# Why are linear schemes group schemes?

According to Nitsure's Construction of Hilbert and Quot Schemes (p. 16), if $$S$$ is Noetherian, a \textit{linear scheme over $$S$$} is a scheme of the form $$V=\operatorname{Spec}\operatorname{Sym}_{\mathsf{O}_{S}}\mathsf{Q}$$ for some coherent sheaf $$\mathsf{Q}$$ over $$S$$.

The author makes the claim that this is naturally a group scheme. Why?

Also, how does this generalize vector bundles?

## 3 Answers

This is a consequence of the fact that the symmetric algebra of $$\mathcal{Q}$$ naturally forms a cogroup object in the category of $$\mathcal{O}_S$$-algebras. Namely, this cogroup object has comultiplication map $$S^\cdot \mathcal{Q} \to S^\cdot \mathcal{Q} \otimes_{\mathcal{O}_S} S^\cdot \mathcal{Q}$$ which is determined by the requirement that for sections $$x \in \mathcal{Q}(U)$$, $$x \mapsto 1 \otimes x + x \otimes 1$$. Similarly, the coinverse map $$S^\cdot \mathcal{Q} \to S^\cdot \mathcal{Q}$$ is the map such that for $$x \in \mathcal{Q}(U)$$, $$x \mapsto -x$$; and the coidentity map $$S^\cdot \mathcal{Q} \to \mathcal{O}_S$$ is the map such that for $$x \in \mathcal{Q}(U)$$, $$x \mapsto 0$$. It it straightforward to show that these definitions satisfy the required identities for a cogroup object. And then, taking $$\mathbf{Spec}$$ of a cogroup object will result in a group object in the category of $$S$$-schemes since $$\mathbf{Spec}$$ is contravariant and takes coproducts to products.

As for how to see this as a generalization of vector bundles, the answer by ggg already gives one possibility: if $$\mathcal{Q}$$ is locally free then $$\mathop{\mathbf{Spec}} S^\cdot \mathcal{Q}$$ is in fact a vector bundle. Another possibility: you can show that in fact, $$\mathop{\mathbf{Spec}} S^\cdot \mathcal{Q}$$ is a module over the ring object $$\mathbb{A}^1_S$$ using a similar construction of $$S^\cdot \mathcal{Q}$$ as a comodule over the coring object $$\mathcal{O}_S[t]$$. Thus, for example if $$S$$ is a $$k$$-scheme where $$k$$ is a field, then every element of $$k$$ induces a section $$S \to \mathbb{A}^1_S$$, and using the module structure this induces an endomorphism of $$\mathop{\mathbf{Spec}} S^\cdot \mathcal{Q}$$; it is then not hard to see that this gives every fiber of $$\mathop{\mathbf{Spec}} S^\cdot \mathcal{Q}$$ a structure of $$k$$-module object. So, for example, the closed points of each fiber form a $$k$$-vector space.

• The answer by @ggg is a useful complement to this answer, in that it constructs the group object structure using a Yoneda lemma point of view which is immensely useful to get used to. This answer just makes it a bit more explicit what structure actually results from that construction. – Daniel Schepler Apr 8 at 17:57

The answer to your first question is because $$V$$ represents the group-valued functor taking an $$S$$-scheme $$f: T \to S$$ to $$(f^* Q^\vee)(T)$$, where $$Q^\vee$$ is the dual of $$Q$$.

If $$Q$$ is a locally free $$\mathcal{O}_S$$-module of rank $$n$$, then $$V$$ is locally (over an open cover $$U_i$$ of $$S$$) of the form $$U_i \times \mathbb{A}^n$$, hence a vector bundle.

For reference see EGA II sections 1.3 and 1.7, Stacks, or section 17.1 of Vakil. There's also an exercise about this in Hartshorne.

• Thanks. I accepted Daniel Schepler's answer because I can only accept one, but yours was equally good. – The Thin Whistler Apr 13 at 20:25

Question: "The author makes the claim that this is naturally a group scheme. Why? Also, how does this generalize vector bundles?"

Answer: If $$V:=k\{e\}$$ is a rank one free module over a commutative ring $$k$$ and $$S:=Sym_k^*(V^*)\cong k[x]$$ is the polynomial ring on $$x$$ over $$k$$, there is a canonical map

$$m:k[x]\rightarrow k[x]\otimes_k k[x]\cong k[x,y]$$

defined by $$m(x):=x+y$$ and $$m(f(x)):=f(x+y)$$. This defines a $$k$$-group scheme structure on $$\mathbb{A}^1_k:=Spec(S)$$. If $$k$$ is a field, this group scheme structure corresponds to the fact that the $$k$$-vector space $$V$$ has an underlying additive abelian group: There is an addition operation

$$+: V \times V \rightarrow V$$

defined by $$+(u,v):=u+v$$ making $$(V,+)$$ an abelian group. The $$k$$-group scheme $$\mathbb{A}^1_k$$ is a "scheme version" of this structure. This generalize to any scheme $$X$$ and any algebraic vector bundle $$E$$ on $$X$$: We define $$\mathbb{V}(E^*):=Spec(T)$$ where $$T:=Sym_{\mathcal{O}_X}^*(E^*)$$ is the "sheaf symmetric algebra" of $$E^*$$ and take relative Spec of this. The sheaf $$T$$ is a sheaf of commutative unital $$\mathcal{O}_X$$-algebras and you may take the "relative Spec" of this sheaf to get a scheme

$$\pi: \mathbb{V}(E^*)\rightarrow X.$$

The group scheme $$\mathbb{V}(E^*)$$ is the geometric vector bundle of the finite rank locally trivial sheaf $$E$$. The map $$\pi$$ has the property that for any point (closed or not) $$x\in X$$ there is a canonical isomorphism

$$\pi^{-1}(x) \cong \mathbb{A}^e_{\kappa(x)}$$

where $$e=rk(E)$$ and $$\kappa(x)$$ is the residue field of $$x$$.

• Thanks. I accepted Daniel Schepler's answer because I can only accept one, but yours was equally good. – The Thin Whistler Apr 13 at 20:25