# Generating Irreducible representations of a simple lie algebra with Schur functors

Let $\mathfrak{g}$ be a simple lie algebra over $\mathbb{C}$. Let $Rep(\mathfrak{g})$ denote the category of finite dimensional $\mathfrak{g}$-modules. For every $V \in Rep(\mathfrak{g})$ define $Rep_V(\mathfrak{g}) \subset Rep(\mathfrak{g})$ to be the smallest symmetric monoidal, idempotent complete, abelian subcategory with duals (so, closed under tensor products, retracts, direct sums and duals) which contains $V$.

Question: Does there always exist an irreducible $V$ for which $Rep_V(\mathfrak{g}) \cong Rep(\mathfrak{g})$? When it exists, is there a unique minimal one (in terms of the order on the weights) such $V$ (up to dualizing)? If not is there a unique self-dual such representation? If it doesn't exist, what is the minimal dimensional $V$ (possibly reducible) which satisfies this condition? Is it unique in some sense?

I'm interested in the question for all $\mathfrak{g}$ of type $A,B,C$ and $D$ (the exceptionals are a luxury). I think the standard representation $V$ in the case of type $A$ generates the entire category in this sense so that the answer is positive for this case but i'm not sure about any of the other cases.

• The irrep V won't be close to unique, even for SL_3, where the irrep with highest weight 2omega_1+omega_2 is also a tensor generator. Aug 14, 2018 at 2:09
• @PeterMcNamara Right, at least its the unique minimal one in the sense that $2 \omega_1 + \omega_2 > \omega_1+ \omega_2$ in the partial ordering of the weights. Aug 14, 2018 at 7:11
• @PeterMcNamara Isn't it also minimal in the sense of dimensions $dim V_{\omega_1+\omega_2} \lt dim V_{2 \omega_1 + \omega_2}$? Aug 14, 2018 at 7:18
• @PeterMcNamara Won't even $\omega_1$ suffice there? We get $\omega_2$ in the tensor square as the alternating part (or just by taking the dual, since the OP also required closure under duals). Aug 14, 2018 at 8:32
• Yes omega_1 suffices here. This was mentioned by the OP in their last paragraph, which is why I gave a different example to show non-uniqueness. Aug 14, 2018 at 9:48

You can replace "simple Lie algebra over $\mathbb{C}$" with "simply connected, simple compact Lie group" and the category of representations will be equivalent as a tensor category.

It's a standard result that a representation of a compact group with finite center is a tensor generator if and only if it is faithful, and at least for a simple Lie group, such an irrep is faithful if and only its highest weight generates the weight lattice modulo the root lattice. In particular, one exists if and only if the center is cyclic. There is a unique minuscule fundamental irrep for each nontrivial coset mod the root lattice, which will be a generator if that coset generates the quotient.

• In $A_n$, this means $\bigwedge{}^m\mathbb{C}^n$ for $m$ and $n$ relatively prime.
• In $B_n$, the Spin group has center $C_2$, so the spin representation is the unique minimal tensor generator.
• In $C_n$, the symplectic group has center $C_2$, so the vector representation is the unique minimal tensor generator.
• In $D_{2n+1}$, the center is $C_4$, so either of the two spin representations is a minimal generator.

The only case where the cyclic property fails is $D_{2n}$, where the weight lattice mod the root lattice is the Klein 4-group. There are 3 minuscule fundamentals (given by the vector rep and the two spin reps) but none is a tensor generator.

• Great! This is exactly what I wanted! Is there a good reference for these statements (especially for "faithful iff generates the center" and "tensor generator iff faithful")? Maybe its in Fulton and Harris? In any case just to make sure I understand, why do you state the statement that a representation is a tensor generator iff it is faithful only for compact groups? Does this fail in general for finite dimensional representations of complex connected simple lie groups? Your answer gives the result only for the simply connected ones but it seems weird for this statement to fail in general. Aug 14, 2018 at 17:24

I assume that you mean here full subcategory, not just subcategory. Assuming that, let $V$ be an irreducible representation of $\mathfrak{g}$ of dimension $n$ which is also $\textit{faithfull}$. This means that the resulting lie algebra map $\mathfrak{g}\to \mathfrak{sl}_n$ is injective. We thus get an injective map of associative algebras $$U(\mathfrak{g})\to U(\mathfrak{sl}_n).$$ We will think of this map as inclusion. Notice that due to the PBW basis, $U(\mathfrak{sl}_n)$ is free over $U(\mathfrak{g})$. Let $W$ be any simple representation of $\mathfrak{g}$. It will be enough to show that $W$ is contained in a direct summand of a direct sum of representation of the form $V^{a,b}:=V^{\otimes a}\otimes (V^*)^{\otimes b}$. Consider now the $\mathfrak{sl}_n$ representation $W':=U(\mathfrak{sl}_n)\otimes_{U(\mathfrak{g})} W$. This representation is generated by a single element (since $W$ is), and is therefore a quotient of $U(\mathfrak{sl}_n)$. Since $U(\mathfrak{sl}_n)$ has a filtration by finite dimensional representations which are direct sums of direct summands of $V^{a,b}$, the same is true for $W'$ (using here the semisimplicity of $\mathfrak{sl}_n$). By restricting back to $\mathfrak{g}$, the same is true for $W'$ considered as a representation of $\mathfrak{g}$. But as $\mathfrak{g}$-representations, $W'$ is a subrepresentation of $W$ (because of the PBW-basis). This implies that $W$ is a subrepresentation of a direct summand of $\oplus V^{a_i,b_i}$, which is what we wanted to show.

• Just to make sure I understand it seems we don't really need $V^*$ since we have $\bigwedge^{n-1} V \cong V^*$. Does that seem right to you? Secondly why is $W^{'}$ a subrepresentation of $W$? I don't see how to use the PBW basis here... Aug 14, 2018 at 17:47
• 2. $W$ is a subrepresentation of $W'$ due to the following reason: let $a_1=1,a_2,\ldots$ be a $free$ basis of $U(\mathfrak{sl}_n)$ over $U(\mathfrak{g})$ (which exists by PBW), Then as a $\mathfrak{g}$ representation $W'$ can be written as $a_1\otimes W\bigoplus \oplus_{i=2}^{\infty}a_i\otimes W$. The first part is isomorphic with $W$. Aug 15, 2018 at 9:37
• Both of the answers were great and contributed independently to answer the question. I had to choose 1 answer to accept though so I chose the one where the final solution was layed out. Thanks for your input! Aug 16, 2018 at 6:40