let $\phi : B(l^{p}(X)) \to B(l^{p}(Y)) $ be an isomorphism, does $\phi$ necessarily preserve rank of operators?

Question

let $l^{p}(X)$ and $l^{p}(Y)$ be some $l^p$ function spaces, $B(l^{p}(X))$ and $B(l^{p}(Y)) $ be bounded linear operators on themselves, $\phi : B(l^{p}(X)) \to B(l^{p}(Y)) $ be an isomorphism as Banach algebras, does $\phi$ necessarily preserve rank of operators? i.e. if $T \in B(l^{p}(X))$ is of rank $n$, then $\phi (T)\in B(l^{p}(Y))$ is also of rank $n$.

it seems to me if $X$ and $Y$ are discrete, then it is true, but I don't know how to prove it. what if $X$ and $Y$ are non-discrete and locally compact?

Answer 1

The following is true in general: If $E, F$ are Banach spaces, $\phi: B(E) \to B(F)$ is a Banach algebra isomorphism (in fact, just an algebra isomorphism is enough). Then $\phi$ preserves ranks of operators.

To prove this, we first observe that we can retrieve the ranks of projections from the algebraic structure. Mimicking the case of von Neumann algebras, suppose $p, q$ are two projections in $B(E)$. Then we write $p \leq q$ iff $pq = qp = p$. We also write $p < q$ to mean $p \leq q$ and $p \neq q$. These conditions are clearly preserved by an algebra isomorphism, so $\phi(p) < \phi(q)$ iff $p < q$. On the other hand, we observe that $p < q$ implies $pE \subsetneq qE$. (Though, as opposed to the case of orthogonal projections on a Hilbert space, this is not an equivalent condition.) Moreover, if $V \subset qE$ is a finite-dimensional proper subspace of $qE$, then there exists a projection from $qE$ onto $V$. Let $p$ be the composition of $q$ with this projection onto $V$. Then $pE = V$ and $p < q$. Thus, for any projection $p$,

$$\dim(pE) = \sup\{n \in \mathbb{N}: \exists\text{ nonzero projections }p_1, \cdots, p_n\text{ s.t. }p_1 < \cdots < p_n = p\}$$

Hence, $\text{rank}(p) = \dim(pE) = \dim(\phi(p)F) = \text{rank}(\phi(p))$ for any projection $p$. For a general operator $T$, its rank is simply given by,

$$\text{rank}(T) = \min\{n \in \mathbb{N} \cup \{\infty\}: \text{ there exists a projection }p\text{ of rank }n\text{ s.t. } pT = T\}$$

Therefore, $\text{rank}(T) = \text{rank}(\phi(T))$, for any operator $T$.

Remark: One may also observe that,

$$\text{rank}(T) = \min\{n \in \mathbb{N} \cup \{\infty\}: \text{ there exists a projection }p\text{ of rank }n\text{ s.t. } Tp = T\}$$

is true as well. Thus, the same argument shows that an algebra anti-isomorphism will also preserve ranks of operators.

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A bit off-topic, but to clarify my comments: If we are dealing with Hilbert spaces $H$ and $K$, then an isometric isomorphism $\phi: B(H) \to B(K)$, which may not be multiplicative, will actually also preserve ranks of operators. This is by a result of Kadison, which shows that any such isometric isomorphism is given by a $\ast$-isomorphism or a $\ast$-anti-isomorphism, followed by multiplication on the left by a fixed unitary. Since $\ast$-isomorphisms, $\ast$-anti-isomorphisms, and multiplication by unitaries all preserve ranks, the result easily follows. It would be interesting to see if the same result holds for general Banach spaces instead of just Hilbert spaces.