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Let $A$ be a finite dimensional algebra over an algebraically closed field. I'm trying to better understand the difference between $A$ being representation infinite and $A$ being $\tau$-tilting infinite when $A$ is a string algebra. So, if $A$ is representation finite, then $A$ is also $\tau$-tilting finite.

I'm wondering what is an example of a string algebra $A$ that is representation infinite but $\tau$-tilting finite.

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  • $\begingroup$ Please don't edit the question after receiving a perfectly good answer. It's just not good form. $\endgroup$
    – Dave Benson
    Commented May 11, 2023 at 20:38
  • $\begingroup$ Thanks for reverting. $\endgroup$
    – Dave Benson
    Commented May 11, 2023 at 20:45

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I think the group algebra of a dihedral $2$-group in characteristic two, mod its socle, is an example. The smallest of these is $k\langle x,y\rangle/(x^2,y^2,xy,yx)$. See Plamondon, "$\tau$-Tilting finite gentle algebras are representation finite."

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  • $\begingroup$ So your example is basically the algebra given by one vertex and two loops quiver, with radical square zero relation. But this algebra is representation infinite and hence by the result of Plamondon it is $\tau$-tilting infinite as well. But I'm looking for an example that is rep infinite, but $\tau$-tilting finite. $\endgroup$
    – It'sMe
    Commented May 11, 2023 at 19:46
  • $\begingroup$ No, it isn't gentle. Read the introduction to Plamondon. Second paragraph. $\endgroup$
    – Dave Benson
    Commented May 11, 2023 at 19:56
  • $\begingroup$ But even then I don't think this is going to be $\tau$-tilting finite, because by the result of Demonet, Iyama and Jasso, $\tau$-tilting finite is same as brick-finite (i.e., $\mathrm{mod}-A$ has finitely many bricks $B$ ($\mathrm{End}_{A}(B)\cong\mathbb{K}$), up to isomorphism. But in the above quiver if you take the dimension vector to be $(2)$, then you have an infinite family of non-isomorphic bricks, and hence it is brick infinite, and therefore $\tau$-tilting infinite. $\endgroup$
    – It'sMe
    Commented May 11, 2023 at 20:05
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    $\begingroup$ Those aren't bricks. $\endgroup$
    – Dave Benson
    Commented May 11, 2023 at 20:06
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    $\begingroup$ To expand, for a local algebra the only brick is the one dimensional module. $\endgroup$
    – Dave Benson
    Commented May 11, 2023 at 20:17

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