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Ruy
Ruy
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Given two C*-algebras $A$ and $B$, the maximal tensor product $A\otimes_{max}B$ is bigger than the minimal tensor product $A\otimes_{min}B$ in the sense that there exists an epimorphism $$A\otimes_{max}B \to A\otimes_{min}B,$$ which is the identity map on the respective copies of the algebraic tensor product $A\otimes_{alg}B$.

TheIn case $A$ and $B$ are unital C*-algebras, the freeamalgamated free product $A{*}B$$A{*}_{\mathbb C}B$ is even bigger than $A\otimes_{max}B$ according to the exact same criteria described above (notice that $A{*}B$$A{*}_{\mathbb C}B$ contains a cannonical copy of the vector space $A\otimes_{alg}B$). This is of course due to the fact that, within the free product, the elements of $A$ are not required to commute with those of $B$.

My question is whether $A\otimes_{max}B$ is still maximal if we weaken the condition of commutativity of elements by replacing it with the commutativity of sets.

Question: Suppose that $C$ is a C*-algebra containing copies of $A$ and $B$, such that $C$ coincides with the closed linear span of $$AB = \{ab:a\in A,\ b\in B\}$$ as well as that of $BA$. In symbols $C=\overline{\hbox{span}}\,AB=\overline{\hbox{span}}\,BA$. Suppose moreover that there exists a *-homomorphism $$\varphi:C\to A\otimes_{max}B$$ such that $\varphi(ab) = a\otimes b$, for all $a$ in $A$, and all $b$ in $B$. Is $\varphi$ necessarily an isomorphism?

Given two C*-algebras $A$ and $B$, the maximal tensor product $A\otimes_{max}B$ is bigger than the minimal tensor product $A\otimes_{min}B$ in the sense that there exists an epimorphism $$A\otimes_{max}B \to A\otimes_{min}B,$$ which is the identity map on the respective copies of the algebraic tensor product $A\otimes_{alg}B$.

The free product $A{*}B$ is even bigger than $A\otimes_{max}B$ according to the exact same criteria described above (notice that $A{*}B$ contains a cannonical copy of the vector space $A\otimes_{alg}B$). This is of course due to the fact that, within the free product, the elements of $A$ are not required to commute with those of $B$.

My question is whether $A\otimes_{max}B$ is still maximal if we weaken the condition of commutativity of elements by replacing it with the commutativity of sets.

Question: Suppose that $C$ is a C*-algebra containing copies of $A$ and $B$, such that $C$ coincides with the closed linear span of $$AB = \{ab:a\in A,\ b\in B\}$$ as well as that of $BA$. In symbols $C=\overline{\hbox{span}}\,AB=\overline{\hbox{span}}\,BA$. Suppose moreover that there exists a *-homomorphism $$\varphi:C\to A\otimes_{max}B$$ such that $\varphi(ab) = a\otimes b$, for all $a$ in $A$, and all $b$ in $B$. Is $\varphi$ necessarily an isomorphism?

Given two C*-algebras $A$ and $B$, the maximal tensor product $A\otimes_{max}B$ is bigger than the minimal tensor product $A\otimes_{min}B$ in the sense that there exists an epimorphism $$A\otimes_{max}B \to A\otimes_{min}B,$$ which is the identity map on the respective copies of the algebraic tensor product $A\otimes_{alg}B$.

In case $A$ and $B$ are unital C*-algebras, the amalgamated free product $A{*}_{\mathbb C}B$ is even bigger than $A\otimes_{max}B$ according to the exact same criteria described above (notice that $A{*}_{\mathbb C}B$ contains a cannonical copy of the vector space $A\otimes_{alg}B$). This is of course due to the fact that, within the free product, the elements of $A$ are not required to commute with those of $B$.

My question is whether $A\otimes_{max}B$ is still maximal if we weaken the condition of commutativity of elements by replacing it with the commutativity of sets.

Question: Suppose that $C$ is a C*-algebra containing copies of $A$ and $B$, such that $C$ coincides with the closed linear span of $$AB = \{ab:a\in A,\ b\in B\}$$ as well as that of $BA$. In symbols $C=\overline{\hbox{span}}\,AB=\overline{\hbox{span}}\,BA$. Suppose moreover that there exists a *-homomorphism $$\varphi:C\to A\otimes_{max}B$$ such that $\varphi(ab) = a\otimes b$, for all $a$ in $A$, and all $b$ in $B$. Is $\varphi$ necessarily an isomorphism?

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Ruy
Ruy
  • 2.3k
  • 10
  • 19

Maximality of the maximal tensor product of C*-algebras

Given two C*-algebras $A$ and $B$, the maximal tensor product $A\otimes_{max}B$ is bigger than the minimal tensor product $A\otimes_{min}B$ in the sense that there exists an epimorphism $$A\otimes_{max}B \to A\otimes_{min}B,$$ which is the identity map on the respective copies of the algebraic tensor product $A\otimes_{alg}B$.

The free product $A{*}B$ is even bigger than $A\otimes_{max}B$ according to the exact same criteria described above (notice that $A{*}B$ contains a cannonical copy of the vector space $A\otimes_{alg}B$). This is of course due to the fact that, within the free product, the elements of $A$ are not required to commute with those of $B$.

My question is whether $A\otimes_{max}B$ is still maximal if we weaken the condition of commutativity of elements by replacing it with the commutativity of sets.

Question: Suppose that $C$ is a C*-algebra containing copies of $A$ and $B$, such that $C$ coincides with the closed linear span of $$AB = \{ab:a\in A,\ b\in B\}$$ as well as that of $BA$. In symbols $C=\overline{\hbox{span}}\,AB=\overline{\hbox{span}}\,BA$. Suppose moreover that there exists a *-homomorphism $$\varphi:C\to A\otimes_{max}B$$ such that $\varphi(ab) = a\otimes b$, for all $a$ in $A$, and all $b$ in $B$. Is $\varphi$ necessarily an isomorphism?