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I'm reading Gortz's Algebraic Geometry Theorem 11.51 and I'm stuck trying to understand this statement:

Excerpt

There is an erratum: in the definition of $\mathcal{E}^{\lambda}$, we should repace $d$ by $d_i$.

Why is the underlined statement true? What does the "uniqueness of $\mathcal{E}^{\lambda}$" exactly mean?

Assume that $\bigoplus_{d_i \ge \lambda} \mathcal{O}_X(d_i) \cong \mathcal{E}^{\lambda} \cong \bigoplus_{e_i\ge \lambda}\mathcal{O}_X(e_i)$. We first show that $d_1=e_1$.

  • Case 1) $d_1 > e_1$ : In this case, note that:

$$\begin{align}0&\neq\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X(d_1), \mathcal{E}^{\lambda}) \\&= \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X(d_1), \bigoplus_{e_i \ge \lambda}\mathcal{O}_X(e_i))\\&\overset{?}{=} \bigoplus_{e_i \ge \lambda} \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X(d_1), \mathcal{O}_X(e_i))\\&=0\end{align}$$

(The last equality is true since $d_1 > e_1 \ge e_2 \ge \cdots $ )

  • Case 2) $d_1 < e_1$ : In this case, similarly, note that

$$\begin{align}0& \neq \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X(e_1), \mathcal{E}^{\lambda}) \\&= \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X(e_1), \bigoplus_{d_i \ge \lambda}\mathcal{O}_X(d_i))\\&\overset{?}{=} \bigoplus_{d_i \ge \lambda} \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X(e_1), \mathcal{O}_X(d_i))\\&=0\end{align}$$

So $d_1=e_1$. And this is a point that I stuck. How can we further show that $d_2 = e_2$, $d_3=e_3$, $\cdots$? Can anyone help?

FShrike
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Plantation
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    Unfortunately I can't help with your question, but I can help with the formatting. Use \begin{align},\end{align} to stop the text running off the page, and when using \overset it's advisable to use both sets of braces: \overset{?}{=} is what I had to insert to avoid some formatting issues – FShrike Dec 27 '22 at 10:27
  • As general advice, do not shy away from induction. I don't know what $d_i,e_i$ are, but probably to prove $d_i=e_i$ you can mimic your proof that $d_1=e_1$ (possibly assuming that $d_k=e_k$ for all $k<i$) – FShrike Dec 27 '22 at 10:29
  • @FShrike : Thanks for comment about formatting. For the second comment, for the direct sums $\bigoplus_{d_i \ge \lambda} \mathcal{O}X(d_i) \cong \mathcal{E}^{\lambda} \cong \bigoplus{e_i\ge \lambda}\mathcal{O}X(e_i)$, $d_i \ge \lambda$ is taken as the subsequence $d_1 \ge d_2 \ge \cdots d{n_0} \ge \lambda > d_{n_0 +1}$ and $e_i \ge \lambda$ is taken as some sequence $e_1 \ge e_2 \ge \cdots e_{n_0} \ge \lambda$ . – Plantation Dec 27 '22 at 10:48

1 Answers1

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Why is $\mathcal{E}^{\lambda}$ uniquely defined?

I claim that $\mathcal{E}^{\lambda}$ is in fact the intersection of the kernels of all maps $\mathcal{E} \rightarrow \mathcal{O}_X(t)$ with $t < \lambda$ (in particular, that doesn’t depend on the given decomposition of $\mathcal{E}$).

The inclusion $\subset$ is clear, because all maps $\mathcal{O}_X(d_i) \rightarrow \mathcal{O}_X(t)$ with $d_i \geq \lambda > t$ are zero.

For the reverse inclusion, we have a map $\mathcal{E} \rightarrow \bigoplus_{d_i < \lambda}{\mathcal{O}_X(d_i)}$ with kernel $\mathcal{E}^{\lambda}$.

Now, $\mathrm{rk}\,\mathcal{E}^{\lambda}-\mathrm{rk}\,\mathcal{E}^{\lambda+1}$ is, by definition of $\mathcal{E}^{\lambda}$, the number of $i$ such that $d_i=\lambda$. Since it depends only on $\mathcal{E}$, it proves uniqueness.

Aphelli
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  • Now, It seems to me somewhat diffcult to follow. :) 1) By the intersection of the kernels of all $\mathcal{E} \to \mathcal{O}X(t)$, you mean $ker(\mathcal{E} \to \prod{t<\lambda}(\mathcal{E}/ker(\mathcal{E}\to\mathcal{O}X(t)))$, as Gortz's Book p.174? 2) How far did you prove the 'reverse inclusion'? why $\mathcal{E}^{\lambda} = ker(\mathcal{E} \to \bigoplus{d_i < \lambda}\mathcal{O}_X(d_i))$ implies the reverse inclusion? 3) Why you mentioned final statement? You wrote, "~, it proves uniqueness." What this statement means? Uniqueness of $d_1 \ge d_2 \ge \cdots \ge d_n$? – Plantation Dec 27 '22 at 12:43
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  • no, I mean, the intersection of the kernels of literally every possible linear map $\mathcal{E} \rightarrow \mathcal{O}_X(t)$ (with $t < \lambda$) – that’s why it depends only on $\mathcal{E}$ and not the decomposition. 2) because $\mathcal{E}^{\lambda}$ is the kernel of the map I consider, it contains the intersection of kernels I was talking about. The other inclusion (the “forward” one) is easy. 3) the datum of $|{i,,d_i=\lambda}|$ for all integers $\lambda$ determines the $(d_1\geq \ldots \geq d_n)$.
    • – Aphelli Dec 27 '22 at 14:39
  • Thanks. I guess that, as the intersection of the kernels, you mean that for each open set $U\subseteq X$, $\cap(ker(\mathcal{E} \to \mathcal{O}_X(t))(U) := \cap ker(\mathcal{E}(U) \to \mathcal{O}_X(t)(U))$ (the intersection is over $t<\lambda$)? And, for comment for 3), is there a way to show the uniqueness of the $(d_1 \ge \cdots \ge d_n)$ from the datums $| {i, d_i = \lambda}| $ easily? By using the induction? What can we easily prove by catching some point? I could show the uniqueness for the case $n=2$ by brutal force(?)-case by case-argument. If needed, I'll upload my proof. – Plantation Dec 28 '22 at 02:33
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    I mean, in fact, the sheaf $U \longmapsto \bigcap_{t<\lambda, f: \mathcal{E} \rightarrow \mathcal{O}_X(t)}{\ker{f(U)}}$. 3) is basically obvious but you can use induction if you’re not convinced – the maximum $\lambda$ for which some $d_i$ is $\lambda$ is $d_1$, and then you consider the $d_2 \geq \ldots \geq d_n$. – Aphelli Dec 28 '22 at 10:13
  • O.K. For the notation $f$ in the sheaf you defined, does it means arbitrary $\mathcal{O}X$-module homomorphism $\mathcal{E} \to \mathcal{O}_X(t)$? Or projection map $\mathcal{E}:= \bigoplus{i=1}^{n}\mathcal{O}_X(d_i) \to \mathcal{O}_X(t)$? – Plantation Dec 28 '22 at 10:53
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    No, it means absolutely every possible $\mathcal{O}_X$-linear map $\mathcal{E} \rightarrow \mathcal{O}_X(t)$. – Aphelli Dec 28 '22 at 11:30
  • O.K. Thanks! Perhaps, if you have time, can you see my other question? ('Existence part' of the classification of vector bundles on $\mathbb{P}_k^{1}$ ) : https://math.stackexchange.com/questions/4606295/classification-of-vector-bundles-of-rank-n-on-mathbbp-k1-gortzs-alge. – Plantation Dec 28 '22 at 11:36