Am I right in assuming that one cannot define an antipode for $M\_q(n)$ the bi-algebra of $nXn$ quantum matrices? If so, does anyone know a proof?



Antipode for  quantum matrices.
I think a proof could be provided like this.


Claim: Let $H$ be a bi-algebra. Then an antipode S on H, if it exists, is unique.


Proof: If H is a bialgebra, then the set of linear maps from H to H inherits an algebra structure given by f\*g(x)=f(x\_1)g(x\_2), where \Delta(x)=x\_1\ot x\_2 is Sweedler's notation. This is an associative algebra structure, as can be easily checked. Moreover the axioms of the antipode assert that S is a left and right inverse to the identity map id:H-->H.


Now a general fact about associative algebras is that inverses are unique when the exist:
if xy=id and zx=id, y=(zx)y = z(xy)=z. //


Now suppose that Mat\_q(n) has an antipode, call it S'. Then, this extends to a map of its localization O\_q(GL\_n), so long as S'(det\_q)\neq 0[GAP - see below]. This is guaranteed, though, because S' was assumed to be bijective in order to be an antipode on Mat\_q(n). Okay, so now you have two antipodes on O\_q(GL\_n), but the previous claim implies they are equal. This implies that S' which you started with originally required inverting the quantum determinant in order to be defined in the first place, which is a contradiction, since quantum determinant is not invertible in Mat\_q(n). //


By the way, Mat\_q(n) is a bialgebra in two ways. First of all, it is a bialgebra coming from a coproduct coming viewing matrices as a monoid under composition. It is also, however, a bialgebra with respect to a different coproduct, coming from regarding matrices as a group under addition. The latter notion also quantizes, and so Mat\_q(n) becomes what is called a "braided bialgebra". This means it's not a bialgebra in the usual sense, but it is a bialgebra in the braided tensor category of U\_q(gl\_n)-modules, which just means that when you write down the axiom for compatibility of product and coproduct, you realize that there are some tensor flips that need to be replaced with braiding. If you regard it in this sense, then you would expect it to have a (braided) antipode, and indeed it does.


All of this can be read, for instance in Majid's "Foundations of Quantum Groups", or also Klymik and Schüdgen's "Quantum Groups and Their Representation Theory".


