This looks like a (slightly) non-additive version of Grothendieck's theory of
"extensions panachées" (SGA 7/I, IX.9.3). There he considers objects (in some
abelian category) $X$ together with a filtation $0subseteq X_1subseteq
X_2subseteq X_3=X$. In the first version he also fixes (just as one does for
extensions) isomorphisms $Prightarrow X_1$, $Qrightarrow X_2/X_1$ and
$Rrightarrow X_3/X_2$. However, in the next version he fixes the isomorphism
class of the two extensions $0rightarrow Prightarrow X_2rightarrow
Qrightarrow0$ and $0rightarrow Qrightarrow X_3/X_1rightarrow Rrightarrow0$
so that if $E$ is an extension of $P$ by $Q$ and $F$ is an extension of $Q$ by
$R$, then the category $mathrm{EXTP}(F,E)$ has as objects filtered objects $X$
as above together with fixed isomorphisms of extensions $Erightarrow X_2$ and
$Frightarrow X_3/X_1$ and whose morphisms are are morphisms of $X$'s preserving
the given structures. The morphisms of $mathrm{EXTP}(F,E)$ are necessarily
isomorphisms so we are dealing with a groupoid. Similarly for objects $A$ and
$B$ $mathrm{EXT}(B,A)$ is the groupoid of extensions of $B$ by $A$.
Grothendieck then shows that $mathrm{EXTP}(F,E)$ is a torsor over
$mathrm{EXT}(R,P)$ (in the category of torsors, Grothendieck had previously
defined this notion). The action on objects of an extension $0rightarrow
Prightarrow Grightarrow Rrightarrow0$ is given by first taking the pullback
of it under the map $X/X_1rightarrow R$ and then using the obtained action by
addition on extensions of $P$ by $F$. To more or less complete the picture,
there is an obstruction to the existence of an object of $mathrm{EXTP}(F,E)$:
We have that $E$ gives an element of $mathrm{Ext}^1(Q,P)$ and $F$ one of
$mathrm{Ext}^1(R,Q)$ and their Yoneda product gives an obstruction in
$mathrm{Ext}^2(P,Q)$.
The case at hand is similar (staying at the case of $n=3$ and with the caveat
that I haven't properly checked everything): We choose fixed isomorphisms with
$K_2$ and a given central extension and with $K_3/K_1$ and another given central
extension (assuming that we have three groups $P$, $Q$ and $R$ as before)
getting a category $mathrm{CEXTP}(F,E)$ of central extensions. We shall shortly
modify it but to motivate that modification it seems a good idea to start with
this. We get as before an action of $mathrm{CEXT}(R,P)$ on
$mathrm{CEXTP}(F,E)$ as we can pull back central extensions just as before. It
turns however that the action is not transitive. In fact we can analyse both the
difference between two elements of $mathrm{CEXTP}(F,E)$ and the obstructions
for the non-emptyness of it by using the Hochschild-Serre spectral sequence. To
make it easier to understand I use a more generic notation. Hence we have a
central extension $1rightarrow Krightarrow Grightarrow G/Krightarrow1$ and
an abelian group $M$ with trivial $G$-action. There is then a succession of two
obstructions for the condition that a given central extension of $M$ by $G/K$
extend to a central extension of $M$ by $G$. The first is $d_2colon
H^2(G/K,M)rightarrow H^2(G/K,H^1(K,M))$, the $d_2$-differential of the H-S
s.s. Now, we always have a map $H^2(G/K,M)rightarrow H^2(G/K,H^1(K,M))$ given
by pushout of $1rightarrow Grightarrow G/Krightarrow1$ along the map
$Krightarrow mathrm{Hom}(K,M)=H^1(K,M)$ given by the action by conjugation of
$K$ on the given central extension of $M$ by $K$ (equivalently this map is given
by the commutator map in that extension). It is easy to compute and identify
$d_2$ but I just claim that it is equal to that map by an appeal to the What Else
Can It Be-principle (which works quite well for the beginnings of spectral
sequences with the usual provisio that the WECIB-principle only works up to a
sign).
This means that we can cut down on the number of obstructions by redefining
$mathrm{CEXTP}(F,E)$. We add as data a group homomorphism $varphicolon
K_3/K_1rightarrowmathrm{Hom}(Q,P)$ that extends $Qrightarrow
mathrm{Hom}(Q,P)$ which describes the conjugation action on $K_2$ and only look
the elements of $mathrm{CEXTP}(F,E)$ for which the action is the given
$varphi$ to form $mathrm{CEXTP}(F,E;varphi)$. Now the action of
$mathrm{CEXT}(R,P)$ on $mathrm{CEXTP}(F,E;varphi)$ should make
$mathrm{CEXTP}(F,E;varphi)$ a
$mathrm{CEXT}(R,P)$-(pseudo)torsor. Furthermore, there is now only a single
obstruction for non-emptyness which is given by $d_3colon H^2(R,M)rightarrow
H^3(P,M)$.
Going to higher lengths there are two ways of proceeding in the original
Grothendieck situation: Either one can look at the the two extensions of one
length lower, one ending with the next to last layer (i.e., $X_{n-1}$) and the
other being $X/X_1$. This reduces the problem directly to the original case
(i.e., we look at filtrations of length $n-2$ on $Q$). One could instead look at
the successive two-step extensions and then look at how adjacent ones build up
three-step extensions and so on. This is essentially an obstruction theory point
of view and quickly becomes quite messy. An interesting thing is however the
following: We saw that in the original situation the obstruction for getting a
three step extension was that $ab=0$ for the Yoneda product of the two twostep
filtrations. If we have a sequence of three twostep extensions whose three step
extensions exist then we have $ab=bc=0$. The obstruction for the existence of
the full fourstep extension is then essentially a Massey product $langle
a,b,crangle$ (defined up to the usual ambiguity). The messiness of such an
iterated approach is well-known, it becomes more and more difficult to keep
track of the ambiguities of higher Massey products. The modern way of handling
that problem is to use an $A_infty$-structure and it is quite possible (maybe
even likely) that such a structure is involved.
If we turn to the current situation and arbitrary $n$ then the first approach
has problems in that the midlayer won't be abelian anymore and I haven't looked
into what one could do. As for the second approach I haven't even looked into
what the higher obstructions would look like (the definition of the first
obstruction on terms of $d_3$ is very asymmetric).
No comments:
Post a Comment