# What is the group of automorphisms of $l^{\infty}$?

What is the group of automorphisms of $l^{\infty}$?

I think it would be the permutations of the integers. Is this right?

• The automorphism group of $l^\infty(X)$ will depend on what $X$ is. Also, what structure do you require automorphisms to preserve? May 2, 2016 at 1:42
• $l^{\infty}(\mathbb{Z})$. The automorphisms have to preserve the $W^{*}$-algebra structure. May 2, 2016 at 1:49
• So do you require that the automorphisms are weak-star continuous? (This might come for free, but I haven't thought deeply about it) May 2, 2016 at 4:50
• @YemonChoi, no it doesn't come for free. Every homeo of $\beta\mathbb{Z}$ will give an auto, but only these which will preserve $\mathbb{Z}$ will give rise to weak* continuous ones. May 2, 2016 at 10:07
• @user89334: yes it does come for free. Every homeo of $\beta \mathbb{Z}$ comes from a permutation of the integers, and every C*-automorphism of $l^\infty(\mathbb{Z})$ is an order-isomorphism, hence is normal, hence is weak*-continuous. May 2, 2016 at 11:55

This is to record what is probably the easiest proof (that the answer is "yes"), following a remark by Nik Weaver.

Noting that $\ell^\infty(\mathbb{Z})\simeq C(\beta\mathbb{Z})$, using Banach-Stone, we see that every auto gives a homeo of $\beta\mathbb{Z}$. But since $\mathbb{Z}$ is the subset of isolated points, every homeo preseves it, and since it is dense the homeo is determined by its restriction to $\mathbb{Z}$.

• Equivalently, but maybe slightly easier, (in the sense that one doesn't have to use the correspondence $\ell^\infty(\mathbb{Z})\cong C(\beta\mathbb{Z})$) one notices that every automorphism must permute the minimal projections of $\ell^\infty(\mathbb{Z})$ which are in one-to-one correspondence with $\mathbb{Z}$ May 3, 2016 at 1:20
• @CalebEckhardt: that is nicer. On the other hand, identifying with $C(\beta\mathbb{Z})$ allows us to invoke the Banach-Stone theorem and say that every linear isometry of $l^\infty(\mathbb{Z})$ with itself has this form. May 6, 2016 at 1:35
• @CalebEckhardt, you're argument seems to suggest that every isometry of $\ell^\infty$ would come from a permutation, which is incorrect: take the multiplication map with a function $\mathbb{Z}\to\{-1,1\}$ (or to $S^1$ over $\mathbb{C}$). May 6, 2016 at 6:08

Yes, the group of automorphisms of $\ell^\infty(\mathbb{Z})$ preserving the $W^*$-algebra structure is the group of permutations of $\mathbb{Z}$.

On the one hand, any permutation of $\mathbb{Z}$ acts as an automorphism of $\ell^\infty(\mathbb{Z})$.

On the other hand, any $W^*$-algebra automorphism of $\ell^\infty(\mathbb{Z})$ sends characters - that is, homomorphisms $\chi: \ell^\infty(\mathbb{Z}) \to \mathbb{C}$ - to characters, so it defines a permutation of the characters. By the Gelfand-Naimark correspondence, there is a compact Hausdorff space $\beta\mathbb{Z}$ called the Stone-Cech compactification of $\mathbb{Z}$ such that

$$\ell^\infty(\mathbb{Z}) \cong C(\beta\mathbb{Z})$$

Points of $\beta\mathbb{Z}$ are the same as characters $\chi: \ell^\infty(\mathbb{Z}) \to \mathbb{C}$. Every point $n \in \mathbb{Z}$ defines a character $\chi$ by

$$\chi(f) = f(n)$$

Not all characters are of this form, as pointed out by Yemon Choi and Uri Bader. However, the weak-$\ast$-continuous characters are. If $\alpha$ is a $W^*$-algebra automorphism of $\ell^\infty(\mathbb{Z})$ and $\chi$ is a weak-$\ast$-continuous character, $\chi \circ \alpha$ is again weak-$\ast$-continuous. So, every $W^*$-algebra automorphism of $\ell^\infty(\mathbb{Z})$ defines a permutation of the integers.

The concept of 'weak-$\ast$-continuous character' is a bit fancy, so I like Nik Weaver's simpler argument below. Just as the integers give the weak-$\ast$-continuous characters on $\ell^\infty(\mathbb{Z})$, they give the isolated points in $\beta\mathbb{Z}$ - but this is easier to understand, and easier to prove. By the functorality of the Gelfand-Naimark correspondence any automorphism of $\ell^\infty(\mathbb{Z})$ comes from a homeomorphism of $\beta\mathbb{Z}$, and these clearly map isolated points to isolated points. So, any automorphism of $\ell^\infty(\mathbb{Z})$ gives a permutation of the integers.

• In the Cstar context, the characters of $\ell^\infty({\bf Z})$ are NOT all given by evaluation at integers -- rather, they correspond to points of the Stone-Cech compactification. I suspect that in your second paragraph you want to restrict attention to weak-star continuous characters, which are all of the form you state. May 2, 2016 at 4:51
• Indeed, characters supported on the boundary of $\mathbb{Z}$ in the Stone Ceck compactification will not be weak* continuous. May 2, 2016 at 5:57
• Thanks, you're right of course! Let me fix my answer just to avoid confusion. May 5, 2016 at 18:19

For a standard Lebesgue space $(X,\mu)$ the autmorphism group of the W* algebra $L^\infty(X,\mu)$ is the group of measure class preserving automorphisms of $(X,\mu)$ - the group of a.e defined measurable map $\sigma:X\to X$ s.t $\sigma_*\mu$ and $\mu$ are equivalent, modulo the relation of being identical a.e. (more generally: the category of standard Lebesgue spaces and (the op of) the category of commutative W* algebras with separable predual are equivalent).

Apply this general fact in the case $(\mathbb{Z},\text{counting})$ and get the answer: yes, every automorphism is given by a permutation.

• More specifically: every C*-algebra automorphism will come from a self homeo of $\beta\mathbb{Z}$. Those which are weak* cont. will preserve the measure class induced by the predual $\ell^1(\mathbb{Z})$, which is supported on $\mathbb{Z}$. Hence this part will be preserved. Since it is dense, the homeo will be defined uniquely by its restriction to $\mathbb{Z}$. May 2, 2016 at 6:07
• Here is another way to see it: every auto of $\ell^{\infty}$ gives an isometry of the unique predual $\ell^1$. Now use Banach-Lamperti which classifies these, and check which dual isometry of these preserves the algebra structure on $\ell^\infty$. May 2, 2016 at 6:25
• Again, it is a general fact that any C*-automorphism of a von Neumann algebra is automatically weak* continuous. May 2, 2016 at 11:57
• @NikWeaver sure, you are right and in fact it is too esay to see it here: $\mathbb{Z}$ consists of the isolated points in $\beta\mathbb{Z}$... May 2, 2016 at 13:23