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## The representation theory of the additive group scheme

In this post we’ll describe the representation theory of the additive group scheme $\mathbb{G}_a$ over a field $k$. The answer turns out to depend dramatically on whether or not $k$ has characteristic zero.

Preliminaries over an arbitrary ring

(All rings and algebras are commutative unless otherwise stated.)

The additive group scheme $\mathbb{G}_a$ over a base ring $k$ has functor of points given by the underlying abelian group functor

$\displaystyle \mathbb{G}_a(-) : \text{CAlg}(k) \ni R \mapsto (R, +) \in \text{Ab}.$

This functor is represented by the free $k$-algebra $k[x]$ together with the Hopf algebra structure given by the comultiplication

$\displaystyle \Delta(x) = x \otimes 1 + 1 \otimes x$

and antipode $S(x) = -x$, and so it is an affine group scheme whose underlying scheme is the affine line $\mathbb{A}^1$ over $k$. From a Hopf algebra perspective, this Hopf algebra is special because it is the free Hopf algebra on a primitive element.

A representation of $\mathbb{G}_a$ can be described in a few equivalent ways. From the functor of points perspective, we first need to describe the functor of points perspective on a $k$-module: a $k$-module $V$ has functor of points

$\displaystyle V(-) : \text{CAlg}(k) \ni R \mapsto R \otimes_k V \in \text{Ab}$

making it an affine group scheme if $V$ is finitely generated projective (in which case its ring of functions is the symmetric algebra $S(V^{\ast})$ on the dual of $V$) but not in general, for example if $k$ is a field and $V$ is an infinite-dimensional vector space. Note that if $V = k$ we recover $\mathbb{G}_a$, and more generally if $V = k^n$ we recover $\mathbb{G}_a^n$.

A representation of $\mathbb{G}_a$ over $k$ is, loosely speaking, a polynomial one-parameter group of automorphisms of a $k$-module. The simplest nontrivial example is

$\mathbb{G}_a \ni x \mapsto \left[ \begin{array}{cc} 1 & x \\ 0 & 1 \end{array} \right]$

which defines a nontrivial action of $\mathbb{G}_a$ on $k^2$.

Formally, we can define an action of $\mathbb{G}_a$ on a $k$-module $V$ as an action of the functor of points of the former on the functor of points of the latter which is linear in the appropriate sense. Explicitly, it is a natural transformation with components

$\displaystyle \eta_R : \mathbb{G}_a(R) \times V(R) \to V(R)$

such that each $\eta_R$ defines an $R$-linear action of the group $\mathbb{G}_a(R) \cong (R, +)$ on the $R$-module $V(R) \cong R \otimes_k V$ in a way which is natural in $R$. So we can think of the parameter $x$ in the one-parameter group above as running over all elements of all $k$-algebras $R$.

$\eta$ can equivalently, by currying, be thought of as a natural transformation of group-valued functors from $\mathbb{G}_a$ to the functor

$\displaystyle \text{Aut}(V)(-) : \text{CAlg}(k) \ni R \mapsto \text{Aut}_R(V \otimes_k R)$

even though the latter is generally not representable by a scheme, again unless $V$ is finitely generated projective; note that if $V = k^n$ then $\text{Aut}(V)$ is an affine group scheme, namely the general linear group $GL_n$.

The advantage of doing this is that we can appeal to the Yoneda lemma: because $\mathbb{G}_a$ is representable, such natural transformations correspond to elements of the group

$\text{Aut}_{k[x]}(V \otimes_k k[x]) \subseteq \text{End}_{k[x]}(V \otimes_k k[x]) \cong \text{Hom}_k(V, V \otimes_k k[x])$

at which point we’ve finally been freed of the burden of having to consider arbitrary $R$. Writing down the conditions required for a map $V \to V \otimes_k k[x]$ to correspond to an element of $\text{Aut}_{k[x]}(V(k[x]))$ giving a homomorphism of group-valued functors $\mathbb{G}_a \to \text{Aut}(V)$, we are led to the following.

Definition-Theorem: A representation of $\mathbb{G}_a$ over $k$ is a comodule over the Hopf algebra of functions $\mathcal{O}_{\mathbb{G}_a} \cong k[x]$.

This is true more generally for representations of any affine group scheme.

Let’s get somewhat more explicit. A comodule over $k[x]$ is in particular an action map $\alpha : V \to V \otimes_k k[x]$. Isolating the coefficient of $x^n$ on the RHS, this map breaks up into homogeneous components

$\displaystyle \alpha_n : V \to V \otimes_k x^n$

which each correspond to an element of $\text{End}_k(V)$, which we’ll call $M_n$. The entire action map can therefore be thought of as a power series

$\displaystyle M(x) = M_0 + M_1 x + M_2 x^2 + \dots$

and the condition that $\alpha$ is part of a comodule structure turns out to be precisely the condition that 1) $M(0) = M_0$ is the identity, 2) that we have

$\displaystyle M(x + y) = M(x) M(y)$

as an identity of formal power series, and 3) that for any $v \in V$, $M_i v = 0$ for all but finitely many $i$. Which of these is nonzero can be read off from the value of $\alpha(v)$, which takes the form

$\displaystyle \alpha(v) = M(x)(v) = \sum_{i \ge 0} M_i v \otimes x^i$.

Equating the coefficients of $x^i y^j$ and of $x^j y^i$ on both sides of the identity $M(x + y) = M(x) M(y)$ (the two coefficients are equal on the LHS, hence must be equal on the RHS) gives

$\displaystyle {i + j \choose i} M_{i+j} = M_i M_j = M_j M_i$

from which it follows in particular that the $M_i$ commute. This is necessary and sufficient for us to have $M(x + y) = M(x) M(y)$, so we can say the following over an arbitrary $k$:

Theorem: Over a ring $k$, representations of $\mathbb{G}_a$ can be identified with modules $V$ over the divided power algebra

$\displaystyle k[M_1, M_2, \dots] / \left( {i + j \choose i} M_{i + j} = M_i M_j \right)$

which are locally finite in the sense that for any $v \in V$ we have $M_i v = 0$ for all but finitely many $i$. Given the action of each $M_i$, the corresponding action of $\mathbb{G}_a$ is

$\displaystyle \sum_i M_i x^i : V \mapsto V \otimes_k k[x]$.

If $k$ is torsion-free as an abelian group, for example if $k = \mathbb{Z}$, then the divided power algebra can be thought of as the subalgebra of $(k \otimes \mathbb{Q})[x]$ spanned by $\frac{x^i}{i!}$, where the isomorphism sends $M_i$ to $\frac{x^i}{i!}$. This is because the key defining relation above can be rewritten $(i + j)! M_{i + j} = (i! M_i) (j! M_j)$ if there is no torsion.

These representations should really be thought of as continuous representations of the profinite Hopf algebra given as the cofiltered limit over the algebras spanned by $M_1, M_2, \dots M_n$ for each $n$, with the quotient maps given by setting all $M_i$ past a certain point to zero. This profinite Hopf algebra is dual to the Hopf algebra $k[x]$ of functions on $\mathbb{G}_a$, and we can more generally relate comodules over a coalgebra $C$ to modules over a profinite algebra dual to it in this way under suitable hypotheses, for example if $k$ is a field.

In characteristic zero

If $k$ is a field of characteristic zero, or more generally a $\mathbb{Q}$-algebra, then the divided power algebra simplifies drastically, because we can now divide by all the factorials running around. Induction on the relation ${i + j \choose i} M_{i + j} = M_i M_j$ readily gives

$\displaystyle M_i = \frac{M_1^i}{i!}$

and we conclude the following.

Theorem: Over a $\mathbb{Q}$-algebra $k$, representations of $\mathbb{G}_a$ can be identified with endomorphisms $M_1 : V \to V$ of a $k$-module which are locally nilpotent in the sense that, for any $v \in V$, we have $M_1^i v = 0$ for all but finitely many $i$. Given such an endomorphism, the corresponding action of $\mathbb{G}_a$ is

$\displaystyle \exp (M_1 x) : V \to V \otimes_k k[x]$.

The corresponding profinite Hopf algebra is the formal power series algebra $k[[\partial]]$ in one variable, with comultiplication $\Delta(\partial) = \partial \otimes 1 + 1 \otimes \partial$. If $k$ is a field of characteristic zero, this means we can at least classify finite-dimensional representations of $\mathbb{G}_a$ using nilpotent Jordan blocks.

Example. The representation $\left[ \begin{array}{cc} 1 & x \\ 0 & 1 \end{array} \right]$ we wrote down earlier corresponds to exponentiating the Jordan block $\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$.

Example. Any coalgebra has a “regular representation,” namely the comodule given by its own comultiplication. The regular representation of $\mathbb{G}_a$ is the translation map

$\displaystyle \mathbb{G}_a \ni x \mapsto (f(y) \mapsto f(y+x)) \in \text{Hom}_k(k[y], k[y] \otimes_k k[x])$.

and the corresponding locally nilpotent endomorphism is the derivative $f \mapsto \partial f$ (hence the use of $\partial$ above). More generally, locally nilpotent derivations on a $k$-algebra $R$ correspond to actions of $\mathbb{G}_a$ on the affine $k$-scheme $\text{Spec } R$.

The regular representation has subrepresentations given by restricting attention to the polynomials of degree at most $n$ for each $n$. These subrepresentations are finite-dimensional, and are classified by the $(n+1) \times (n+1)$ nilpotent Jordan block, which follows from the fact that they have a one-dimensional invariant subspace given by the constant polynomials.

In positive characteristic

In positive characteristic the binomial coefficient ${i + j \choose i}$ in the relation ${i + j \choose i} M_{i + j} = M_i M_j$ is sometimes zero, so things get more complicated. For starters, the same induction as before shows that if $k$ is a field of characteristic $p$, or more generally an $\mathbb{F}_p$-algebra, we have

$\displaystyle M_i = \frac{M_1^i}{i!}, i < p$.

However, when we try to analyze $M_p$, we find that we can put no constraint on it in terms of smaller $M_i$ but instead that $M_1^p = 0$. In fact we have $M_i^p = 0$ for all $i$, because the identity

$\displaystyle M(x + y) = M(x) M(y)$

gives by induction

$\displaystyle M(px) = M(x)^p = \sum_{i \ge 0} M_i^p x^{ip} = 1$

(here we use the fact that we know the $M_i$ commute). So things are not determined just by $M_1$; we at least also need to know $M_p$. Note that $M_1^p = 0$ guarantees that the exponential $\exp (M_1 x)$ still makes sense in characteristic $p$, because it’s no longer necessary to divide by factorials divisible by $p$.

Continuing from here we find that

$\displaystyle M_{i_0 + i_1 p} = \frac{M_1^{i_0}}{i_0!} \frac{M_p^{i_1}}{i_1!}, i_0, i_1 < p$

using the fact that ${ip \choose p} M_{ip} = M_p^i$ and ${ip \choose p}$ is not divisible by $p$ for $i < p$ by Kummer’s theorem, as well as Lucas’s theorem which gives more precisely ${ip \choose p} \cong i \bmod p$. This gets us up to $p^2 - 1$ and then when we try to analyze $M_{p^2}$ we find that we cannot relate it to any of the smaller $M_i$, because ${p^2 \choose i}$ is always divisible by $p$ (using either Kummer’s or Lucas’s theorems), and that $M_p^p = 0$, which we already knew.

The general situation is as follows. We can write $M_n$ as a product of $M_i, i < n$ precisely when we can find $1 \le i \le n-1$ such that ${n \choose i}$ is nonzero $\bmod p$. Kummer’s theorem implies that this is possible precisely when $n$ is not a power of $p$, from which it follows that each $M_n$ is a product of $M_{p^i}$. We can be more precise as follows. Write $n$ in base $p$ as

$\displaystyle n = i_0 + i_1 p + \dots + i_k p^k.$

Then induction gives

$\displaystyle {i_0 + i_1 p + \dots \choose i_0, i_1 p, \dots } M_n = M_{i_0} M_{i_1 p} \dots M_{i_k p^k}$

where the LHS is congruent to $1 \bmod p$ by a mild generalization of Lucas’s theorem and the RHS can be rewritten as above, giving

$\displaystyle M_n = \frac{M_1^{i_0}}{i_0!} \frac{M_p^{i_1}}{i_1!} \dots \frac{M_{p^k}^{i_k}}{i_k!}$.

This is enough for the $M_n$ to satisfy every relation defining the divided power algebra, and so we conclude the following.

Lemma: Over an $\mathbb{F}_p$-algebra $k$, the divided power algebra can be rewritten

$\displaystyle k[M_1, M_p, \dots]/(M_1^p, M_p^p, \dots)$.

Corollary: Over an $\mathbb{F}_p$-algebra $k$, representations of the additive group scheme $\mathbb{G}_a$ correspond to modules $V$ over the algebra $k[M_1, M_p, \dots]/(M_1^p, M_p^p, \dots)$ which are locally finite in the sense that for all $v \in V$, we have $M_{p^i} v = 0$ for all but finitely many $i$. Given the action of each $M_{p^i}$, the corresponding action of $\mathbb{G}_a$ is

$\displaystyle \exp (M_1 x + M_p x^p + \dots) : V \mapsto V \otimes_k k[x]$.

As before, we can equivalently talk about continuous modules over a suitable profinite Hopf algebra.

Example. We can classify all nontrivial $2$-dimensional representations over a field $k$ of characteristic $p$ as follows. There is some smallest $i$ such that the action of $M_{p^i}$ is nonzero. In a suitable basis, $M_{p^i}$ acts by a nilpotent Jordan block $J = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$, and all the other $M_{p^j}$ commute with it. A calculation shows that this implies that each $M_{p^j}$ must also have the form $\left[ \begin{array}{cc} 0 & c_j \\ 0 & 0 \end{array} \right]$ for some scalars $c_j$, and hence our representation has action map

$\displaystyle \exp \left( J (x^{p^i} + c_{i+1} x^{p^{i+1}} + \dots) \right) = \left[ \begin{array}{cc} 1 & x^{p^i} + c_{i+1} x^{p^{i+1}} + \dots \\ 0 & 1 \end{array} \right]$.

Contrast to the case of characteristic zero, where there is only one isomorphism class of nontrivial $2$-dimensional representation (given by $M_1 = J$).

Example. Consider again the regular representation given by the translation action of $\mathbb{G}_a$ on itself:

$\displaystyle \mathbb{G}_a \ni x \mapsto (f(y) \mapsto f(y+x)) \in \text{Hom}_k(k[y], k[y] \otimes_k k[x])$.

We find as before that $M_1 = \partial$ is the derivative. Now, in characteristic $p$ the derivative of a polynomial is well-known to have the curious property that $\partial^p = 0$; among other things this means that translation is no longer given by just exponentiating the derivative. However, since over $\mathbb{Z}$ we have

$\displaystyle \partial^p y^n = n(n-1) \dots (n-(p-1)) y^{n-p}$

we have, over $\mathbb{Z}$ and hence over any $k$, a well-defined differential operator

$\displaystyle \frac{\partial^p}{p!} y^n = {n \choose p} y^{n-p}$

(for $n \ge p$; otherwise zero) and this differential operator must be $M_p$; similarly for higher powers of $p$ we have

$\displaystyle M_{p^k} y^n = \frac{\partial^{p^k}}{(p^k)!} y^n = {n \choose p^k} y^{n-p^k}$

(for $n \ge p^k$; otherwise zero, as above).

Endomorphisms

Actually we should have expected an answer like this all along, or at least we should have known that we would also need to write down representations involving powers of Frobenius in addition to the obvious representations of the form $\exp (M_1 x)$. This is because in characteristic $p$, the Frobenius map gives a nontrivial endomorphism $\mathbb{G}_a \to \mathbb{G}_a$, and the pullback of any representation along an endomorphism is another representation. Pulling back along the Frobenius map sends $\exp (Mx)$ to $\exp (Mx^p)$. More generally, any sum of powers of the Frobenius map gives a nontrivial endomorphism of $\mathbb{G}_a$ as well.

The fact that this sort of thing doesn’t happen in characteristic zero means that $\mathbb{G}_a$ can’t have any non-obvious endomorphisms like the Frobenius map there. In fact we can say the following.

Proposition: Over a ring $k$, endomorphisms $\mathbb{G}_a \to \mathbb{G}_a$ are in natural bijection with additive polynomials, namely polynomials $f(x) \in k[x]$ such that $f(0) = 0$ and $f(x + y) = f(x) + f(y)$.

Proof. This is mostly a matter of unwinding definitions. By the Yoneda lemma, maps $\mathbb{G}_a \to \mathbb{G}_a$ (on underlying affine schemes, maps $\mathbb{A}^1 \to \mathbb{A}^1$) correspond to elements of $\mathbb{G}_a(k[x]) \cong k[x]$, hence to polynomials $f(x)$, and the condition that such a map corresponds to a group homomorphism is precisely the condition that $f(0) = 0$ and $f(x + y) = f(x) + f(y)$. $\Box$

Now let’s try to classify all such polynomials. Writing $f(x) = \sum f_i x^i$, equating the coefficient of $x^i y^j$ on both sides as before gives

$\displaystyle {i + j \choose i} f_{i + j} = \begin{cases} 0 & \mbox{if } i, j \ge 1 \\ f_i & \mbox{if } j = 0 \\ f_j & \mbox{if } i = 0 \end{cases}$

If $k$ is torsion-free, and in particular if $k$ is a $\mathbb{Q}$-algebra, then this implies that $f_n = 0$ for $n \ge 2$ (by setting $n = i + j$ where $i, j \ge 1$), and since $f(0) = f_0 = 0$ the only possible nonzero coefficient is $f_1$. So $f$ is scalar multiplication by the scalar $f_1$.

On the other hand, if $k$ is an $\mathbb{F}_p$-algebra, then it can again happen as above that $f_n$ doesn’t vanish if ${n \choose i}$ is always divisible by $p$, which as before happens iff $n$ is a power of $p$. In this case $f$ is a linear combination

$\displaystyle f(x) = \sum_i f_{p^i} x^{p^i}$

of powers of the Frobenius map, which is clearly an endomorphism. In summary, we conclude the following.

Proposition: If $k$ is torsion-free, the endomorphism ring of $\mathbb{G}_a$ is $k$ acting by scalar multiplication. If $k$ is an $\mathbb{F}_p$-algebra, the endomorphism ring of $\mathbb{G}_a$ is $k[F]$, with $F$ acting by Frobenius.

The endomorphisms generated by Frobenius can be used to write down examples of affine group schemes which only exist in positive characteristic. For example, the kernel of Frobenius is the affine group scheme $\alpha_p \cong \text{Spec } k[x]/x^p$ with functor of points

$\displaystyle \alpha_p : \text{CAlg}(k) \ni R \mapsto \{ r \in R : r^p = 0 \} \in \text{Ab}$.

### 5 Responses

1. This is a cool post; thanks!

I’m used to seeing divided power algebras appear as cohomology rings of various spaces, e.g. ΩS^3. Do you know if this perspective of modules over divided power algebras as 𝔾_a-representations leads to something interesting in algebraic topology?

• I have no idea; in general the appearance of divided power algebras in mathematics is mysterious to me, so I was glad to be able to get a handle on this one. I guess they are a sign that there is a dual polynomial algebra lurking around, which I guess here is the homology?

• There’s recently been the philosophy going around that divided powers are a consequence of working over the integers instead of the sphere spectrum. (Note that S –> Z is the group-completion of the \pi_0 projection Fin^\simeq –> \mathbb{N} of commutative monoid spaces, whose path components are exactly crushing symmetric groups.) I wouldn’t be surprised if the spectral affine line enhances to a version of G_a whose representations don’t involve any divided powers.

• > There’s recently been the philosophy going around that divided powers are a consequence of working over the integers instead of the sphere spectrum.

What are some examples which demonstrate this philosophy? I’m guessing one of them is HH(F_p) vs THH(F_p); are there others?

• Yes, certainly HH vs. THH is a prominent example. Another place that divided powers show up is in Koszul duality (the slogan is that “polynomial algebras are Koszul-dual to divided power algebras”), and I vaguely recall convincing myself a long time ago that this is also repaired by working over the sphere. On the other hand, these may secretly be the same example: the Koszul-dual E_1-coalgebra of an E_1-algebra is its factorization homology over the pointed circle. This should be fairly straightforward to check, as the “free algebra” functor is a left adjoint and factorization homology is a colimit.