# Action on group $\operatorname{Ext}^i(\mathcal{L}, \mathcal{M})$ by scalar multiplication

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.

• 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.
– abx
Commented May 18, 2020 at 15:20
• @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" Commented May 18, 2020 at 15:29

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.

• 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. Commented May 18, 2020 at 15:40
• This is a good exercise. For instance, you can easily deduce this from the above exact sequence. Commented May 18, 2020 at 16:22
• 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 Commented May 19, 2020 at 8:32
• ...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... Commented May 19, 2020 at 9:02
• The functor $Hom(-,-)$ has two arguments, you can derived with respect to any of these, and the results $Ext^i(-,-)$ are known to be isomorphic. Therefore, if you are interested in $Ext^1(L,M)$ you start with an exact sequence $$0 \to M \to E \to L \to 0$$ and either apply $Hom(L,-)$ to get a morphism $Hom(L,L) \to Ext^1(L,M)$, or apply $Hom(-,M)$ to get $Hom(M,M) \to Ext^1(L,M)$. Commented May 19, 2020 at 10:08

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?

• 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. Commented Jun 24, 2020 at 4:52
• @Sasha: I have adjust my elaboration of your argument above to your notations. Hope, it's correct now? Commented Jun 24, 2020 at 13:24