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So it is well known that when you tensor together two induced modules for an algebraic group $\nabla(\lambda) \otimes \nabla(\mu)$ that the result has a filtration by other induced modules, (I.e. it has a good filtration.). It is also well known that the sections can be determined by the character formula, $$ \mathrm{ch}(\nabla(\lambda) \otimes \nabla(\mu)) = \sum_{ \nu \mbox{ a weight of }\nabla(\mu)} \chi( \lambda + \nu) $$ where $\chi(\lambda)$ is the (formal) character of $\nabla(\lambda)$.

Can anyone give me a reference for this and/or a name for this character formula? I've seen it variously called "Brauer's character formula" or the "Littlewood-Richardson Rule" which is correct? (I can't seem to find it in Jantzen, I haven't tried anywhere else yet.)

Thanks for your help.

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Related: mathoverflow.net/questions/21875/… –  B. Bischof Aug 28 '12 at 13:50
    
Also, just to be clear, have you read this summary? encyclopediaofmath.org/…;. –  B. Bischof Aug 28 '12 at 13:56
    
A more exact reference (in spite of its header) has more complete information: mathoverflow.net/questions/85593/ Unfortunately, the clumsy search mechanism on MO makes it too time-consuming to track down all the old questions which are relevant. This one for instance can be located fastest by searching for 'Klimyk', even though Brauer's formula is older. Searching for 'Brauer' will turn up lots of interesting but irrelevant stuff. –  Jim Humphreys Aug 28 '12 at 15:33
    
@ Bischof: I had a look at the webpage - it's not in there. Also the mathoverflow question you reference is not so helpful for my question. I really just need a reference for the known formula, I know how to do the actual calculation! –  A Parker Aug 28 '12 at 15:39
    
@ humphreys. Thanks, this is helpful. I gather from that page that it's called the littlewood richardson rule in type A and could be called Klimyk's Formula or Klimyk's rule for Lie algebras or algebraic groups? I've never heard it called that before! I did find a reference in Jantzen from which it can be derived: RAGS: II lemma 5.8 (b) –  A Parker Aug 28 '12 at 15:44

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