St. Petersburg Mathematical Journal, 19. cilt,495-765. sayfalarAmerican Mathematical Society, 2008 |
Kitabın içinden
8 sonuçtan 1-3 arası sonuçlar
Sayfa 1006
... RIEMANN - HILBERT PROBLEM FOR A TREE We shall discuss the problem of actually computing the solution of the Riemann- Hilbert problem for a dessin . Clearly , calculating the inverses that go into the Wronskians in the solution given ...
... RIEMANN - HILBERT PROBLEM FOR A TREE We shall discuss the problem of actually computing the solution of the Riemann- Hilbert problem for a dessin . Clearly , calculating the inverses that go into the Wronskians in the solution given ...
Sayfa 1010
... equation . From this formula we can recover the solution of the Riemann - Hilbert problem for the tree given at the end of the preceding section . Whereas the leading coefficients of D2 and D3 are the discriminants of the corresponding ...
... equation . From this formula we can recover the solution of the Riemann - Hilbert problem for the tree given at the end of the preceding section . Whereas the leading coefficients of D2 and D3 are the discriminants of the corresponding ...
Sayfa 1013
... Hilbert problem for the tree of P has a solution of order at most two . Then , by Theorem 3 , the tree of P is a ... Riemann - Hilbert problem , Aspects Math . , vol . E22 , Friedr . Vieweg & Sohn , Braunschweig , 1994. MR1276272 ( 95d ...
... Hilbert problem for the tree of P has a solution of order at most two . Then , by Theorem 3 , the tree of P is a ... Riemann - Hilbert problem , Aspects Math . , vol . E22 , Friedr . Vieweg & Sohn , Braunschweig , 1994. MR1276272 ( 95d ...
İçindekiler
A Brudnyi and D Kinzebulatov Uniform subalgebras of L on the unit | 495 |
N A Vavilov Can one see the signs of structure constants? | 519 |
Waldemar Hołubowski A new measure of growth for groups and algebras | 545 |
Telif Hakkı | |
27 diğer bölüm gösterilmiyor
Diğer baskılar - Tümünü görüntüle
Sık kullanılan terimler ve kelime öbekleri
algebra arrangements assume automorphism base belong called character closed coefficients commutative complete condition Consequently consider constant construction contains continuous corresponding curve defined definition denote determined direct elementary elements equal equation equivalent example exists extension fact field Figure finite fixed formula function given gives Hence homogeneous homomorphism ideal implies inequality integer introduce inverse irreducible isomorphism Lemma linear Math Mathematical matrices means module Moreover multiplication natural normal Note objects obtain operator particular permutation points polynomial positive present prime problem projective Proof properties Proposition prove relations Remark representation respectively result ring root satisfies scheme sequence similar simple solution space statement structure subgroup subset suffices Suppose Theorem theory tree values variety vector weight