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$\DeclareMathOperator\sl{\mathfrak{sl}}\newcommand\hw{\mathrm{hw}}\newcommand\lw{\mathrm{lw}}$Consider $\mathfrak{g}=\sl_2(\mathbb{C})$. Fix $\lambda,\mu\in\mathbb{C}\setminus \frac{1}{2}\mathbb{Z}_{\geq 0}$, and we may assume that $\lambda+\mu\in\mathbb{C}\setminus \frac{1}{2}\mathbb{Z}_{\geq 0}$.

For the vector space $M^{\hw}(\lambda)$ with formal basis $\{v_{\lambda-j}: j\in\mathbb{Z}_{\geq 0} \}$ we define action of linear operators $E$, $F$ and $H$ as $$ H(v_{\lambda-j})=2(\lambda-j)v_{\lambda-j}, $$ $$ E(v_{\lambda-j})=j v_{\lambda-j+1}, $$$$ F(v_{\lambda-j})=(2\lambda-j)v_{\lambda-j-1}. $$

For the vector space $M^{\lw}(\mu)$ with formal basis $\{w_{\mu+j}: j\in\mathbb{Z}_{\geq 0} \}$ we define action of linear operators $E$, $F$ and $H$ as $$ H(w_{\mu+j})=-2(\mu-j)w_{\mu+j}, $$ $$ E(w_{\mu+j})=(2\mu-j) w_{\mu+j+1}, $$$$ F(w_{\mu+j})=j w_{\mu+j-1}. $$

Here $M^{\hw}(\lambda)$ is a Verma module of the highest weight $\lambda$, and $M^{\lw}(\mu)$ is a Verma module of the lowest weight $\mu$.

For $\lambda_1,\lambda_2\in\mathbb{C}\setminus \frac{1}{2}\mathbb{Z}_{\geq 0}$ such that $\lambda_1+\lambda_2\in\mathbb{C}\setminus \frac{1}{2}\mathbb{Z}_{\geq 0}$ it is easy to obtain that $$ M^{\hw}(\lambda_1)\otimes M^{\hw}(\lambda_2)=\bigoplus\limits_{J=0}^\infty M^{\hw}(\lambda_1+\lambda_2-J), $$ and similarly for the tensor product decomposition of two Verma modules of the lowest weight.

The question is to find how to decompose $M^{\hw}(\lambda)\otimes M^{\lw}(\mu)$. The complication which emerges is that one apparently runs into direct integration, since Casimir operator seems to have a continuous spectrum.

EDIT: The resulting module has to have a countable basis, yes. But if I take the Casimir operator $C$ and act with it and comultiplication $\Delta$ on a weight vector of a form $u=\sum\limits_{i,j=0}^\infty a_{i,j} v_{\lambda-i}\otimes w_{\mu+j}$, $$ \Delta C (u)= A u $$ gives a system of equations on coefficients $\{a_{i,j}\}$ and the eigenvalue A. Even though this system can be separated into subsystems, where we fix $i-j=\alpha$, $\alpha\in\mathbb{Z}$, unfortunately, there appears to be no conditions restricting the eigenvalue $A$(as opposed to a system of equations in the case of $M^{\hw}(\lambda_1)\otimes M^{\hw}(\lambda_2)$). Hence, I assume, one should integrate over it somehow. I apologize in advance if I am mistaken.

Any hints on how to proceed, any references in this direction will be appreciated!

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    $\begingroup$ Sorry for my naive comment : how can the Casimir have continuous spectrum since a basis of that tensor product is countable? $\endgroup$ Commented Aug 16 at 12:27
  • $\begingroup$ @NicolasHemelsoet Dear Nicolas Hemelsoet, thank you very much for your question! I have also asked this question myself when approaching the problem, hence, decided to add the answer to it to my original post. $\endgroup$
    – clouds
    Commented Aug 16 at 12:55
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    $\begingroup$ Actually, this tensor product won't contain any lowest weight vector. So it can't contains any $M^{lw}(\nu)$. Similarly, it won't contain any $M^{hw}(\nu)$, so you won't be able to express this tensor product in term of Verma modules. $\endgroup$ Commented Aug 16 at 13:19
  • $\begingroup$ @NicolasHemelsoet True! My naive guess is that it should be something known as dense modules. See, for example, Volodymyr Mazorchuk "Lectures on $\frak{sl}_2(\mathbb{C})$-modules", Section 3.3. But even if it is true, I don't know how to rigorously obtain this result. $\endgroup$
    – clouds
    Commented Aug 16 at 13:50
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    $\begingroup$ The theory of direct integrals works well in the context of unitary representations on Hilbert spaces. May I suggest rephrasing the question in the context of unitary representations on Hilbert spaces of the Lie group $\widetilde{SL(2,R)}$ (the universal cover of $SL(2,R)$), after imposing the additional constraint that 𝜆 and 𝜇 are negative real numbers (to ensure unitarity)? Note that the tensor product of Hilbert spaces is not the algebraic tensor product – it is a completion of the algebraic tensor product. I am very interested in the answer to that question. $\endgroup$ Commented Aug 16 at 21:25

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