Is $T(V) \rtimes T(V^* \otimes V)$ a bialgebra?

Question

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!

Answer 1

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.