# (geometric/intuitive) interpretation of ext

In my current work I have to deal a lot with ext-groups (of modules). I feel kind of familar with the formalism, e.g. the connection between n-th extensions and ext. But I don't have a feeling about the meaning of $Ext$. Is there a informal/intuitive (geometric) interpretation of $Ext(M,N)$ in terms of Morphisms $M \to N$?

For example, how far is the following interpretation away from beeing right?

Consider a free resolution $\cdots \to F_n \to \cdots \to F_1 \to F_0 \to M$ of $M$ does $Ext^i(M,N)$ tells me something about the morphisms in the i-th syzygy $M_i$? e.g consists $Ext^1(M,N)$ of the morphisms of the module generated by the relations of the generators of $M$ modulo the ones, which come from the trivial relations?

I'm mostly interested in the case of $\mathcal{O}_X$-Modules for (toric) varieties or $\mathbb{C}[S]$-Algebras for a semi-groups $S$.

best regards, Johannes

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This was crossposted from MO. In the future, when you post your question in multiple fora please provide links to the other posts - as you can imagine, it would be frustrating for someone to put time into answering your question here, only to see hear from you that you'd already gotten the solution elsewhere. –  Zev Chonoles Jan 4 '13 at 11:11

You specifically ask about Ext in terms of syzygies. Consider modules over a ring $R$, let $F_* \to M$ be a projective resolution with differential $d_*$, write $\Omega^n (M)$ for the $n$th syzygy $\ker d_{n-1}$ in this resolution. Then $$\operatorname{Ext}_R^n(M,N) \cong \hom_R(\Omega^n(M), N) / X$$ (isomorphism of abelian groups) where $X$ denotes the space of morphisms factoring through $\Omega^n(M) \hookrightarrow F_n$. In special circumstances you can assume $X=0$, e.g. $N$ simple, $F_*$ a minimal resolution. A good place to learn about this stuff in more detail is Benson's Representations and Cohomology I.

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2mt_: thnx, something like this is exactly what I was looking for. can you give me a source or idea of the proof of this statement? –  Johannes Jan 4 '13 at 18:25
I gave you a source in my last sentence. The idea is that in $\hom_R(F_*,N)$, a morphism $f$ is in the kernel of the differential iff $f \circ d = 0$ iff $f$ kills $\operatorname{im} d$, in which case $f$ induces a well-defined map on $F_n / \operatorname{im} d_{n+1} = F_n/\ker d_n \cong \operatorname{im} d_n = \Omega^n(M)$. –  mt_ Jan 4 '13 at 18:50

As mentioned on MO, this http://www.math.wayne.edu/~isaksen/Expository/carrying.pdf gives a wonderful elementary viewpoint.

I have another, simple explanation using matrices that explicitly shows why cocycles arise.

Suppose we consider representations of a group $G$ over a field $K$. Suppose we have two one-dimensional representations $\chi_1,\chi_2$ (also known as "characters"). We'd like to find extensions of $\chi_2$ by $\chi_1$, i.e. representations $V$ such that there is an exact sequence $$1 \to \chi_1 \to V \to \chi_2 \to 1.$$

Now $V$ will automatically be two-dimensional (over $K$), so we can write the representation as $\rho:G \to \mathrm{GL}_2(K)$. Choose a basis adapted to the subspace $\chi_1$, so that $\rho(g)$ for $g \in G$ has a matrix representation $$\begin{pmatrix} \chi_1(g) & f(g)\\ 0 & \chi_2(g) \end{pmatrix}$$

where $f$ is some function $G \to K$. Notice that if $f$ is always zero, then $V$ is the trivial extension (i.e. the product of the two one-dimensional representations), and conversely, if $V$ is trivial, then choosing an appropriate basis makes the function $f$ equal to $0$. In other words, this upper right matrix entry governs the non-triviality of the extension.

Now, the fact that $\rho(gh)=\rho(g)\rho(h)$ implies that $$\begin{pmatrix} \chi_1(gh) & f(gh)\\ 0 & \chi_2(gh) \end{pmatrix} = \begin{pmatrix} \chi_1(g) & f(g)\\ 0 & \chi_2(g) \end{pmatrix} \begin{pmatrix} \chi_1(h) & f(h)\\ 0 & \chi_2(h) \end{pmatrix} = \begin{pmatrix} \chi_1(gh) & \chi_1(g)f(h)+\chi_2(h)f(g)\\ 0 & \chi_2(gh) \end{pmatrix}$$

Therefore, $f(gh)= \chi_1(g)f(h)+\chi_2(h)f(g)$. One can (and should) think of this as a cocycle condition for this map $G \to K$.

In particular, if $\chi_1=\chi_2=1$, then $f$ is a homomorphism $G \to K$.

Intuitively, this explains why representations of a finite group over a field of characteristic prime to the order of the finite group should be semisimple; the only homomorphisms from that finite group to the additive group of the field are trivial.

Exercise: Think about how an isomorphism between two extensions gives a coboundary making the two cocycles cohomologous.

Note that one can cook up cocycles for arbitrary finite-dimensional representations; the block structure would just look a bit different. If we examine $\mathrm{Ext}(K,M)$, where $K$ is the trivial one-dimensional representation and $M$ has degree $m$, then an extension will be a representation of $G$ into $(m+1) \times (m+1)$ matrices, with a block of $m$ in the upper left corresponding to $M$. The "upper right" will be a column of length $m$. This makes sense - the target of the mapping out of $G$ should be into $M$. In fact, this is a cocycle for the first group cohomology of $M$. More generally, a cocycle for an element of $\mathrm{Ext}^1(N,M)$ is a map from $G$ into $\mathrm{Hom}_K(N,M)$. This makes sense, for if $N$ has degree $n$, then the upper-right entry should be an $m \times n$ matrix, which should represent an element of $\mathrm{Hom}_K(N,M)$.

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"Is there a informal/intuitive (geometric) interpretation of $\text{Ext}(M,N)$ in terms of Morphisms $M \longrightarrow N$?"

There is an interpretation of $\operatorname{Ext}$ groups exactly as some kind of morphisms $M \to N$. For this you consider instead of abelian category $A$ (of modules or sheaves of modules) derived category $D(A)$, then $$\text{Ext}^i(M,N)=\text{Hom}_{D(A)}(M,N[i]).$$ This is nothing more than interpretation of $\operatorname{Ext}$ in terms of resolutions, but sometimes it could simplifies calculations (for example, definition of product is obvious) and it looks (at least for me) like more conceptually clear definition.

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thnx, I heared of it. but I did not remember. That's a very nice formula, but now you have to give me an intuition about derived categories ;) –  Johannes Jan 4 '13 at 18:23