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Suppose $\mathcal{C}$ is a dg category (over some base) with all colimits. We say that $X\in \mathcal{C}$ is a generator if $\mathcal{C}$ is equivalent to $\operatorname{End}_\mathcal{C}X$-modules (via the Yoneda functor). My question: are there some "niceness" conditions we can impose on the pair $(\mathcal{C}, X)$ which guarantee that all deformations of $X$ (i.e. all fibers of a suitably perfect sheaf of objects of $\mathcal{C}$ over a curve with special fiber $X$, say) are still generators?

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  • $\begingroup$ Rigidity is sufficient. $\endgroup$
    – Sasha
    Commented Dec 4, 2016 at 18:00
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    $\begingroup$ You mean non-existence of nontrivial deformations? Thanks :) $\endgroup$
    – Dmitry Vaintrob
    Commented Dec 4, 2016 at 18:04
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    $\begingroup$ Could you explain or give a reference for what do you mean by a sheaf of objects of $\mathcal C$ here? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 4, 2016 at 19:00
  • $\begingroup$ I am open to interpretation. But a simple working definition on an affine variety $\text{Spec}(R)$ would be a compact object in the category of objects of $\mathcal{C}$ with $R$-action. $\endgroup$
    – Dmitry Vaintrob
    Commented Dec 4, 2016 at 21:37

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Let me give an example (which you are probably aware of) that shows how a generator may become a non-generator after a deformation.

Let $\mathcal C$ be the derived category of $\mathbb{P}^1$ and $X = \mathcal{O} \oplus \mathcal{O}(-1)[1]$. This is clearly a generator. On the other hand, we can deform the sum to an extension $$ 0 \to \mathcal{O}(-1) \to \mathcal{O} \to \mathcal{O}_P \to 0, $$ where $P \in \mathbb{P}^1$ is a point, and the sheaf $\mathcal{O}_P$ is definitely not a generator.

I think a similar trick can be used in a much more general situation, so I do not believe there is a reasonable condition (besides the rigidity of $X$) ensuring that the property "being a generator" is open.

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    $\begingroup$ Thanks! This is a great example that I should have thought of before posting. I suppose it shows that the dga $\text{End}^*_{\mathbb{P}^1}(\mathcal{O}_P)$ is a deformation of $\text{End}^*_{\mathbb{P}^1}(\mathcal{O}\oplus \mathcal{O}(-1)[1]),$ but they are not deformations in the opposite order (which is why I was curious about this) $\endgroup$
    – Dmitry Vaintrob
    Commented Dec 5, 2016 at 20:35

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