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If $\varphi: A\to B$ is a (norm-continuous, unital, involutive) homomorphism of $C^*$-algebras, then the image $\varphi(A)$ is closed in $B$ and therefore is a $C^*$-algebra with the $C^*$-norm induced from $B$ (G.Murphy, 3.1.6).

If in addition $A$ is a von Neumann algebra, will $\varphi(A)$ be a von Neumann algebra as well?

Sergei Akbarov
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1 Answers1

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No. The "why" depends on your definition of von Neumann algebra.

If you define von Neumann algebra the way it is usually used, it is not an intrinsic notion: both the double commutant and the ultraweak operator topology depend on the environment (i.e. $B(H)$). In that case, take a non-type I factor, say a II$_1$, and take an irreducible representation. Then the image will be dense in some $B(H)$, but it cannot be everything (because in a II$_1$ factor the identity is finite, for instance). So the image is not a von Neumann algebra.

The above shows that the usual definition of von Neumann algebra has its caveats. An intrinsic definition can be given: a C$^*$-algebra that admits a Banach pre-dual. More concretely this means that a von Neumann algebra is a C$^*$-algebra that admits an isometric surjective embedding onto the dual of a Banach space.

So in this second case, if $\varphi$ is injective, its image will be a von Neumann algebra. When $\varphi$ is not injective, it is easy to come up with a counterexample: you can take $A=B(H)$ and $\varphi$ the quotient map onto the Calkin algebra, which is not a von Neumann algebra (in an intrinsic way: it is not C$^*$-isomorphic to any von Neumann algebra).

Martin Argerami
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  • I thought, this question becomes senseless, if the definition depends on the environtment... I didn't understand, why is the answer "yes" in the second case? – Sergei Akbarov Nov 19 '14 at 21:03
  • Because if two C$^*$-algebras are isomorphic and one embeds isometrically in a dual, you can use the isomorphism to get an embedding of the other algebra in the dual. So with the intrinsic definition of von Neumann algebra, $\varphi(A)$ is a von Neumann algebra. – Martin Argerami Nov 19 '14 at 21:07
  • Excuse me, perhaps you are speaking about the situation where $\varphi$ is injective? I meant that it is not necessarily injective. Or I didn't understand something? – Sergei Akbarov Nov 19 '14 at 21:13
  • Yes, I though that you were saying injective. But it doesn't change things too much. I have edited the answer. – Martin Argerami Nov 19 '14 at 21:24
  • Ah, OK... Could you, please, give a reference with the explanation of why Calkin is not von Neumann? – Sergei Akbarov Nov 19 '14 at 21:36
  • I don't know an easy reference. Calkin's original work already shows (I think, I have never read the paper) that the Calkin algebra contains increasing bounded nets of selfadjoints with no supremum (there is an argument here: http://mathoverflow.net/questions/172073/why-do-the-projections-in-the-calkin-algebra-not-form-a-lattice ). Brown-Douglas-Fillmore show that the Calkin algebra has a masa that is not generated by its projections. – Martin Argerami Nov 19 '14 at 21:56
  • Yes, thank you! – Sergei Akbarov Nov 19 '14 at 21:59
  • Glad I could help. – Martin Argerami Nov 19 '14 at 22:08
  • I have a question about the first example you show here, In order to get a contraction by using the finite property of 1, I guess we need to make sure that $H$ is infinite dimensional. The question is, how to maker sure that the irreducible representation is infinite dimensional? I don't think it is a simple claim. – Yanyu Apr 10 '22 at 00:45
  • A factor is simple, so any representation is isometric. So an infinite-dimensional factor (all but those of type I$_n$, $n\in\mathbb N$) does not have finite-dimensional representations. This applies in particular to II$_1$-factors, which are the ones I mentioned in the answer. – Martin Argerami Apr 10 '22 at 00:55
  • Thanks for your help! I think I understand it now! In addition to that, In order to show the fact that Calikin algebra is not a von Neumann algebra, I think we could consider the invertible elements here. We know that for a von Neumann algebra, the invertible elements will be connected. However, In the Calikin algebra, the invertible elements will be Fredholm operator, which is definitely not connected since $\pi_0 (B(H)/K(H))=\mathcal{Z}$, characterized by the Fredholm index – Yanyu Apr 10 '22 at 07:28
  • Although I don't think the claim will be true for any factor, we need this von neumann algebra to be finite, like you write here. https://math.stackexchange.com/questions/329429/von-neumann-algebra-factors. The reason why we need this is that we hope the factor to be simple in $C^$ algebra sense instead of $W^$ algebra sense, since we the representation may not be noraml – Yanyu Apr 10 '22 at 08:30
  • To be more specific, we need this claim since we only know that the representation is norm continuous instead of normal. If we only know that $M$ is a factor, then what we do know is that any normal representation of $M$ should be isometric instead of any representation. – Yanyu Apr 10 '22 at 08:54
  • In addition to that, I notice that for the first example, we know that $\pi(M)$ will be isometrically isomorphic to $M$, which means that $\pi(M)$ is actually isometrically isomorphic to a dual of a Banach space. Notice that we know $\pi(M)$ is also a $C^*$ algebra, hence $\pi(M)$ is actually a von Neumann algebra by Sakai's definition. Although $\pi(M)$ is $S.O.T.$ dense in $B(H)$ and not equal to $B(H)$, this only means that $\pi(M)$ won't be a von Neumann subalgebra of $B(H)$. However, for an appropriate Hilbert space $K$, we will have the fact $\pi(M)\subset B(K)$ is a von Neumann algebra. – Yanyu Apr 10 '22 at 11:11