The equality $\operatorname{Spec}\operatorname{Fun}\mathfrak{g}^* = \mathfrak{g}^*$

affine-schemesalgebraic-geometrylie-algebrasschemes

I don't know how many of my hyphotesis will be needed, maybe this is a more general fact.

Suppose $\mathfrak{g}$ is a complex simple Lie algebra, and let $\mathfrak{g}^*$ be its dual (as a vector space). Define the following spaces:

  • $\operatorname{Fun}\mathfrak{g}^* $ is the space of regular functions on $\mathfrak{g}^*$, i.e. $\operatorname{Sym}\mathfrak{g}$, the symmetric algebra of $\mathfrak{g}$
  • $\operatorname{Inv}\mathfrak{g}^* = (\operatorname{Fun}\mathfrak{g}^*)^{\mathfrak{g}}$ is the space of invariants with respect to the coadjoint action of $\mathfrak{g}$ on $\mathfrak{g}^* $
  • $\mathcal{P} =\operatorname{Spec}\operatorname{Inv}\mathfrak{g}^*$

Now to the actual question: the book I'm currently reading states "the inclusion $\operatorname{Inv}\mathfrak{g}^* \to \operatorname{Fun}\mathfrak{g}^* $ gives rise to a surjective map $p\colon \mathfrak{g}^* \to \mathcal{P}$."

I can't understand why: I know that given an homomorphism of rings $A \to B$ you have a morphism of schemes $\operatorname{Spec}B \to \operatorname{Spec}A$, so if the equality I mention in the title is true then it would be ok. But I don't see why that is, and I suspect it's not: how is $\mathfrak{g}^*$ even a scheme? Should I identify it with the affine space $\mathbb{A}^{\dim \mathfrak{g}}$ by choosing a basis?

Best Answer

Let $k$ be a field and let $V$ be a $k$-vector space. Consider the scheme $\mathbb A_V$ over $k$ given by (via the functor-of-points viewpoint), for any $k$-algebra $A$, $$\mathbb A_V(A):=V\otimes_kA,$$ where $V\otimes_kA$ denotes taking the tensor product, then looking at the underlying set. This is the scheme which deserves to be called $V$, since the $k$-valued points recovers the set of points of $V$.

This provides a functor $\mathbb A_V\colon k\mathrm{-Alg}\to\mathrm{Sets}$, which is representable (as readily seen by choosing a basis) by an affine scheme $\mathrm{Spec}(B)$. We hope to find a more canonical description of the $k$-algebra $B$.

Now, the claim in your original question essentially asks to prove $\mathbb A_V$ can be represented by $\mathrm{Spec}(\mathrm{Sym}V^*)$, i.e., that there is a canonical bijection $$\hom_{k\mathrm{-Alg}}(\mathrm{Sym}V^*,A)\cong \mathbb A_V(A).$$ But this is true since: $$\hom_{k\mathrm{-Alg}}(\mathrm{Sym}V^*,A)\cong\hom_{k\mathrm{-Vect}}(V^*,A)\cong V\otimes_kA.$$ The first isomorphism is the observation that $\mathrm{Sym}\colon k\mathrm{-Vect}\to k\mathrm{-Alg}$ is the left adjoint to the forgetful functor. Explicitly, given a linear homomorphism $f\colon V^*\to A$, one can associate the ring homomorphism $\mathrm{Sym}V^*=\bigoplus_{n\ge0}\mathrm{Sym}^nV^*\to A$ sending $v_1\otimes\cdots \otimes v_n$ to $f(v_1)\cdots f(v_n)$.

The second isomorphism is standard, and sends an element $v\otimes a\in V\otimes_kA$ to the $k$-linear homomorphism $V^*\to A:v^\vee\mapsto \langle v^\vee,v\rangle a$.

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