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This is one of small the unsettled questions I had in my senior project. I want to prove for type $D$ we have $R(T)$ is a free module over $R(G)$ by finding a basis. I think we should have,$R(G)\cong R(g)$, $R(T)\cong R(h_{g})$, but since $SO_{6}$ is not simply connected this probably does not work and I have to "ascend" to spin groups, but I do not know how.

Define the representation ring of a lie algebra to be the formal sums of its characters, it is not hard to show that $$R(su_{4})\cong \mathbb{Z}[x+y+z+w,xy+yz+zx+wz+wy+wz,xyz+yzw+xzw+xyw]/(xyzw-1)$$ and $$R(h_{su_{4}})\cong \mathbb{Z}[x,y,z,w]/(xyzw-1)$$

a typical basis of $R(h_{su_{4}})$ over $R(su_{4})$ consists of $x^{i}y^{j}z^{k}, 0\le i\le 3, o\le j\le 2, 0\le k\le 1$.

I proved that the weight lattice of $su_{4}$ and $so_{6}$ are isomorphic, and their Weyl group are both isomorphic to $S_{4}$. So $R(h_{so_{6}})$ should be a free module over $R(so_{6})$ with rank 24 as well. But I found I could not use this to find a basis for $R(h_{so_{6}})$ over $R(so_{6})$, because we have:

$$R(so_{6})\cong \mathbb{Z}[x+y+z+x^{-1}+y^{-1}+z^{-1},x^{\frac{1}{2}}y^{\frac{1}{2}}z^{\frac{1}{2}}+x^{\frac{1}{2}}y^{-\frac{1}{2}}z^{-\frac{1}{2}}+x^{-\frac{1}{2}}y^{-\frac{1}{2}}z^{\frac{1}{2}}+x^{-\frac{1}{2}}y^{\frac{1}{2}}z^{-\frac{1}{2}},x^{-\frac{1}{2}}y^{-\frac{1}{2}}z^{-\frac{1}{2}}+x^{\frac{1}{2}}y^{-\frac{1}{2}}z^{\frac{1}{2}}+x^{-\frac{1}{2}}y^{\frac{1}{2}}z^{\frac{1}{2}}+x^{\frac{1}{2}}y^{\frac{1}{2}}z^{-\frac{1}{2}}]$$

the first is the standard representation with weights $\pm L_{i}$, the second and the third are the spin representations one obtain from clifford algebra or "ascend" to spin groupgroup(can be found at Fulton&Harris, Chapter 23.2 or here). As one commentator noted I am not clear about the relationship between $R(so_{6})$ and $R(h_{so_{6})$. R(h_{so_{6}})$. and $$R(h_{so_{6}})\cong \mathbb{Z}[x,y,z,x^{-1},y^{-1},z^{-1}]$$ because we know the two diagonal submatrices in$so_{6}$must be skew-symmetric. From$A+D^{T}=0$we conclude$T$is isomorphic to$S^{1}\times S^{1}\times S^{1}$. Thus we conclude this. I thought it would be a simple change of variable to prove the two cases are just the same, but I found the isomorphism between$R(so_{6})$and$R(su_{4})$does not extend nicely to an isomorphism between$R(h_{so_{6}})$and$R(h_{su_{4}})$. So I believe I must be confused. My advisor suggested me that maybe there is some subtly in$Spin_{6}$, but I still do not know how to estbalish an isomorphism or to find the basis right away. 6 added explanation. This is one of small the unsettled questions I had in my senior project. I want to prove for type$D$we have$R(T)$is a free module over$R(G)$by finding a basis. I think we should have,$R(G)\cong R(g)$,$R(T)\cong R(h_{g})$, but since$SO_{6}$is not simply connected this probably does not work and I have to "ascend" to spin groups, but I do not know how. Define the representation ring of a lie algebra to be the formal sums of its characters, it is not hard to show that $$R(su_{4})\cong \mathbb{Z}[x+y+z+w,xy+yz+zx+wz+wy+wz,xyz+yzw+xzw+xyw]/(xyzw-1)$$ and $$R(h_{su_{4}})\cong \mathbb{Z}[x,y,z,w]/(xyzw-1)$$ a typical basis of$R(h_{su_{4}})$over$R(su_{4})$consists of$x^{i}y^{j}z^{k}, 0\le i\le 3, o\le j\le 2, 0\le k\le 1$. I proved that the weight lattice of$su_{4}$and$so_{6}$are isomorphic, and their Weyl group are both isomorphic to$S_{4}$. So$R(h_{so_{6}})$should be a free module over$R(so_{6})$with rank 24 as well. But I found I could not use this to find a basis for$R(h_{so_{6}})$over$R(so_{6})$, because we have: $$R(so_{6})\cong \mathbb{Z}[x+y+z+x^{-1}+y^{-1}+z^{-1},x^{\frac{1}{2}}y^{\frac{1}{2}}z^{\frac{1}{2}}+x^{\frac{1}{2}}y^{-\frac{1}{2}}z^{-\frac{1}{2}}+x^{-\frac{1}{2}}y^{-\frac{1}{2}}z^{\frac{1}{2}}+x^{-\frac{1}{2}}y^{\frac{1}{2}}z^{-\frac{1}{2}},x^{-\frac{1}{2}}y^{-\frac{1}{2}}z^{-\frac{1}{2}}+x^{\frac{1}{2}}y^{-\frac{1}{2}}z^{\frac{1}{2}}+x^{-\frac{1}{2}}y^{\frac{1}{2}}z^{\frac{1}{2}}+x^{\frac{1}{2}}y^{\frac{1}{2}}z^{-\frac{1}{2}}]$$ the first is the standard representation with weights$\pm L_{i}$, the second and the third are the spin representations one obtain from clifford algebra or "ascend" to spin group. As one commentator noted I am not clear about the relationship between$R(so_{6})$and$R(h_{so_{6})$. and $$R(h_{so_{6}})\cong \mathbb{Z}[x,y,z,x^{-1},y^{-1},z^{-1}]$$ because we know the two diagonal submatrices in$so_{6}$must be skew-symmetric. From$A+D^{T}=0$we conclude$T$is isomorphic to$S^{1}\times S^{1}\times S^{1}$. Thus we conclude this. I thought it would be a simple change of variable to prove the two cases are just the same, but I found the isomorphism between$R(so_{6})$and$R(su_{4})$does not extend nicely to an isomorphism between$R(h_{so_{6}})$and$R(h_{su_{4}})$. So I believe I must be confused. My advisor suggested me that maybe there is some subtly in$Spin_{6}\$, but I still do not know how to estbalish an isomorphism or to find the basis right away.