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I have a couple of questions about the notations & their meaning used in "The Geometry of Moduli Spaces of Sheaves" by Huybrechts & Lehn, in Example 2.2.2 (page 38):

$V$ is assumed to be a be a finite dimensional vector space over field $k$. Let $0 \leq r \leq dim(V)$. the Grassmann functor is defined in the text as $\underline{Grass}(V,r):(Sch/k)^o \rightarrow (Sets)$ that associates every $k$-scheme $S$ of finite type to the set of all sub sheaves $\mathfrak{U} \subset \mathcal{O}_S \otimes V$ such that the quotient $ F = (\mathcal{O}_S \otimes V)/\mathfrak{U} $ is locally free of rank $r$.

For each $r$-dim linear subspace $W \subset V$ we consider the sub functor $\mathcal{G}_W$ of $\underline{Grass}(V,r)$, that maps every $k$-scheme S to those locally free quotients $F$ for which the canonical composition $\mathcal{O}_S \otimes W \rightarrow \mathcal{O}_S \otimes V \rightarrow F$ is an isomorphism and therefore it induces a splitting of the inclusion $W \subset V$ (...splitting as what and in which category?)

From this we can conclude that $\mathcal{G}_W$ is represented by by an affine space $G_W \subset \text{Spec}(S^*\text{Hom}(V,W)^{\vee})$, "corresponding to homomorphisms that split the inclusion map $W \subset V$" ???

Questions:

Firstly (possibly a stupid question) what is the scheme $\text{Spec}(S^*\text{Hom}(V,W)^{\vee})$ concretly? my guess would be that $S^*\text{Hom}(V,W)^{\vee}$ is nothing by the symmetric algebra $Sym(Hom(V, W)^{\vee})$, is this true? If my guess is correct then I might suppose that $\text{Spec}(S^*\text{Hom}(V,W)^{\vee})$ is the scheme that represents the functor $T : (Sch) \to (Sets)$ assigning $$S \mapsto H^0(S, \mathcal{O}_S \otimes_{\mathcal{O}_{\mathbb{Z}}} Sym(Hom(V,W)^{\vee}))$$ Does it make sense ? Is there a more concrete description of $\text{Spec}(S^*\text{Hom}(V,W)^{\vee})$ ?

And why gives $\mathcal{O}_S\otimes V \to \mathcal{O}_S\otimes W$ an $S$-point of $\text{Spec}(S^*\text{Hom}(V,W)^{\vee})$ as stated in the text?

Some background: The notation "$S$-point" of a (affine) scheme means that via Yoneda embedding we interpret this scheme as a functor $(Sch/k) \to (Sets)$ given by $S \mapsto \text{Spec}(S^*\text{Hom}(V,W)^{\vee})(S)= Hom(S, \text{Spec}(S^*\text{Hom}(V,W)^{\vee})$. Why is $\mathcal{O}_S\otimes V \to \mathcal{O}_S\otimes W$ an element/"point" of it?

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  • $\begingroup$ I don't have time to make this a detailed answer, so here are a few remarks. A) Indeed, $S^*$ is the symmetric algebra. Moreover, $\mathrm{Spec}(S^*\mathrm{Hom}(V,W)^{\vee})$ is a coordinate-free way to get the affine scheme $\mathbb A^{\dim V\dim W}$ corresponding to the vector space $\mathrm{Hom}(V,W)^{\vee})$ and the claim is that $\mathcal G_W$ is the sub-space corresponding to those homomorphisms inducing a splitting of vector spaces. And the very rough idea why is that the splitting (of vector bundles) given by $\mathcal O\otimes W\to \mathcal O\otimes V\to F$ (to be cont'd) $\endgroup$
    – Ben
    Commented Sep 26, 2019 at 17:25
  • $\begingroup$ (cont.) is really a family of splittings of vector spaces parametrised by $S$. Does that help? $\endgroup$
    – Ben
    Commented Sep 26, 2019 at 17:25
  • $\begingroup$ a bit, but not fully. what I don't understand why every homomorphism $\phi:\mathcal{O}_S\otimes V \to \mathcal{O}_S\otimes W$ correspond exactly to the set $Hom(S, Sym(Hom(V,W)^{\vee})$? first of all what is $\mathcal{O}_S\otimes W$? if we set $n:= dim(V), m:= dim(W)$ then $\mathcal{O}_S\otimes V \simeq \mathcal{O}_S^n, \mathcal{O}_S \otimes W \simeq \mathcal{O}_S^m$ and $\phi: \mathcal{O}_S^n \to \mathcal{O}_S^m$ is fully determined by the images of $e_1,e_2,...,e_n$ (= free "base" of $\mathcal{O}_S^n$) in $\mathcal{O}_S^m$ with free base $f_1,...,f_m$. $\endgroup$
    – user705174
    Commented Sep 27, 2019 at 18:37
  • $\begingroup$ therefore for every point $s \in S$ we obtain equatons $\phi_s(e_i)= \sum_{k=1}^n a_{i,j;s} e_k $ in $\mathcal{O}_{S,s}^m$. Since $Hom(V,W)$ is a vector space of dimension $n \cdot m$ has canonical base $e_i \otimes f_j$ every morphism $\phi$ is determined by a "vector" $\sum_{i,j} ^{nm} c_{ij} e_i \otimes f_j$ with $c_{ij} \in \mathcal{O}_S(S)$ and therefore every morphism $\phi$ is $\endgroup$
    – user705174
    Commented Sep 27, 2019 at 18:38
  • $\begingroup$ determined by "linear" component of $ \mathcal{O}_S \otimes_Z Sym(Hom(V,W)^{\vee})$. why do we need "higher graded elements" like $(e_1 \otimes f_1) \wedge (e_2 \otimes f_2) \in \mathcal{O}_S \otimes_Z Sym(Hom(V,W)^{\vee})$ do describe a morphism $\phi: \mathcal{O}_S\otimes V \to \mathcal{O}_S\otimes W$ ? $\endgroup$
    – user705174
    Commented Sep 27, 2019 at 18:38

1 Answer 1

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Let me concentrate on your first question; this should clarify the authors' claim. We'll see if that's enough for you to figure out the rest.

Even though later we will be interested in the rather specific $k$-vector space $\hom(V,W)$ of linear maps, for now, it is conceptually easier to consider any finite-dimensional $k$-vector space $V$. I like to think of it as a vector bundle over $\mathrm{Spec}(k)$. And a vector bundle (thought of as a sheaf) is ought to have a "total space" – a scheme $|V|$ over $k$ whose sections correspond to the elements of $V$, universally. Meaning that for every $k$-scheme $X$, the $k$-morphisms $X\to |V|$, being the same as the sections of the pull-back $|V|\times_kX$, should be the global sections of the pulled back vector bundle $V\otimes_k\mathcal{O}_X$, i.e., $V\otimes_k\mathcal{O}_X(X)$. For short, we want $\hom_k(X,|V|) = V\otimes_k\mathcal{O}_X(X)$. I claim that this is solved by $\mathrm{Spec}(S^\bullet V^\vee)$. In fact, $$\begin{align*} \hom_k(X, \mathrm{Spec}(S^\bullet V^\vee))&=\hom_{k\text{-alg}}(S^\bullet V^\vee,\mathcal{O}_X(X))\\ &\cong\hom_{k\text{-vect}}(V^\vee,\mathcal O_X(X))\\ &\cong V\otimes_k\mathcal O_X(X), \end{align*}$$ where the bottom isomorphism comes from the natural map $V\otimes_k\mathcal O_X(X)\to \hom_{k\text{-vect}}(V^\vee,\mathcal O_X(X))$, mapping a homogeneous element $v\otimes f$ to the homomorphism $(\varphi\mapsto \varphi(v)\cdot f)\in \hom_{k\text{-vect}}(V^\vee,\mathcal O_X(X))$. It's an isomorphism since $V$ is finite-dimensional.

Returning to $\hom(V,W)$ and its associated affine scheme $|\hom(V,W)| = \mathrm{Spec}(S^\bullet \hom(V,W)^\vee)$: Let $U\subset\hom(V,W)$ be the affine subspace consisting of those linear maps $V\to W$ which restrict to the identity on $W$; equivalently, "that split the inclusion map $W\subset V$". Moreover, for every $k$-algebra $\mathcal O_X(X)$ it makes sense to define $U\otimes_k\mathcal O_X(X)\subset \hom(V,W)\otimes_k \mathcal O_X(X)$ in an obvious way and there exists an affine sub-scheme $\mathcal U\subset |\hom(V,W)|$ such that via the above isomorphisms, $\hom(X,\mathcal U) = U\otimes_k \mathcal O_X(X)$. (I'll leave the details up to you.)

What the authors claim is simply that via the indicated map $\mathcal G_W\to |\hom(V,W)|$, $\mathcal G_W$ is isomorphic to $\mathcal U$. (Let me know in the comments if you need more clarifications or more hints towards the proof.)

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  • $\begingroup$ I think I understand the point. $|V|$ might be thought as be "that what it represent" and and if we find another scheme as $\mathrm{Spec}(S^\bullet V^\vee)$ in our case that represents the same functors as induced by $V$, then Yoneda lemma tells that thy coinside up to isomorphism. hope that is it. Thank you very much for your time and help! $\endgroup$
    – user705174
    Commented Oct 5, 2019 at 17:08

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