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Let $X$ be a proper scheme over field $k$ and $\mathcal{L}, \mathcal{M}$ two invertible $\mathcal{O}_X$-modules. Then $Hom_{\mathcal{O}_X}(\mathcal{L}, \mathcal{M}) \cong Hom_{\mathcal{O}_X}(\mathcal{O}_X, \mathcal{M}\otimes \mathcal{L}^{\vee}) \cong H^0(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$.

Therefore derived functors coinside as well as we assumed $X$ sufficiently nice:

$\operatorname{Ext}^i(\mathcal{L}, \mathcal{M}) \cong H^i(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$.

The right hand side has $ H^i(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ a natural structure of a $k$ vector space, therefore we can talk about subspaces, multiplication by scalars form $k$ and the whole other basic linear algebra stuff.

On the other hand the Abelian group $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ has an interpretation as set of all extension classes

$$0 \to \mathcal{L} \to ? \to \mathcal{M} \to 0$$

where two classes are considered in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ as equal if there exist commutative diagram between the two exact sequences such that the vertical arrows between $\mathcal{L}$ and $\mathcal{M}$ are identities and the middle vertical arrow a isomorphism of $\mathcal{O}_X$-modules.

QUESTION 1: by $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M}) \cong H^1(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ the Ext-group is also endowed with structure of a $k$ vector space and I'm asking if there is a nice description how two extension classes in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M}) $ differ from each other /or related to each other if their corresponding elements in $H^1(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ differ by a multiplication by a scalar $a \in k^*$:

in other words if

$$0 \to \mathcal{L} \to E_1 \to \mathcal{M} \to 0$$

and

$$0 \to \mathcal{L} \to E_2 \to \mathcal{M} \to 0$$

are two representers of two extension classes in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ and the vectors $v_{E_1}$ and $v_{E_2} \in H^1(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ lie on the same line $k \cdot v_{E_1}$:

i.e. there exist a $a \in k^*$ with $v_{E_2}=a \cdot v_{E_1}$, is there a meaningful contruction between $E_1$ and $E_2$ in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ relating them to each other in dependence of $a$?

In other words how the two exact sequences of $E_1$ and $E_2$ are in this case related to each other in sophisticated way reflecting that their corresponding vectors in $v_{E_1}$ and $v_{E_2} \in H^1(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ are only differ by a scalar.

Or more generally, how the action of $k$ on by scalar multiplication $H^1(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ can be transfered to an action on the exact sequences representing extension classes from $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$?

QUESTION 2:

How to see that the $0$ in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ (the neutral element of this Abelian group) corresponds to the class of splitting extension

$$0 \to \mathcal{L} \to \mathcal{L} \oplus \mathcal{M} \to \mathcal{M} \to 0$$

I often saw in comments/ remarks on this issue that people just say 'that's because the two objects are canonical' from both viewpoints: in a vector space as well extension classes.

But I nowhere found a "clean" constructive argument why this identification is true diving in explicit machinery how thegroup elements of the Ext^1 group are identified with extension classes.

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    $\begingroup$ The category of $\mathscr{O}_X$-modules is a $k$-linear abelian category, so $\operatorname{Ext}^i(\mathscr{L},\mathscr{M}) $ is a $k$-vector space by definition. Your questions would be a better fit at MSE. $\endgroup$ – abx May 18 at 15:20
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    $\begingroup$ @abx: Please, read the question a bit more carefully. I understand that $\operatorname{Ext}^i(\mathscr{L},\mathscr{M})$ abstractly inherits from $H^i(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ the $k$-vector space structure. Well, abstractly that's of course clear. The point is that I want see what explicitly happens with a representing element of extension class when the inherited $k$-action acts on it. ie how it "deforms" from the initial representing element. That is, "what happens inside", not "if something happens in well defined way" $\endgroup$ – MortyPB May 18 at 15:29
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Remark. Exact sequence $0 \to L \to E \to M \to 0$ corresponds to $Ext^1(M,L)$, not to $Ext^1(L,M)$.

Q1. $a \in k^\times$ acts on $Ext^1(L,M)$ via pullback along $a:L \to L$ or via pushout along $a: M \to M$.

Q2. There are two options: either one can check that the split sequence is the neutral element for the addition, or that in the long exact sequence $$ 0 \to Hom(L,M) \to Hom(L,L \oplus M) \to Hom(L,L) \to Ext^1(L,M) $$ the element $1_L \in Hom(L,L)$ goes to 0.

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  • $\begingroup$ Do you know a recommendable reference where it is explaned why this inherited action by scalar multiplication induce exactly the pullback resp pushout. Up to now I nowhere found a homological algebra book treating exactly this point. $\endgroup$ – MortyPB May 18 at 15:40
  • $\begingroup$ This is a good exercise. For instance, you can easily deduce this from the above exact sequence. $\endgroup$ – Sasha May 18 at 16:22
  • $\begingroup$ Another remark: probably your hint on second way to show that that $L \oplus M$ corresponds to $0$ shold be beginn with applying $Hom(-,L)$ instead of $Hom(L,-)$ to the sequence, right? I learned this topic on that way that in general if $0 \to L \to X \to M \to 0$ is a ex sequence then it's corresponding class arrise as image of $id_L$ of the boundary map after application of $Hom(-,L)$ and forming the long exact sequence $\endgroup$ – MortyPB May 19 at 8:32
  • $\begingroup$ ...and a question on your hint on the verification about what kind of action is induced by $k$: Essentially this follows simply from the fact that the map $Hom(L,L) \to Ext^1(L,M)$ in compatible the $k$-multiplication; ie it's a $k$-morphism. That was the spice in your hint, right? More precisely, in addition we have to exploit the naturality of the boundary map... $\endgroup$ – MortyPB May 19 at 9:02
  • $\begingroup$ more concretly we write an arbitrary sequence $0 \to L \to E \to M \to 0$ draw a vertical arrow $a: L \to L$ (=the multiplication by $a \in k$) and obtain pullback $a^*E$. Thus we obtain a commutative diagram between the sequence $0 \to L \to E \to M \to 0$ and $0 \to L \to a^*E \to M \to 0$ above conncted by the vertical maps. Then we apply $Hom(-,L)$ and compare the classes in $Ext^1(M,L)$ in the related commutative square. I think that is it? $\endgroup$ – MortyPB May 19 at 9:09
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That is we start with an arbitrary extension $0 \to M \to e_2 \to L \to 0$ represented by the class of the image $\Phi_{e_2}:=\delta(id_L)$ with respect the delta-map in lower row in second diagram below and it's pullback extension $e_2$ in the upper row. Now we want determine that the extension $\overline{e_1}$ is represented by multiplication $a \cdot \Phi_{e_2} =: \Phi_{e_1}$.

We apply $Hom(L,-)$ to diagram

$$ \require{AMScd} \begin{CD} 0 @> >> M @> >> e_1 @>a^{-1} >> L @> >> 0\\ @VVV @VVV @VVV @VV\cdot{a}V \\ 0 @> >> M @> >> e_2 @> >> L @> >> 0 \end{CD} $$

and obtain

$$ \require{AMScd} \begin{CD} Hom(L, E) @> >> Hom(L,L) @>\delta >> Ext(L,M) @> >> \\ @VVV @VV\cdot{a}V @VVV \\ Hom(L,\overline{E}) @> >> Hom(L,L) @>\delta >> Ext(L,M) @> >> \end{CD} $$

That's a diagram of $k$-vector spaces. As you explaned in the answer the extension $e_1$ is forced to be the pullback of $e_2$: i.e. $e_1= a^*e_2$. $k$-linearity and commutativity of the maps imply $a \cdot \Phi_{e_2}=a \cdot \delta(id_L) = \delta(a \cdot id_L) = \Phi_{e_1}$. So $e_1=a e_2$. Is this the correct result of the $k^*$ action by scalar multiplication on $Ext(L,M)$? Or do I have somewhere implemented your hints on my question 1) in wrong way?

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    $\begingroup$ In the first diagram take both lines to be the original extension (with the arrow $E \to L$ in the top line modified by $a^{-1}$) and the vertical arrows to be 1, 1, and $a$. Note that the diagram commutes and is the pullback diagram (pullback of the bottom line with respect to the right vertical arrow). Let $e_1$ and $e_2$ be the extensional classes of the lines. Consider the second diagram, the vertical arrows on it are induced by the morphisms of the second argument, hence they are $1$, $a$, $1$. The commutativity of the second square then reads as $ae_2 = e_1$. This is what you need. $\endgroup$ – Sasha Jun 24 at 4:52
  • $\begingroup$ @Sasha: I have adjust my elaboration of your argument above to your notations. Hope, it's correct now? $\endgroup$ – MortyPB Jun 24 at 13:24

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