Finitistic dimension conjecture

An ordinary vector can be described by a few independent coordinates. When vectors also carry algebraic operations, their generators may satisfy relations, and those relations may themselves have relations. For one fixed finite-dimensional algebra, can the shortest descriptions that finish after finitely many such steps ever require arbitrarily many steps?

The objects being described are finitely generated modules: vector-like spaces on which the algebra acts. The permitted building pieces are projective modules, which can occur as direct summands of free coordinate modules. The conjecture asks for one finite ceiling on the shortest finite projective resolutions for that algebra. Modules needing an infinite resolution are left out, and the ceiling may change with the algebra.[1]

A module can be presented by generators and relations, but resolving the relations may require further layers. For one fixed finite-dimensional algebra, must there be a ceiling on the number of layers in every shortest resolution that is finite? An ordinary vector space $K^2$, viewed as a module over the field $K$, already is free and has projective dimension zero; this easy case illustrates the count, not the general conjecture.

A free module $A^m$ consists of finite coordinate lists with entries in $A$. A projective module is a direct summand of a free module. A projective resolution maps a projective object $P_0$ onto $M$, maps $P_1$ onto the remaining kernel, and continues until no further kernel remains. The shortest possible finite length is $\operatorname{pd}_A(M)$. To define the little finitistic dimension, let $M$ range over finitely generated modules of finite projective dimension, and take

$$\operatorname{findim}(A)=\sup_M\operatorname{pd}_A(M).$$

The conjecture requires this supremum to be finite for every finite-dimensional $A$. It does not assert a single bound for all algebras or finite projective dimension for every module.[1]

Bounds are available under special assumptions on the algebra or on a related module. The missing result is a bound for every finite-dimensional algebra. A construction with an infinite resolution would not refute this claim; a counterexample would need finite shortest resolutions of unbounded lengths within one algebra.[1]

The cited preprint obtains finite bounds for opposite endomorphism algebras of particular τ-tilting modules when the underlying algebra has finite representation type or is minimal representation infinite. Its depth and delooping estimates explain where those bounds come from. These hypotheses do not cover every finite-dimensional algebra, and an infinite delooping level alone would not disprove finite finitistic dimension.[1]

Relations determine how algebraic pieces can fit together, not just how many coordinates they have. A finite ceiling would also limit the levels at which extension data can remain nonzero for the modules in question. It would turn many individually finite calculations into a family with a common stopping level, while leaving modules with infinite resolutions outside the guarantee.[1]

Extension groups record ways of joining modules, with higher groups recording longer extension chains. If a module $M$ has a projective resolution of length $b$, then $\operatorname{Ext}_A^j(M,N)=0$ for every $j>b$ and every module $N$: the resolution has no terms in those higher degrees from which to compute them. A finite $\operatorname{findim}(A)$ would give a common such cutoff for every finitely generated $M$ of finite projective dimension. The value would therefore control more than the chosen presentation; it would limit where nontrivial extension information can persist across that entire class of representations. This does not make modules of infinite projective dimension disappear.[1]