I have a conjecture concerning how "tightly" two equivalent $n$-fold extensions of modules over an algebraic group over a field might
be "linked". I suspect that the 
question has been already studied in at least some depth, and a
positive or negative answer (or partial positive or negative answer) would either way be greatly helpful.


**Background.** Let $G$ be an algebraic group over a field $k$. Let
$M$ and $N$ be finite dimensional (as $k$-vector spaces)
$G$-modules (that is, representations of $G$ over $k$), and let


$\xi:0 \rightarrow N \rightarrow X\_{n-1} \rightarrow \ldots
\rightarrow X\_0 \rightarrow M \rightarrow 0$


$\nu: 0 \rightarrow N \rightarrow Y\_{n-1} \rightarrow \ldots
\rightarrow Y\_0 \rightarrow M \rightarrow 0$


be two $n$-fold extension of $M$ by $N$, where all of the $X\_i$
and $Y\_i$ are also finite dimensional. Call the extensions $\xi$
and $\nu$ "linked" if there is a "linkage" between them, that is,
a commutative diagram


\begin{array}{cccccccccccccc}
\xi: 0 &\to & N & \to & X\_{n-1} & \to & \ldots & \to & X\_0 & \to &
M & \to & 0 \\ \newline & & \wr\downarrow & & \downarrow & & & &
\downarrow & & \wr\downarrow \\ \newline
\nu: 0 &\to & N & \to & Y\_{n-1} & \to & \ldots & \to & Y\_0 & \to & M & \to & 0 \\
\end{array}


Call $\xi$ and $\nu$ "equivalent" if they are equivalent with
respect to the equivalence relation generated by "linked", that
is, if there is a finite chain of linkages leading from $\xi$ to
$\nu$. (For those of you more accustomed to other ways of thinking of the $\text{Ext}$ functor, this is the same as saying that $\xi = \nu$ as elements of $\text{Ext}^n\_{G(k)}(M,N)$.) 


**The conjecture:** There is a positive integer $m$ (perhaps depending on $G$, $k$,
$n$, $M$ and $N$, but *not* on the particular extensions $\xi$ and
$\nu$) such that, if $\xi$ and $\nu$ are equivalent, then they are
linked via a linkage of length no greater than $m$.


**Remarks.** To illustrate what is going on here, consider three $n$-fold
extensions $\xi, \tau$ and $\nu$ of $M$ by $N$, and suppose we
have a commutative diagram


\begin{array}{cccccccccccccc}
\xi: 0 &\to & N & \to & X\_{n-1} & \to & \ldots & \to & X\_0 & \to &
M & \to & 0 \\ \newline & & \wr\downarrow & & \downarrow & & & &
\downarrow & & \wr\downarrow \\ \newline \tau: 0 &\to & N & \to &
Z\_{n-1} & \to & \ldots & \to & Z\_0 & \to & M & \to & 0 \\
\newline
& & \wr\uparrow & & \uparrow & & & & \uparrow & & \wr\uparrow \\
\newline
\nu: 0 &\to & N & \to & Y\_{n-1} & \to & \ldots & \to & Y\_0 & \to &
M & \to & 0 \\ \newline
\end{array}


Then $\xi$ and $\nu$ are equivalent, since there is a length-$2$ chain
of linkages between them, namely $\xi \rightarrow \tau \leftarrow
\nu$. But it is by no means guaranteed that there is an ACTUAL
linkage between $\xi$ and $\nu$, that is, a commutative diagram


\begin{array}{cccccccccccccc}
\xi: 0 &\to & N & \to & X\_{n-1} & \to & \ldots & \to & X\_0 & \to &
M & \to & 0 \\ \newline & & \wr\downarrow & & \downarrow & & & &
\downarrow & & \wr\downarrow \\ \newline \nu: 0 &\to & N & \to &
Y\_{n-1} & \to & \ldots & \to & Y\_0 & \to & M & \to & 0 \\
\newline
\end{array}


The conjecture asks: suppose that all you know is that there
is *some* finite chain of linkages leading from $\xi$ to $\nu$, of
perhaps very huge length. Does there then also exist a chain of
linkages leading from $\xi$ to $\nu$ of length no greater than
$m$? (again, $m$ perhaps depending on the group, the field, the
length $n$ of the extensions, and the modules $M$ and $N$, but *not* on the particular extensions $\xi$ and $\nu$ [this would be utterly trivial of course]).


Note that this conjecture is trivially true for the case $n=1$.
This is because a linkage between $1$-fold extensions


\begin{array}{cccccccccccccc}
\xi: 0 &\to & N & \to & X\_{0} & \to & M & \to & 0 \\ \newline & &
\wr\downarrow & & \wr\downarrow & & \wr\downarrow \\ \newline
\nu: 0 &\to & N & \to & Y\_{0} & \to & M & \to & 0 \\
\end{array}


is necessarily an isomorphism of exact sequences (that is, the
second map of the linkage is also invertible). Thus you can take
any finite length chain of linkages, compose them, and you have a
single linkage. Thus, for $n = 1$, no matter $G$, $k$, $M$ or $N$, you can take in all cases $m=1$.


I also suspect that the conjecture is true for some further rather
trivial examples. For example, let $G = G\_0$, where $G\_0$ is the
trivial algebraic group over $k$ (that is, when $G$-modules are
nothing more than vector spaces over $k$, and morphisms between
them are nothing more than $k$-linear maps), let $n=2$, and let
$M$ and $N$ both be the vector space $k$. Consider then the
following $2$-fold extension of $k$ by $k$:


\begin{array}{cccccccccccc}
\xi: 0 &\to & k & \to & k^{100} & \to & k^{99} & \to & k & \to & 0 \\
\end{array}


It can be shown straightforwardly that this extension is
equivalent, via a single linkage, to


\begin{array}{cccccccccccc}
\tau: 0 &\to & k & \to & k^2 & \to & k & \to & k & \to & 0 \\
\end{array}


But so also can $\nu$ be linked to $\tau$, where $\nu$ is the
extension


\begin{array}{cccccccccccc}
\nu: 0 &\to & k & \to & k^{200} & \to & k^{199} & \to & k & \to & 0 \\
\end{array}


Since all extensions of $k$ by $k$ essentially looks like this,
this seems to say that, for $G = G\_0$, $n=2$, and $M = N = k$, we
can always take $m=2$.


The truth or falsity of these results have real implications towards my current research, and being much less than an expert in cohomology, they have been torturing me for some
time. I have no idea whether this result is in
general true, nor under what conditions it may be true.


Any partial answers (such as "if $G$ is this, $k$ is that, $n$ is this, and
if $M$ and $N$ are this and that, then the conjecture is
true/false") are very much welcome. For reasons I won't go into,
a sharp estimate of such a chain length bound $m$ is by no means required for my
purposes; all I need to know is whether or not such an $m$ exists.


Thanks so much for any advice or help.



Can the 'linkages' between equivalent extensions of modules of an algebraic group be taken to have bounded length?
The linkage bound is 2.


If the algebraic group is simple, say over an algebraically closed $k$,
then one has the following lemma. 


Lemma. If $V$, $W$ are finite dimensional,
there is an $m$ depending on $V$, $W$, so that if $St\_n$ 
is the $n$-th Steinberg module with $n\geq m$ the natural map
 $Ext^i(V,W)\to Ext^i(V,W\otimes
St\_n\otimes St\_n)$ vanishes for $i\geq 1$.


See for instance my 1977 
[joint paper with Cline, Parshall and Scott](http://resolver.sub.uni-goettingen.de/purl?GDZPPN002093200). So this gives a way to kill extension classes while staying within
the category of finite dimensional representations.
Now it seems one may argue as in Sasha's answer, dualized.


Indeed consider a representative $E$ of an element of $Ext^1(V,W)$.
Let us write $W\otimes St\_n \otimes St\_n$ as $I\_n(W)$.
For $n$ large the lemma implies that $E$ is the Yoneda composite $P\circ f$
of $P:0\to W\to I\_n(W)\to I\_n(W)/W\to 0$ with an element 
$f:V\to I\_n(W)/W$. Now if one has a representative of $Ext^i(V,W)$ with $i>1$, 
write it as
Yoneda composite $E\circ F$ of an 
$E\in Ext^1(Z,W)$ and an $F\in Ext^{i-1}(V,Z)$. 
Applying the previous result one gets a
linkage map from $E\circ F$ towards a Yoneda composite $P\circ Q$ of the representative $P:0\to W\to I\_n(W)\to I\_n(W)/W\to 0$
with a representative $Q=f\circ F$ of an element of $Ext^{i-1}(V, I\_n(W)/W)$. 
So we use that there is a linkage from $(P\circ f)\circ F$ towards 
$P\circ(f\circ F)$. 
One repeats this untill one has a linkage map from the original $E\circ F$
towards a Yoneda 
composite $R\circ S$ of an extension 
$R:0\to W\to I\_{n\_1}(W)\to I\_{n\_2}(X\_1)\to \cdots I\_{n\_{i-2}}(X\_{i-2})\to
I\_{n\_{i-2}}(X\_{i-2})/X\_{i-2}\to 0$ [ depending only on $W$ ] with an element $S$ of $Ext^1(V,I\_n(X\_{i-2})/X\_{i-2})$.
Given a second representative of the same class of $Ext^i(V,W)$ one may repeat the construction and if the $n\_i$ are
big enough [ meaning they grow fast enough ] one ends up with the same class of $Ext^1(V,I\_n(X\_{i-2})/X\_{i-2})$,
by a dimension shift argument. So the linkage is bounded by 2 as in 
Sasha's answer.


For reductive groups over an algebraically closed $k$ the Lemma still holds,
interpreted appropriately. For nonreductive groups one may take an exhaustive 
filtration $k\subset M\_1\subset M\_2\cdots$ of the coordinate ring $k[G]$ by finite dimensional submodules [ for the action by left translation ]
and replace $St\_n \otimes St\_n$ with $M\_n$ in the argument above.
The argument needs to be modified a bit. For instance the lemma is no longer available. 
But the limit over $n$ of the $Ext^i(V,W\otimes M\_n)$
vanishes and that is enough to construct all the needed maps. 


