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?

  • $\begingroup$ Here's a hint: en.wikipedia.org/wiki/Von_Neumann_algebra#Factors $\endgroup$
    – Aweygan
    Sep 3 '19 at 14:14
  • $\begingroup$ 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 $III$? $\endgroup$
    – math112358
    Sep 3 '19 at 14:53
  • $\begingroup$ Neither of those types of factors are separable or finite. $\endgroup$
    – Aweygan
    Sep 3 '19 at 15:19
  • $\begingroup$ @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. $\endgroup$ 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 $d$.

  • $\begingroup$ 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 $[x]_D,[y]_D \in D(A)$? $\endgroup$
    – math112358
    Sep 4 '19 at 3:20
  • $\begingroup$ 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 $q\sim q+p$. $\endgroup$ Sep 4 '19 at 4:36
  • $\begingroup$ ,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 $? $\endgroup$
    – math112358
    Sep 4 '19 at 6:19

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