# the $K_0$ group of von Neumann algebra factor of type $II_{\infty}$ and type $III$

Suppose $$M$$ is a is a von Neumann algebra factor of type II$$_{\infty}$$, and $$N$$ is a is a von Neumann algebra factor of type III. I have no idea how to prove that $$K_0(M)=K_0(N)=0$$.

What are the definitions of von Neumann algebra factor of type II$$_{\infty}$$ and von Neumann algebra factor of type III?

• Here's a hint: en.wikipedia.org/wiki/Von_Neumann_algebra#Factors Sep 3 '19 at 14:14
• there is a statement :factors that are separable or finite, two projections are equivalent if and only if they have the same trace. I can only prove the above conclusion when $M$ is a type $II_1$ factor.Would you mind showing me the proof when $M$ is type $II_{\infty}$ or type $III$? Sep 3 '19 at 14:53
• Neither of those types of factors are separable or finite. Sep 3 '19 at 15:19
• @Aweygan: every infinite-dimensional von Neumann algebra is not separable (in norm); so no one uses that terminology. A von Neumann algebra is separable when it is separable in one of the weaker topologies (wot, sot, etc.). When $H$ is separable, any von Neumann subalgebra of $B(H)$ is separable. Sep 3 '19 at 17:38

What happens is that the Grothendieck group of a semigroup that has an "infinity" is always trivial. This is because $$\infty+d=\infty+c$$ for any $$c,d$$, so $$(\infty,\infty)\sim(c,d)$$. Both type II$$_\infty$$ and type III factors have infinite projections, so the above applies.
When the algebra is non-separable, we can still do the above. There will be infinite projections of different cardinalities, so it is enough to choose an infinity that is greater than both $$c$$ and $$d$$.
• Let S be an abelian semigroup,the Grothendieck goup $G(S)=\{r_S(x)-r_S(y):x,y\in S\}$,where $r_S$ is the Grothendieck map.We have the following conclusion:$r_S(x)=r_S(y)$iff $x+z=y+z$ for some $z\in S$.Let $S=D(A),K_0(A)=G(D(A))$,for each $p\in P_{\infty}(A)$,let $[p]_D$ denote the equivalence class containing $p$.If $q$ is an infinite projection,why can we view $[q]_D$ as $\infty$,in other words,why $[x]_D+[q]_D=[y]_D+[q]_D$ for any $[x]_D,[y]_D \in D(A)$? Sep 4 '19 at 3:20
• That's where you need to know a bit about von Neumann algebras and projections. If $p\leq 1-q$, $p\preceq q$, and $q$ is infinite, then $q\sim q+p$. Sep 4 '19 at 4:36
• ，I only know the fact that allprojections in a factor can be compared and the definition of infinite projections,why there exists a projection p such that $p\leq1-q$ and $p\preceq q$? Sep 4 '19 at 6:19