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It's a big, famous, hard problem in operator algebras to determine if the von Neumann algebras $L(F_2)$ and $L(F_3)$ are isomorphic, or not. Here $F_n$ is the free group on n generators and $L(F_n)$ is the weak-operator-topology closure of the group algebra $\mathbb C[F_n]$ acting naturally on the Hilbert space $\ell^2(F_n)$.

I presume it must be known if the algebras $\mathbb C[F_2]$ and $\mathbb C[F_3]$ are isomorphic or not. But from casually asking a few algebraists, I've never had any luck in finding this out (I admit to not working very hard on this!) I'm guessing some (co)homology theories must help...? What about for replacing $\mathbb C$ by a more general ring?

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  • $\begingroup$ How far does K-theory for C*-algebras get you with this sort of problem? Can it distinguish between the reduced group C*-algebras of $F_2$ and $F_3$, for example? $\endgroup$
    – Paul Siegel
    Commented May 23, 2010 at 12:52
  • $\begingroup$ @Paul: yes, I think. Can't remember who did the calculation (Pimsner? Voiculescu?) but this is probably covered in the "usual" introductions to K-theory of C*-algebras. Maybe also in Davidson's book? $\endgroup$
    – Yemon Choi
    Commented May 23, 2010 at 13:03
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    $\begingroup$ Yes, Pimsner and Voiculescu showed that the $K_1$ group of the reduced C*-algebra of $F_n$ is $\mathbb{Z}^n$ in "K-groups of reduced crossed products by free groups", 1982. It doesn't seem to be in Davidson's C* book, but it follows from Problem 10.11.11 (g) and (h) on page 91 of Blackadar's K-theory book, 2nd ed. $\endgroup$
    – Jonas Meyer
    Commented May 24, 2010 at 2:55

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Well, yes. Imagine that you have an algebra $A$ over $\mathbb{C}$ and you want to find out whether it is $\mathbb{C}[F_2]$ or $\mathbb{C}[F_3]$. Pick any one-dimensional $A$-module $M$ and compute $\operatorname{Ext}^1_A(M,M)$. If $A=\mathbb{C}[F_2]$, you'll get a $2$-dimensional vector space over $\mathbb{C}$, while if $A=\mathbb{C}[F_3]$, you'll get a $3$-dimensional vector space.

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    $\begingroup$ Just to expand very slightly on Leonid's answer: ${\rm Ext}^n_A([C,C)$, where $A=kG$ and $C$ is the one-dimensional module with trivial action, is the $n$th cohomology group of G (over ground field k) or from another point of view the $n$th Hochschild cohomology group of $kG$ with coefficients in $C$, $\endgroup$
    – Yemon Choi
    Commented May 23, 2010 at 13:09
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    $\begingroup$ BTW, something like the early parts of Brown's "Cohomology of groups" should (IIRC) have an explanation why you get 2 and 3 (the ranks of the groups) as the dimension of 1st cohomology. To a (geometric) group theorist, free groups are nice. $\endgroup$
    – Yemon Choi
    Commented May 23, 2010 at 13:22
  • $\begingroup$ Well, that was easy (perhaps I shouldn't ask people interested only in finite-dimensional algebras...) $\endgroup$
    – Matthew Daws
    Commented May 23, 2010 at 14:12
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    $\begingroup$ Since $K(F_n,1)$ is the wedge of $n$ circles, the first cohomology of $F_n$ with trivial coefficients over any base ring $R$ is free $R$-module of rank $n$, by elementary topology. $\endgroup$
    – Victor Protsak
    Commented May 27, 2010 at 9:43
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One has $Hom({\mathbb C}[F_n],{\mathbb C}) = ({\mathbb C}^{\times})^n$ with the obvious topology. (Here, $Hom$ denotes the space of $\mathbb C$-linear homomorphisms.) This of course uses a little bit more than only the algebra structure, but every ${\mathbb C}$-linear isomorphism would preserve the topology on the space of $\mathbb C$-linear representations. Since the spaces $({\mathbb C}^{\times})^n$ are not homeomorphic for different $n$, the claim follows. The same applies to the maximal group $C^{\star}$-algebra of $F_n$. One has $Hom(C^{\star}(F_n),{\mathbb C}) = (S^1)^n$, where one considers only $\star$-homomorphisms.

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  • $\begingroup$ Could you add a bit more detail about how you are defining the topology on the character space in such a way that purely algebraic isomorphism of the group algebras induces homeomorphisms of the character spaces? (Obviously if the isomorphism of algebras extends up to some Banach completion then I agree this is trivial, so I am probably just overlooking something obvious...) $\endgroup$
    – Yemon Choi
    Commented Aug 16, 2010 at 20:07
  • $\begingroup$ I was thinking about the topology of pointwise convergence on elements in ${\mathbb C}[F_n]$. $\endgroup$
    – Andreas Thom
    Commented Aug 16, 2010 at 20:12
  • $\begingroup$ Another way of putting it is that every countably dimensional algebra carries a canonical locally convex topology which is called the fine topology. It is a topological algebra with this topology. Moreover, every ${\mathbb C}$-linear map into any other locally convex vector space is continuous. For finitely generated algebra, the pointwise convergence on the generators gives another way of defining the topology on the $Hom$-space. $\endgroup$
    – Andreas Thom
    Commented Aug 16, 2010 at 20:19

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