von Neumann algebra factors

I have two basic questions on von Neumann algebras :

1) If a von Neumann algebra $M$ is simple (only trivial ideals), is it a factor (i.e. $M\cap M'=\mathbb C \cdot 1_M$ ?).

2) If the reduced group algebra of a discrete group $C_r(\Gamma)$ (i.e. the completion of $C[\Gamma]$ with respect to the norm induced by the left regular representation) is simple, is it true that the associated von Neumann algebra $L(\Gamma)=C_r(\Gamma)''$ is a factor ? Thanks a lot, sorry fot these naive questions.

• Hi, welcome to math.SE! The main difficulty with your post is that it contains a lot of notation and terms which most people reading will not know (or remember). It would be really helpful if you provided the definitions, or else links to definitions. The ones that don't make sense to me here in the context are "$Cid$", $M'$, "is it a factor,"associated algebra $C_r(\Gamma)''$". Mar 13, 2013 at 14:50
• Hello, sorry I didn't know how much I had to write there... I am going to complete it. I have one question with latex, how do I input the set of complex numbers, I have tried $\C$ but it does not seem to work... ?
– vonN
Mar 13, 2013 at 14:55
• \Bbb C should do the trick Mar 13, 2013 at 16:34

The answer to your first question is yes.

Suppose that $M$ is not a factor; it must contain a nontrivial central projection $p \in M \cap M'$. But then $pMp$ is a nontrivial ideal in $M$. So if $M$ is simple, it must be a factor.

The answer to your second question is also yes.

Let $\Gamma$ be a countably infinite discrete group. Suppose that the reduced group $C^*$-algebra $C^*_r(\Gamma)$ is simple. Then the group $\Gamma$ must be i.c.c.. That is, each of its conjugacy classes different from $\{1\}$ is infinite. But then the group von Neumann algebra $L(\Gamma)$ is a $II_1$-factor.

"Simplicity implies i.c.c." follows from Proposition 3 in http://arxiv.org/abs/math/0509450, as noted in Section X of that paper.

That "i.c.c. implies factor" goes back as far as Murray and von Neumann.

• Thanks a lot, it is very clear. Since you were kind enough to answer these two easy questions, I may ask one last : What would be a good reference for you to learn von Neumann algebras ? I have some notes from Jones and I would like something else to read so that I can understand more...
– vonN
Mar 14, 2013 at 10:02
• Three things: 1. This might be better as a separate question; 2. Ask your professor/whoever you are working with for advice (each book will cover both too much material and not enough of what are you interested in); 3. This list looks pretty good. Mar 14, 2013 at 10:55
• 1. ok 2.It is pure curiosity so there is be no "perfect" reference, I wanted that someone specialist gives me some advices 3.Exactly what I needed, thanks.
– vonN
Mar 14, 2013 at 12:38

To complement Tom's answer, it is interesting to know that the following statements are equivalent:

• $$M$$ is simple as a von Neumann algebra (i.e., no sot-closed ideals)

• $$M$$ is simple as a C$$^*$$-algebra (i.e., no norm-closed ideals)

• When $$M$$ is finite, it is a factor

Indeed, if $$M$$ is not a factor then its centre is a nontrivial von Neumann algebra and so it has a nontrivial projection $$p$$; thus $$pM$$ is a nontrivial (sot-closed, norm-closed) ideal of $$M$$, so $$M$$ is not simple neither as a von Neumann algebra nor as a C$$^*$$-algebra. This shows that either of the first two assertions implies the last one.

That C$$^*$$-simplicity implies von Neumann simplicity is trivial. So the only nontrivial assertion is that a factor is simple as a C$$^*$$-algebra. This follows from Dixmier's Approximation Theorem (see, for instance, section 8.3 in Kadison-Ringrose), which says that for any $$x\in M$$, there exists $$y\in M\cap M'$$ such that $$y\in\overline{\text{conv}\,\{uxu^*:\ u\in\mathcal U(M)\}}.$$ So if you take a norm closed ideal $$J\subset M$$ and $$M$$ is a factor, take $$x\in J$$ nonzero, and now by Dixmier's Approximation Theorem there exists $$\lambda I$$ that belongs to the norm-closure of the convex hull of the unitary orbit of $$x$$, which is in $$J$$. So $$I\in J$$ and thus $$J=M$$ (it is easy to see that $$\lambda\ne0$$ by using the faithful trace).

Edit: This last argument requires the $$M$$ is finite (or a similar condition) to guarantee that $$\lambda$$ above is not zero. For instance if you take $$M=B(H)$$, which has a proper ideal, and $$x$$ a rank-one projection, then the $$y$$ from Dixmier's Theorem is zero. Thanks Søren for noticing.