Let Δ be an indecomposable root system in a real inner product space E, and suppose that Φ is a simple system of roots in Δ, with respect to an ordering of E. If Φ = { α 1 , … , α l }, prove that α 1 + ⋯ + α l ∈ ΔI know that any positive root γ may be written as a sum of simple roots, and furthermore that every partial sum is itself a root, but I am unsure if that will help me or not. Any hints to get me started?

Question

Let   Δ be an indecomposable root system in a real inner product space   E, and suppose that   Φ is a simple system of roots in   Δ, with respect to an ordering of   E. If   Φ  =  {      α    1    ,  …  ,      α    l    }, prove that      α    1    +  ⋯  +      α    l    ∈  ΔI know that any positive root   γ may be written as a sum of simple roots, and furthermore that every partial sum is itself a root, but I am unsure if that will help me or not. Any hints to get me started?

gozaderaradiox5 · Accepted Answer

First, I would let   Φ be a base for an indecomposable root system, and choose labels for the elements of   Φ according to the following algorithm:Suppose you have chosen       Φ    k    :=  {      α    1    ,  …  ,      α    k    } such that the corresponding root system       Δ    k   is indecomposable (when   k  =  1 this is automatic). Then, there exists       α          k      +      1        ∈  Φ such that       Φ          k      +      1       is a base for an indecomposable root system       Δ          k      +      1      . Indeed, otherwise      s    p    a    n    (      Φ    k    )  ⊥      s    p    a    n    (  Φ  ∖      Φ    k    )  ,so   Δ is not indecomposable, a contradiction.This is enough to obtain strict inequality since, by construction, we must have   (      α    j    ,      α    l    )  &amp;lt;  0 for some   j  &amp;lt;  l.

Lucian Maddox · Accepted Answer

Great, but maybe there is another solution?

Let <mi mathvariant="normal">Δ be an indecomposable root system in a real inner

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