Let $V$ be a finite-dimensional $\mathfrak{sl}_2$-module. There is a standard base $\{e,f,h\}$ in $\mathfrak{sl}_2$, I use standard notation ($h$, for instance, is the diagonal matrix with $1$ and $-1$ on the diagonal). The character of $V$ is a Laurent polynomial $$\chi_V(z)=\sum_{m\in \mathbb{Z}}\mathrm{dim}V(m)z^m,$$ where $V(m)=\mathrm{Ker}(h-m\cdot Id_V)$. Suppose $U$ is another finite-dimensional representation of $\mathfrak{sl}_2$. I would like to prove that $\chi_{V\otimes U}=\chi_V\cdot \chi_U$.
Okay, thus one has the special condition for coefficients, i.e. $$\mathrm{dim}(V\otimes U)(m)=\sum_{\lambda+\mu=m}\mathrm{dim}V(\lambda)\cdot \mathrm{dim}U(\mu),$$ really?
I guess we are to prove the isomorphism $$(V\otimes U)(m)\simeq\bigoplus_{\lambda+\mu=m}V(\lambda)\otimes U(\mu)?$$
Could you please help me to do that? The only thing I understand is that there is an obvious embedding $V(\lambda)\otimes U(\mu)\subset (V\otimes U)(m)$ for all $\mu,\lambda$ such that $\mu+\lambda=m$.
It's almost the same as Jyrki's comment, but you don't need to argue with dimensions: If $V$ is an ${\mathfrak s}{\mathfrak l}_2({\mathbb C})$-module for which you have found some direct sum decomposition $V = \bigoplus\limits_{n\in{\mathbb Z}} V^{\prime}_n$ into ${\mathbb C}$-subspaces such that $V^{\prime}_n$ consists of $h$-eigenvectors of eigenvalue $n$, then this must be the $h$-eigenspace decomposition of $V$ by the linear algebra fact that eigenspaces for different eigenvalues are linearly independent.
In particular, if $V$ and $U$ have decompositions into $h$-eigenspaces $V(\lambda)$ and $U(\mu)$, respectively, then $$V\otimes U = \bigoplus\limits_{n\in{\mathbb Z}}\left[\bigoplus\limits_{\lambda+\mu=n} V(\lambda)\otimes U(\mu)\right]$$ by the additivity of the tensor product, and since $\bigoplus\limits_{\lambda+\mu=n} V(\lambda)\otimes U(\mu)$ is contained in the $h$-eigenspace of $V\otimes U$ for the eigenvalue $n$, the above allows to conclude $(V\otimes U)(n)=\bigoplus\limits_{\lambda+\mu=n} V(\lambda)\otimes U(\mu)$.