Characters group and cocharacters group Hom duality

Question

Let $T$ be an algebraic torus over $\mathbb{C}$. For brevity denote $C = \mathbb{C}^\times = \mathbb{G}_{m,\mathbb{C}}$ the multiplicative subgroup of $\mathbb{C}$. Define character group by $$ X^*(T) = \operatorname{Hom}(T,C) = \{ f : T \to C \mid f \mbox{ is homomorphism} \} $$ and cocharacter group $$ X_*(T) = \operatorname{Hom}(C,T) = \{ f^\vee : C \to T \mid f^\vee \mbox{ is homomorphism} \} $$ Then

$\operatorname{Hom}( X^*(T), \mathbb{Z} ) = X_*(T) $

I ask:

1. How to prove this fact by elementary or straight forward methods?
2. The equality implies that each homomorphism from $X^*(T)$ to $\mathbb{Z}$ defines a cocharacter $\chi^\vee \in X_*(T)$? I want to find this cocharacter expliticly.
3. What is the converse statement - i.e. taking Hom on $X_*(T)$ to get $X^*(T)$?

Thank you.

Answer 1

As you might expect, this is all rather formal:

Adhering to the functorial spin on algebraic geometry - and varying your notation slightly, if only to make it easier to type - define the $\mathbb C$ algebraic group $G$ by $$ G(A)=\mathbb G_m ( A) = A^*,$$ where $A^*$ is set of the invertible elements of the $\mathbb C$-algebras $A$.

We can also write $G= \mathbb G_m = \mathop{\rm Spec} R$, with $R=\mathbb C[T,T^{-1}]$. In fact, the identification is canonical (i.e. identify $a\in G(A) = A^*$ for any $\mathbb C$-algebra $A$, with $\lambda_a:T\mapsto a\in A$, for $\lambda_a\in {\rm hom}_{\rm alg}(R,A)$ - so the element of $T\in G(R)$ corresponds to the identity homorphism – i.e. $T\to T$).

We get, in the category of algebraic groups, after a bit more formalism, perhaps, that $${\rm hom} (G,G) = \mathbb Z,\tag{*}$$ canonically, with the identity homomorphism on the left identified with $1$ on the right – at the level of algebras, $T \mapsto T^n$ on the left corresponds to $n\in \mathbb Z$ on the right.

However, if you prefer, identify $G$ with $G(\mathbb C) = \mathbb C^*$, as you have done - the only point that matters here is $(*)$. This is all window dressing to be able to talk about the $n$ in $z \mapsto z^n$, for $n\in \mathbb Z$.

In any event, for any algebraic group $T$ one certainly has a canonical pairing $$X_*(T) \otimes X^*(T) \to {\rm hom} (G,G) = \mathbb Z.$$ That is, $$ (f^\vee, f) \mapsto f\circ f^\vee.$$

We would like to see that the pairing is perfect if $T$ is a torus.

If $T= G^n$, then $X^*(T) = \mathop {\rm hom} (G^n,G) = \mathbb Z^n$, and $X_*(T) =\mathop {\rm hom} (G,G^n) = \mathbb Z^n$. So, if $a = (a_1\cdots, a_n) \in \mathbb Z^n = X_*(T)$, and $b= (b_1,\cdots,b_n)\in \mathbb Z^n=X^*(T)$, the pairing is $$ (a,b) \mapsto \sum b_k \,a_k, $$ and this is clearly perfect.

[With your identification, the co-character is $z\mapsto (z^{a_1},\cdots, z^{a_n})$, the character is $(z_1,\cdots, z_n)\mapsto z_1^{b_1} \cdots z_n^{b_n}$, and the pairing gives us $$z \mapsto z^{\sum{b_k a_k}}.]$$

But, by definition, in the general case, if $T$ is a torus, there is an isomorphism $\lambda \colon T \xrightarrow{\sim} G^n$, although the choice is not necessarily canonical.

Still: $$(f^\vee , f ) \mapsto f\circ f^\vee = f \circ\left( \lambda^{-1} \circ \lambda \right) \circ f^\vee = \left(f\circ \lambda^{-1}\right) \circ \left( \lambda \circ f^\vee\right),$$ which brings us back to the previous case.