \otimes ((v^* \otimes I_{(1)})(u))_{(2)} I_{(1)} ) ) \otimes (I_{(1)} \sharp (I_{(2)} \otimes w)). \quad (1) \end{align} The right hand side is \begin{align} & \Delta( 1 \sharp (v^* \otimes w) ) \Delta( u \sharp 1) \\ & = ( (1 \sharp (I_{(2)} v^* \otimes I_{(1)}) ) \otimes (I_{(1)} \sharp (I_{(2)} \otimes w)) ) ( (u_{(1)} \sharp (I_{(2)} \otimes u_{(2)} I_{(1)}) ) \otimes (I_{(1)} \sharp (I_{(2)} \otimes 1)) ) \\ & = ( (1 \sharp (I_{(2)} v^* \otimes I_{(1)}) ) (u_{(1)} \sharp (I_{(2)} \otimes u_{(2)} I_{(1)}) ) \otimes ( (I_{(1)} \sharp (I_{(2)} \otimes w)) (I_{(1)} \sharp (I_{(2)} \otimes 1)) ) \\ & = ( (I_{(2)} v^* \otimes I_{(1)})( u_{(1)} ) \sharp (I_{(2)} \otimes I_{(1)}) (I_{(2)} \otimes u_{(2)} I_{(1)}) ) \otimes ( I_{(1)} (I_{(2)} \otimes I_{(1)})(I_{(1)}) \sharp (I_{(2)} \otimes w) (I_{(2)} \otimes 1) ) \\ & = ( (I_{(2)} v^* \otimes I_{(1)})( u_{(1)} ) \sharp (I_{(2)} I_{(2)} \otimes I_{(1)} u_{(2)} I_{(1)} ) \otimes ( I_{(1)} (I_{(2)} \otimes I_{(1)})(I_{(1)}) \sharp (I_{(2)} I_{(2)} \otimes w) ). \quad (2) \end{align} But it seems that (1) and (2) do not equal?

My question is: are there some multiplication and comultiplication in $T(V) \rtimes T(V^* \otimes V)$ such that $T(V) \rtimes T(V^* \otimes V)$ is a bialgebra? Any references, comments will be greatly appreciated!

$\endgroup$
1
  • $\begingroup$ The general construction that one might attempt to use is the Majid--Radford biproduct $B\rtimes H$ of a bialgebra $B$ which lives in the category of modules over another Hopf algebra $H$. It seems to me that in your example the coaction of $T(V)$ on $T(V^*\otimes V)$ does not make the bialgebra structure a morphism of comodules. It this case their construction would not apply. $\endgroup$ Jun 21, 2016 at 14:43

1 Answer 1

1
$\begingroup$

As @Zahlendreher said, the good language is that of bialgebras and comodule algebras so that you can do the bicross-product.

Identifying $V^*\otimes V\cong(End(V))^*=:C$ then $V$ is a right $C$-comodule, because $V$ is clearly an $End(V)$ left module. The coaction that you use ("$v_i\mapsto \sum_k v_k\otimes E_{ki}^*$") is precisely this one.

Then $B:=T(C)$ is a bialgebra because $C$ is a coalgebra, and $V$ is a $B$-comodule because it is a $C$-comodule. Then, $A:=T(V)$ is not a $C$-comodule but it is a $B$-comodule, moreover it is a $B$-comodule-algebra.

In general, if $B$ is a bialgebra and $A$ a $B$-comodule algebra then $A\# B$ is just an algebra. If you want it to be a bialgebra using the bicros-product construction then you need $A=T(V)$ to be a bialgebra in the Yetter-Drinfel'd category associated to $B=T(End(V)^*)$. I think this is a quite extra structure. I could be surprised if you can do it without considering some quotient of $T(C)$ associated to that extra structure. If you consider $A=TV$ as coalgebra with deconcatenation it doesn't work.. you get "infinitesimal bialgebras" in the sense of Marcelo Aguiar.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.