# Vector space structure on extensions of vector bundles

Let $$F, G$$ be vector bundles on a scheme over the field of complex numbers. I know that the set $$V:=Ext^1(F, G)$$ has the structure of an additive group given by the Baer sum of extensions. But $$V$$ also has the vector space structure. Namely, if $$a\in\mathbb{C}^*$$ and $$\xi\in Ext^1(F, G)$$ is given by $$0\longrightarrow G\stackrel{g}\longrightarrow E\stackrel{f}\longrightarrow F\longrightarrow 0$$ is it true that $$a\cdot\xi$$ is given by an extension $$0\longrightarrow G\stackrel{g}\longrightarrow E\stackrel{a\cdot f}\longrightarrow F\longrightarrow 0?$$

• Isn't $a \cdot \xi$ always equivalent to $\xi$? So your multiplication acts trivially on the set of equivalence classes of extensions. Dec 14, 2019 at 15:17
• Yes, it is equivalent, but in general $a\cdot\xi$ and $\xi$ are different points in the vector space $Ext^1(F, G)$. So my question is what is the right nonzero map in the sequence given by the extension $a\cdot\xi$.
– mark
Dec 14, 2019 at 15:26
• What's $0 \cdot \xi$? Dec 14, 2019 at 16:56
• It's the trivial extension $G\to G\oplus F\to F.$
– mark
Dec 14, 2019 at 17:13
• If $\varepsilon$ denotes the trivial extension, is $2 \cdot \varepsilon$ also the trivial extension? Dec 14, 2019 at 17:22

In fact, $$\operatorname{Ext}^1(F,G)$$ admits two vector spaces structures. One which comes from $$F$$ and another one which comes from $$G$$. But you can show that these two structures are identical here.

The construction of the first one is as follow : let $$a\in\mathbb{C}$$ and consider the multiplication by $$a$$ map $$a:F\to F$$. You can form the pullback if $$E\to F$$ along $$a$$. This gives a commutative diagram : $$\require{AMScd} \begin{CD} 0@>>> G@>>> E\times_{F,a} F@>>> F@>>>0\\ @.@|@VVV@VVaV\\ 0@>>> G@>>> E@>>>F@>>>0 \end{CD}$$ If $$a=0$$, then $$E\times_{F,a} F$$ is isomorphic to $$G\oplus F$$. Now if $$a\neq 0$$, then $$a$$ is an isomorphism so $$E\times_{F,a} F$$ is isomorphic to $$E$$. Modulo this isomorphism, the pullback diagram looks like : $$\require{AMScd} \begin{CD} 0@>>> G@>>> E @>a^{-1}f>> F@>>>0\\ @.@|@|@VVaV\\ 0@>>> G@>>> E@>>f>F@>>>0 \end{CD}$$

Hence, the extension $$a\xi$$ is actually given by $$0\longrightarrow G\longrightarrow E\overset{a^{-1}f}\longrightarrow F\longrightarrow 0$$

For sake of completness, lets have a look at the other possible construction : we use instead the structure on $$G$$. Let $$a\in\mathbb{C}$$ and consider the multiplication by $$a$$ as a map $$a:G\to G$$. We can form the pushout diagram of $$G\to E$$ along $$a$$. This gives a commutative diagram : $$\require{AMScd} \begin{CD} 0@>>> G@>>> E @>>> F@>>>0\\ @.@VaVV@VVV@|\\ 0@>>> G@>>> G\coprod_{a,G} E @>>>F@>>>0 \end{CD}$$ Again, if $$a=0$$ then $$G\coprod_{a,G} E=G\oplus F$$ and the extension splits. If $$a\neq 0$$, then $$G\coprod_{a,G} E\simeq E$$ and modulo this isomorphism the pushout looks like : $$\require{AMScd} \begin{CD} 0@>>> G@>g>> E @>>> F@>>>0\\ @.@VaVV@|@|\\ 0@>>> G@>a^{-1}g>> E @>>>F@>>>0 \end{CD}$$ So for the second vector space structure, the class $$a\xi$$ is reprensented by $$0\longrightarrow G\overset{a^{-1}g}\longrightarrow E\longrightarrow F\longrightarrow 0$$

It remains to show that these two structures are identical, but this follows from the commutativity of :

$$\require{AMScd} \begin{CD} 0@>>> G@>a^{-1}g>> E @>f>> F@>>>0\\ @.@|@VaVV@|\\ 0@>>> G@>>g> E @>>a^{-1}f>F@>>>0 \end{CD}$$