Errata to Invariant Theory with Applications January 28, 2017

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1 Invarant Theory wth Applcatons Jan Drasma and Don Gjswjt http: // verson of 7 December 2009 Errata and addenda by Darj Grnberg The followng s a haphazard lst of errors I found n Invarant Theory wth Applcatons by Jan Drasma and Don Gjswjt. 16. Errata Page 5, 1.1: Replace Clearly, the elements of V are regular of degree by Clearly, the elements of V are regular functons and are homogeneous of degree. Page 7, 1.3: dscrbed descrbed. Page 8, Example 1.3.2: althought although. Page 8, Example 1.3.3: wth the same exponent wth the same coeffcent. Page 9, proof of Proposton 1.4.1: Replace To each c = c 1,..., c n C n by To each c = c 1,..., c n ) C n. Page 9, proof of Proposton 1.4.1: In 1.8), the entres n the last column should be c n, c n 1,..., c 2, c 1 not c n, c n 1,..., c 2, c 1 ). Page 9, proof of Proposton 1.4.1: Replace shows that χ Ac t) = t n + c n 1 t n c 1 t + c 0 by shows that χ Ac t) = t n + c 1 t n c n 1 t + c n. Page 9, Exercse 1.4.2: Replace that χ Ac t) = t n + c n 1 t n c 1 t + c 0 by that χ Ac t) = t n + c 1 t n c n 1 t + c n. Page 9, proof of Proposton 1.4.1: In 1.9), replace s 1 A c ), s 2 A c ),..., s n A c ) by s 1 A c ), s 2 A c ),..., s n A c )). Page 9, proof of Proposton 1.4.1: Replace dense n O Mat n C)) by dense n Mat n C). Ths mstake appears twce.) Page 9, Exercse 1.4.3: Replace nonzero egenvalues by egenvalues. Page 10, Exercse 1.4.3: Replace dstnct and nonzero by nonzero. 1

2 Page 10, Exercse 1.4.3: It mght be worth notcng that the fact you are mentonng about the Vandermonde determnant s a consequence of Lemma below usng the well-known fact that the determnant of a square matrx equals the determnant of ts transpose). Page 15, Theorem 2.2.9: You msspell Sylvester as Sylverster. Page 15, proof of Theorem 2.2.9: Remove the comma n Snce, Ã contans. Page 15, proof of Theorem 2.2.9: You wrte: t follows that Bez f ) has rank 2k + r. How does ths follow? I only see that Bez f ) has rank 2k + r. Page 20: Replace every element T U V by every element t U V. Page 20: Replace for T to zero by for t to zero. Page 21: wth of k-tensors wth k-tensors. Page 23: so that the v α, α = k a bass of V should be so that the v α wth α = k form a bass of S k V. Page 23: Replace π v 1 v k ) by π v 1 v k ). Page 23, Exercse : Replace v v v by v v v. Page 24, Exercse 3.1.4: You should requre that at least one of U and V s fnte-dmensonal. Page 24, Exercse 3.1.4: Replace somorhsm by somorphsm. Page 25: Replace so that g h f ) = hg) f by so that g h f ) = gh) f. Page 25, Example 4.0.8: Replace G module by G-module. Page 26, Example 4.0.9: Replace G module by G-module. Page 27, proof of Proposton 4.0.7: v v) = v v) = gv gv). g G gv gv) should be g G Page 28, Lemma 4.1.1: Replace G modules by G-modules. Page 28, 4.1: of the somorphsm classes of G-modules should be of the somorphsm classes of rreducble G-modules. Page 29, Exercse 4.1.2: Remove the superscrpt G. 2

3 Page 31, Lemma 5.0.9: Dxon s Dckson s. Page 32, proof of Hlbert s Bass Theorem: Dxon s Dckson s. Page 32: In 5.2), add a whtespace before for all f V 1. Page 32: Ths a G-module morphsm Ths s a G-module morphsm. Page 33, Exercse : wth zero coeffcent should be wth constant coeffcent equal to 0. Page 33, Exercse : I am wonderng whether you really mean subalgebra here and not graded subalgebra. Page 34, proof of Theorem 5.1.1: Replace β <= G by β G. Page 34, proof of Theorem 5.1.1: Replace p j = f α z α 1 1 zα n n. α =j f α z α 1 1 xα n n by p j = α =j Page 34, proof of Theorem 5.1.1: You wrte: Recall that p j s a polynomal n p 1,..., p G. Dd you actually prove ths anywhere? Ths s a partcular case of the followng fact: In the polynomal rng C [x 1, x 2,..., x n ], each S n - nvarant polynomal f C [x 1, x 2,..., x n ] S n can be wrtten as a polynomal n the Newton polynomals p 1, p 2,..., p n. 1 Ths s probably worth statng as an exercse n Chapter 2. Page 37, proof of the weak Nullstellensatz: Replace f k,ξ := x 1,..., x n 1, ξ) by f k,ξ := f k x 1,..., x n 1, ξ). Page 37, proof of the weak Nullstellensatz: Replace all three k sgns =1 by k sgns. j=1 Page 37: Nulstellensatz Nullstellensatz. Page 39, proof of Theorem : Replace the k sgn by a k sgn. =1 j=1 Page 41, Lemma 6.2.6: Replace from C [Y] C [X] by from C [Y] to C [X]. 1 The proof of ths fact s easy: By Theorem 2.1.1, t suffces to show that the s 1, s 2,..., s n are polynomals n p 1, p 2,..., p n. In other words, t suffces to show that s k s a polynomal n p 1, p 2,..., p n for each k {1, 2,..., n}. But ths easly follows by strong nducton over k ndeed, 2.18) gves a way to wrte each s k for k {1, 2,..., n} as a polynomal n p 1, p 2,..., p n, provded that s 1, s 2,..., s k 1 have already been wrtten n ths form). 3

4 Page 41, proof of Lemma 6.2.8: are a regular maps are regular maps. Page 42, Example 6.3.3: Replace act on the W by act on the vector space W. Page 43, Theorem 6.3.4: In property 4, replace φ : Z C m by φ : Z C m. Page 43, proof of Theorem 6.3.4: In the proof of property 3, replace φ : Z U by φ : Z C m. Page 47, proof of Theorem : Replace Hence w s n the null-cone N V by Hence w s n the null-cone N W. Page 49: Replace Let W d=0 W d be a drect sum by Let W = d=0 W d be a drect sum. Page 49: In 8.1), replace V and V d by W and W d, respectvely. Page 49, Example : Replace H C [x 1,..., x n ]) by H C [x 1,..., x n ], t). Page 50, Theorem 8.1.1: Replace of a fnte group by of a fnte group G. Page 50, proof of Theorem 8.1.1: In 8.5), replace tr L d g)) by t d tr L d g)). Page 50, proof of Theorem 8.1.1: Replace lets fx by let s fx. Page 50, proof of Theorem 8.1.1: Replace the nner sum tr L d g)) d=0 by the nner sum t d tr L d g)). d=0 Page 50, proof of Theorem 8.1.1: You wrte: Pck a bass x 1,..., x n of V that s a system of egenvectors for L 1 g). It s worth justfyng why such a bass exsts. Namely, you are usng the apocryphal theorem from lnear algebra that says that f U s a fnte-dmensonal C-vector space, and f α s an element of GL U) havng fnte order, then α s dagonalzable. You are applyng ths theorem to U = V and α = L 1 g), whch s allowed because the element L 1 g) of GL V ) has fnte order snce the element g of G has fnte order). Ths s not a dffcult argument, but I don t thnk t s obvous enough to be entrely left to the reader.) Page 50, proof of Theorem 8.1.1: Replace for a system by form a system. Page 50, proof of Theorem 8.1.1: On the frst lne of the computaton 8.7), replace 1 + λ n t + λ n t 2 + ) by 1 + λ n t + λ 2 nt 2 + ). 4

5 Page 50, proof of Theorem 8.1.1: On the frst lne of the computaton 8.8), replace tr L d g)) by t d tr L d g)). Page 50, proof of Theorem 8.1.1: On the thrd lne of the computaton 8.8), replace deti ρ g) t by det I ρ g) t). Page 51, 8.2: Replace whch u and v, dffer by whch u and v dffer. Page 51, 8.1: It s worth pontng out that you use the word code to mean lnear code. Page 52: Furhermore Furthermore. Page 53, Theorem 8.2.6: Replace x 4 y 4) by x 4 y 4) 4. Page 57, Example 9.1.8: Replace k l by Page 59, 9.2: Replace Consder the map λ : G GL[C [ x j, 1/ det x) ] ) by Consder the map λ : G GL C [ x j, 1/ det x) ]). k. l Page 65, Exercse : Replace larger enough by large enough. Page 65, proof of Proposton : Replace standard bass C 2 by standard bass of C 2. Page 65, proof of Proposton : Replace nduced bass of S d V) by nduced bass of S k V). Page 65, proof of Proposton : Replace d λ) x y k by d λ) x y k. Page 65, proof of Proposton : You clam that d 0 and every d wth c = 0 are nonzero polynomals wth λ. I would suggest explanng why they are nonzero. Namely, the polynomal d 0 s nonzero because d 0 = c λ y k and because not all c are 0); meanwhle, the polynomals d wth c = 0 are nonzero because they satsfy d 0) = c = 0.) Page 66, proof ) of Proposton : Replace Then for every the vector µ 0 µ k 0 µ 1 u = λ c x y k belongs to U by Then for every µ ) µ 0 {µ 0, µ 1,..., µ k } the vector µ k 0 µ 1 u = λ c x y k wth λ = µ 2 ) belongs to U. Page 66, proof of Proposton : In 10.4), replace S d End S 2 V) C )) by S d S 2 V) C ). Page 66, Exercse : Replace SL 2 C) module by SL 2 C)-module. 5

6 Page 68, 11.1: Replace t dentfes the space End V k) S k wth of symmetrc tensors by t dentfes the space End V k) S k space End V) k) S k of symmetrc tensors. End V) k) S k wth the Page 68, 11.1: Replace Applyng the followng theorem to H = S n by Applyng the followng theorem to H = S k. Page 69, proof of Theorem : represatons representatons. Page 69, proof of Theorem : You wrte: By complete reducblty, the map U ) d) G S d U ) G s surjectve. Actually, you don t need to use complete reducblty here: The projecton map π : U ) d S d U, u 1 u 2 u d u 1 u 2 u d has a G-equvarant secton namely, the lnear map ψ : S d U U ) d, Hence, the restrcton u 1 u 2 u d 1 d! σ S d u σ1) u σ2) u σd). U ) d) G S d U ) G of the map π to the G- nvarants has a secton as well namely, the restrcton of the secton ψ : S d U U ) d to the G-nvarants). Therefore, ths restrcton s surjectve. I lke ths argument more not just because t avods the use of complete reducblty, but also because t s more general t works for any subgroup G of GL n, ncludng those for whch the representatons nvolved fal to be completely reducble). Page 70, proof of Theorem : If d = k should be If k = d k. Page 74: a fx a stochastc we fx a stochastc. Page 75, 12.3: Replace formal lnear combnatons of the alphabet V by formal lnear combnatons of the alphabet B. Page 75, 12.3: Replace Next we defne a polynomal map ψ T : rep T) p leaft) by Next we defne a polynomal map Ψ T : rep T) p leaft) V p. There were two typos here: ψ T should be Ψ T, and the V p was mssng.) Page 76, 12.3: I suppose that p leaft) f p) should be p leaft) f p). 6