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What are all the complex finite dimensional linear representation of $GL(N,\mathbb{C})$?

We already know all the complex finite dimensional linear representation of SU(N).

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  • $\begingroup$ No extra hypotheses? You don't want it to be algebraic? Analytic? Smooth? Continuous? $\endgroup$ – Qiaochu Yuan Nov 20 '12 at 19:59
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    $\begingroup$ And what does this have to do with geometric Langlands? $\endgroup$ – Qiaochu Yuan Nov 20 '12 at 20:00
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    $\begingroup$ I see that you've asked a number of other questions, many appropriate for this site. This particular question is more appropriate for Google. $\endgroup$ – Allen Knutson Nov 20 '12 at 20:04
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    $\begingroup$ Finite-dimensional does not imply analytic and smooth. We can use wild automorphisms of $\mathbb{C}$ that permute a transcendence basis. If you don't like this, there are also the non-analytic representations $A\mapsto |\det(A)|$ and $A\mapsto \left[ \begin{array}{cc} 1 & \log |\det A|\\ 0 & 1\end{array}\right]$. $\endgroup$ – Richard Stanley Nov 20 '12 at 21:43
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    $\begingroup$ The context of the question is unclear, but, ... At least probably one wants to assume continuity of the repn. Continuity certainly does imply smoothness, because even in infinite-dimensional (continuous) repns, Garding's argument proves density of smooth vectors, which in a finite-dimensional repn must be everything. A refinement of such an argument proves that real analytic vectors are dense in "nice" repns, and finite-dimensional are nice enough so that all vectors are real-analytic. But, as Richard S. notes, certainly not complex-analytic. Then it's about highest weights? Clarify? $\endgroup$ – paul garrett Nov 21 '12 at 3:38

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