Part of one of my calculations involves (the innocent looking) expression $\sum_{\alpha\in\Sigma} (\alpha,\alpha)$ for simple Lie algebras. I have two methods of calculating it -- which don't agree. I'm pretty sure that the first one is wrong, but I don't know why. Any help is welcome (which is why I posted here)!

First my starting 'facts' (see e.g. the free book by Cahn or the comprehensive Knapp):

Given a simple Lie algebra $\mathfrak{g}$ and a basis to the Cartan subalgebra $\{h_i\ ,\ i=1\ldots,r\}$, the components of the roots are defined by $$[h_i,e_\alpha] = \alpha_i e_\alpha$$

The Killing form restricted to the Cartan subalgebra is (Knapp Cor (2.24): but with index ridden notation) $$ g_{ij}=\mathrm{tr}h_i h_j = \sum_{\alpha\in\Sigma}\alpha_i\alpha_j $$ The inner product on the root space is defined via the Killing form (Knapp eqn (2.28)): $$(\alpha,\beta) = \alpha^i\beta_i = \alpha_i g^{ij} \beta_j=\mathrm{tr}(h_\alpha h_\beta)\ ,\qquad g^{ij}g_{jk}=\delta^i_k$$

So we get our first (and probably wrong) way of calculating: $$\sum_{\alpha\in\Sigma} (\alpha,\alpha) = g^{ij}\sum_{\alpha\in\Sigma} \alpha_i\alpha_j = g^{ij}g_{ij} = \sum_i \delta^i_i = \mathrm{rnk} \mathfrak{g} $$

The second method is to just enumerate and sum over all roots. For a simply laced Lie algebra this is easy, because all roots have the same length $(\alpha,\alpha)=l$: $$\sum_{\alpha\in\Sigma} (\alpha,\alpha) = l\sum_{\alpha\in\Sigma} 1 = l\ (\mathrm{dim}\mathfrak{g}-\mathrm{rnk}\mathfrak{g}) $$

These results are not compatible...

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    $\begingroup$ I'm confused, what is $\alpha^i$ and $g^{ij}$, are you using Einstein summation notation? Also, shouldn't there be $\beta_i$ in your formula with $(\alpha, \beta)$? $\endgroup$ – Steven Sam Feb 24 '10 at 6:40
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    $\begingroup$ Won't you know which one is right by testing them on an example where the two expressions disagree? $\endgroup$ – Qiaochu Yuan Feb 24 '10 at 7:00
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    $\begingroup$ Qiaochu: The last formula is correct. The number of roots of a semisimple Lie algebra is its dimension minus the dimension of a Cartan subalgebra. $\endgroup$ – Steven Sam Feb 24 '10 at 7:11
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    $\begingroup$ @Sam - Yes, I'm using Einstein summation convention. $g^{ij}$ is the inverse of the matrix of the Killing form and is used to raise the index on $\alpha^i:=g^{ij}\alpha_j$. I fixed the typo with $\beta$. $\endgroup$ – Simon Feb 24 '10 at 9:22
  • $\begingroup$ @Qiaochu - I agree. The last formula is correct (for simply laced Lie algebras - not for $b_n$, $c_n$, etc). But I don't know why the first formula is incorrect. Where is the wrong step? $\endgroup$ – Simon Feb 24 '10 at 9:24

The reason for the discrepancy and to that the first calculation is not correct is that indeed the Cartan-Killing metric is a metric on the weight space, but it is not the metric with respect to which the root lengths are defined, (it is up to a scalar multiple). The root lengths are defined with the Euclidian metric in the weight space, and since the primitive weights are not orthogonal, this metric is not diagonal in the primitive weight basis, but it is different from the Cartan-Killing metric by a scalar multiple. A properly normalized metric tensor is given in table 7 of the following review article by Slansky for all simple Lie algebras. The normalization convention in this article is taken by fixing the root lengths of simply laced algebras to 2. One can easily check, for example, using the Cartan matrices of table 6 that the squared lengths of the simple roots of An are 2, or the length of the roots of B2 is 1,2 respectively.

  • $\begingroup$ Hi David, I really appreciate your help, but I'm not sure I understand your answer. In the basis defined by a choice of the Cartan subalgebra, the Cartan-Killing metric is used to define the innerproduct for roots. Slansky never defines this explicitly, but most texts do, eg Wybourne or Knapp. If you express the roots/weights using Dynkin labels (ie using the basis of primitive weights), then the matrices given in Slansky's table 7 are the correct metric tensors. But I'm not sure where that comes into this calculation. Could you be perhaps more specific in which step/definition is wrong? $\endgroup$ – Simon Feb 25 '10 at 5:34
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    $\begingroup$ Hi Simon, The normalization of the root lengths is a semisimple Lie algebra is arbitrary. So one can use the Cartan-Killing metric to define the root lengths, but then they will not be as the convention used by you (1-for An) or Slansky (2-for An). You can check that even for sL2 there will be a discrepancy between the two methods. Here is an article in which the computation of the scaling parameter (which is basically proportional to the Coxeter number is given: rose-hulman.edu/~brought/Epubs/preprints/killing.pdf $\endgroup$ – David Bar Moshe Feb 25 '10 at 8:30
  • $\begingroup$ Hi David, I did not assume the root lengths were one - I set them to some unknown $l$ that I thought was arbitrary. I was convinced by someone else to check the simple cases of sl2 and sl3 and it turned out in both computations the two methods matched. This is confirmed by the discussion in the paper you linked to. Thanks for your help (and the useful links). $\endgroup$ – Simon Feb 25 '10 at 15:01

It turns out that I was wrong and both methods give correct results. In fact using the two results you can reproduce the scaling parameters for the simple Lie algebras given in the Broughton paper linked to by David.

My mistake was in assuming that the length of the roots is always arbitrary. It's well known that the choice of the invariant bilinear on the algebra is unique up to a scale. The restriction of the bilinear onto the Cartan subalgebra is dualized to give the metric on the root space and thus passes on the choice of scale to the roots.

But I had already chosen the Killing form as my bilinear - a necessary ingredient in the first method I presented. Thus the length of my roots was fixed.

Direct construction of a few low rank cases (something that I should have done before posting) confirmed that everything works out ok.

As a quick check and application, we can reproduce the scaling factor given by Broughton for $A_n$. We have $\mathrm{rnk}(A_n)=n$ and $\mathrm{dim}(A_n)=(n+1)^2-1$, so $l=\frac{\mathrm{rnk}(A_n)}{\mathrm{dim}(A_n)-\mathrm{rnk}(A_n)}=\frac{1}{n+1}$.


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