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.