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Let $\mathfrak{g}$ be a simple finite-dimensional Lie algebra over $\mathbb{C}$ and let $U_q(\hat{\mathfrak{g}})$ be the corresponding quantum affine algebra (here $q$ is not a root of unity). We know that, if $V$ is a finite-dimensional $U_q(\hat{\mathfrak{g}})$-module, then $W:=V\otimes V^*$ has the trivial simple module as submodule, that is, we have a monomorphism $\mathbb{C}\hookrightarrow W$.

Are there sufficient conditions to ensure that this is the only nontrivial submodule of $W$? Maybe if $W$ is highest-weight... I don't know. Thanks in advance for answers and references. For my purposes, we can assume that $V$ is simple, but some more general answers are also welcome...

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    $\begingroup$ My intuition says the answer is no and that the situation is even worse: that the trivial module is not just a submodule but even a direct summand (so that there exist some other (not necessarily simple) module $U$ such that $W = \mathbb{C} \oplus U$. If this is true, there must be a projection operator from $W$ to its trivial submodule with kernel $U$. I think I can guess what this projection operator is (so proving that this guess is correct would settle the question) but I'm to tired now to check that it is indeed a $U_q(\hat{\mathfrak{g}})$ map. I'll type my guess in the next comment $\endgroup$
    – Vincent
    Nov 25, 2019 at 22:15
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    $\begingroup$ Note that $W$ has an interpretation as the space of all linear maps (not just $U_q(\hat{\mathfrak{g}})$-maps but all of them) from $V$ to $V$. I believe the trivial module is sitting in there as the scalar multiples of the identity map. If this is true, then there is a natural projection operator $T: W \to \mathbb{C}I$ given by $x \mapsto \frac{1}{n} Trace(x)$. Clearly this is a linear projection operator (the second condition just says that $T \circ T = T$). So, as I said, what needs to be checked (or debunked) is that it is a $U_q(\hat{\mathfrak{g}})$-map. I leave that to you $\endgroup$
    – Vincent
    Nov 25, 2019 at 22:18
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    $\begingroup$ O where I wrote $n$ I meant $\dim V$. I was inadvertently thinking of my personal favorite module $V$ whose dimension just happens to be $n$ $\endgroup$
    – Vincent
    Nov 25, 2019 at 22:20
  • $\begingroup$ @Vincent Thank you for the comments. I think this is not the case, because the category of finite-dimensional $U_q(\hat{\mathfrak{g}})$-modules is not semissimple (and not braided), as in case of quantum group $U_q(\mathfrak{g})$ and in classical setting (for $U(\mathfrak{g})$). In these cases your argument works. $\endgroup$
    – cl4y70n____
    Nov 25, 2019 at 22:41
  • $\begingroup$ Hmmm perhaps you are right, but be careful: having a semi-simple category is MUCH stronger than what I am claiming. There EVERY subrepresentation is a direct summand, I only claim that this very special submodule of these very special representations are. In other words: in the examples you give my argument works, BUT we do not even NEED my argument there since we can just say 'semi-simplicity' and be done with it, without looking into traces etc. So I still think there is something to check here. $\endgroup$
    – Vincent
    Nov 26, 2019 at 8:27

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