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For a semisimple Lie algebra $\mathfrak{g}$, a highest weight module $V(\lambda)$ with highest weight weight $\lambda$ has the property that every submodule $W$ of $V(\lambda)$ is a direct sum of the weight spaces of $W$.

Now consider quantum affine algebra $U_q(\hat{\mathfrak{g}})$. Let $V(\lambda)$ be the highest weight $U_q(\hat{\mathfrak{g}})$-module with highest weight weight $\lambda$. Do we still have the property that every submodule $W$ of $V(\lambda)$ is a direct sum of the weight spaces of $W$?

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Yes. See Proposition 3.2.1 in Jin Hong and Seok-Jin Kang's book Introduction to Quantum Groups and Crystal Bases.

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  • $\begingroup$ @Christopher, but it seems that proposition is for quantum groups not quantum affine algebras. $\endgroup$
    – Jianrong Li
    Commented Apr 10, 2012 at 1:51
  • $\begingroup$ @user, The name quantum group means different things to different people, but in the book I referenced, quantum groups are the same thing as quantum (or quantized) enveloping algebras, i.e., algebras defined by certain generators and relations coming from the Cartan datum of the underlying Lie algebra. If these aren't the same objects as you are interested in, then it might be helpful if you could edit your question to be more specific about what you mean by a quantum affine algebra. $\endgroup$
    – Christopher Drupieski
    Commented Apr 10, 2012 at 2:40
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    $\begingroup$ @user, Also, the book I referenced permits the Cartan datum to correspond to a generalized Cartan matrix, so the quantum enveloping algebras constructed there include those corresponding to affine Lie algebras. $\endgroup$
    – Christopher Drupieski
    Commented Apr 10, 2012 at 2:44
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    $\begingroup$ @Chris: It's probably useful to comment also on whether or not the nature of the parameter q matters at this point (indeterminate vs. root of unity). I guess it doesn't matter for the initial formalism involving highest weight modules and submodules. $\endgroup$
    – Jim Humphreys
    Commented Apr 10, 2012 at 17:27

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