Is $R(su_{4})\cong R(so_{6})$?

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(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 $R(h_{so_{6}})$.

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.

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I have worked with $R(G)$ quite a bit, but since I've never seen the notation before, what is $R(h_G)$ for a group/lie algebra $G$? –  ARupinski Apr 13 '12 at 2:16
@ARupinski: $h$ is the Cartan subalgebra of $G$, to distinguish the two rings I use $R_{h_{G}}$. –  Kerry Apr 13 '12 at 2:30
Hmm, shouldn't $R(h_{so_6})$ contain $R(so_6)$ as a subring? –  Gjergji Zaimi Apr 13 '12 at 3:44
Yes, that is what I am expecting. But I do not know what is wrong so I really need some help. –  Kerry Apr 13 '12 at 3:56
The first one is the standard representation, the second and the third are the spin representations. –  Kerry Apr 13 '12 at 3:57
It is a standard fact that $Spin(6)$ and $SU(4)$ (hence $so_{6}$ and $su_{4}$) are isomorphic. An easy way to see this is to observe that $SU(4)$ is simply connected and acts on the exterior square of $\mathbb{C}^4$, preserving a (complex) quadratic form (coming form exterior squaring and $det=1$) and a real structure (a conjugate-linear involution, hodge star followed by conjugation) for which the quadratic form is definite. Then the equality of dimensions (and easy calculation of the kernel) shows that $SU(4)\to SO(6)$ is a double covering.
From a higher standpoint, you may also observe that the Dynkin diagrams for $D_3$ and $A_3$ are the same, hence the maximal compact subgroups of the corresponding simply connected complex Lie groups are isomorphic.