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As a character of an algebraic group is defined to be any morphism (of algebraic groups) from this group to $\mathbb{G}_m$, how to prove:

$\mathbb{G}_a$ has no nontrivial characters.

If I take the ground field $K=\mathbb{C}$, and define a map $\phi: \mathbb{G}_a \rightarrow \mathbb{G}_m$, $a+bi \mapsto $exp$(a+bi)$, why isn't $\phi$ a character of $\mathbb{G}_a$? (Why isn't it a morphism of algebraic groups?)

Many thanks.

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What exactly is $\mathbb{G_m}$? –  Thelonius Sep 16 '12 at 2:29

4 Answers 4

up vote 9 down vote accepted

One way to establish that there is no nontrivial homomorphism of algebraic groups from $\mathbb{G}_a$ to $\mathbb{G}_m$ is via the Jordan decomposition: every element $x$ in a linear algebraic group may uniquely be written as the product of a semisimple element $x_s$ and a unipotent element $x_u$, and this decomposition is preserved by homomorphisms of algebraic groups. See for instance Theorem 6 of these notes of Ryan Reich.

Now we can realize $\mathbb{G}_a$ as the subgroup of matrices $\{ \left[ \begin{array}{cc} 1 & a \\ 0 & 1 \end{array} \right] , \ a \in K \}$,

which shows that every element of $\mathbb{G}_a$ is unipotent. On the other hand, every element of $\mathbb{G}_m$ is semisimple. Since an element which is both unipotent and semisimple is the identity, the only homomorphism from $\mathbb{G}_a$ to $\mathbb{G}_m$ is the trivial one.

As for your example with the exponential map: that's not an algebraic function! Remember that in dealing with homomorphisms of algebraic groups, we restrict to maps which are locally given by polynomials.

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Thank you very much! This answer is really detailed and impressing! Many thanks! –  ShinyaSakai Jul 22 '11 at 18:40

If ${\Bbb G}_a$ had a non-trivial character, then there would be a systematic (in fact, functorial) way to produce invertible elements in any ring $R$ (or maybe in any $R_0$-algebra, where $R_0$ is the ring of definition of the character) from its additive structure.

This would be quite strange........

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Thank you very much. Now I understand why I am wrong. –  ShinyaSakai Jul 22 '11 at 18:45

You might be interested in this answer of mine from MO. Over a base ring $R$ which contains $\mathbb{Q}$, I show that the hom's from $\mathbb{G}_a$ to $\mathbb{G}_m$ are in canonical bijection with the nilpotent elements of $R$. In particular, if $R$ is itself a field, then the only nilpotent is $0$, giving the trivial hom.

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I am really very very lucky to see admirable and helpful teachers here. Thank you very much, Mr Speyer. This answer gives me a deeper look into this question. –  ShinyaSakai Jul 22 '11 at 18:57

Since these are affine algebraic groups, there is an equivalent ring-theoretic statement. Namely, a character over a commutative ring $R$ is an $R$-Hopf algebra map from $R[x,x^{-1}]$ to $R[y]$, where the comultiplication is $x \mapsto x \otimes x$ on the left and $y \mapsto y \otimes 1 + 1 \otimes y$ on the right.

To be an $R$-algebra map, $x$ has to go to an invertible element of $R[y]$. We will write it as $f(y) = \sum_{i=0}^n r_i y^i$, where $r_0 \in R^\times$. Taking tensor square is a functor on algebras, so we get a map $$R[x,x^{-1}] \otimes_R R[x,x^{-1}] \to R[y] \otimes_R R[y] = R[y,z]$$ that in particular yields the assignment: $$x \otimes x \mapsto f(y) f(z) = \sum_{j=0}^n \sum_{k=0}^n r_j r_k y^j z^k.$$ To have a Hopf algebra map, it is necessary and sufficient that this be equal to the image of $f(y)$ under the comultiplication map, which is $$\sum_{i=0}^n \sum_{j=0}^i \binom{i}{j} r_i y^j z^{i-j}.$$ Comparing the coefficients of $y^j z^k$ yields the system of equations $r_jr_k = \binom{j+k}{j} r_{j+k}$, where $r_m = 0$ for $m > n$.

We find that $r_0 = 1$ (since $r_0$ is an invertible idempotent), and all $r_i$ for $i \geq 1$ are nilpotent. Since the complex numbers have no nonzero nilpotents, all characters defined over $\mathbb{C}$ are trivial.

A little more work shows that there is a functor from schemes to sets that takes the spectrum of a ring $R$ to the character group of the additive group over $R$. This functor is formally represented by the multiplicative formal group $\widehat{\mathbb{G}_m}$, whose reduced subscheme is $\operatorname{Spec} \mathbb{Z}$. This is an example of Cartier duality where one has to allow formal schemes to take duals of commutative affine groups.

Regarding the exponential function, it does not yield a map from $R[x,x^{-1}]$ to $R[y]$, but in characteristic zero, there is a formal exponential map to the completion $R[[y]]$, which is the coordinate ring of the formal additive group $\widehat{\mathbb{G}_a}$. This is given by sending $x$ to $\exp(y) = \sum_{i = 0}^\infty \frac{y^i}{i!}$, and it is straightforward to check that the assignment $r_i = 1/i!$ satisfies the system of equations $r_j r_k = \binom{j+k}{j} r_{j+k}$. In positive characteristic, you can't divide by $i!$, but there is a formal exponential to a formal divided power series algebra, which yields a sort of PD-additive group. In other words, the exponential does yield a homomorphism when restricted to an infinitesimal neighborhood of the identity, but that homomorphism cannot be extended to a homomorphism (in group schemes) on the whole additive group.

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Thank you for giving me a deeper insight into this quesiton. Thank you very much. –  ShinyaSakai Jul 27 '11 at 0:02

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