MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
@BS: I understand. This is very helpful. My main obstacle is I do not know how to establish this isomorphism on the representation ring level, despite the proof you offered. I will try to think about this harder. – Kerry Apr 13 '12 at 16:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.