A question on the root systems of simple Lie algebras in the 90 degree case - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:18:05Z http://mathoverflow.net/feeds/question/60935 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60935/a-question-on-the-root-systems-of-simple-lie-algebras-in-the-90-degree-case A question on the root systems of simple Lie algebras in the 90 degree case K McKenzie 2011-04-07T12:57:58Z 2011-06-17T04:29:19Z <p>I've been taking a look at simple Lie algebras for particle physics and I've found myself wondering about the following question. It can be shown that the adjacent roots in a root diagram corresponding to a simple Lie algebra must always lie at either 0 ,30, 45, 60 or 90 degrees (if the rank $\geq2$). In all cases but the 90 degree case (forgetting the trivial 0 degree case), the relative lengths of the roots are uniquely determined given the angle between them. But it turns out that in the 90 degree case there are no constraints whatsoever on their relative lengths. So my first question is: are there infinitely many Lie algebras corresponding to the 90 degree case, one for every possible ratio in the lengths of the roots? Or is there only one algebra, which just doesn't care about the relative lengths of its roots?</p> <p>Any help would be greatly appreciated! (Ps I am an extremely mediocre graduate physicist, so please go easy on me!)</p> http://mathoverflow.net/questions/60935/a-question-on-the-root-systems-of-simple-lie-algebras-in-the-90-degree-case/60936#60936 Answer by Jim Humphreys for A question on the root systems of simple Lie algebras in the 90 degree case Jim Humphreys 2011-04-07T13:15:33Z 2011-04-07T13:15:33Z <p>If you look only at a <em>simple</em> Lie algebra, no two "adjacent" simple roots in the Dynkin diagram can form a right angle: being joined by at least one edge forces a different angle. In the simple case there is no ambiguity about relative lengths of roots, but of course in a direct sum of simple Lie algebras the different simple ideals involved are not directly related. </p> <p>By the way, the nontrivial angles between adjacent simple roots are actually <em>obtuse</em> rather than acute angles. </p> http://mathoverflow.net/questions/60935/a-question-on-the-root-systems-of-simple-lie-algebras-in-the-90-degree-case/60992#60992 Answer by Ben Webster for A question on the root systems of simple Lie algebras in the 90 degree case Ben Webster 2011-04-07T20:12:39Z 2011-04-07T20:12:39Z <p>The length of a root is not a notion attached to a Lie algebra on its own, but to a <a href="http://en.wikipedia.org/wiki/Quadratic_Lie_algebra" rel="nofollow">nondegenerate quadratic Lie algebra</a>: a Lie algebra with a <strong>choice</strong> of invariant inner product. If instead of the question you did asked, you had said</p> <blockquote> <p>are there infinitely many quadratic Lie algebras corresponding to the 90 degree case, one for every possible ratio in the lengths of the roots? </p> </blockquote> <p>the answer would have been "Yes!" In fact, there is a quadratic Lie algebra for each choice of non-zero length for the two roots. The underlying Lie algebras are, of course, isomorphic, but the forms are different.</p> http://mathoverflow.net/questions/60935/a-question-on-the-root-systems-of-simple-lie-algebras-in-the-90-degree-case/68028#68028 Answer by Lucas Seco for A question on the root systems of simple Lie algebras in the 90 degree case Lucas Seco 2011-06-17T04:29:19Z 2011-06-17T04:29:19Z <p>Ben, I also asked myself that same question and the notion that made it clear for me was that of an isomorphism between root systems: it is a linear map that sends all the roots of one system to all the roots of the other system, preserving the Cartan-Killing numbers of the corresponding roots (see Humphreys' book, Section 9.2)</p> <p>The beauty is that such an isomorphism does not need to be an isometry! So the the infinitely many two-dimensional root systems of the 90$^o$ case are all isomorphic between themselves, in particular all isomorphic to $A_1 \times A_1$.</p> <p>In terms of the corresponding Lie algebras, you have also infinitely many, all of them isomorphic to ${\frak sl}(2) \times {\frak sl}(2)$.</p> <p>PS - It can be shown that an isomorphism between two <strong><em>irreducible</em></strong> root systems must be conformal: it must scale the metrics by a constant factor. In particular, every automorphism of an irreducible root system is an isometry. (Ask me if you need a proof of this, it is very simple but it is not in Humphreys' book, I think.)<br> The irreducibility here is crucial since an $A_1 \times A_1$ root system with distinct root lengths admit an automorphism which is not an isometry.</p>