# Naive questions on double duals of von Neumann algebras

Let $M$ be a von Neumann algebra. I try to understand the double dual $M^{**}$ which is also an "abstract" von Neumann algebra. I denote by $i_M \colon M \to M^{**}$ the canonical map. It seems to me that there exists a central projection $e \in M^{**}$ such that $M=eM^{**}$.

1) Is $i_M$ a unital normal injective $*$-homomorphism ? My feeling is that the answer is no. But what are the correct algebraic properties of this map?

2) Sometimes in the litterature, I see the map $Q \colon M \to M^{**}$, $x \mapsto ex$. What is the connection between $Q$ and $i_M$? I believe that $Q(x)=ei_M(x)$ and that $Q(x)\not= i_M(x)$.

3) In the litterature, I see the formula $M^{**}=M \oplus M_*^\perp$. It seems to me that the identification of $M$ in $M^{**}$ is not for the map $i_M$. Can you confirm ?

There are two different objects.

• $M^{**}$, the double dual of $M$.

• $M''$ the universal enveloping algebra of $M$.

The algebra $M''$ is usually constructed by taking all (unitary equivalence classes of) representations and taking the direct sum. Or, by taking all (unitary equivalence classes of) states and using their GNS representations: $$\tag1 M''=\overline{\bigoplus_{f\in S(M)} \pi_f(M)}^{wot}$$

It turns out that $M^{**}$ and $M''$ are isometrically isomorphic as Banach spaces and, more importantly, homeomorphic when we put the weak$^*$-topology on $M^{**}$ and the $\sigma$-weak topology on $M''$.

One can use the isomorphism between $M^{**}$ and $M''$ to define a product on $M^{**}$; this product agrees with the Arens product.

Because of the above identification, it is very common for people to talk about the double dual but work on the enveloping von Neumann algebra.

It seems to me that there exists a central projection $e\in M^{∗∗}$ such that $M=eM^{∗∗}$.

Yes (if we are talking about $M''$). The projection $e$ corresponds, in $(1)$, to the summands given by the normal states.

Is $i_M$ a unital normal injective ∗-homomorphism ? My feeling is that the answer is no. But what are the correct algebraic properties of this map?

As written, the question doesn't make sense, because $M^{**}$ has no algebraic structure per se. So no, $i_M$ is not multiplicative. If one induces a product on $M^{**}$ via the isomorphism with $M''$, then $i_M$ is multiplicative.

Sometimes in the literature, I see the map $Q:M\to M^{∗∗}$, $x\mapsto ex$. What is the connection between $Q$ and $i_M$? I believe that $Q(x)=ei_M(x)$ and that $Q(x)\ne i_M(x)$.

C$^*$ and von Neumann algebra people do not work on the double dual, they work on $M''$ where they have a natural structure. So equation that holds is that $Q=e\,\pi_u$, where $\pi_u$ is the universal representation. One may abuse notation and write $Q=e\,i_M$, in the sense that if $\rho:M^{**}\to M''$ is the isometric isomorphism mentioned above, we have $$\rho\circ i_M=\pi_u.$$ So $Q=e(\rho\circ i_M)$.

In the literature, I see the formula $M^{∗∗}=M\oplus M^\perp_∗$. It seems to me that the identification of $M$ in $M^{∗∗}$ is not for the map $i_M$. Can you confirm ?

This is related to $(1)$. You can decompose the direct sum into two parts:

• the summands corresponding to normal states; this gives you a faithful representation of $M$ that is $\sigma$-weakly continuous, so the image is isomorphic with $M$

• the summands corresponding to non-normal states (singular), that are usually denoted by $M_*^\perp.$

So $M^{**}=M\oplus M_*^\perp$ looks like a bit of abuse of notation.

• I do not see clearly why $\hat{xy} = \hat{x} \hat{y}$ implies that $\phi(xy) = \phi(x) \, \phi(y)$. It is clear that $(\widehat{x \, y})(\phi) = \phi(x y)$ but why does the product of $\hat{x} \hat{y}$ evaluated in $\phi$ equate $\phi(x)\phi(y)$? If the argument were true, wouldn't it also hold for $M = M_n(\mathbb{C})$? In that case $i_M$ is the identity but no linear functional is multiplicative. – Adrián González-Pérez May 1 '18 at 16:37
• In any case, even if the map $i_M$ is a $\ast$-homomorphism it wouldn't be normal. Otherwise by Goldstine Theorem its image would be weak-$\ast$ dense inside $M^{\ast \ast}$, but the image of a normal $\ast$-homorphism has to be weak-$\ast$ closed. That will get a contradiction unless $M = M^{\ast \ast}$. – Adrián González-Pérez May 1 '18 at 16:41
• Yes, you are right. I was trying to emphasize that $i_M$ is not multiplicative. I'll just delete that. – Martin Argerami May 1 '18 at 16:43
• Thank you. However, note that if $A$ ia a Banach algebra then we can equip $A^{**}$ with two canonical products (Arens products) such that $i_A \colon A \to A^{**}$ is a homomorphism for both products. In the case where $A=M$ is a von Neumann algebra, the two products coincide (Arens regularity). – Zouba May 1 '18 at 19:00
• I don't know a reference. The way I think about it, the normal states give rise to normal representations. The normal representations preserve the $\sigma$-weak topology, so the image of the sum is spatially isomorphic to $M$; when you map a von Neumann algebra via a non-normal representation, you get $\pi(M)\subset B(H_\pi)$, but $\pi(M)$ is not a von Neumann algebra, in the sense that it is not equal to its double commutant.So the part coming from the non-normal states puts $M$ in $M''$ in a way that it is not a von Neumann algebra. – Martin Argerami May 2 '18 at 21:41