Antipode for  quantum matrices. 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?
 A: If I remember correctly, the antipode for GL_q(2) (or GL_q(n)) is a sort of "inverse matrix" defined using the inverse of the quantum determinant det_q((a, b \ c, d)) = ad - q*bc.
Since the antipode is unique when it exists, restricted to GL_q (which is dense in M_q) it should coincide with this one, which is not well defined in M_q.
Another way, assume the antipode is defined and take classical limit q-->1, then you would have an antipode in the bialgebra of coordinates of usual matrices, but that is not possible since M_n is not a group scheme.
A: 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".
