From: Dave Rusin Date: Tue, 28 Jul 1998 11:02:15 -0500 (CDT) To: mjoao@alf1.cii.fc.ul.pt Subject: Re: Vector spaces Newsgroups: sci.math.research The answer to your question is "yes" if "automorphism" means a sufficiently narrow class of objects. I think it must, although I didn't succeed in proving that; perhaps you can finish the proof. But the real question is in classifying the automorphisms Z of J, not so much investigating their relationship with End(V). In article you write: >Let V be a vector space (finite or infinite dimension) and let J be the >set of all endomorphisms of V whose rank is at most one. (That is, if k >belongs to J then Dim( k(V) ) leq 1). So J is the set of elements of the form ( v tensor phi) in (V tensor V*) where V* is the dual space. If V is finite dimensional, we can pick a basis and then write J as the set of matrices v1 * v2' for arbitrary column vectors v1, v2. In both cases, the constituents of each element of J are only defined up to a common scalar. This describes J as a set in a vector space. The multiplicative structure is given by (v tensor phi) . (w tensor psi) = (phi(w)).v tensor psi, or in matrix terms, (v1*v2') . (v3*v4') = (v2 dot c3)v1 * v4' >Denote by End(V) the semigroup of all endomorphisms of V. That's M_n(R) in the finite-dimensional-with-basis case, of course. In the case of a general vector space I suppose you will want to restrict to continuous endomorphisms? (Here and in the definition of J.) >The set J is closed under composition of endomorphisms and hence it is a >semigroup. In fact it is a right ideal in the sense that for all f in >End(V) and every k in J, we have that fk belongs to J. > >So, for every f in End(V), we can define a mapping as follows: > >G_f : J --> J ; G_f (k) = fk. So G_f( v tensor phi ) = f(v) tensor phi . In the matrix case this says G_M ( v1 * v2' ) = (M v1) * v2' for each square matrix M. >On the other hand, as J is a semigroup, we can fix Z, an automorphism of >the semigroup J. It seems to me this is where your question really lies. You need to ask, what _are_ the automorphisms Z of the semigroup J ? I believe the correct answer is, they are of the form (v tensor phi) -> ( L(v) tensor (phi o L^(-1)) ). for some automorphism L of V; in matrix terms this says they are of the form (v1 * v2') -> (A v1) * ((A^(-1))'v2)' for some invertible matrix A. I _can_ show that the automorphisms must have the form (v tensor phi) -> (F(v) tensor G(phi) ) for some _functions_ F : V -> V, G : V^* -> V^*; I can prove more too, but I didn't see how to prove F must be _linear_. However, let me assume this to be true. >Now, for every f in End(V) define a mapping as follows: > >F_f : J --> J ; F_f (k) = Z (f Z^-1 (k)). > >(Z^-1 is the inverse of the automorphism Z). >Observe that Z^-1 (k) is an element of J and hence f Z^-1(k) is the >composition of two endomorphisms of V. So this one would be F_f ( v tensor phi ) = ( F(f(F^(-1)(v))) tensor G^(-1)(G(phi)) ) and of course the second factor is just phi; assuming F is linear, the first factor is the image of v under the linear map F o f o F^(-1). >So, my question is this: for every f in End(V) is there a g in End(V) such >that F_f = G_g ? So this seems to be "yes": g = F^(-1) o f o F is also linear. Thus the work remaining is to show the automorphisms of J have the desired form. Here's a sketch of a partial proof (you'll notice I blithely assume all real values are nonzero, etc.) Suppose we are given an automorphism of J. We may write it in the form (v tensor phi) -> (F(v,phi) tensor G(v,phi)), where F and G are only well-defined up to a common scalar multiple; in particular, their images [F(v,phi)] in P^(V) and [G(v,phi)] in P^1(V*) _are_ well-defined. If this is indeed a homomorphism of algebras, then from the product law (v tensor phi) . (w tensor psi) = (phi(w)).v tensor psi we see [G(w,psi)] = [G((phi(w).v, psi)] which means, since w and v are arbitrary, that [G(w,psi)] depends on psi alone. Writing the product law as (v tensor phi) . (w tensor psi) = v tensor (phi(w)).psi we see likewise that [F(v,phi)] depends only on v. Picking representative vectors in each of the lines [F(v)] and [G(psi)] we then see the automorphism has the form v tensor phi -> ( F(v) tensor t(v,phi).G(phi) ) where, in order to be well-defined, we need for every c some constants with F(c v) = lambda(c,v,phi) F(v) G(1/c phi) = 1/lambda(c,v,phi) t(v,phi)/t(cv,phi/c) G(phi) The first equation forces lambda to be independent of phi and by symmetry it is also independent of v. Then the defining property of lambda makes it a homomorphism R^*-> R^*; if we assume continuity we have lambda(c) = c^r for some real r . Next let's get rid of the t's. Using the product law again, we get (F(v) tensor t(v,phi) G(phi)) . (F(w) tensor t(w,psi) G(psi)) = (F( phi(w).v ) tensor t(phi(w).v,psi) G(psi) ); the left side becomes t(v,phi) G(phi)( F(w) ) . F(v) tensor t(w,psi) G(psi) = t(v,phi) G(phi)( F(w) ) t(w,psi). F(v) tensor G(psi) so the multiplicativity is satisfied iff t(v,phi) G(phi)( F(w) ) t(w,psi). F(v) = t(phi(w).v,psi) F( phi(w).v ) Both sides are multiples of the same vector; we get this equation among scalars: G(phi)( F(w) ) t(v,phi) t(w,psi) = t(phi(w).v,psi) ( phi(w) )^r Fix a pair (w0, phi0) with phi0(w0)=1; then if G(phi0)( F(w0) ) = c, this equation shows t(v,psi) = c t(v,phi0) t(w0, psi) for all v, psi. So if we re-express our automorphism as (v,phi) -> (F'(v) tensor t'(v,psi) G'(psi) ) where G'(phi) = G(phi)*t(w0,phi)*c and F'(v) = F(v)*t(v,phi0), then t'(v,phi) = t(v,phi)/t(v,phi0)/t(w0,phi) = 1 is _constant_. So we have shown that our automorphism may be written in the desired form (v tensor phi) -> (F(v) tensor G(w)) for some maps F:V->V, G:V*->V* which satisfy the additional property that there is a constant r with (*)... G(phi)( F(w) ) = (phi(w))^r for all phi and w. From this I would like to show that F must be linear. Actually I think it's then also true that r=1, G is linear, and G = F^(-1)^* (that is, G(phi) = phi o F^(-1) ), although as the earlier part of the proof shows, this information is not necessary for your purposes. Note that F need _not_ be linear if the above equation (*) is only assumed to hold for a _single_ phi; e.g. if V = R^2 and F(w1,w2) = (w1^3, w2^3) then (*) holds (with r=3) for both phi=projection to 1st coordinate and phi=projection to second coordinate, as long as G(phi)=phi for these two functionals phi. So to prove of the linearity of F we will have to use the fact that (*) is to hold for _all_ phi. I don't see how to do this. dave