User:Jallotta/Yoneda Extensions

In mathematics, Yoneda Extensions are a method used in homological algebra to define higher Ext groups without using derived functors. It is a useful method of explicitly computing Ext groups because one does not need a projective or injective resolution.

Definition and computation
For any abelian category $$\mathcal{C}$$ and $$A,B$$ two objects in $$\mathcal{C}$$, define


 * $$ \operatorname{Ext}_\mathcal{C}^0(A,B) = \operatorname{Hom}_\mathcal{C}(A,B).$$

We define $$ \mathrm{Ext}_\mathcal{C}^1(A,B) $$ as the group of extensions of A by B modulo an equivalence relation as follows. Given two extensions $$ E, E' $$ we say they are equivalent if there exists a morphism of short exact sequences with equality on the ends. In other words a morphism $$ E \to E' $$ that makes the following diagram commute.

INSERT COMMUTATIVE DIAGRAM

The group of $$n$$-extensions of $$A$$ by $$B$$ modulo equivalence is defined to be $$ \mathrm{Ext}_\mathcal{C}^n(A,B) $$. For an equivalence of $$n$$-extensions, it is often useful to use intermediate stages. So two extensions are equivalent if there is a commutative diagram of the following form.

INSERT COMMUTATIVE DIAGRAM

When it is clear which category is being used, the subscript will usually be dropped and it is written as $${\mathrm{Ext}}^n(A,B)$$.

Group law
Ext functors derive their name from the relationship to extensions. Given $$R$$-modules $$A$$ and $$B$$, there is a bijective correspondence between equivalence classes of extensions
 * $$0\rightarrow B\rightarrow C\rightarrow A\rightarrow 0$$

of $$A$$ by $$B$$ and elements of
 * $$\operatorname{Ext}_R^1(A,B).$$

Given two extensions
 * $$0\rightarrow B\rightarrow C\rightarrow A\rightarrow 0$$ and
 * $$0\rightarrow B\rightarrow C'\rightarrow A\rightarrow 0$$

we can construct the Baer sum, by forming the pullback $$\Gamma$$ of $$C\rightarrow A$$ and $$C'\rightarrow A$$. We form the quotient $$Y=\Gamma/\Delta$$, with $$\Delta=\{(-b,b):b\in B\}$$. The extension
 * $$0\rightarrow B\rightarrow Y\rightarrow A\rightarrow 0$$

thus formed is called the Baer sum of the extensions $$C$$ and $$C'$$.

The Baer sum ends up being an abelian group operation on the set of equivalence classes, with the extension
 * $$0\rightarrow B\rightarrow A\oplus B\rightarrow A\rightarrow 0$$

acting as the identity.

Ext in abelian categories
This identification enables us to define $$\operatorname{Ext}^1_{\mathcal{C}}(A,B)$$ even for abelian categories $$\mathcal{C}$$ without reference to projectives and injectives. We simply take $$\operatorname{Ext}^1_{\mathcal{C}}(A,B)$$ to be the set of equivalence classes of extensions of $$A$$ by $$B$$, forming an abelian group under the Baer sum. Similarly, we can define higher Ext groups $$\operatorname{Ext}^n_{\mathcal{C}}(A,B)$$ as equivalence classes of n-extensions


 * $$0\rightarrow B\rightarrow X_n\rightarrow\cdots\rightarrow X_1\rightarrow A\rightarrow0$$

under the equivalence relation generated by the relation that identifies two extensions


 * $$0\rightarrow B\rightarrow X_n\rightarrow\cdots\rightarrow X_1\rightarrow A\rightarrow0$$ and


 * $$0\rightarrow B\rightarrow X'_n\rightarrow\cdots\rightarrow X'_1\rightarrow A\rightarrow0$$

if there are maps $$X_m\rightarrow X'_m$$ for all $$m$$ in $$1,2,..,n$$ so that every resulting square commutes.

The Baer sum of the two n-extensions above is formed by letting $$X_1$$ be the pullback of $$X_1$$ and $$X'_1$$ over $$A$$, and $$X_n$$ be the pushout of $$X_n$$ and $$X'_n$$ under $$B$$. Then we define the Baer sum of the extensions to be
 * $$0\rightarrow B\rightarrow X_n\rightarrow X_{n-1}\oplus X'_{n-1}\rightarrow\cdots\rightarrow X_2\oplus X'_2\rightarrow X_1\rightarrow A\rightarrow0.$$

Properties of Ext
The Ext functor exhibits some convenient properties, useful in computations.


 * $$\operatorname{Ext}^i_{\mathrm{Mod}_R}(A,B)=0$$ for $$i>0$$ if either $$B$$ is injective or $$A$$ is projective.


 * The converse also holds: if $$\operatorname{Ext}^1_{\mathrm{Mod}_R}(A,B)=0$$ for all $$A$$, then $$\operatorname{Ext}^i_{\mathrm{Mod}_R}(A,B)=0$$ for all $$A$$, and $$B$$ is injective; if $$\operatorname{Ext}^1_{\mathrm{Mod}_R}(A,B)=0$$ for all $$B$$, then $$\operatorname{Ext}^i_{\mathrm{Mod}_R}(A,B)=0$$ for all $$B$$, and $$A$$ is projective.


 * $$\operatorname{Ext}^n_{\mathrm{Mod}_R}(\bigoplus_\alpha A_\alpha,B)\cong\prod_\alpha\operatorname{Ext}^n_{\mathrm{Mod}_R}(A_\alpha,B)$$


 * $$\operatorname{Ext}^n_{\mathrm{Mod}_R}(A,\prod_\beta B_\beta)\cong\prod_\beta\operatorname{Ext}^n_{\mathrm{Mod}_R}(A,B_\beta)$$

Ring structure and module structure on specific Exts
One more very useful way to view the Ext functor is this: when an element of $$\operatorname{Ext}^n_{\mathrm{Mod}_R}(A,B)$$ is considered as an equivalence class of maps $$f: P_n\rightarrow B$$ for a projective resolution $$P_*$$ of $$A$$ ; so, then we can pick a long exact sequence $$Q_*$$ ending with $$B$$ and lift the map $$f$$ using the projectivity of the modules $$P_m$$ to a chain map $$f_*: P_*\rightarrow Q_*$$ of degree -n. It turns out that homotopy classes of such chain maps correspond precisely to the equivalence classes in the definition of Ext above.

Under sufficiently nice circumstances, such as when the ring $$R$$ is a group ring, or a k-algebra, for a field k or even a noetherian ring k, we can impose a ring structure on $$\operatorname{Ext}^*_{\mathrm{Mod}_R}(k,k)$$. The multiplication has quite a few equivalent interpretations, corresponding to different interpretations of the elements of $$\operatorname{Ext}^*_{\mathrm{Mod}_R}(k,k)$$.

One interpretation is in terms of these homotopy classes of chain maps. Then the product of two elements is precisely the composition of the corresponding representatives. We can choose a single resolution of $$k$$, and do all the calculations inside $$\operatorname{Hom}_{\mathrm{Mod}_R}(P_*,P_*)$$, which is a differential graded algebra, with homology precisely $$\operatorname{Ext}_{\mathrm{Mod}_R}(k,k)$$.

Another interpretation, not in fact relying on the existence of projective or injective modules is that of Yoneda splices. Then we take the viewpoint above that an element of $$\operatorname{Ext}^n_{\mathrm{Mod}_R}(A,B)$$ is an exact sequence starting in $$A$$ and ending in $$B$$. This is then spliced with an element in $$\operatorname{Ext}^m_{\mathrm{Mod}_R}(B,C)$$, by replacing
 * $$\rightarrow X_1\rightarrow B\rightarrow 0$$ and $$0\rightarrow B\rightarrow Y_n\rightarrow$$

with
 * $$\rightarrow X_1\rightarrow Y_n\rightarrow$$

where the middle arrow is the composition of the functions $$X_1\rightarrow B$$ and $$B\rightarrow Y_n$$.

These viewpoints turn out to be equivalent whenever both make sense.

Using similar interpretations, we find that $$\operatorname{Ext}_{\mathrm{Mod}_R}^*(k,M)$$ is a module over $$\operatorname{Ext}^*_{\mathrm{Mod}_R}(k,k)$$, again for sufficiently nice situations.

Interesting examples
If $$\mathbb ZG$$ is the integral group ring for a group $$G$$, then $$\operatorname{Ext}^*_{\mathrm{Mod}_{\mathbb ZG}}(\mathbb Z,M)$$ is the group cohomology $$H^*(G,M)$$ with coefficients in $$M$$.

For $$\mathbb F_p$$ the finite field on $$p$$ elements, we also have that $$H^*(G,M)=\operatorname{Ext}^*_{\mathrm{Mod}_{\mathbb F_pG}}(\mathbb F_p,M)$$, and it turns out that the group cohomology doesn't depend on the base ring chosen.

If $$A$$ is a $$k$$-algebra, then $$\operatorname{Ext}^*_{\mathrm{Mod}_{A\otimes_k A^{op}}}(A,M)$$ is the Hochschild cohomology $$\operatorname{HH}^*(A,M)$$ with coefficients in the module M.

If $$R$$ is chosen to be the universal enveloping algebra for a Lie algebra $$\mathfrak g$$, then $$\operatorname{Ext}^*_{\mathrm{Mod}_R}(R,M)$$ is the Lie algebra cohomology $$\operatorname{H}^*(\mathfrak g,M)$$ with coefficients in the module M.