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Let $ R $ be a local Noetherian Gorenstein domain. Suppose a module $ M $ fits into an exact sequence $$ 0 \rightarrow K \rightarrow R^n \rightarrow M \rightarrow 0 $$ Then we write $ K = tM $. A module $ M $ is defined to be Cohen-Macaulay if $ M $ can be written as $ t(tN) $ for some module $ N $. I understand that this is not the standard definition but I'd like to work with this definition for reasons below.

Suppose $ M $ itself isCohen-Macaulay. In almost every case I've seen, $ M $ may actually be written as $ t(tM) $. Here's a worked out example:

Let $ R = \mathbb{C} [[x,y,z]] / (xy - z^3) $. Consider $ M_1 = (x,z) $, the ideal generated by $ x $ and $ z $ in $ R $. Then we have a sequence $$ 0 \rightarrow (x, z^2) \rightarrow R^2 \rightarrow (x,z) \rightarrow 0 $$ where the first map is $ f \rightarrow ( \frac{-zf}{x}, f) $ (This is well defined, as $ f = xp + z^2q \implies \frac{-zf}{x} = -zp - \frac{z^3q}{x} = -zp - yq $) and it can be checked easily that this is exact. Similarly, we have $$ 0 \rightarrow (x, z) \rightarrow R^2 \rightarrow (x,z^2) \rightarrow 0 $$ where the first map now is $ f \rightarrow ( \frac{-z^2f}{x}, f) $. These two computations show that $ M_1 = t(tM_1) $. And also show that if $ M_2 = (x,z^2) $, then $ M_2 = t(tM_2) $.

I've heard that this is no coincidence and has connections with mirror symmetry. Can I know the reason behind why this is happening?

Edit : I goofed up in the definition, now it should be clear.

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    $\begingroup$ $tM$ is far from being well-defined — any module with a surjective map to $K$ qualifies. So your "$t(tM)$" is not correctly defined. $\endgroup$
    – abx
    Commented Jan 12, 2022 at 20:32
  • $\begingroup$ But I never said it is well-defined, it's just a notation. $\endgroup$
    – Cranium Clamp
    Commented Jan 12, 2022 at 20:53
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    $\begingroup$ To spell out concretely then, $ M $ is C-M if there exist two sequences $ K \rightarrow R^n \rightarrow N \rightarrow 0 $ and $ M \rightarrow R^m \rightarrow K \rightarrow 0 $. I don't see any problems here but correct me otherwise. $\endgroup$
    – Cranium Clamp
    Commented Jan 12, 2022 at 20:57
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    $\begingroup$ Why confuse the definition of CM? The object you define has already a name - second syzygy module. $\endgroup$
    – Mohan
    Commented Jan 13, 2022 at 2:54
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    $\begingroup$ I guess that it might be more reasonable to write something like $K\in T(M)$ where $T(M)$ is the collection of $\ker(P\to M)$ and $P\to M$ run through all surjections from finite free modules, and you write $T(T(M))$ for the collection of $T(N)$ for $N$ running through all modules in $T(M)$. $\endgroup$
    – Z. M
    Commented Jan 13, 2022 at 7:33

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