Posted by Jason Polak on 23. August 2016 · Write a comment · Categories: group-theory · Tags: ,

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Let $k$ be a commutative ring. Let $\G_a$ be group functor $\G_a(R) = R$ and $\G_m$ be the group functor $\G_m(R) = R^\times$, both over the base ring $k$. What are the homomorphisms $\G_a\to \G_m$? In other words, what are the characters of $\G_a$? This depends on the ring, of course!

The representing Hopf algebra for $\G_a$ is $k[x]$. And, the representing Hopf algebra for $\G_m$ is $k[x,x^{-1}]$. Homomorphisms $\G_a\to \G_m$ correspond to Hopf algebra maps $k[x,x^{-1}]\to k[x]$. Such a map is a $k$-algebra homomorphism that satisfies the additional conditions for being a Hopf algebra homomorphism.

What are these conditions in this case? First, a $k$-algebra homomorphism will $k[x,x^{-1}]\to k[x]$ is given by an invertible polynomial $a\in k[x]$. That already puts some serious conditions on the base ring. For this map to be a Hopf algebra map, the element $a= a_0 + a_1x + \cdots + a_nx^n\in k[x]$ has to satisfy some pretty special requirements. To state them, for a Hopf algebra $A$ let us use the following notation for the maps that give $A$ its Hopf algebra structure:

  • Comultiplication: $\Delta: A\to A\otimes_k A$
  • Counit: $\epsilon: A\to k$
  • Coinverse: $S:A\to A$.

Then, the conditions for $a$ to give a Hopf algebra map are given by the following conditions in the Hopf algebra $k[x]$:

  1. $\Delta(a) = a\otimes a$
  2. $\epsilon(a) = 1$
  3. $S(a) = a^{-1}$

Now, in $k[x]$, the counit is determined by the formula $\epsilon(x) = 0$. Therefore, by condition (2), we see that the constant term of $a$ must actually be $1$. And of course we already know that $a$ is invertible. So if $k$ is a integral domain, then $a=1$ and the only homomorphism $\G_a\to \G_m$ is actually the trivial one, given on points by sending everything to $1$.

Actually, the same is true in many cases even if $k$ is not an integral domain. That’s because in $k[x]$, we know that $S(x) = -x$. Therefore, by requirement (3), we see that if $a_n^2 = 0$. Therefore, if nontrivial homomorphisms exist, then $k$ must have nilpotent elements. Indeed, if $a_1^2 = 0$ then one possibility is $a = 1 + a_1x$. The reader should check (1)-(3) to make sure that they hold.

In fact, if $k$ has characteristic zero and $c$ is nilpotent then $a = \exp(c)$, the “exponential series” map gives a homomorphism too, which is just a generalisation of the example $a= 1 + a_1x$. Indeed, $\exp(c)$ makes sense whenever $c$ is nilpotent and $k$ has characteristic zero. Condition (3) follows from the formula $\exp(a)\exp(-a) = \exp(0) = 1$. Condition (2) follows from the formula $\exp(0) = 1$. What about condition (1)? If we choose an isomorphism $k[x]\otimes_k k[x] \cong k[x,y]$, then (3) becomes the formula $\exp(x + y) = \exp(x)\exp(y)$.

Are all Hopf algebra homomorphisms $k[x,x^{-1}]\to k[x]$ given by $x\mapsto \exp(c)$ for some nilpotent $c\in k$? No! Of course, the exponential series doesn’t work in general in positive characteristic. Of course it does work sometimes: if $c$ is nilpotent of order $\ell$ and the characteristic of $k$ is greater than $\ell$, for example.

But take even the case when $a$ is of the form $a = 1 + a_1x + a_2x^2$. Working out conditions (1)-(3) gives the following conditions on the coefficients of $a$:

  • $a_2^2 = 0$
  • $2a_2 – a_1^2 = 0$
  • $a_1a_2 = 0$.

If $2\not=0$ in $k$ then this just reduces to the exponential series for an element $a_1$ such that $a_1^3 = 0$. However, if $2 = 0$ then we just get the conditions $a_2^2 = a_1^2 = a_1a_2= 0$. And indeed, one can find a ring $k$ and elements $a_1,a_2\in k$ that satisfy these conditions: take $k = \Z/2[t]/t^2$ and take $a_1 = a_2 = t$. Or, $k= \Z/2[t]/t^2\times \Z/2[t]/t^2$ and $a_1 = (t,0)$ and $a_2 = (0,t)$. Neither of these comes from an exponential series.

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