I study Nitin Nitsure's paper Construction of Hilbert and Quot Schemes (arXiv:math/0504590) and have some problems with the content of imposed universal property (F) in the section "Use of Flattening Stratification" (p 26).

Reading the imposed universal property (F) following parts confuse me:

Question 1: About the composition map $h$. First of all I guess the author had $W_Y \otimes_{O_Y} \operatorname{Sym}^r V_Y$ in mind and not $W \otimes_{O_S} \operatorname{Sym}^r V$, or not?

Let assume that. But why is $\pi_Y^* W_Y \otimes_{O_Y} \operatorname{Sym}^r V_Y = \pi_Y^* \pi_{Y*} E_Y$ and not $\pi_Y^* \pi_{Y*} E_Y(r)$?

I think that this identification should arise in same way as in the construction of

$\alpha:$ $ \frak{Quot} $ $^{\Phi, L}_{E/X/S} \to $ $\frak{Grass} $ $(W \otimes_{O_S} \operatorname{Sym}^r V, \Phi(r))$

from the section "Embedding Quot into Grassmannian" (the previous part on same page). Let introduce abbreviations for these functors:

We set $G:=$ $\frak{Grass}$ $(W \otimes_{O_S} \operatorname{Sym}^r V, \Phi(r))$ and $Q:=$ $\frak{Quot}$ $^{\Phi, L}_{E/X/S}$. The explicit construction of $\alpha$ is given there.

I assume that the author hasn't changed notations when passing to "Use of Flattening Stratification" and therefore it should still be $\pi_Y^* W_Y \otimes_{O_Y} \operatorname{Sym}^r V_Y = \pi_Y^* \pi_{Y*} E_Y(r)$?

Question 2: Based on insights from Question 1 I think that the cokernel of $h$ should be $q: E_Y(r) \to F$ because $h$ should be $\pi_Y^* K_Y \to E_Y(r)$. Then I come to conclusion that the universal property (F) on $F$ should be read as follows:

(UP) $F(-r)$ (ie twist of $F$ by $O_Y(-r)$ and not $F$ itself !) is flat over $Y$ with its Hilbert polynomial on all fibers equal to $\Phi$ if and only if $\phi : Y \to T$ factors via $T' \to T$.

But then the proof shows an embedding of $ \frak{Quot} $ $^{\Psi, L}_{E(-r)/X/S} $ with $\Psi(X) := \Phi(X-r)$ and not $ \frak{Quot} $ $^{\Phi, L}_{E/X/S} $. That's a problem.

The reason I think so is the definition of relative representability, see Tag 0023 from the Stacks Project.

Reparaphrasing this we have to show for fixed $f \in G(Y) \cong \operatorname{Hom}(Y, \operatorname{Grass}(W \otimes_{O_S} \operatorname{Sym}^r V, \Phi(r)))$ that $h_T \times_{f, G, \alpha} Q(Y)$ is the set of pairs $(\phi, q) \in h_T(Y) \times Q(Y)$ where $q: E_Y \to D$ is the quotient with $G(\phi)(f)=\alpha(q)$.

In this sense we have to show that this is representable. Now what does this condition mean? By construction $ G(\phi)(f)= f_Y: W_Y \otimes_{O_Y} \operatorname{Sym}^r V_Y \to \phi^* \mathcal{J}$ and for a quotient $q: E_Y \to D \in Q(Y)$ the image $\alpha(q)$ is by construction of $\alpha$ exactly $\pi_{Y*} E_Y(r) \to \pi_{Y*} D(r) $ and the requirement is that $\alpha(q)= f_Y$.

Therefore I think that the correct formulation in the last sentence of universal (F) should be as in (UP). Is that true or do I make a thinking error and misunderstand the construction?

Question 3: I not understand why if we take at the end $T:= \operatorname{Grass}(W \otimes_{O_S} \operatorname{Sym}^r, \Phi(r))$ then the corresponding locally closed subscheme $T' \subset T$ represents the functor $\frak{Quot}$ $^{\Phi, L}_{E/X/S}$ "by construction". Precisely, by the construction of what? Of $\alpha$?


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