Algebraic Peter-Weyl theorem in the case of $G=SL_2$. The algebraic Peter-Weyl theorem says that for a linear reductive group $G$ we have $\mathbb{C}[G] = \oplus_{V} V \otimes V^* $, where $V$ runs over the set of all non-isomorphic irreducible representations of $G$. 
I would like to know this result explicitly in the case of $G=SL_2$. We have $\mathbb{C}[SL_2] = \mathbb{C}[a,b,c,d]/(ad-bc-1)$. How to write $\mathbb{C}[SL_2] = \oplus_{V} V \otimes V^* $ explicitly (write the representations $V$ explicitly)? Thank you very much.
 A: Let $V$ be an irreducible left $G$-module. Via the isomorphism $V \otimes V^* \cong \mathrm{End}_\mathbf{C}(V)$ we obtain an injection (by the double centralizer theorem)
$$V \otimes V^* \cong \mathrm{End}_\mathbf{C}(V) \rightarrow \mathbf{C}[G]$$ sending an endomorphism $\phi$ to the function $g \mapsto \mathrm{trace}(\phi g^{-1})$, where $g$ is regarded as an endomorphism of $V$. One checks that this is a morphism of $G$-bimodules, where for $g,h \in G$ the notation $g \phi h$ is self-explanatory and on a function $f \in \mathbf{C}[G]$ we have
$$(gfh)(x)=f(g^{-1}xh^{-1}) \quad \hbox{for all $g,h,x \in G$}.$$ These are the inclusions in the Peter-Weyl theorem, so your question amounts to computing them explicitly for some realization of the irreps of $\mathrm{SL}_2$. As noted in the comments, the irreps of $\mathrm{SL}_2$ are indexed by non-negative integers $\lambda$, and the corresponding irrep $L(\lambda)$ is isomorphic to the space of homogeneous polynomials in two variables of degree $\lambda$. A geometric realization of the corresponding inclusions follows in the next paragraph. 
There is a surjective (orbit) map $G=\mathrm{SL}_2(\mathbf{C}) \rightarrow \mathbf{C}^2 \setminus \{0 \}$ defined by
$$\left(\begin{matrix} a & b \\ c & d \end{matrix} \right) \mapsto  \left(\begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix}\right) =\left( \begin{matrix} a \\ c \end{matrix}\right).$$ This induces an isomorphism $G/U \cong \mathbf{C}^2 \setminus \{0 \}$, and hence an isomorphism of function spaces $$\mathbf{C}[\mathbf{C}^2 \setminus \{0 \}]=\mathbf{C}[x,y] \rightarrow \mathbf{C}[G/U]=\mathbf{C}[G]^U.$$ The image of the space of polynomials of degree $\lambda$ in $\mathbf{C}[G]$ is a copy of the $\lambda+1$-dimensional irrep of $G$. 
Now we extend this to obtain all functions on $G$. There is a basis of $L(\lambda)$ given by the monomials $x^i y^{\lambda-i}$ indexed by integers $0 \leq i \leq \lambda$. Given two integers $0 \leq i,j \leq \lambda$ the corresponding matrix unit $E_{ij} \in \mathrm{End}(L(\lambda))$ goes to the function
$$f_{ij}(g)=\mathrm{trace}(E_{ij} g^{-1})=(g^{-1})_{ij},$$ where we regard $g$ as a matrix acting on $L(\lambda)$ and the subscript means take its $i,j$th matrix coefficient. Now for the matrix
$$g=\left( \begin{matrix} a & b \\ c & d \end{matrix} \right)$$ we see that $g^{-1}$ acts on a monomial $x^j y^{\lambda-j}$ via
$$g^{-1}(x^j y^{\lambda-j})=(ax+by)^j(cx+dy)^{\lambda-j}$$ in which the coefficient of $x^i y^{\lambda-i}$ is 
$$f_{ij}(g)=(g^{-1})_{ij}=\sum_{k+\ell=i} {j \choose k} {\lambda-j \choose \ell} a^k b^{j-k} c^\ell d^{\lambda-j-\ell} $$
I don't know of a nicer formula for a basis for the corresponding space $V \otimes V^*$. But I can say this: if you filter the coordinate ring by total degree in $a,b,c,d$ and use the relation $ad-bc$ to rewrite the resulting functions, then upon taking the associated graded you get the span of all monomials $a^ib^jc^k d^\ell$ of total degree $\lambda$ and not divisible by $bc$ (because you have rewritten these in terms of $ad$ plus lower degree terms). 
A: The function $V \otimes V^* \to \mathbb{C}[G]$ is given by $(v, \lambda) \mapsto (g \mapsto \lambda(gv))$. Let $G = SL_2$ and write the coordinate on $SL_2$ as $$ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right). $$
As shown in the post, $Span_{\mathbb{C}}\{c, d\}$ is a representation of $G$. Let $V = Span_{\mathbb{C}}\{c, d\}$ and take $e_1 = c = (1, 0)^T$, $e_2 = d = (0, 1)^T$. Then $V^* = Span_{\mathbb{C}}\{e_1^*, e_2^*\}$, where $e_i^*(e_j)=\delta_{ij}$. Let $v = e_1$, $\lambda=e_1^*$. Then $(e_1, e_1^*) \mapsto (g \mapsto e_1^*(ge_1))$. Suppose that $$ g = \left( \begin{matrix} g_{11} & g_{12} \\ g_{21} & g_{22} \end{matrix} \right). $$ $e_1^*(ge_1) = e_1^*(g_{11}, g_{21})^T = e_1^*(g_{11} e_1 + g_{21} e_2) = g_{11}$. Therefore $(e_1, e_1^*) \mapsto x_{11}$, where $x_{11}$ is the function on $SL_2$ whose value on a matrix $g$ in $SL_2$ is the $(1,1)$ entry of $g$. Similarly $(e_i, e_j^*) \mapsto x_{ji}$. We have $x_{11}x_{22} - x_{12} x_{21} = 1$ since $x_{ij}$'s are functions on $SL_2$.
