Is $R(su_{4})\cong R(so_{6})$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T21:23:13Z http://mathoverflow.net/feeds/question/93926 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93926/is-rsu-4-cong-rso-6 Is $R(su_{4})\cong R(so_{6})$? Changwei Zhou 2012-04-13T01:15:43Z 2012-05-10T09:26:09Z <p>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. </p> <p>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 $$R(h_{su_{4}})\cong \mathbb{Z}[x,y,z,w]/(xyzw-1)$$ </p> <p>a typical basis of $R(h_{su_{4}})$ over $R(su_{4})$ consists of $x^{i}y^{j}z^{k}, 0\le i\le 3, o\le j\le 2, 0\le k\le 1$. </p> <p>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:</p> <p>$$R(so_{6})\cong \mathbb{Z}[x+y+z+x^{-1}+y^{-1}+z^{-1},x^{\frac{1}{2}}y^{\frac{1}{2}}z^{\frac{1}{2}}+x^{\frac{1}{2}}y^{-\frac{1}{2}}z^{-\frac{1}{2}}+x^{-\frac{1}{2}}y^{-\frac{1}{2}}z^{\frac{1}{2}}+x^{-\frac{1}{2}}y^{\frac{1}{2}}z^{-\frac{1}{2}},x^{-\frac{1}{2}}y^{-\frac{1}{2}}z^{-\frac{1}{2}}+x^{\frac{1}{2}}y^{-\frac{1}{2}}z^{\frac{1}{2}}+x^{-\frac{1}{2}}y^{\frac{1}{2}}z^{\frac{1}{2}}+x^{\frac{1}{2}}y^{\frac{1}{2}}z^{-\frac{1}{2}}]$$</p> <p>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&amp;Harris, Chapter 23.2 or <a href="http://books.google.com/books?id=6GUH8ARxhp8C&amp;printsec=frontcover&amp;dq=fulton+harris&amp;hl=en&amp;sa=X&amp;ei=ALmIT_rvAcTL0QH-_dH1CQ&amp;ved=0CDgQ6AEwAA#v=onepage&amp;q=fulton%2520harris&amp;f=true" rel="nofollow">here</a>). As one commentator noted I am not clear about the relationship between $R(so_{6})$ and $R(h_{so_{6}})$. </p> <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/93926/is-rsu-4-cong-rso-6/93963#93963 Answer by BS for Is $R(su_{4})\cong R(so_{6})$? BS 2012-04-13T15:05:40Z 2012-04-13T15:05:40Z <p>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.</p> <p>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.</p>