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Let $A$ be a finite dimensional algebra. The big finitistic dimension of $A$ is

$$\operatorname{FinDim}(A)=\sup\{\operatorname{pd}(M)\mid M\in \text{Mod-}A \text{ and } \operatorname{pd}(M)<\infty\},$$ where $\operatorname{pd}(M)$ is the projective dimension of $M$, $\text{Mod-}A$ is the category of right $A$-modules.

Problem: Does there exist a finite dimensional algebra $A$ such that $\operatorname{FinDim}(A)=\infty?$

http://cn.arxiv.org/abs/1804.09801 (see Definition 4.1.)

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  • $\begingroup$ A more interesting question (although not very mathematical) would be whether one thinks that the conjecture is true. I think it is false. The opinion seems to be roughly split 50-50 on true-false when I asked several people working on it. $\endgroup$ – Mare Oct 9 '18 at 12:32
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You quoted a very recent arxiv article, where it is stated that this is an open problem. No progress has been made towards this open problem since then.

The only big progress in the questions sourounding the finitistic dimension conjecture was made 1992 in https://link.springer.com/article/10.1007/BF02100610 where it was proven that the first finitistic dimension conjecture is wrong.

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    $\begingroup$ Just to back up what Mare says: I'm the author of the article linked to in the OP, and since writing it I have spoken to, or had email exchanges with, most of the people who know most about the finitistic dimension conjecture(s). I'm pretty sure I would have heard if anybody had found a counterexample to the big finitistic conjecture. And I haven't. $\endgroup$ – Jeremy Rickard Oct 9 '18 at 12:06
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    $\begingroup$ @JeremyRickard Do you believe the finitistic dimension conjecture is true (big or small)? $\endgroup$ – Mare Oct 9 '18 at 12:34
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    $\begingroup$ No. But not with very strong conviction. I wouldn't be very surprised if they turned out to be true. $\endgroup$ – Jeremy Rickard Oct 9 '18 at 12:50

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