3
$\begingroup$

this is just a basic linear algebra question, which I do not have a idea.

Suppose that we have a group homomorphism $\phi:GL_n(\mathbb{R})\rightarrow GL_m(\mathbb{R})$. Is there always a functor $F:Vect_n \rightarrow Vect_m$ where for any $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$, $F(f)=\phi(f)$? Clearly if there is a canonical choice of a basis for each vector space then we can define a functor. But I am thinking of defining a functor in an intrinsic way, like $V^{\otimes m}$, $\textrm{Sym}^nV$, $\Lambda^n V$ etc.. Is it always possible to construct $F$ is this way?

Thank you.

$\endgroup$
1
  • 2
    $\begingroup$ What is $\text{Vect}_n$? Anyway, the keyword you want is "Schur functor." IIRC it is necessary and sufficient that $\phi$ is algebraic. Without this condition you get problems with representations like irrational powers of the absolute value of the determinant. $\endgroup$ Oct 28, 2013 at 7:31

1 Answer 1

4
$\begingroup$

If we assume $Vect_n$ means the category whose objects are $n$-dimensional real vector spaces and whose morphisms are $\mathbb{R}$-linear maps, then your question doesn't quite make sense, because the notation $\phi(f)$ requires $f$ to be invertible. If instead we assume $Vect_n$ means the core of the previous category, i.e., morphisms are isomorphisms of real vector spaces, then a positive answer to your question follows from global choice. This is essentially your reasoning involving a basis: we may define a functor on a skeleton of the source category and choose a noncanonical extension.

The usual method for expressing naturality of a representation is to describe it as the restriction of an endofunctor on the category of all finite dimensional real vector spaces, instead of restricting to a single dimension. The representations you get are precisely those defined by polynomial equations (i.e., algebraic representations), and the functors are direct sums of Schur functors. (Edit: A previous version of this paragraph restricted to additive endofunctors, but these are uniquely determined by what happens to the one dimensional vector space, and hence a bit too restrictive.)

In general, you can get many other representations. For example, we may choose a discontinuous automorphism $\psi$ of $GL_1(\mathbb{R})$, and take the tensor product of any nice map $GL_n \to GL_m$ with the composite of determinant with $\psi$, to get a rather awful object. Even with continuous homomorphisms, one can get things like $A \mapsto \begin{pmatrix} 1 & s \log |\det A | \\ 0 & 1 \end{pmatrix}$ for real numbers $s$, so I doubt there is a good classification.

$\endgroup$
6
  • $\begingroup$ Just endofunctors, I think. Most of them are not additive. $\endgroup$ Oct 28, 2013 at 17:52
  • $\begingroup$ I think I believe this statement over $\mathbb{R}$ but I'm a little confused about other fields. First, weird things happen in positive characteristic involving divided powers. Second, even in characteristic zero it seems that the category of additive endofunctors is the category of representations of the underlying field $k$ into the rings $M_n(k)$ (all regarded as rings, not $k$-algebras). In order for the stated equivalence to hold every such representation must be conjugate to the diagonal embedding but I don't see how to rule out the possibility of pathological embeddings. $\endgroup$ Oct 28, 2013 at 17:58
  • $\begingroup$ For a simple example, complex conjugation gives an endofunctor of finite-dimensional complex vector spaces which is not naturally isomorphic to the identity (the only Schur functor which it could possibly be), e.g. by taking traces. $\mathbb{R}$ is somewhat special in that it has no nontrivial endomorphisms. $\endgroup$ Oct 28, 2013 at 18:53
  • $\begingroup$ Thanks for your answer, Carnahan. My question was not well-posed, and your interpretation seems to be good. $\endgroup$
    – ECortez
    Oct 30, 2013 at 1:36
  • $\begingroup$ Also, I think so, that it should be endofunctors, and additive endofunctors are just $M_n(k)$s. I am wondering there is any other notion of naturallity, that containing things like tensor product. Nevertheless, thank everyone for indicating Schur functors. $\endgroup$
    – ECortez
    Oct 30, 2013 at 1:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.