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Consider an algebraic quantum group $(A, \Delta)$ in the sense of Van Daele, i.e. a regular multiplier Hopf $^*$-algebra with a positive left integral $\varphi$ and a positive right integral $\psi$.

It is well-known that we can make the following construction:

• Embed $A\subseteq B(L^2(A, \psi)).$
• Define $M= A''\subseteq B(L^2(A, \psi))$.
• Show that $M$ with an extension of the coproduct $\Delta$ becomes a von Neumann algebraic locally compact quantum group in the sense of Kustermans and Vaes.

See for instance Van Daele's recent survey paper Algebraic quantum groups and duality II for details about this construction.

In this case, we can show that the right invariant nsf weight $\widetilde{\psi}$ extends $\psi$. I'm interested in the following: can we show (using elementary means) that a right-invariant nsf weight $\widetilde{\psi}: M_+\to [0, \infty]$ extending $\psi: A\cap M^+ \to \mathbb{C}$ is necessarily unique? Obviously, this is true since in the general theory of locally compact quantum groups since a right invariant nsf weight is uniquely determined (up to constant). I am interested to see a more elementary approach in this case though.

Note that if $A$ is discrete, this is fairly easy, since a positive element in $M$ is a $\sigma$-strong limit of an increasing net of positive elements in $A$, so normality of the weight does the trick.

Consider an algebraic quantum group $(A, \Delta)$ in the sense of Van Daele, i.e. a regular multiplier Hopf $^*$-algebra with a positive left integral $\varphi$ and a positive right integral $\psi$.

It is well-known that we can make the following construction:

• Embed $A\subseteq B(L^2(A, \psi)).$
• Define $M= A''\subseteq B(L^2(A, \psi))$.
• Show that $M$ with an extension of the coproduct $\Delta$ becomes a von Neumann algebraic locally compact quantum group in the sense of Kustermans and Vaes.

See for instance Van Daele's recent survey paper Algebraic quantum groups and duality II for details about this construction.

In this case, we can show that the right invariant nsf weight $\widetilde{\psi}$ extends $\psi$. I'm interested in the following: can we show (using elementary means) that a right-invariant nsf weight $\widetilde{\psi}: M_+\to [0, \infty]$ extending $\psi: A\cap M^+ \to \mathbb{C}$ is necessarily unique? Obviously, this is true since in the general theory of locally compact quantum groups since a right invariant nsf weight is uniquely determined (up to constant). I am interested to see a more elementary approach in this case though.

Note that if $A$ is discrete, this is fairly easy, since a positive element in $M$ is a $\sigma$-strong limit of an increasing net of positive elements in $A$, so normality of the weight does the trick.

Thanks in advance for your help!

Consider an algebraic quantum group $(A, \Delta)$ in the sense of Van Daele, i.e. a multiplier Hopf $^*$-algebra with a positive left integral $\varphi$ and a positive right integral $\psi$.

It is well-known that we can make the following construction:

• Embed $A\subseteq B(L^2(A, \psi)).$
• Define $M= A''\subseteq B(L^2(A, \psi))$.
• Show that $M$ with an extension of the coproduct $\Delta$ becomes a von Neumann algebraic locally compact quantum group in the sense of Kustermans and Vaes.

See for instance Van Daele's recent survey paper Algebraic quantum groups and duality II for details about this construction.

In this case, we can show that the right invariant nsf weight $\widetilde{\psi}$ extends $\psi$. I'm interested in the following: can we show (using elementary means) that a right-invariant nsf weight $\widetilde{\psi}: M_+\to [0, \infty]$ extending $\psi: A\cap M^+ \to \mathbb{C}$ is necessarily unique? Obviously, this is true since in the general theory of locally compact quantum groups since a right invariant nsf weight is uniquely determined (up to constant). I am interested to see a more elementary approach in this case though.

Note that if $A$ is discrete, this is fairly easy, since a positive element in $M$ is a $\sigma$-strong limit of an increasing net of positive elements in $A$, so normality of the weight does the trick.

Consider an algebraic quantum group $(A, \Delta)$ in the sense of Van Daele, i.e. a regular multiplier Hopf $^*$-algebra with a positive left integral $\varphi$ and a positive right integral $\psi$.

It is well-known that we can make the following construction:

• Embed $A\subseteq B(L^2(A, \psi)).$
• Define $M= A''\subseteq B(L^2(A, \psi))$.
• Show that $M$ with an extension of the coproduct $\Delta$ becomes a von Neumann algebraic locally compact quantum group in the sense of Kustermans and Vaes.

See for instance Van Daele's recent survey paper Algebraic quantum groups and duality II for details about this construction.

In this case, we can show that the right invariant nsf weight $\widetilde{\psi}$ extends $\psi$. I'm interested in the following: can we show (using elementary means) that a right-invariant nsf weight $\widetilde{\psi}: M_+\to [0, \infty]$ extending $\psi: A\cap M^+ \to \mathbb{C}$ is necessarily unique? Obviously, this is true since in the general theory of locally compact quantum groups since a right invariant nsf weight is uniquely determined (up to constant). I am interested to see a more elementary approach in this case though.

Note that if $A$ is discrete, this is fairly easy, since a positive element in $M$ is a $\sigma$-strong limit of an increasing net of positive elements in $A$, so normality of the weight does the trick.