Download An Introduction to Homological Algebra by Aaron Marcus PDF

Download An Introduction to Homological Algebra by Aaron Marcus PDF

By Aaron Marcus

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Example text

Fundamental root systems 33 the Jacobi identity and the fact that [ep eyJ =l= 0 yield that either a) [n ... [ei, ei b) [([... [ei, ei,1 ... ei,J, [... [ej, ej,1 ... ej,_lJ], ej,] =l= O. 2] • • • ei,J eit], [ ... [ej, ej,J ... eit- 1 ]] =l= 0, or In case a), we have lX = f3' + y', where f3' = f3 + lXj" y' = Y - lXiI are positive roots; in case b), lX' = lX - lXiI is a root, and lX = lX' + lXiI' In either case the assertion follows by inductive hypothesis. It follows from Lemma 2 and the fact that the roots span the dual space of SJ (otherwise we should have 0 =l= hE SJ with lX (h) = 0 for all roots lX, from which h is in the center of ~) that the simple roots span the dual space of SJ.

By Lemma 2, 2(lX(h) - 1) = -2, or 2lX(h) = 0, a contradiction to (he" hp) =1= O. Thus the lemma is proved. 10. Let the hypotheses be as in Lemma 9, and let lX and {3 be roots, {3 O. Then not all of lX, lX + {3, lX + 2{3, lX + 3{3, lX + 4{3 are roots. Thus not all lX + k {3 are roots. Namely, if one of the above sequence is zero, then by Lemma 7 either lX = 0, lX = -{3, or lX = (3(P = 5), so that either lX + 2{3, lX + 3{3, or lX + {3 is 2{3, which is not a root. If none is zero, and if all are roots, then (lX + 2(3) ± lX cannot be roots by Lemma 5, nor can (lX + 4(3) ± (lX + 2(3).

Thus let ~ have prime characteristic p;;;; 3. By (2), if v is as in the statement, v (ee)j+l, if not zero, belongs to the characteristic root {3 + j + 1 of he; hence v (ee)i+l = O. Also, j < P - 1; thus we have an integer k, 0 ~ k ~ ~ j < P - 1, with v (ea)k =1= 0, v (ee)k+l = O. By analogy with (2), v Ie he = ({3 - 1) v Ie, from which v Ie = 0 by assumption. = k + 1 yields 2{3 = -k, since 0 < k + 1 ~ P - 1 and since v (ee)k =1= o. Thus the lemma will be proved once we show k = f. Suppose k < j; then let 0 =1= u E )8, u he = ({3 + k + 1) u.

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