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 $$0\longrightarrow G\stackrel{g}\longrightarrow E\stackrel{a\cdot f}\longrightarrow F\longrightarrow 0?$$

  • $\begingroup$ Isn't $a \cdot \xi$ always equivalent to $\xi$? So your multiplication acts trivially on the set of equivalence classes of extensions. $\endgroup$
    – Magma
    Dec 14, 2019 at 15:17
  • $\begingroup$ 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 $a\cdot\xi$. $\endgroup$
    – mark
    Dec 14, 2019 at 15:26
  • $\begingroup$ What's $0 \cdot \xi$? $\endgroup$
    – Magma
    Dec 14, 2019 at 16:56
  • $\begingroup$ It's the trivial extension $G\to G\oplus F\to F.$ $\endgroup$
    – mark
    Dec 14, 2019 at 17:13
  • $\begingroup$ If $\varepsilon$ denotes the trivial extension, is $2 \cdot \varepsilon$ also the trivial extension? $\endgroup$
    – Magma
    Dec 14, 2019 at 17:22

1 Answer 1


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 :

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .