- #1
bartadam
- 40
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I have a six dimensional lie algebra. I checked it was semi-simple by checking if the killing form was not invertible. I found a set of two (I think) maximally commuting elements which I called H1 and H2 and found their matrix representation by calculating the structure constants.
I have put these matrices into maple and calculated
H=aH_1+bH_1
and calculated the eigenvalues of H and got four (as expected) eigenvalues as follows,...
\sqrt{2}i(a+b),\ -\sqrt{2}(a+b),\ \sqrt{2}(a-b)\ and\ -\sqrt{2}i(a-b)
So I think the commutators are as follows
[\stackrel{\rightarrow}{H},E^{\pm \alpha}]=\pm\sqrt{2}i(1,1)E^{\pm\alpha}
[\stackrel{\rightarrow}{H},E^{\pm \beta}]=\pm\sqrt{2}i(1,-1)E^{\pm\beta}
plus all the others which follow from cartan weyl procedure. Where \stackrel{\rightarrow}{H}=(H_1,H_2)
This makes the algebra A1+A1=D2 as it has two orthogonal root spaces of A1.
Am I doing this properly?
Unfortunately I have not calculated the generators E^{\alpha} to check the above because the algebra in the basis which I derived it is is such that calculating these is horrifically tedious if you understand me. Are the generators just going to be the dot product of the eigenvectors with the original basis?
I have put these matrices into maple and calculated
H=aH_1+bH_1
and calculated the eigenvalues of H and got four (as expected) eigenvalues as follows,...
\sqrt{2}i(a+b),\ -\sqrt{2}(a+b),\ \sqrt{2}(a-b)\ and\ -\sqrt{2}i(a-b)
So I think the commutators are as follows
[\stackrel{\rightarrow}{H},E^{\pm \alpha}]=\pm\sqrt{2}i(1,1)E^{\pm\alpha}
[\stackrel{\rightarrow}{H},E^{\pm \beta}]=\pm\sqrt{2}i(1,-1)E^{\pm\beta}
plus all the others which follow from cartan weyl procedure. Where \stackrel{\rightarrow}{H}=(H_1,H_2)
This makes the algebra A1+A1=D2 as it has two orthogonal root spaces of A1.
Am I doing this properly?
Unfortunately I have not calculated the generators E^{\alpha} to check the above because the algebra in the basis which I derived it is is such that calculating these is horrifically tedious if you understand me. Are the generators just going to be the dot product of the eigenvectors with the original basis?