How to generally characterize $\operatorname{Ext}^n$?

Question

There's this well known characterization of the first $\operatorname{Ext}$ group of two objects in a abelian category:

There is an isomorphism between $\operatorname{Ext}^1(B,A)$ and the group of equivalence classes of short exact sequences of the form $A \hookrightarrow \_\_ \twoheadrightarrow B$. If $A$ and $B$ are abelian groups, these are called the group extensions of $B$ by $A$.

Can we say anything crisp like this to characterize $\operatorname{Ext}^2$? Or $\operatorname{Ext}^n$ in general? It looks like we can, but I am unfamiliar with $n$-extensions and the construction of the Baer sum of two chains.

Answer 1

There is this description in terms of extensions of length $n$, but as far as I know it's not useful for anything; in any case I've never used it for anything. Here are three descriptions that are actually useful:

• $\text{Ext}^n(A, B)$ is the set of maps from $A$ to $B[n]$ (the chain complex which is $B$ concentrated in degree $n$) in the derived category. This is maybe the easiest way to see that for all $A, B, C$ there are natural maps $\text{Ext}^n(A, B) \otimes \text{Ext}^m(B, C) \to \text{Ext}^{n+m}(A, C)$ (it's just given by composition in the derived category); I think thinking about this map in terms of extensions is a pointless headache.
• $\text{Ext}^n(A, B)$ classifies extensions of $A$ by $B[n-1]$; one way to say this is that it classifies chain complexes $C$ fitting into a short exact sequence $0 \to B[n-1] \to C \to A \to 0$, up to a suitable equivalence relation.
• $\text{Ext}^n(A, B)$ is the $n^{th}$ homology of a chain complex $\text{RHom}(A, B)$ called the derived hom. In many situations it's not enough to consider the Ext groups and one must consider this entire chain complex, for example if $A = B$ derived endomorphisms have a dg algebra structure containing more information than just the corresponding graded algebra given by taking homology.

Answer 2

You already found a description via equivalence classes of $n$-extensions in the link. This is described in more detail in some homological algebra books like those of Weibel and Hilton and Stammbach. In case the abelian category has some more properties, for example it is a module category, one can define the syzygy $\Omega^n(M)$ of a module $M$ as the $n^\text{th}$ kernel in a projective resolution of $M$. In this case one has $\operatorname{Ext}^n(B,A)=\operatorname{Ext}^1(\Omega^{n-1}(B),A)$ and one has a nice description of $\operatorname{Ext}^n$ via the description of $\operatorname{Ext}^1$ using short exact sequences. Another nice description is possible using derived categories and their $\operatorname{Hom}$, see for example the last chapter of the book by Weibel, or the book by Gelfand and Manin.

Answer 3

For the sake of keeping this post self-contained, I should include a description of how to characterize $\operatorname{Ext}^n$ as $n$-extensions as outlined in this section on Wikipedia.

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For two objects $A$ and $B$ in an abelian category, define an $n$-extension of $B$ by $A$, as an exact sequence $$ 0 \to A \to X_1 \to \dotsb \to X_n \to B \to 0 \,. $$ Then we want to say two $n$-extensions are equivalent if there are maps $\{f_1, \dotsc,f_n\}$ such that every square in this diagram commutes: $$ \require{AMScd} \begin{CD} 0 @>>> A @>>> X_1 @>>> \dotsb @>>> X_n @>>> B @>>> 0\\ @.@V{\mathrm{id}}VV @V{f_1}VV @. @V{f_n}VV @V{\mathrm{id}}VV \\ 0 @>>> A @>>> Y_1 @>>> \dotsb @>>> Y_n @>>> B @>>> 0\\ \end{CD} $$

This relation is clearly reflexive and transitive. Showing that it is symmetric, and can indeed be regarded as an honest equivalence relation takes some work (look for a proof of the short-five lemma to help). Under this relation, there is a set bijection between equivalence classes of $n$-extensions and $\operatorname{Ext}^n(B,A)$.

We can strengthen this result by imposing a group structure on this set of equivalence classes and saying that it and $\operatorname{Ext}^n(B,A)$ are isomorphic as groups. Let $X_n \times_B Y_n$ be the pullback of $X_n$ and $Y_n$ over $B$, and let $X_1 +_A Y_1$ be the pushout of $X_1$ and $Y_1$ over $A$. Define the Baer sum of the two $n$-extensions above as $$ 0 \to A \to X_1 +_A Y_1 \to X_2 \oplus Y_2 \to \dotsb \to X_{n-1} \oplus Y_{n-1} \to X_n \times_B Y_n \to B \to 0 \,. $$ This is another $n$-extension of $B$ by $A$. The Baer sum imposes a group structure on the set of equivalence classes of $n$-extensions, the identity element being the class of split $n$-extensions with a canonical representative

$$ X_i = \underbrace{A \oplus \dotsb \oplus A}_{\binom n i \text{ copies}} \oplus \underbrace{B \oplus \dotsb \oplus B}_{\binom n {n-i+1} \text{ copies}} \;, $$

making it isomorphic to the group $\operatorname{Ext}^n(B,A)$.