On the automorphism group of a linear algebraic monoid
Proceedings of the American Mathematical Society
Let 5 be a connected regular monoid with zero. It is shown that an automorphism of S is inner if and only if it sends each idempotent of S to a conjugate idempotent. In the language of semigroup theory, the automorphism group of S maps homomorphically into the automorphism group of the finite lattice of J-classes of S, and the kernel of this homomorphism is the group of inner automorphisms of S. In particular, if the Jj-classes of S are linearly ordered, then every automorphism of 5 is inner.
... sm of 5 is inner. Throughout this paper Z+ will denote the set of all positive integers and K an algebraically closed field. i3ïtn( K ) denotes the monoid of all « X « matrices over K. GL(n, K) denotes the group of units of 91tn(/v), and SL(n, K) the group of matrices of determinant 1 in zJUJ^K). We will follow the notation and terminology of [2,4] concerning linear algebraic monoids. Let S be a connected algebraic monoid with group of units G. By an automorphism of S is meant a semigroup automorphism a of 5 such that both a and a"1 are polynomial maps. An automorphism a of S is inner if there exists g E G such that a(a) = g~lag for all a OE S. We let ^(S) denote the finite lattice of all regular ^-classes of S, and E(S) the partially ordered set of all idempotents of 5. It follows from the work of the author [4, 5] and Renner [7, 8] that 5 is regular if and only if the closure of the radical of G is a Clifford semigroup. In particular, if S has a zero then S is regular if and only if G is a reductive group. Theorem 1. Let S be a connected regular monoid with zero and o an automorphism of S. Then a is an inner automorphism of S if and only if o(J) = J for all J E %(5) (i.e. a(e) is a conjugate of e for all e E E(S)). Proof. Suppose a(J) for all J E %(5). We must show that a is inner.