Let $k$ be a field, $\mathsf{Vect}$ the category of finite dimensional vector spaces, and $\mathsf{C} = Fun(\mathsf{Vect},\mathsf{Vect})$ the abelian category of pointed endofunctors (sending $0$ to $0$). The question has 2 parts, one for characteristic $0$ fields and the other for characteristic $p>0$.

Assume $char(k)=0$

Definition: An endofunctor $F \in \mathsf{C}$ is called polynomial if it can be expressed as a sum

$$F(V) = \bigoplus_{n \in \mathbb{N}}P(n) \otimes_{k[S_n]}V^{\otimes n}$$

Where $P(n)$ are $k$-linear representations of the symmetric group s.t. $P(n)=0$ for $n \gg 0$.


1. Is there a characterization of the polynomial functors among all endofunctors?

2. What's an example of a non-polynomial functor in $\mathsf{C}$?

3. Is $\mathsf{C}$ a semi-simple category? If so is there an explicit description of it? (hopefully identifying it with something built out of categories of representations).

Assume $char(k) = p \gt 0$

I think that in this case $\mathsf{C}$ will probably never be semi-simple (my intuition being that the exact sequence $0 \to \bigwedge^2 \to \otimes^{2} \to S^2 \to 0$ doesn't seem to be split in general) and so there looks to be some interesting (in my opinion) homological algebra going on here.

Question: Is $\mathsf{C}$ well understood in this case? By which I mean that there's a full list of all isomorphism classes and all Ext groups can be calculated in principle.

And finally, where can I read more about this topic?

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    $\begingroup$ Example of a non-polynomial functor: the exterior algebra $V \mapsto \bigwedge V$ (note that this is finite-dimensional). It can be expressed as a sum $\bigoplus P(n) \otimes_{k[S_n]} V^{\otimes n}$, but the $P(n)$ are all nonzero; in fact they equal the sign representation. (You probably need characteristic $0$ for this description of the $P(n)$.). To see that it's not polynomial, look at the growth of the dimension. $\endgroup$ May 16, 2017 at 21:17
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    $\begingroup$ The "short exact sequence" you write down in the char p case only makes sense for $r=2$. $\endgroup$ May 16, 2017 at 21:31
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    $\begingroup$ @Qfwfq Determinant isn't even a functor - what do you do with maps between spaces of different dimension? $\endgroup$ May 16, 2017 at 23:13
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    $\begingroup$ @SaalHardali: it's the natural quotient map $u \otimes v \mapsto u \wedge v$. Wedge is not alternating tensors; it's the quotient of $T^2$ by the space spanned by vectors of the form $x \otimes x$. $\endgroup$ May 17, 2017 at 3:13
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    $\begingroup$ In characteristic two one sees that your definition of polynomial endofunctors is not very good: In the exact sequence $0\to \wedge^2\to\otimes^2\to S^2\to 0$ the $\wedge^2$ does not satisfy your definition. The definitions by Friedlander and Suslin are better. Also note that their category is no full subcategory of $Fun(\rm Vect,Vect)$. $\endgroup$ May 22, 2017 at 8:53

1 Answer 1


There are a couple of equivalent ways to characterise polynomial functors. One is to say that $F$ is a polynomial functor of degree $n$ if the function $$\hom(U, V)\to \hom(F(U), F(V))$$ is polynomial of degree n. A second, equivalent formulation goes via cross-effects: A functor is polynomial of degree $n$ if its $n+1$-th cross-effect vanishes. As far as I know, these definitions were first introduced in the classic paper

Eilenberg, Samuel; MacLane, Saunders, On the groups $H(\Pi,n)$. II, Ann. Math. (2) 60, 49-139 (1954). ZBL0055.41704.

In this paper the concept of cross-effect was introduced, and the equivalence of the two definitions proved.

In characteristic $0$, the category of polynomial functors is equivalent to the direct sum of representation categories of symmetric groups. In particular, it is semi-simple. This seems to be due to MacDonald

Macdonald, I.G., Polynomial functors and wreath products, J. Pure Appl. Algebra 18, 173-204 (1980). ZBL0455.18002.

In positive characteristic, the category is not semi-simple. The following papers introduced the systematic study of functor categories between vector spaces over $\mathbb F_p$, and in particular related them to modules over the Steenrod algebra

Henn, Hans-Werner; Lannes, Jean; Schwartz, Lionel, The categories of unstable modules and unstable algebras over the Steenrod algebra modulo nilpotent objects, Am. J. Math. 115, No.5, 1053-1106 (1993). ZBL0805.55011.

Kuhn, Nicholas J., Generic representations of the finite general linear groups and the Steenrod algebra. I, II, and III, Am. J. Math. 116, No.2, 327-360 (1994). ZBL0813.20049.

There has been a lot of work on the homological algebra in this category, with striking applications in particular by Friedlander and Suslin to the homology of general linear group (Edit: for this application one needs the category of strict polynomial functors - a variant of the "naive" notion of polynomiality). However I don't think there is an effective procedure known for calculating the ext groups between general polynomial functors. For an up to date survey of the subject you may want to consult, for example, the following book

Franjou, Vincent, Touze, Antoine Lectures on functor homology., , Proceedings of the conference on functor homology, Nantes, France, April 2012. Cham: Birkh\"auser/Springer (ISBN 978-3-319-21304-0/hbk; 978-3-319-21305-7/ebook). Progress in Mathematics 311, 7-39 (2015). ZBL1338.18011.

Lastly let me mention that Kuhn also proved that the some results about the category of polynomial functors between vector spaces over different characteristics is in fact semi-simple. See his comment below.

Kuhn, Nicholas J., Generic representation theory of finite fields in nondescribing characteristic, Adv. Math. 272, 598-610 (2015). ZBL1354.18001.

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    $\begingroup$ Friedlander and Suslin distinguish between polynomial functors, defined in terms of cross effects, and strict polynomial functors, defined in terms of polynomial maps hom(U,V)→hom(F(U),F(V)). $\endgroup$ May 17, 2017 at 8:29
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    $\begingroup$ I did not find the proof of equivalence of the two notions in the paper of Eilenberg and MacLane. $\endgroup$ May 19, 2017 at 7:59
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    $\begingroup$ But I found a counterexample in the Habilitation thesis of Antoine Touzé. It goes like this: Tensor each complex vector space with the bimodule $M=\Bbb C$ with ordinary multiplication on the right, but multiplication by complex conjugate on the left. $\endgroup$ May 19, 2017 at 9:23
  • $\begingroup$ I had in mind Theorem 9.11 in their paper. It seems to say exactly that the two notions are equivalent - did I misread it? I am sorry, I don't understand the counterexample yet. Is it giving an a functor that is polynomial according to one of the notions but not the other? $\endgroup$ May 19, 2017 at 12:15
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    $\begingroup$ Greg has slightly messed up his description of my `nondescribing characteristic' result. I show that if $\mathbb F$ is a finite field, and $k$ is a field of different characteristic, then the abelian category of functors from $\mathbb F$--vector spaces to $k$--vector spaces splits as the product over $n$ of the categories of $k[GL_n(\mathbb F)]$--modules. This is only semisimple if $k$ has characteristic 0. I also note that NO nonconstant functors in this category are polynomial. $\endgroup$ Jun 15, 2017 at 21:57

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