I am trying to understand Hochschild homology, in particular the Hochschild–Kostant–Rosenberg theorem. As far as I understand this result gives an isomorphism between the algebraic (Kähler) differential forms of a smooth commutative $k$-algebra and the Hochschild homology. This clearly implies that the homology of the complex is infinite-dimensional, which seems strange to me. How can Hochschild homology be a good invariant of an algebra if the dimensions of its homology groups are all infinite-dimensional?

Edit: Based on the answers below, perhaps the interesting invariant of HH is the length of the complex, not the dims of the cohomology groups?

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    $\begingroup$ The K"ahler differentials of a smooth K-algebra are finitely generated. $\endgroup$
    – the L
    Commented Sep 6, 2020 at 14:28
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    $\begingroup$ Kahler forms are a module over the algebra and their “size” should be understood in terms of rank as a module. Have you worked through the example of affine space? $\endgroup$ Commented Sep 6, 2020 at 14:28
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    $\begingroup$ If $A$ has finite dimension over a field $K$, the (co)homology is that of a complex with finite-dimensional spaces, hence $HH_n(A)$ has finite dimension (bounded above by $\dim(A)^n$). $\endgroup$
    – YCor
    Commented Sep 6, 2020 at 14:42
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    $\begingroup$ BTW you write in the title "is interesting" and in the text "is a good invariant". Both terms are subjective, but I'm not sure they have the same meaning. That HH is interesting does not a priori mean that it's a good invariant (assuming "good invariant" means "efficient in distinguishing non-isomorphic algebras"); there are other good reasons it can be "interesting". $\endgroup$
    – YCor
    Commented Sep 6, 2020 at 14:46
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    $\begingroup$ Regarding your thoughts about length: for the polynomial ring in d variables (or affine space if you are ore geometrical) the-result-which-serves-as-the-prototype-for the HKR theorem is an identification of HH_n(R,R) as an R-module with the n'th exterior power of $\Omega_{R/k}$ as an R-module, and if $M$ is a free R-module of rank $d$ then $\bigwedge_R^n M =0$ whenever $n\geq d+1$ $\endgroup$
    – Yemon Choi
    Commented Sep 6, 2020 at 22:15


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