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Hey Everyone

Let $A$ be an algebra over a field (group rings $k[G]$ for group cohomology, the Steenrod Algebra). We want to compute, say, $Ext_A(k,k)$, so let $F_*\to k$ be an $A$-free resolution. Elements of $Ext^j_A(k,k)$ can be represented as Yoneda extensions, essentially exact sequence $k\to M_1 \to ... \to M_{j} \to k$, which we denote $\mathcal{X}$, and a chain map map $F_*\to \mathcal{X}$ with the cocycle, $F_{j+1}\to k$, on one side and the identity map $k\to k$ on the other. Using pushouts one can, given a cocycle, generate $\mathcal{X}$, but the modules in $\mathcal{X}$ can be rather large.

My question is then this. Is there a way to either generate a "small" extension (measured however you'd like, perhaps the size of an extension should be the largest dimension over $k$ in any bi-digree) representing a cocycle, or given the a Yoneda extension, is there a way to make some sort of local change to it that can make it "smaller" (in which case we can search for a "local optimum").

This question is motivated as a workaround to the problem presented by Bob Bruner's in


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The Steenrod Algebra is not of finite type. – Fernando Muro Nov 25 '12 at 22:12
Oops, Why on earth did I write finite type... I must be typing without thinking. – Joseph Victor Nov 26 '12 at 6:58
trivial, unless $G$ itself is graded. – Joseph Victor Nov 26 '12 at 18:12
To be honest, I didn't mean to write "graded algebra of finite type". I fixed the finite type part. – Joseph Victor Nov 26 '12 at 18:47

Assume that a projective resolution $P \to k$ over $A$ is given (as far as I understand, Bruner is able to compute such a $P$ in a reasonable amount of time) and that $f: P_n \to k$ is a cocycle. Than there is an extension $\mathcal{X}$ representing the class of $f$ that is as small as $P$ is.

In more detail: Let $L := d_n(\ker(f)) \le P_{n-1}$ where $d_n: P_n \to P_{n-1}$. Then the class of $f$ is represented by the extension $$\mathcal{X}:\qquad 0 \to k \to P_{n-1}/L \to P_{n-2} \to \cdots \to P_0 \to k \to 0$$ (cf. page 9 of the paper Benson, Carlson: Projective resolutions and Poincaré duality complexes. Can be found on: and holds for each augmented $k$-algebra $A$).

Usually one will try to choose $P$ as small as possible. If $A$ is Artinian (which holds for instance if $A=kG$ with $G$ finite) then each f.g. $A$-module has a projective cover. Hence $k$ has a minimal projective resolution which is a natural candidate for $P$.

If $A$ is only known to be Noetherian, one can construct a "small" resolution by computing a minimal generating set for $\ker(d_n)$ as $A$-module (let it have $k_n$ elements) and by choosing $P_{n+1} := A^{k_n}$ with the obvious map $P_{n+1} \twoheadrightarrow \ker(d_n) \hookrightarrow P_n$.

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Thanks Ralph. In the case of group cohomology, in playing around I was able to make $\mathcal{X}$ much smaller than the bar resolution, which obviously is not minimal. However, I was hoping to make \mathcal{X} smaller than the resolution being used even if that resolution was itself minimal (in the sense of the algorithm from the last paragraph). Is this something that can be done? – Joseph Victor Nov 26 '12 at 7:05

You ask for a way to make local changes to decrease the size of the extension. Here are some observations for length two:

If $M \cap \operatorname{im} \iota = \{0\}$ then

$$ 0 \to k \stackrel \iota \to A \stackrel j \to B \stackrel \pi \to k \to 0$$ is equivalent to $$ 0 \to k \to A/M \to B/j(M) \to k \to 0$$ via the obvious maps. Equally if $N \leq B$ and $\pi|_N \neq 0$ then the sequence is equivalent to $$0 \to k \to j^{-1}(N) \to N \to k \to 0$$

I imagine that using these is enough to reduce the size considerably in many cases.

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I tried something like this but game up. I don't think these are still exact. Pick some 1-d $N$ which maps onto $k$, then $j^{-1}(N)=0$ but exactness, but $k\to 0$ is not injective. Am I missing something? – Joseph Victor Nov 27 '12 at 3:26
$j$ has a kernel equal to $\operatorname{im} \iota$, so $j^{-1}(N)$ contains $\operatorname{im} \iota$ and is never zero. In the case you describe, the sequence splits. – M T Nov 27 '12 at 8:42
Good point, mt. These are clearly exact. – Joseph Victor Nov 27 '12 at 18:37

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