Suppose that we have the following commutative diagram of graded Lie algebras

$$\begin{array} A 0& {\longrightarrow} & C_n & {\longrightarrow} & A_{n+1} &{\longrightarrow} & A_n & {\longrightarrow} &0 \\ & & \downarrow{} & &\downarrow{}& &\downarrow{}\\ 0& {\longrightarrow}&D_n & \stackrel{}{\longrightarrow} &B_{n+1} & \stackrel{}{\longrightarrow} & B_n & {\longrightarrow}&0 & \end{array}$$ for all $n\in\mathbb{Z}^+$, where the both rows are split exact sequences and every vertical map is onto.

My question is, suppose that we know $A_n$, $C_n$, $D_n$ for all $n\in\mathbb{Z}^+$ and $B_1$, do we have enough information to uniquely determine up to isomorphism every $B_n$ (likely by induction)?

Here by "know", I mean that we can write $A_n$ (similarly $C_n, D_n, B_1$) as the free Lie algebra modulo some known relations $$B_n=L[S_n]/I_n,$$ where $L[S_n]$ is the free Lie algebra generated by a set $S_n$ and $I_n$ is 2-sided Lie ideal generated by some known relations.

If the answer to my question is positive, then is there a systematic way to determine the structure of $B_n$?


It looks to me like it is true that $B_n$ is uniquely determined. Let $K_n$ be the kernel of the map $A_n\to B_n$ and let $E_n$ be the kernel from $C_n$ to $D_n$. Then we have an exact sequence $0\to E_n\to K_{n+1}\to K_n$. $E_n$ is known and $K_n$ is known inductively. That means that $K_{n+1}$ can be determined as the subspace of $A_{n+1}$ generated by all preimages of $K_n$ and images f $D_n$. Therefore $B_{n+1}=A_{n+1}/K_{n+1}$ is uniquely determined.

  • $\begingroup$ Thank you for your answer! May I clarify two things? 1. Why do we have the exact sequence $0\to D_n\to K_{n+1}\to K_n\to 0$? 2. You mentioned that $K_n$ is known inductively (from the exact sequence?). If we can find $K_n$ inductively from the exact sequence $0\to D_n\to K_{n+1}\to K_n\to 0$, then why not finding $B_n$ directly from the exact sequence $0\to D_n\to B_{n+1}\to B_n\to 0$? $\endgroup$ – Zuriel Nov 22 '14 at 7:28
  • $\begingroup$ @Zuriel: I've sketched the inductive step. Assume $K_n$ is known, and then this argument tells you how to find $K_{n+1}$. (Try it out with $n=1$.) $\endgroup$ – Cheerful Parsnip Nov 22 '14 at 11:50
  • $\begingroup$ @Zuriel: the reason you can't do this directly with $B_{n+1}$ is that an exact sequence with known first and last term does not determine the middle term! However, it does when we are looking at subspaces of a known exact sequence. $\endgroup$ – Cheerful Parsnip Nov 22 '14 at 11:51
  • $\begingroup$ @Zuriel: also, there was a typo! The kernel should have been $E_n$. $\endgroup$ – Cheerful Parsnip Nov 22 '14 at 11:56
  • 1
    $\begingroup$ If you have an exact sequence $0\to A\to B\to C\to 0$ which includes in an exact sequence $0\to A'\to B'\to C'\to 0$, then $B<B'$ is the set of all preimages of elements of $C$ and images of $A$. $\endgroup$ – Cheerful Parsnip Nov 24 '14 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.