By Wu X.

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Hence it leaves irreducibles invariant, since they are determined up to isomorphism by their character according to the theorem. 3 Let M ∈ Suppose: m -mod and N ∈ N indn+m M; (i) indm+n mn M nm N m+n ind (ii) M N appears in resm+n mn M mn Then indm+n mn M n -mod be irreducible modules. N with multiplicity one. N is irreducible. Proof Suppose for a contradiction that K = indm+n N is reducible. mn M Then we can find a proper irreducible submodule S, and set Q = K/S. By Frobenius reciprocity, M N appears in resm+n m n Q with non-zero multiplicity.

1(i). 3 Let M ∈ n -mod and N ∈ n+1 -mod be irreducible modules, and a ∈ F . Then f˜a M N if and only if e˜ a N M. 4). 1(i). 4 Let M N ∈ n -mod be irreducible. Then f˜a M f˜a N if and only if M N . Similarly, providing a M a N > 0, e˜ a M e˜ a N if and only if M N . 1 The map ch K n -mod →K n -mod is injective. Crystal operators 50 Proof We need to show that the characters of the irreducible modules in n -mod are linearly independent in K RepI n . Proceed by induction on n, the case n = 0 being trivial.

Now t = hd = 0, as required. d (iii) We remind that h denotes the map with h t = n → th . Given (i), it is straightforward to check that f is a homomorphism of -bimodules. To see that it is an isomorphism, it suffices by dimension n to show that it is injective. Suppose t lies in the kernel, then f t h = ht = 0 for all h ∈ n . Hence t = 0 by (ii). Now we extend this result to n. 2 Define a linear map fw = for each f ∈ n n →d by f d−1 w if w ∈ dS , otherwise, 0 w ∈ Sn . Then: -bimodules; (i) is a homomorphism of (ii) the map f n is an isomorphism of → Hom n d n h→h -bimodules.

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