# Can the 'linkages' between equivalent extensions of modules of an algebraic group be taken to have bounded length?

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.

• In your diagrams, it looks like the vertical maps at the ends of your linkage diagrams are merely isomorphisms. To be clear, I think these maps need to be equalities, and not just isomorphisms. – Christopher Drupieski Nov 25 '11 at 0:18
• Also, your examples for the trivial group are wrong, as exact sequences have trivial Euler characteristic. So $\xi$ is no exact sequence. – Wilberd van der Kallen Nov 25 '11 at 8:46

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. 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.

Take a projective resolution $P_\bullet$ of $M$ and let $K = Ker(P_{n-2} \to P_{n-3})$. Since $P_i$ are projective there is an isomorphism $Ext^n(M,N) = Ext^1(K,N)$. Let $0 \to N \to E_\xi \to K \to 0$ be the short exact sequence corresponding to the class of $\xi$. Then it follows that there is a link form $$0 \to N \to E_\xi \to P_{n-2} \to \dots \to P_0 \to M$$ to your $\xi$. Since $E_\xi \cong E_\nu$ the same argument show that there is a link from the above extension to your $\nu$. So, the length of a linkage is bounded by 2.

Thanks for the reply, but there are two problems with this (which is not your fault, I should have been clearer). Firstly, I am working with modules over an algebraic group, for which there are not necessarily enough projectives. And yes, I had tried to make the above argument work for injective resolutions (of which there are enough), but have run into some issues. Do you see a way to make your argument work using an injective resolution?

The second problem (and I didn't make this clear) is that I was really hoping to take the chain of linkages to be between extensions comprised of only finite dimensional modules, and I don't see that an algebraic group has enough finite dimensional injectives. If I could find a length-$2$ linkage between an infinite dimensional extension comprised of injectives, that would be worth something to me, but to be able to take them to be finite dimensional would be much better.

Of course, this raises a question that I hadn't even considered: given that $\xi$ and $\nu$ are equivalent, does there even necessarily exist a chain of linkages from $\xi$ to $\nu$ using extensions comprised of only finite dimensional modules, let alone one of bounded length?

• Next time indicate whose reply you are responding to. – Wilberd van der Kallen Nov 24 '11 at 19:31
• In general, an algebraic group may not have any finite-dimensional rational injective modules. – Christopher Drupieski Nov 25 '11 at 0:17